Linear Algebra Done Right
Solutions Manual
Sheldon Axler
7 July 1997
© 1997 Sheldon Axler

This solutions manual provides answers to all the exercises in my book
Linear Algebra Done Bight. It is distributed (without charge) only to
instructors who are using Linear Algebra Done Right as a textbook. If you
are such an instructor, please contact me (preferably via e-mail) to obtain
the appropriate version of the solutions manual. This version of the
solutions manual is intended to accompany the second edition (1997) of Linear
Algebra Done Right.
This solutions manual is copyrighted by me, but instructors using Linear
Algebra Done Right as a textbook have permission to modify and/or abridge
this solutions manual and to distribute copies only to students enrolled in
their courses (if students are charged for copies, the price may not exceed the
cost of materials and reproduction). If you want to modify and/or abridge
this solutions manual, you can request (via e-mail) the ЖЩК file from me
(or from Springer if I am no longer around).
Some instructors may not want to distribute any of these solutions to
students; other instructors may want to save class time by distributing to
students solutions to some or even many of the exercises. Obviously students
should not sec the solution to an exercise until they have attempted the
exercise themselves. Please do not distribute solutions needlessly, especially
if there is a danger that many solutions will get passed down to next year's
class. I will not distribute this solutions manual to students, so if neither
you nor colleagues in your department make copies for students, you can be
reasonably confident that your students are working the exercises without
benefit of these solutions.
This solutions manual has not been subjected to the same amount of
scrutiny as the book, so errors are more likely. I would be grateful for
information about any errors that you notice. If you know nicer solutions
to any of the exercises than the solutions given here, please let me know so
that I can improve future versions of this solutions manual.
Please check my web site for errata arid other information about Linear
Algebra Done Right. I welcome comments about either the book or the
solutions manual.
Have fun!
Sheldon Axler
Mathematics Department
San Francisco State University
San Francisco, CA 94132, USA
e-mail: axlerQmath.sfsu.edu
www home page: http://math.sfsu.edu/axler

Chapter 1
Vector Spaces
1. Suppose α and b are real numbers, not both 0. Find real numbers с and d
such that
l/(a + Ы) = с + di.
SOLUTION: Multiplying both the numerator and the denominator of the
left side of the equation above by α - Ы gives
a — bi ,.
—z—^ = c + dt.
Thus we must have
α — b
C = 2 ■ 13. and d = 2 ■ 1.2'
a2 + br az + bz
because α and b are not both 0, we are not dividing by 0.
Comment: Note that these formulas for с and d are derived under the
assumption that a + bi has a multiplicative inverse. However, we can forget
about the derivation and verify (using the definition of complex
multiplication) that
{а + ы)Ьть^--Лт^ = ^
which shows that every nonzero complex number does indeed have a
multiplicative inverse.
1

Chapter 1. Vector Spaces
Show that
-1 + y/3i
2
is a cube root of 1 (meaning that its cube equals 1).
SOLUTION: Using the definition of complex multiplication, we have
l + \/3i\2 -l-y/3i
(™r =
Thus
/-1 + у/Зг\3 = /-1-у/Зг\ /-1 + у/Зг\
= 1.
Prove that - (-υ) = υ for every υ € V.
Solution: Let ν G V. By the definition of additive inverse, we have
ν + (-v) = 0.
The additive inverse of —v, which by definition is —(—v), is the unique vector
that when added to -v gives 0. The equation above shows that ν has this
property. Thus -(-v) = v.
Comment: Using 1.6 twice leads to another proof that -(-υ) = υ.
However, the proof given above uses only the additive structure of V, whereas a
proof using 1.6 also uses the multiplicative structure.
Prove that if α € F, υ G V, and av = 0, then α = 0 or υ = 0.
Solution: Suppose that α G F, v G V, and
av = 0.
We want to prove that α = 0 or ν = 0. If α = 0, then we are done. So
suppose that α φ 0. Multiplying both sides of the equation above by 1/a
gives

3
Chapter 1. Vector Spaces
The associative property shows that the left side of the equation above equals
Iv, which equals v. The right side of the equation above equals 0 (by 1.5).
Thus υ = 0, completing the proof.
5. For each of the following subsets of F3, determine whether it is a subspace
ofF3:
(a) {(xi, x2,x3) € F3 : xi + 2x2 + 3x3 = 0};
(b) {(Х1,х2,хз) € F3 : xi + 2x2 + 3x3 = 4};
(c) {(хьхг.хз) G F3 : xix2X3 = 0};
(d) {(xi,x2,X3)GF3:xi = 5x3}.
Solution: (a) Let
U = {(χι,x2, хз) G F3 : χι + 2x2 + 3x3 = 0}.
To show that U is a subspace of F3, first note that (0,0,0) € U, so U is
nonempty.
Next, suppose that (x\,X2txz) G U and (уъУ2,Уз) G U. Then
xi + 2x2 + Зхз = 0
yi + 2y2 + 3y3 = 0.
Adding these equations, we have
(*i +Vi) + 2(x2 + y2) + 3(x3 + уз) = 0,
which means that (xi + yi,X2 + 2/2,^3 + Уз) G U. Thus U is closed under
addition.
Next, suppose that (х1,Х2,хз) G U and α € F. Then
xi + 2x2 + Зхз = 0.
Multiplying this equation by a, we have
(αχι) + 2(ахг) + З(ахз) = 0,
which means that (ах1,ахг,ахз) € U. Thus U is closed under scalar
multiplication.
Because U is a nonempty subset of F3 that is closed under addition and
scalar multiplication, U is a subspace of F3.

4 Chapter 1. Vector Spaces
(b) Let
C/ = {(xi,x2,:c3)€F3:xi+2x2 + 3x3 = 4}.
Then (4,0,0) € U but 0(4,0,0), which equals (0,0,0), is not in U. Thus U
is not closed under scalar multiplication. Thus U is not a subspace of F3.
(c) Let
U = {(хьхг.хз) € F3 : X1X2X3 = 0}.
Then (1,1,0) € U and (0,0,1) € U, but the sum of these two vectors, which
equals (1,1,1), is not in U. Thus U is not closed under addition. Thus U is
not a subspace of F3.
(d) Let
U = {(хьхг.хз) € F3 :χχ = 5x3}.
To show that U is a subspace of F3, first note that (0,0,0) € U, so U is
nonempty.
Next, suppose that (хьХ2,хз) € U and (2/1,2/21 Уз) € U. Then
X\ = 5Хз
У1 = 52/3.
Adding these equations, we have
xi +yi = 5(х3+2/з),
which means that (xi + 2/i,X2 + 2/2,хз + Уз) € U. Thus U is closed under
addition.
Next, suppose that (хьХ2)Хз) € U and α € F. Then
xi = 6x3.
Multiplying this equation by a, we have
αχι = 5(ахз),
which means that (ах1,ахг,ахз) € U. Thus U is closed under scalar
multiplication.
Because U is a nonempty subset of F3 that is closed under addition and
scalar multiplication, U is a subspace of F3.

Chapter 1. Vector Spaces
Give an example of a nonempty subset U of R2 such that U is closed under
addition and under taking additive inverses (meaning —u € U whenever
и € J7), but U is not a subspace of R2.
SOLUTION: Let U = {(τη,ή) : m and η are integers}. Then clearly
U is closed under addition and under taking additive inverses. However,
(1,1) € U but j(l, 1), which equals (^, 5), is not in U, so U is not closed
under scalar multiplication. Thus U is not a subspace of R2.
Of course there are also many other examples.
Give an example of a nonempty subset U of R2 such that U is closed under
scalar multiplication, but U is not a subspace of R2.
Solution: Let U be the union of the two coordinate axes in R2. More
precisely, let
U = {(χ, 0) : χ € R} U {(0, у) : у € R}.
Then clearly U is closed under scalar multiplication. However, (1,0) and
(0,1) are in U but their sum, which equals (1,1) is not in U, so U is not
closed under addition. Thus U is not a subspace of R2.
Of course there are also many other examples.
Prove that the intersection of any collection of subspaces of V is a subspace
of V.
Solution: Suppose {UQ}aer is a collection of subspaces of V; here Г
is an arbitrary index set. We need to prove that Паег ^<*> which equals the
set of vectors that are in Ua for every a € Г, is a subspace of V.
The additive identity 0 is in UQ for every а € Г (because each Ua is a
subspace of V). Thus 0 € ПаеГ^а· ^n particular, По€Г^а 'IS a nonempty
subset of V.
Suppose u, υ € Паег Ua· Then u, v. € UQ for every a € Г. Thus
u + v € UQ for every а € Г (because each UQ is a subspace of V). Thus
и + υ € Паег U<*· Thus Паег ^ is closed under addition.
Suppose и € Паег Ua and а € F. Then и € Ua for every а € Г. Thus
au € Ua for every а € Г (because each Ua is a subspace of V). Thus
au € Паег U<*· Thus Паег U<* ls closed under scalar multiplication.
Because Паег UQisa. nonempty subset of V that is closed under addition
and scalar multiplication, Паег U<* 1S a subspace of V.
COMMENT: For many students, the hardest part of this exercise is
understanding the meaning of an arbitrary intersection of sets. Instructors who

б
Chapter 1, Vector Spaces
do not want to deal with this issue should change the exercise to "Prove that
the intersection of any finite collection of subspaces of V is a subspace of V."
Many students will then prove that the intersection of two subspaces of V
is a subspace of V and use induction to get the result for finite collections
of subspaces.
9. Prove that the union of two subspaces of V is a subspace of V if and only if
one of the subspaces is contained in the other.
Solution: Suppose U and W are subspaces of V such that UuW is a
subspace of V. We will use proof by contradiction to show that U dW ox
W С U. Suppose that our desired result is false. Then U <£ W and W gLU.
This means that there exists и € U such that и £ W and there exists w € W
such that w £ U. Because и and w are both in C/U W, which is a subspace
of V, we can conclude that и + w € U U W. Thus u + tu€C/oru + tu€ W.
First consider the possibility that u+w € C/. In this case w, which equals
(u + w) + (—it), would be in the sum of two elements of U and hence we
would have w € U, contradicting our assumption that w £U.
Now consider the possibility that u+w € W. In this case u, which equals
(u + w) + (—го), would be in the sum of two elements of W and hence we
would have и € W, contradicting our assumption that и ф W.
The two paragraphs above show that u + w £ U and u + w £ W,
contradicting the final sentence of the first paragraph of this solution. This
contradiction completes our proof that U С W or W С U.
The other direction of this exercise is trivial: if we have two subspaces
of V", one of which is contained in the other, then the union of these two
subspaces equals the larger of them, which is a subspace of V.
10. Suppose that U is a subspace of V. What is U + U1
Solution: By definition, U+U = {u+υ :u,veU}. Clearly U С U+U
because if u € 17, then и equals и + 0, w*hich expresses и as a sum of two
elements of U. Conversely, U + U CU because the sum of two elements of U
is an element of U (because U is a subspace of V). Conclusion: U + U = U.
11. Is the operation of addition on the subspaces of V commutative?
Associative? (In other words, if J7i, С/г, Us are subspaces of V, is J7i +U2 = U2 + U1I
Is (Ux + U2) +U3 = Ui+ (U2 + J73)?)
Solution: Suppose U\, С/г, С/3 are subspaces of V.
A typical element of C/i +C/2 is a vector of the form щ +u2, where щ € U\
and u2 € U2. Because addition of vectors is commutative, u\ + u2 equals

Chapter 1. Vector Spaces
u2 + щ, which is a typical element of U2 + U1. Thus C/i + C/2 = C/2 + C/i. In
other words, the operation of addition on the subspaces of V is commutative.
A typical element of (C/i + U2) + Щ is a vector of the form (щ + u2) + из,
where щ G C/i, иг G C/2, and из € C/3. Because addition of vectors is
associative, (ui + иг) + из equals щ + (иг + из), which is a typical element
of C/i + (C/2 + C/3). Thus (C/i + C/2) + C/3 = C/i + (C/2 + C/3). In other words,
the operation of addition on the subspaces of V is associative.
Does the operation of addition on the subspaces of V have an additive
identity? Which subspaces have additive inverses?
Solution: The subspace {0} is an additive identity for the operation
of addition on the subspaces of V. More precisely, if C/ is a subspace of V,
then U + {0} = {0} + U = U.
For a subspace U of V to have an additive inverse, there would have to
be another subspace W of V such that U + W = {0}. Because both U and
W are contained in C/ + W, this is possible only if U = W = {0}. Thus {0}
is the only subspace of V that has an additive inverse.
Prove or give a counterexample: if C/i, С/г, W are subspaces of V such that
Ui + W = C/2 + W,
then C/i = C/2.
Solution: To construct a counterexample for the assertion above,
choose V to be any nonzero vector space. Let C/i = {0}, C/2 = V, and
W = V. Then C/j + W and C/2 + W are both equal to V, but C/i 7ε C/2.
Of course there are also many other examples.
Suppose U is the subspace of ^(F) consisting of all polynomials ρ of the
form
p(z) =az? +bz5,
where a, b G F. Find a subspace W of V(F) such that "P(F) = U φ W.
Solution: Let W be the set of all polynomials (with coefficients in F)
whose z2-coefficient and ^-coefficient both equal 0. Then every polynomial
in V(F) can be written uniquely in the form ρ+ 9, where ρ G U and q &W.
Thus "P(F) = U φ W.

8
Chapter I. Vector Spaces
Comment: There are other possible choices for W that give a correct
solution to this exercise, but the choice for W made above is certainly the
most natural one.
15. Prove or give a counterexample: if U\, U2, W are subspaces of V such that
V = UX@W and V = U2@W,
then Ui = U2.
SOLUTION: To construct a counterexample for the assertion above, let
V = F2, let J7i = {(χ,Ο) : χ € F}, let U2 = {(0,y) : у € F}, and let
W = {(z,z):zeF}. Then
F2 = Ui®W and F2 = J72©H^
as is easy to verify, but J7i φ U2.
Of course there are also many other examples.

Chapter 2
Finite-Dimensional
Vector Spaces
1. Prove that if (υι,..., vn) spans V, then so does the list
(v\ -v2,V2 -υ3,...,υη_ι - vn,vn)
obtained by subtracting from each vector (except the last one) the following
vector.
Solution: Suppose (v\,. ..,vn) spans V. Let υ € V. To show that
υ € span(vi - щ, щ — vz, ■ ■ ■, υη_ι - vn, vn), we need to find αϊ,..., αη € F
such that
ν = αι(υι - v2) + а2(г>2 - v3) + ...αη_ι(υη-ι - vn) +anvn.
Rearranging terms of the equation above, we see that we need to find
αϊ,..., an € F such that
(a) υ = aw + (a2 - ai)v2 + (a3 - a2-)vz + ■ ■ ■ + (an - α„_ι)υη.
Because (v\,..., vn) spans V, there exist b\,..., bn € F such that
(b) υ = b\Vi + b2V2 + 63^3 Η 1- bnvn-
Comparing equations (a) and (b), we see that (a) will be satisfied if we
choose αϊ to equal b\ and then choose a2 to equal 62 + o-ι and then choose
аз to equal 63 + a2, and so on.
9

10 Chapter 2. Finite-Dimensional Vector Spaces
2. Prove that if (i>i,..., vn) is linearly independent in V, then so is the list
(i>i -V2,v2 -ι>3,...,ι>η-ι -νη,υη)
obtained by subtracting from each vector (except the last one) the following
vector.
Solution: Suppose (i>i,..., vn) is linearly independent in V. To prove
that the list displayed above is linearly independent, suppose oi,...,a„€F
are such that
a\(vi -v2) + a2(v2 - i>3) + ■ ■ ·+ αη_ι(ι>η-ι - νη) + α„ι>η = 0.
Rearranging terms, the equation above can be rewritten as
a-ivi + (a2 -αι)υ2 + (α3 - a2)v3 + ■·· + (an- θη-{)υη = 0.
Because (υι,..., vn) is linearly independent, the equation above implies that
αϊ =0
a2 - αϊ = 0
α3 - α2 = 0
о-п -о,п-\ =0.
The first equation above tells us that αϊ = 0. That information, combined
with the second equation, tells us that a2 = 0. That information, combined
with the third equation, tells us that a3 = 0. Continue in this fashion,
getting αϊ = -■· = an = 0. Thus (vi - v2,V2 - v3,. ..,υη_ι - νη,νη) is
linearly independent.
3. Suppose (υι,..., vn) is linearly independent in V and w € V. Prove that if
(v\ +w,...,vn+w) is linearly dependent, then w G span(vi,...,υη)-
Solution: Suppose {v\ + w,... ,vn + w) is linearly dependent. Then
there exist scalars ab ... ,an, not all 0, such that
αι(υι +w) Η Ι-αη(υη + го) = 0.
Rearranging this equation, we have
αιΐ>ι Η 1- OnVn = -(ai + 1- dn)w.

11
Chapter 2. Finite-Dimensional Vector Spaces
If αϊ Η \-an were 0, then the equation above would contradict the linear
independence of (υι,..., vn)· Thus αϊ + · · · + an φ 0. Hence we can divide
both sides of the equation above by -(αϊ + ··· + αη), showing that w G
span(vi,...,t;n).
4. Suppose m is a positive integer. Is the set consisting of 0 and all polynomials
with coefficients in F and with degree equal to m a subspace of V(F)?
Solution: The set consisting of 0 and all polynomials with coefficients
in F and with degree equal to m is not a subspace of ^(F) because it is not
closed under addition. Specifically, the sum of two polynomials of degree m
may be a polynomial with degree less than m. For example, suppose m = 2.
Then 7 + 4z + 5z2 and 1 + 2z - 5z2 are both polynomials of degree 2 but
their sum, which equals 8 + 6z, is a polynomial of degree 1.
5. Prove that F°° is infinite dimensional.
Solution: For each positive integer m, let е™, be the element of F°°
whose mth coordinate equals 1 and whose other coordinates equal 0:
em = (0I... ,0,1,0,...).
ΐ
mth coordinate
Then (ei,..., em) is a linearly independent list of vectors in F°°, as is easy
to verify. This implies, by the marginal comment attached to 2.6, that F°°
is infinite dimensional.
6. Prove that the real vector space consisting of all continuous real-valued
functions on the interval [0,1] is infinite dimensional.
Solution: Let V denote the real vector space of all continuous real-
valued functions on the interval [0,1). For each positive integer m, the list
(l,x, ...,xm) is linearly independent in V (because if αο,...,α„, G R are
such that
an + αχχ Λ 1- Omim = 0
for every χ G [0,1], then the polynomial above has infinitely many roots
and hence all its coefficients must equal 0). This implies, by the marginal
comment attached to 2.6, that V is infinite dimensional.

12 Chapter 2. Finite-Dimensional Vector Spaces
7. Prove that V is infinite dimensional if and only if there is a sequence
vi,V2,... of vectors in V such that (v\,...,vn) is linearly independent for
every positive integer n.
Solution: First suppose that V is infinite dimensional. Choose v\
to be any nonzero vector in V. Choose V2,vs,... by the following
inductive process: suppose that v\,..., υη-1 have been chosen; choose any vector
vn € V such that vn £ span(ui,.. .,υη_ι)—because V is not finite
dimensional, span(ui,... ,υη_ι) cannot equal V so choosing vn in this fashion is
possible. The linear dependence lemma (2.4) implies that (υι,..., vn) is
linearly independent for every positive integer n, as desired.
Conversely, suppose there is a sequence υι, иг,... of vectors in V such
that (vi,..., vn) is linearly independent for every positive integer n. This
implies, by the marginal comment attached to 2.6, that V is infinite
dimensional.
8. Let U be the subspace of R5 defined by
U = {(11,12,13,14^5) € R5 : χι = 3x2 and хз = 7x4}.
Find a basis of U.
Solution: Obviously
U = {(3X2,X2,7X4,X4,X5) : X2,X4,X5 G R}.
From this representation of U, we see easily that
((3,1,0,0,0), (0,0,7,1,0), (0,0,0,0,1))
is a basis of U.
Of course there are also other possible choices of bases of U.
9. Prove or disprove: there exists abasis (ро,Р\,Р2,Рз) of ^(F) such that none
of the polynomials ро,Р\,Р2,Рз has degree 2.
Solution: Define Ро,РьР2,Рз е ^(F) by
PoOO = 1.
Pi{z) = z,
P2(z) = z2 + z3,
Pz{z) = zz.

13 Chapter 2. Finite-Dimensional Vector Spaces
None of the polynomials ро,РьР2)Рз has degree 2, but (рсьРьРг.Рз) is a
basis of ^(F), as is easy to verify.
Of course there are also other possible choices of bases of ^(F) without
using polynomials of degree 2.
10. Suppose that V is finite dimensional, with dim V = n. Prove that there
exist one-dimensional subspaces U\,..., Un of V" such that
V = Ux φ · · · Φ Un.
Solution: Let (vi,...,vn) be a basis of V. For each j, let Uj equal
span(vj); in other words, Uj = {avj : a € F}. Because (v\,..., vn) is a basis
of V, each vector in V can be written uniquely in the form
axvi + ■ · ■ + anvn,
where αϊ,..., αη € F (see 2.8). By definition of direct sum, this means that
V = Ux®---®Un.
11. Suppose that V is finite dimensional and U is a subspace of V such that
dimJ7 = dim V. Prove that U = V.
Solution: Let (ui,...,ti„) be a basis of U. Thus η = dim[7, and
by hypothesis we also have η = dim V. Thus (ui,..., un) is a linearly
independent (because it is a basis of U) list of vectors in V with length
dimV". From 2.17, we see that (щ,...,ип) is a basis of V. In particular
every vector in V is a linear combination of (u\,... ,Un)· Because each
Uj € U, this implies that U = V.
12. Suppose that ρο,Ρι, · · · ,pm are polynomials in Vm(F) such that Pj(2) = 0
for each j. Prove that (ρο,Ρι, · · · ,Pm) is not linearly independent in Vm(F).
Solution: Because py(2) = 0 for each j, the constant polynomial 1 is
not inspan(po,... ,pm). Thus (prj,-..,Pm) is not a basis of Vm(F). Because
(poi · · ■ iPm) is a list of length m + 1 and Vm(F) has dimension m + 1, this
implies (by 2.17) that (po,... ,pm) is not linearly independent.
13. Suppose U and W are subspaces of R8 such that dimC/ = 3, dimW = 5,
and U + W = R8. Prove that U П W = {0}.
Solution: We know (from 2.18) that
dim(J7 + W) = dimU + dim W - dim(J7 П W).

14 Chapter 2. Finite-Dimensional Vector Spaces
Because dim(J7 + W) = 8, dim С/ = 3, and dimH^ = 5, this implies that
dim(C7 nW^) = 0. Thus U П W = {0}.
14. Suppose that U and W are both five-dimensional subspaces of R9. Prove
that UnW φ {Ъ}.
Solution: Using 2.18 we have
9 > dim(C7 + W)
= dim U + dim W - dim{U П W)
= lO-dim(tfnW).
Thus dim(C/ П W) > 1. In particular, U П W φ {0}.
15. You might guess, by analogy with the formula for the number of elements
in the union of three subsets of a finite set, that if Όγ,ϋι,£/3 are subspaces
of a finite-dimensional vector space, then
dimit/i + U2 + Uz)
= dim U\ + dim U2 + dim £/3
- dim(Ui П U2) - dim(C/i П Uz) - dim(C72 П Uz)
+ dim(C7i П U2 П U3).
Prove this or give a counterexample.
Solution: To give a counterexample, let V = R2, and let
U\ ={(x,0) :x€R},
^2 = {(0,y):y€R},
UZ = {(x,x) :x€ R}.
Then Ui + U2 + Uz = R2, so dim(C7i + U2' + Uz) = 2. However,
dim Ui = dim U2 = dim C/3 = 1
and
dim(C/i П U2) = dim(C7i П Щ) = dim(C72 П Uz) = dim(C/i П U2 П Uz) = 0.
Thus in this case our guess would reduce to the formula 2 = 3, which
obviously is false.
Of course there are also many other examples.

15
Chapter 2. Finite-Dimensional Vector Spaces
16. Prove that if V is finite dimensional and U\,...,Um are subspaces of V,
then
dim(J7i + · · · + Um) < dim U\ 4 + dim Um.
Solution: For each j = l,...m, choose a basis for Uj. Put these
bases together to form a single list of vectors in V. Clearly this list spans
U\ -\ \-Um. Hence the dimension of U\ Η fJ7m is less than or equal to the
number of vectors in this list (by 2.10), which equals dim J7i -\ l· dim Um.
In other words,
dim(J7i Η 1- Um) < dim U\ Λ 1- dim Um.
17. Suppose V is finite dimensional. Prove that if U\,..., Um are subspaces of V
such that V = U\ φ · · · φ Um, then
dim V = dim J7i Η \- dim Um.
Comment: This exercise deepens the analogy between direct sums of
subspaces and disjoint unions of subsets. Specifically, compare this exercise
to the following obvious statement: if a finite set is written as a disjoint
union of subsets, then the number of elements in the set equals the sum of
the number of elements in the disjoint subsets.
Solution: Suppose that J7i,..., Um are subspaces of V such that V =
U\ θ · · · θ Um. For each j = 1,... m, choose a basis for Uj. Put these bases
together to form asingle list В of vectors in V. Clearly В spans J7i Η \-Um,
which equals V. If we show that В is also linearly independent, then it will
be a basis of V. Thus the dimension of V will equal the number of vectors B.
In other words, we will have
dim V = dim U\ Η + dim J7m,
as desired.
We still need to show that В is linearly independent. To do this, suppose
that some linear combination of В equals 0. Write this linear combination
as щ -\ 1- ιι,η, where we have grouped together the terms that come from
the basis vectors of J7i and called their sum «i, and similarly up to 1%.
Thus we have
Щ + ■ · · + Щп = 0,
where each Uj € Uj. Because V — U\ φ · - · φ J7m, this implies that each Uj
equals 0. Because each Uj is a linear combination of our basis of Uj, all the

16 Chapter 2. Finite-Dimensional Vector Spaces
coefficients in the linear combination defining щ must equal 0. Thus all the
coefficients in our original linear combination of В must equal 0. In other
words, В is linearly independent, completing our proof.

Chapter 3
Linear Maps
1. Show that every linear map from a one-dimensional vector space to itself is
multiplication by some scalar. More precisely, prove that if dim V = 1 and
Τ G £(V, V), then there exists α G F such that Tv = av for all υ G V.
Solution: Suppose dim V = 1 and Τ € C(V, V). Let и be any nonzero
vector in V. Then every vector in V is a scalar multiple of u. In particular,
Tu = au for some α G F.
Now consider a typical vector υ G V. There exists b G F such that
ν = bu. Thus
Tv = T(bu)
= bT(u)
= b(au)
= a(bu)
= av.
2. Give an example of a function /: R2 —> R such that
f(av) = af{y)
for all α G R and all ν G R2 but / is not linear.
Solution: Define /: R2 -> R by
/(s,y) = (s3 + y3)1/3.
Then f(av) — af(v) for all α G R and all ν G R2. However, / is not linear
because /(1,0) = 1 and /(0,1) = 1 but
17

18 Chapter 3. Linear Maps
/((1,0)+ (0,1)) =/(1,1)
7^/(1,0) + /(0,1).
Of course there are also many other examples.
Comment: This exercise shows that homogeneity alone is not enough
to imply that a function is a linear map. Additivity alone is also not enough
to imply that a function is a linear map, although the proof of this involves
advanced tools that are beyond the scope of this book.
3. Suppose that V is finite dimensional. Prove that any linear map on a sub-
space of V can be extended to a linear map on V. In other words, show that
if U is a subspace of V and S € C(U, W), then there exists Τ € C(V, W)
such that Tu = Su for all и €U.
Solution: Suppose U is a subspace of V and S € C(U, W). Let
(щ,..., ιι,η) be a basis of U. Then (ui,..., ttm) is a linearly independent list
of vectors in V, and so can be extended to a basis (щ,..., Um, v\,..., vn)
of V (by 2.12). Define Τ € C(V, W) by
Т(<цщ +...0^11^. + biVi + ...bnvn) = aiSui Η {-a^Sum.
Then Tu = Su for all и € U.
COMMENT: Defining T: V -> W by
Tv={Sv if* €17;
\0 \[у$и.
does not work because this map is not linear.
4. Suppose that Γ is a linear map from V to F. Prove that if и € V is not in
nullT, then
V = nullTe {au: a € F}.
Solution: Suppose и € V is not in nullT. If a € F and au € nullT,
then 0 = T(au) = aTu, which implies that α = 0 (because Tu φ 0). Thus
nullTn{au:a€F} = {0}.
If ν € V, then

19
Chapter 3. Linear Maps
V={V-¥aU)
Tv
Tu
Note that Τ (υ — |^ιι) = Tv-lj^Tu = 0. Thus the equation above expresses
an arbitrary vector υ € V as the sum of a vector in null Γ and a scalar
multiple of u. Hence V = null Τ + {au : a € F}. Using 1.9, we conclude
that V = null Τ φ {au : a € F}.
Suppose that Τ € C(V, W) is injective and (v\,..., υη) is linearly
independent in V. Prove that {Tv\,...,Tvn) is linearly independent in W.
Solution: To show that {Tv\,..., Tvn) is linearly independent, suppose
αϊ,..., On € F are such that
αιΓυι +··· + αηΤνη = 0.
Because Γ is a linear map, this equation can be rewritten as
Γ(αιυι +···+αηυη) = 0.
Because Τ is injective, this implies that
a\V\ -\ Ι-αηυη = 0.
Because {v\,..., vn) is linearly independent, the equation above implies that
a\ = '" = α„ = 0. Thus {Tv\,...,Tvn) is linearly independent.
Prove that if S\,...,Sn are injective linear maps such that S\...Sn makes
sense, then S\ ...Sn is injective.
Solution: Suppose that Si,...,Sn are injective linear maps such that
Si...Sn makes sense (which means that the domains of S\,..., Sn are such
that Si... Sn is well defined). Suppose υ is.a vector in the domain of S\... Sn
(which equals the domain of Sn) such that
{Sl...Sn)v = 0.
To show that S\... Sn is injective, we need to show that ν = 0 (see 3.2). To
do this, rewrite the equation above as
Si((S2...Sn)v) = 0.
Because Si is injective, this implies that

20 Chapter 3. Linear Maps
(S2...Sn)v = 0.
The same argument, now applied to the equation above, shows that
{S3...Sn)v = 0.
Repeat this process until reaching the equation Snv = 0, which implies
(because Sn is injective) that υ = 0, as desired.
7. Prove that if (vi,...,vn) spans V and Τ € C(V,W) is surjective, then
(Tvu...,Tvn) spans W.
Solution: Suppose that {v\,..., vn) spans V and Τ € £(V, W) is
surjective. Let w € W. Because Τ is surjective, there exists ν ζ. V such that
ϋΓυ = tu. Because (υι,..., vn) spans V, there exist αϊ,..., αη € F such that
ν = aivi Η Ι-αηυη.
Applying Τ to both sides of this equation, we get
Tv = axTvi + · · · + OnTvn.
Because Tv = w, the equation above implies that w € span(!Tt;i,..., Tvn).
Because w was an arbitrary vector in W, this implies that {Tv\,..., Tvn)
spans W.
8. Suppose that V is finite dimensional and that Τ € C(V, W). Prove that
there exists a subspace U of V such that U П null Γ = {0} and range Τ =
{TuiueU}.
Solution: There exists a subspace U of V such that
V = mniT®U;
this follows from 2.13 (with null Γ playing the role of U and U playing the
role of W).
From the definition of direct sum, we have U Π null Γ = {0}.
Obviously range Γ D {Tu : и € U}. To prove the inclusion in the other
direction, suppose ν € V. Then there exist w € null Γ and u' €U such that
υ = w + u'.
Applying Τ to both sides of this equation, we have Tv = Tw + Tu' = Tu'.
Thus Tv € {Tu : и € 17}. Because υ was an arbitrary vector in V (and thus
Tv is an arbitrary vector in range T), this implies that

21
Chapter 3. Linear Maps
ranger С {74i :u€C/}.
Thus rangeX1 = {Tu : и € 17}, as desired.
9. Prove that if Τ is a linear map from F4 to F2 such that
null Τ = {(χι, X2, яз, £<t) € F4 : X! = 5x2 and X3 = 7x4},
then Τ is surjective.
Solution: Suppose Τ € £( FA, F2) is such that null Γ is as above.
Then ((5,1,0,0), (0,0,7,1)) is a basis of nullT, and hence dim null Γ = 2.
From 3.4 we have
dim range Τ = dim F4 - dim null Τ
= 4-2
= 2.
Because range Γ is a two-dimensional subspace of R2, we have range Τ = R2.
In other words, Τ is surjective.
10. Prove that there does not exist a linear map from F5 to F2 whose null space
equals
{(χι,Χ2)^3)^4)^5) e F5 : χι = 3x2 and X3 = x4 = x5}.
Solution: Suppose U is the subspace of F5 displayed above. Then
((3,1,0,0,0), (0,0,1,1,1)) is a basis of U, and hence dim U = 2.
If Γ € £(F5, F2) then from 3.4 we have
dim null Τ = dim F5 — dim range Τ
= 5 - dim range Τ
>3
>dimJ7,
where the first inequality holds because range Г С F2. The inequality above
shows that if Г € £(F5, F2), then null Γ φ U, as desired.
11. Prove that if there exists a linear map on V whose null space and range are
both finite dimensional, then V is finite dimensional.
Solution: Suppose there exists a linear map Τ from V into some
vector space such that null Τ and range Τ are both finite dimensional. Thus

22
Chapter 3. Linear Maps
there exist vectors щ,...,!!^ € V and w\,.. .,wn € range Τ such that
{u\,...,Uj„) spans null Γ and (w\,... ,wn) spans range T. Because each
Wj € range T, there exists Vj € V such that tuj = Tvj.
Suppose υ € V". Then Γυ € range Γ, so there exist b\,...,bn € F such
that
Tv = b\Wi Η \r bnwn
= biTvi + ■ ■ ■ + bnTvn
= T(blVl + ...bnvn).
The equation above implies that T(v — b\Vi — ... — bnvn) = 0. In other
words, υ -b\V\ - ··· - bnvn € nullT. Thus there exist αϊ,...,am € F such
that
υ - bivi bnvn = сцщ Η 1- amum.
The equation above can be rewritten as
ν = aim -{ 1- amum + Mi -\ 1- bnvn.
The equation above shows that an arbitrary vector ν € V is a linear
combination of («i,..., um, v\,..., vn). In other words, (u\,..., Um, V\,..., vn)
spans V. Thus V is finite dimensional.
Comment: The hypothesis of 3.4 is that V is finite dimensional (which
is what we are trying to prove in this exercise), so 3.4 cannot be used in this
exercise.
12. Suppose that V and W are both finite dimensional. Prove that there exists
a surjective linear map from V onto W if and only if dim W < dim V.
Solution: First suppose that there exists a surjective linear map Τ
from V onto W. Then
dim W = dim range Τ
= dim V - dim null Τ
< dim V,
where the second equality comes from 3.4.
To prove the other direction, now suppose that dim W < dim V. Let
(tui,..., wm) be a basis of W and let (v\,. ..,vn) be a basis of V. For
oi,..., α„ € F define Γ(αιυι Η 1- anvn) by

23
Chapter 3. Linear Maps
T{axvx + · · · + anvn) = a^wi +■·■ + OmWm.
Because dim W < dim V, we have m < η and so a„, on the right side of the
equation above makes sense. Clearly Γ is a surjective linear map from V
onto W.
13. Suppose that V and W are finite dimensional and that U is a subspace
of V. Prove that there exists Τ € C(V, W) such that null Τ - U if and only
ifdimt/>dimVr-dimH'.
Solution: First suppose that there exists Τ € C(V, W) such that
nuUT = J7. Then
dimJ7 = dimuullT
= dim V — dim range Τ
> dim V - dim W,
where the second equality comes from 3.4.
To prove the other direction, now suppose that dim U > dim V — dim W.
Let (iti,... ,Um) be a basis of U. Extend to a basis (ui,... ,umii;i,... ,vn)
of V. Let (w\,..., wp) be a basis of W. For a\,...,a„,,b\,...,bn € F define
T(aiui + · · · + anUm + Mi + · · · + bnvn) by
T{aiu\ Η 1- CLmUm + biv\ Η 1- bnvn) = biwi Η \- bnwn.
Because dim W > dim V — dim U, we have p>n and so wn on the right side
of the equation above makes sense. Clearly Τ € C(V, W) and null Γ = U.
14. Suppose that W is finite dimensional and Τ € C(V, W). Prove that Τ is
injective if and only if there exists S € C(W, V) such that ST is the identity
map on V.
Solution: First suppose that Τ is injective. Define S': range Γ —> V
by
S'(Tv) = v;
because Τ is injective, each element of range Γ can be represented in the
form Tv in only one way, so Γ is well defined. As can be easily checked, S'
is a linear map on range T. By Exercise 3 of this chapter, S' can be extended
to a linear map S € C(W, V). If ν € V, then (ST)v = S{Tv) = S'{Tv) = v.
Thus ST is the identity map on V", as desired.

24
Chapter 3. Linear Maps
To prove the implication in the other direction, now suppose that there
exists S € C(W, V) such that ST is the identity map on V. If u, ν e V are
such that Tu = Tv, then
и = {ST){u) = S{Tu) = S(Tv) = (ST)v = ν
and hence и = v. Thus Г is injective, as desired.
15. Suppose that V is finite dimensional and Τ € C(V, W). Prove that Τ is
surjective if and only if there exists S € C(W, V) such that TS is the identity
map on W.
Solution: First suppose that Τ is surjective. Thus W, which equals
rangeT, is finite dimensional (by 3.4). Let (tui,... ,wm) be a basis of W.
Because Τ is surjective, for each j there exists Vj Ε V such that Wj = Tvj.
Define S € C(W, V) by
S(aiioi Η \- OmWm) = a\V\ -\ (- amvm.
Then
{TS){a\Wi -\ \-amWm) = T(cliVi Η l-amt;m)
= αιΓυι +-··+αηχΤυτη
= aitoi Η |-amtum.
Thus TS is the identity map on W.
To prove the implication in the other direction, now suppose that there
exists S € C(W, V) such that TS is the identity map on W. If w e W, then
ш = T(Sw), and hence tu € range T. Thus range Γ = W. In other words, Τ
is surjective, as desired.
16. Suppose that U and V are finite-dimensional vector spaces and that S €
£(V, W), Τ Ε C(U, V). Prove that
dim null ST < dim null S + dim null T.
Solution: Define a linear map T": null ST -> V by T'u = Tu. If
u € null ST, then 5(Ги) = 0, which means that Tu € null S. In other
words, range Τ С null 5. Now
dim null ST = dim null Τ + dim range Γ7
< dim null T' + dim null S
< dim null Τ + dim null S,

25
Chapter 3. Linear Maps
where the first line follows from 3.4 (applied to T1), the second line holds
because range T1 с null S, and the third line holds because of the obvious
inclusion null T1 с null T.
17. Prove that the distributive property holds for matrix addition and matrix
multiplication. In other words, suppose A, B, and С are matrices whose
sizes are such that A(B+C) makes sense. Prove that AB + AC makes sense
and that A(B +C) = AB + AC.
Solution: Because A(B+C) makes sense, В and С must have the same
size. Furthermore, the number of columns of A (let's call this number n)
must equal the number of rows of В and C. All this means that AB + AC
makes sense.
To prove that A(B + C) = AB + AC, just use the definition of matrix
addition, the definition of matrix multiplication, and the usual distributive
property for elements of F. Specifically, let a.jtk, bj,fc, and Cj·^ denote the
entries in row j, column к of A, B, and C, respectively. The entry in row j,
column к of В + С is bj^ + Cjfc. Thus the entry in row j, column к of
A(B + C) is
η
^2ajAbr,k +Cr,fc).
r=l
which equals
η π
У] 0-j,rbr,k + Σ аЗ,г^г,ку
г=1 г=1
which equals the entry in row j, column к of AB + AC, as desired.
18. Prove that matrix multiplication is associative. In other words, suppose A,
B, and С are matrices whose sizes are such that (AB)C makes sense. Prove
that A(BC) makes sense and that (AB)C = A(BC).
Solution: This exercise can be done by a brute force calculation, in
the style of the solution to the previous exercise. Here is a solution that uses
only the associativity of the product of linear maps (which is easy to verify
because composition of functions is clearly associative) and the nice property
that the matrix of the product of two linear maps equals the product of the
matrices of the two linear maps (see 3.11).
Suppose A is an m-by-n matrix, β is an n-by-p matrix, and С is a p-by-g
matrix; the sizes much match up like this in order for (AB)C to make sense.

Let R e £(Fn, Fm), S e C(F", Fn),Te £(F«, F") be such that, with respect
to the standard bases, M(R) = A,M(S) = B,M(T) = C; 3.19 insures that
such linear maps exist. Now
(AB)C = (M(R)M(S))M(T)
= M(RS)M(T)
= M((RS)T)
= M(R(ST))
= M(R)M(ST)
= M(R)(M(S)M(T))
= A(BC).
Suppose Τ Ε £(Fn, Fm) and that
M(T) =
0.1,1
αι,η
im,n
O-rn.,1 · ■■ O-r,
where we are using the standard bases. Prove that
T{x\, . . . , Xn) = (αι,ιΧι Η +a\,nXn,---,<hn,lXl Η Ютп.пЯп)
for every (χι,..., xn) e Fn.
Comment: This exercise shows Τ has the form promised on page 39.
Solution: Let χ = (xi,...,xn) e Fn. Using the standard bases, we
then have
M{Tx) = M{T)M{x)
0.1,1
αι,η x\
am,\ ··· Οτη,η
Xr
ο-ι,ιΧι Η Ι-αι,ηΧη
Οτη,Ι^Ι Η \-<hn,nXn
where the first equality comes from 3.14. The last equation implies that

27
Chapter 3. Linear Maps
Tx = (αι,ιΧι Η 1- αιιηχη,..., amiiXi Η 1- α„,ιηχη),
as desired.
20. Suppose (i>i,...,vn) is a basis of V. Prove that the function T: V —>
Mat(n, 1,F) defined by
Tv = M(v)
is an invertible linear map of V onto Mat(n, 1, F); here M{v) is the matrix
of υ € V with respect to the basis (vi,..., vn)·
Solution: Suppose u, w e V. We can write
и = a\V\ + ■ · · + OnVn and w = b\V\ + · - · + bnvn
for some αι,...,αη,ί>ι,... ,bn EF. Thus
и + w = (αϊ + b\)v\ Η 1- (αη + bn)vn.
Hence
T(u + w) = M{u + w)
=
αϊ +bi
_an+bn
a\
;
. °" .
+
' bi '
;
- b" .
= M(u)+M(w)
= 5
^u + 2
X
which shows that Τ satisfies that additivity property required for linearity.
If с € F, then
cu = cai^i + - ■ - + canvn.
Hence

28
Chapter 3. Linear Maps
T{cu) = M{cu)
ca.\
can
αϊ
= с
On
= cM{u)
= cTu,
which shows that Τ satisfies the homogeneity property required for linearity.
Thus Τ is linear.
If Tu = 0, then αϊ = · · · = α„ = 0, which implies that и = 0. Thus Τ is
iujective.
If Ci,...,c € F, then
T(civi + --- + cnvn) =
Cl
Cn
which implies that Τ is surjective.
Because the linear map Τ is iujective and surjective, it is invertible
(see 3.17).
21. Prove that every linear map from Mat(n, 1,F) to Mat(m, 1,F) is given by
a matrix multiplication. In other words, prove that if
Τ € £(Mat(n, 1, F), Mat(m, 1, F)),
then there exists an m-by-n matrix A such that ТВ = AB for every В €
Mat(n,l,F).
Solution: The vector spaces Mat(n, 1,F) and Mat(m, 1,F) have
obvious bases (consisting of matrices that have 0 in all entries except for a 1
in one entry). Let A be the matrix of Τ with respect to these bases. Note
that if В € Mat(n, 1, F), then M(B) = В and M(TB) = ТВ. Thus

29 Chapter 3. Linear Maps
ТВ = M{TB)
= M(T)M(B)
= AB,
where the second equality comes from 3.14.
22. Suppose that V is finite dimensional and 5,Te £{V)· Prove that ST is
invertible if and only if both S and Τ are invertible.
Solution: First suppose ST is invertible. Thus there exists R e C(V)
such that R(ST) = (ST)R = I. If ν € V is such that Tv = 0, then
υ — Iv
= R(ST)v
= 0.
Because υ was an arbitrary vector in nullT1, this shows that null Γ = {0}.
Thus Τ is injective (by 3.2), and hence Τ is invertible (by 3.21), as desired.
If и € V, then
u = Iu
= (ST)Ru
= S(TRu),
which shows that и € range S. Because и was an arbitrary vector in V, this
implies that range S = V. Thus V is surjective, and hence V is invertible
(by 3.21), as desired.
To prove the implication in the other direction, now suppose that both
S and Τ are invertible. Then
(st)(t-1 s-1) = sctt-^s-1
= SS~l
= 1
and
(T^S'^iST) = T'^S^SW
= T-iT
= 1.

30
Chapter 3. Linear Maps
Thus Τ 1S 1 satisfies the properties required for an inverse of ST. Thus
ST is invertible and (ST)~l = Γ_Ι5_Ι.
23. Suppose that V is finite dimensional and S, Τ € C(V). Prove that ST = I
if and only if TS = I.
Solution: First suppose that
ST = I.
Because / is invertible, the previous exercise implies that S and Τ are both
invertible. Multiply both sides of the equation above by T~l on the right,
getting
Now multiply both sides of the equation above by Τ on the left, getting
TS=I,
as desired.
To prove the implication in the other direction, simply reverse the roles
of S and Τ in the direction we have already proved, showing that if TS = I,
then ST = I.
24. Suppose that V is finite dimensional and Τ € C(V). Prove that Γ is a scalar
multiple of the identity if and only if ST = TS for every S e C{V).
Solution: First suppose that Τ = o.I for some a £ F. Let S € £(V).
Then
ST = S(al)
= aS m
= (al)S
= TS.
To prove the implication in the other direction, suppose now that ST =
TS for all S € £(V). We begin by proving that (υ, Τν) is linearly dependent
for every υ € V. To do this, fix υ € V, and suppose that (υ, Τν) is linearly
independent. Then {ν,Τν) can be extended to a basis (v,Tv,Ui,...,Un)
of V. Define S € C(V) by
S(av + bTv + cibi Η 1- CnU„) = bv.

31
Chapter 3. Linear Maps
Thus S{Tv) = ν and Sv = 0. Thus the equation S(Tv) = T{Sv) becomes
the equation ν = 0, a contradiction because (ν,Τυ) was assumed to be
linearly independent. This contradiction shows that (v, Tv) is linearly
dependent for every υ € V. This implies that for each υ € V \ {0}, there exists
av € F such that
Tv = avv.
To show that Γ is a scalar multiple of the identity, we must show that av
is independent of v. To do this, suppose v, w € V \ {0}. We want to show
that ay = aw. First consider the case where (υ, го) is linearly dependent.
Then there exists 6 € F such that w = bv. We have
aww = Tw
= T(bv)
= bTv
= b(avv)
= avw,
which shows that a„ = aw, as desired.
Finally, consider the case where (v, w) is linearly independent. We have
а„+ш(1> + w) = T{v + w)
= Tv + Tw
= avv +aww,
which implies that
(av+w — av)v + {av+w — aw)w = 0.
Because (υ, го) is linearly independent, this implies that av+w = av and
o-v+w = ciyj, so again we have Oy = dy,, as desired.
25. Prove that if V is finite dimensional with dim V > 1, then the set of non-
invertible operators on V is not a subspace of C(V).
Solution: Suppose that V is finite dimensional with dim V > 1. Let
η = dim V and let (ιί ,..., i>n) be a basis of V. Define S,T Ε C(V) by
S(diVi Η \- OnVn) = CLiVi
and

32 Chapter 3. Linear Maps
T{axv\ Η 1- OnVn) = a2V2 Η 1- α„υη.
Then 5 is not injective because Sv2 = 0 (this is where we use the hypothesis
that dim V > 1), and Τ is not injective because Tv\ = 0. Thus both S and
Τ are not invertible. However, S + Τ equals /, which is invertible. Thus the
set of noninvertible operators on V is not closed under addition, and hence
it is not a subspace of C(V).
Comment: If dim V = 1, then the set of noninvertible operators on V
equals {0}, which is a subspace of C(V).
26. Suppose η is a positive integer and aij € F for i, j = 1,..., n. Prove that
the following are equivalent:
(a) The trivial solution x\ = ■ ■ ■ = xn = 0 is the only solution to the
homogeneous system of equations
η
^2ahkxk = 0
Y^anikxk = 0.
k=i
(b) For every ci,...,c„ € F, there exists a solution to the system of
equations
η
22а1<кХк = Cl
22an>kX.k = c"·
k=\
Note that here we have the same number of equations as variables.
Solution: Define Τ € £(Fn) by
η η
Γ(χι,...,χ„) = (У^а^кХк, ■ · ·, У]ап,кХк)-
k=i k=i
Then (a) above is the assertion that Τ is injective, and (b) above is the
assertion that Τ is surjective. By 3.21, these two assertions are equivalent.

Chapter 4
Polynomials
1. Suppose m and η are positive integers with m < n. Prove that there exists
a polynomial ρ G Pn(F) with exactly m distinct roots.
Solution: Define peVn(1?) by
p{z) = {z- i)n~m+l(z-2)(z-Z)...(z-m).
Then ρ is a polynomial of degree η with exactly m distinct roots (which are
l,...,m).
2. Suppose that z\,..., z,„+i are distinct elements of F and w\,..., wm+\ € F.
Prove that there exists a unique polynomial ρ € Vm(F) such that
p(Zj) = Wj
for j = l,...,m + 1.
Solution: Define Γ: "Pm(F) -* Fm+1 by
Tp= (ρ{ζι),...,ρ(ζτη+ι)).
We need to prove that Τ is injective (which implies that at most one
polynomial ρ satisfies the condition required by the exercise) and surjective (which
implies that at least one polynomial ρ satisfies the condition required by the
exercise).
Clearly Γ is a linear map. If ρ e null T, then
p(Zl) = ■■■=p(zm+i) = 0,
which means that ρ is a polynomial of degree m with at least m + l distinct
roots, which means that ρ = 0 (by 4.3). Thus ρ is injective, as desired.
33

34
Chapter 4. Polynomials
Now
dim range Τ = dim "Pm(F) - dim null Τ
= (m + l)-0
= dimFm+1,
where the first equality comes from 3.4 and the second equality holds because
null Γ = {0}. The last equality above implies that range Τ = Fm+1. Thus
Τ is surjective, as desired.
Comment: Surjectivity of Τ can also be proved by using an explicit
construction. But linear algebra, specifically 3.4, gives us surjectivity easily
once we get injectivity.
3. Prove that if p, q € ^(F), with ρ φ 0, then there exist unique polynomials
s,reV{F) such that
q = sp + r
and degr < degp. In other words, add a uniqueness statement to the
division algorithm (4.5).
Solution: Suppose p, q € V(F), with ρ φ 0. We know from the division
algorithm (4.5) that there exist s,r € V(F), with degr < degp, such that
q = sp + r.
To prove that s and г are unique, suppose that s,f are in ^(F), with
degr < degp and
q = sp + f.
Subtracting the last two equations are rearranging, we have
(s — s)p = r — f.
The right side of the equation above is a polynomial whose degree is less
than degp. If s were not equal to s, then the left side of the equation above
would be a polynomial whose degree is at least degp. Thus we must have
s = s, which, from the equation above, implies that f = r. Thus the choices
of s and r were indeed unique.

35 Chapter 4. Polynomials
4. Suppose ρ € V{C) has degree m. Prove that ρ has m distinct roots if and
only if ρ and its derivative j/ have no roots in common.
Solution: First suppose that ρ has m distinct roots. Because ρ has
degree m, this implies that ρ can be written in the form
p(z) = c(z -\i)...(z- \m),
where X\,...,km are distinct. To prove that ρ and j/ have no roots in
common, we must show that p'(Xj) φ 0 for each j. To do this, fix j. The
expression above for ρ shows that we can write ρ in the form
p(z) = (z-kj)q(z),
where q is a polynomial such that q(Xj) φ 0. Differentiating both sides of
this equation, we have
p'{z) = {z-\j)q'{z)+q{z).
Thus
p'(Ai) = ϊ(λ,)
as desired.
To prove the other direction, we will proved the contrapositive, meaning
that we will prove that if ρ has less than m distinct roots, then ρ and p1
have at least one root in common. To do this, suppose that ρ has less than
m distinct roots. Then for some root λ of p, we can write ρ in the form
p(z) = (z-\)nq(z),
where η > 2 and q is a polynomial. Differentiating both sides of this
equation, we have
p'(z) = (z- X)nq'(z)+n(z - X)n-lq(z).
Thus ρ'(λ) = 0, and so λ is a common root of ρ and p7, as desired.
5. Prove that every polynomial with odd degree and real coefficients has a real
root.
Solution: Suppose that ρ is a polynomial with odd degree and real
coefficients. By 4.14, ρ is a constant times the product of factors of the form

36 Chapter 4. Polynomials
χ — λ and/or χ2 + ax + β, where λ, α, β € R. Not all the factors can be of
the form x2 + ax + β, because otherwise ρ would have even degree. Thus at
least one factor must be of the form χ — λ. Any such λ is a real root of p.
Comment: Here is another proof, using calculus but not using 4.14.
Suppose ρ is a polynomial with odd degree m. We can write ρ in the form
p(x) = ao+ a{x Η h amxm,
where Go, ·.., Ощ G R and am φ 0. Replacing ρ with —ρ if necessary, we
can assume that dm > 0. Now
/ \ m/ a0 , al , am-l , ч
Ρ(*) = * (^Γ + ^ΓΤ+··· + —+ a-)·
This implies that
lim p(x) = -co and lim p(x) = oo.
X—t—OO X-»00
The intermediate value theorem now implies that there is a real number λ
such that ρ(λ) = 0. In other words, ρ has a real root.

Chapter 5
Eigenvalues and Eigenvectors
1. Suppose Τ G C(V). Prove that if U\,..., Um axe subspaces of V invariant
under T, then U\-\ + Um is invariant under T.
Solution: Suppose U\,..., Um are subspaces of V invariant under T.
Consider a vector и Ε U\ + ■■■ + Um. There exist щ G U\,...,Um G Um
such that
и = щ Η 1- um.
Applying Τ to both sides of this equation, we get
Tu = Тщ + ■ ■ ■ + Trim.
Because each Uj is invariant under T, we have Тщ G U\,..., Tum G Um.
Thus the equation above shows that Tu G U\ -\ 1- Um, which implies that
U\ -i + Um is invariant under T.
2. Suppose Τ € C( V). Prove that the intersection of any collection of subspaces
of V invariant under Τ is invariant under Τ.
Solution: Suppose {UQ}Q^r is a collection of subspaces of V invariant
under T\ here Γ is an arbitrary index set. We need to prove that П<*ег UQ>
which equals the set of vectors that are in UQ for every α G Г, is invariant
under T. To do this, suppose и G Паег ^*· Then и G Ua for every α G Г.
Thus Tu G Ua for every α G Γ (because every Ua is invariant under T).
Thus Tu G Паег ^ai which implies that f\aer UQ is invariant under T.
3. Prove or give a counterexample: if U is a subspace of V that is invariant
under every operator on V, then U = {0} or U = V.
37

38
Chapter 5. Eigenvalues and Eigenvectors
Solution: We will prove that if U is a subspace of V that is invariant
under every operator on V, then U = {0} or U = V. Actually we will prove
the (logically equivalent) contrapositive, meaning that we will prove that
if U is a subspace of V such that U Φ {0} and Ό φ V, then there exists
Τ € C(V) such that U is not invariant under T. To do this, suppose U is a
subspace of V such that U φ {0} and U φ V. Choose и € U \ {0} (this is
possible because U φ {0}) and w € V \ U (this is possible because U φ V).
Extend the list (u), which is linearly independent because и Ф 0, to a basis
(u, «i,..., vn) of V. Define Τ € C(V) by
T(au + b\ v\ + ... bnvn) = aw.
Thus Tu = w. Because u€ U but w £ U, this shows that U is not invariant
under T, as desired.
4. Suppose that S, Τ € C(V) are such that ST = TS. Prove that null(T - XI)
is invariant under S for every λ € F.
Solution: Fix λ € F. Suppose υ G пи11(Г - λ/). Then
{Τ - XI){Sv) = TSv -XSv
= STv — XSv
= S(Tv - Xv)
= 0.
Thus Sv € nul\(T - XI). Hence null(r - XI) is invariant under S.
5. Define Τ € £(F2) by
T(w,z) = (z,w).
Find all eigenvalues and eigenvectors of T.
Solution: Suppose λ is an eigenvalue of T. For this particular operator,
the eigenvalue-eigenvector equation T(w,z) = X(w,z) becomes the system
of equations
ζ = Xw
w = Xz.
Substituting the value for ζ from the first equation into the second equation
gives w = X2w. Thus 1 = λ2 (we can ignore the possibility that w = 0

39 Chapter 5. Eigenvalues and Eigenvectors
because if w = 0, then the first equation above implies that ζ = 0). Thus
A = lorA = -l. The set of eigenvectors corresponding to the eigenvalue 1
is
{(го, го) : го € F};
The set of eigenvectors corresponding to the eigenvalue -1 is
{(го, -го) : го € F}.
6. Define Τ € £(F3) by
T{z\,zi,zz) = (2*2,0,5*3 )·
Find all eigenvalues and eigenvectors of T.
Solution: Suppose λ is an eigenvalue of T. For this particular operator,
the eigenvalue-eigenvector equation T{z\,zi,zz) = λ{ζι,ζι,ζζ) becomes the
system of equations
2*2 = Xz\
5*3 = λ*3·
If λ φ 0, then the second equation implies that zi = 0, and the first
equation then implies that z\ = 0. Because an eigenvalue must have a
nonzero eigenvector, there must be a solution to the system above with
*з φ 0. The third equation then shows that λ = 5. In other words, 5 is the
only nonzero eigenvalue of T. The set of eigenvectors corresponding to the
eigenvalue 5 is
{(0,0,*3):^eF}.
If λ = 0, the first and third equations above show that zi = 0 and *з = 0.
With these values for Z2,z$, the equations above are satisfied for all values
of z\. Thus 0 is an eigenvalue of T. The set of eigenvectors corresponding
to the eigenvalue 0 is
{(*i,0,0):*i €F}.
7. Suppose η is a positive integer and Τ € £(Fn) is defined by
T(xi,..., xn) = (χι Η \- xn,..., χι Η 1-1„);
in other words, Τ is the operator whose matrix (with respect to the standard
basis) consists of all 1 's. Find all eigenvalues and eigenvectors of T.

40
Chapter 5. Eigenvalues and Eigenvectors
Solution: Suppose λ is an eigenvalue of T. For this particular
operator, the eigenvalue-eigenvector equation Tx = Xx becomes the system of
equations
x\ + hin = λχι
χι + · · · + xn = Xxn-
Thus
λχι = ■■ · = λχη.
Hence either λ = 0 or x\ = · · · = xn.
Consider first the possibility that λ = 0. In this case all the equations in
the eigenvector-eigenvalue system of equations above become the equation
x\ + · · · + xn = 0. Thus we see that 0 is an eigenvalue of Τ and that the
corresponding set of eigenvectors equals
{(xb...,xn)€Fn:xi +···+χη = 0}.
Now consider the possibility that x\ = ■ · · = xn; let t denote the
common value of xi,... ,xn. In this case all the equations in the eigenvector-
eigenvalue system of equations above become the equation nt = Xt. Hence λ
must equal η (an eigenvalue must have a nonzero eigenvector, so we can take
t φ 0). Thus we see that η is an eigenvalue of Τ and that the corresponding
set of eigenvectors equals
{(xb...,Xn) € Fn :xi =··· =xn}·
Because the eigenvector-eigenvalue system of equations above implies
that λ = 0 or xi = ·■■ = xn, we see that Τ has no eigenvalues other
than 0 and n.
8. Find all eigenvalues and eigenvectors of the backward shift operator Τ €
£(F°°) defined by
T{Z\ ,Z2,Z3,...) = (Z2, Zz,... ).
Solution: Suppose λ is an eigenvalue of T. For this particular
operator, the eigenvalue-eigenvector equation Tz = Xz becomes the system of
equations

41 Chapter 5. Eigenvalues and Eigenvectors
zi = \z\
zz = λζ-ι
Zi = \zz
From this we see that we can choose z\ arbitrarily and then solve for the
other coordinates:
Zl = \Z\
Zz = λΖ2 = λ2 2ι
Z\ = kzz = A Zy
Thus each λ € F is an eigenvalue of Τ and the set of corresponding
eigenvectors is
{(го, Аго, λ2 го, λ3 го...) : го G F}.
9. Suppose Τ € C(V) and dim range Τ = к. Prove that Τ has at most к + 1
distinct eigenvalues.
Solution: Let Ai,...,Am be the distinct eigenvalues of T, and let
ui,...,um be corresponding nonzero eigenvectors. If \j φ 0, then
T(vj/Xj) = vj.
Because at most one of Ai,..., Am equals 0, this implies that at least m— 1 of
the vectors v\,..., vm are in range T. These vectors are linearly independent
(by 5.6), which implies that
77i — 1 < dim range Τ = к.
Thus τη < к + 1, as desired.
10. Suppose Τ G C{V) is invertible and A G F\{0}. Prove that A is an eigenvalue
of Τ if and only if ^ is an eigenvalue of T~l.
Solution: First suppose that A is an eigenvalue of T. Thus there exists
a nonzero vector υ G V such that
Tv = Xv.

42 Chapter 5. Eigenvalues and Eigenvectors
Applying T~l to both sides of the equation above, we get ν = XT~lv, which
is equivalent to the equation T~lv = ^v. Thus j is an eigenvalue of T~l.
To prove the implication in the other direction, replace Τ by T~l and λ
by j and then apply the result from the paragraph above.
11. Suppose S,T € C(V). Prove that ST and TS have the same eigenvalues.
Solution: Suppose that λ € F is an eigenvalue of ST. We want to
prove that λ is an eigenvalue of TS. Because λ is an eigenvalue of ST, there
exists a nonzero vector ν € V such that
{ST)v = Xv.
Now
(TS){Tv) = T(STv)
= T(Xv)
= XTv.
If Τυ φ 0, then the equation above shows that λ is an eigenvalue of TS, as
desired.
If Tv = 0, then λ = 0 (because S(Tv) = Xv) and furthermore Τ is
not invertible, which implies that TS is not invertible (by Exercise 22 in
Chapter 3), which implies that λ (which equals 0) is an eigenvalue oiTS.
Regardless of whether or not Tv = 0, we have shown that λ is an
eigenvalue of TS. Because λ was an arbitrary eigenvalue of ST, we have shown
that every eigenvalue of ST is an eigenvalue of TS.
Reversing the roles of S and T, we conclude that every eigenvalue of TS
is also an eigenvalue of ST. Thus ST and TS have the same eigenvalues.
12. Suppose Τ € C(V) is such that every vector in V is an eigenvector of T.
Prove that Γ is a scalar multiple of the identity operator.
Solution: For each ν € V, there exists α„ € F such that
Tv = CLyV.
Because TO = 0, we can choose Go to be any number in F, but for ν € V"\{0}
the value of av is uniquely determined by the equation above.
To show that Γ is a scalar multiple of the identity, we must show that av
is independent of ν for ν € V \ {0}. To do this, suppose v, w € V \ {0}. We
want to show that av = a^,. First consider the case where (v,w) is linearly
dependent. Then there exists b € F such that w = bv. We have

43 Chapter 5. Eigenvalues and Eigenvectors
OwW = Tw
= T(bv)
= bTv
= b(avv)
= OvW,
which shows that a„ = a™, as desired.
Finally, consider the case where (v, w) is linearly independent. We have
Ov+w(v + w) = T(v + w)
= Tv + Tw
= dvv + aww,
which implies that
(а„+ш - av)v + (α„+ω - aw)w = 0.
Because (v,w) is linearly independent, this implies that av+w = a^ and
dv+xu = аш, so again we have tiy = aw, as desired.
13. Suppose Τ € C(V) is such that every subspace of V with dimension
dimV- 1
is invariant under T. Prove that Γ is a scalar multiple of the identity
operator.
Solution: Suppose that Τ is not a scalar multiple of the identity
operator. By the previous exercise, there exists и € V such that и is not an
eigenvector of T. Thus (u, Tu) is linearly independent. Extend (u, Tu) to a
basis (it, Tu, v\,..., vn) of V. Let
U = span(u, vi,... ,vn).
Then U is a subspace of V and dimJ7 = dimV" — 1. However, U is not
invariant under Τ because и € U but Tu ^ U. This contradiction to our
hypothesis about Τ shows that our assumption that Τ is not a scalar multiple
of the identity must have been false.
14. Suppose S, Τ € C(V) and S is invertible. Prove that if ρ € V(F) is a
polynomial, then
p(STS~l) = Sp(T)S~l.
Solution: First suppose m is a positive integer. Then

44
Chapter 5. Eigenvalues and Eigenvectors
(STS~l)m = (STS-l)(STS-l)...(STS~l)
= ST(S-lS)T(S-lS)...(S~lS)TS-1
=STmS-\
which is our desired equation in the special case when p(z) = zm.
Multiplying both sides of the equation above by a scalar and then summing a finite
number of equations of the resulting form shows that p(STS~l) = Sp(T)S~l
for every polynomial ρ € V(F).
15. Suppose F = C, Τ € C(V), ρ € V(C), and α € С. Prove that α is an
eigenvalue of p(T) if and only if α = ρ(λ) for some eigenvalue λ of T.
Solution: First suppose that α is an eigenvalue of p(T). Thusp(T)-aI
is not injective. Write the polynomial p(z) — a in factored form:
p(z) - a = c(z - λι)... (ζ - km),
where c,Ai,...,Am € C. We can assume that с φ 0 (otherwise ρ is a
constant polynomial, in which case the desired result clearly holds). The
equation above implies that
p(T) -aI = c(T-klI)...(T- kmI).
Because p(T) - al is not injective, this implies that Τ — Xjl is not injective
for some j. In other words, some Xj is an eigenvalue of T. The formula
above for p(z) — a shows that p(kj) — a = 0. Hence α = p(kj), as desired.
For the other direction, now suppose that α = ρ(λ) for some eigenvalue
λ of T. Thus there exists a nonzero vector υ € V such that
Tv =■ kv.
Repeatedly applying Τ to both sides of this equation shows that Tkv =■ kkv
for every positive integer k. Thus
p(7> = p{k)v
= av.
Thus α is an eigenvalue of p(T).
16. Show that the result in the previous exercise does not hold if С is replaced
with R.

45 Chapter 5. Eigenvalues and Eigenvectors
Solution: Define Τ € £(R2) by T(x, y) = (-y, x). Define ρ € V(R) by
p(x) = x2. Then p(T) = T2 = —I, and hence -1 is an eigenvalue of p(T).
However, Τ has no eigenvalues (as we saw on page 78 of the textbook; the
point here is that eigenvalues are required to be real because we are working
on a real vector space), so there does not exist an eigenvalue λ of Γ such
that -1 =ρ(λ).
Of course there are also many other examples.
17. Suppose V is a complex vector space and Τ € C(V). Prove that Τ has an
invariant subspace of dimension j for each j = 1,..., dim V.
Solution: There is a basis {v\,..., udim v) with respect to which Τ has
an upper-triangular matrix (see 5.13). For each j = 1,..., dim V, the span
of {v\,..., vj) is a j-dimensional subspace of V that is invariant under Τ
(by 5.12).
18. Give an example of an operator whose matrix with respect to some basis
contains only O's on the diagonal, but the operator is invertible.
Solution: Let Τ € £(F2) be the operator whose matrix (with respect
to the standard basis) is
0 1'
1 0 "
Obviously this matrix has only O's on the diagonal, but Τ is invertible
(because TT — I, as is clear from squaring the matrix above).
Of course there are also many other examples.
Comment: This exercise and the next one show that 5.16 fails without
the hypothesis that an upper-triangular matrix is under consideration.
19. Give an example of an operator whose matrix with respect to some basis
contains only nonzero numbers on the diagonal, but the operator is not
invertible.
Solution: Define Τ € £(F2) to be the operator whose matrix (with
respect to the standard basis) is
Then T(1,0) = T(0,1) = (1,1), so Γ is not injective, so Γ is not
invertible, even though the diagonal of the matrix above contains only nonzero
numbers.

46
Chapter 5. Eigenvalues and Eigenvectors
Of course there are also many other examples.
20. Suppose that Τ € C{V) has dim V distinct eigenvalues and that S € C{V)
has the same eigenvectors as Τ (not necessarily with the same eigenvalues).
Prove that ST = TS.
Solution: Let η = dimV. There is a basis (υι,...,υη) of V
consisting of eigenvectors of Τ (see 5.20 and 5.21). Letting λχ,..., λη be the
corresponding eigenvalues, we have
Tvj = Xjvj
for each j. Each vj is also an eigenvector of S, so
Svj = a3-vj
for some acj € F.
For each j, we have
(ST)vj = S{Tvj) =■ kjSvj = ctjkjVj
and
{TS)vj = T(Svj) = ajTvj = ajXjvj.
Because the operators ST and TS agree on a basis, they are equal.
21. Suppose Ρ € £(V) and P2 = P. Prove that V = null Ρ θ range P.
Solution: First suppose и € null Ρ Π range P. Then Pu = 0, and
there exists w € V such that и = Pw. Applying Ρ to both sides of the
last equation, we have Pu = Ρ w = Pw. But Pu = 0, so this implies
that Pw = 0. Because и = Pw, this implies that и = 0. Because и was an
arbitrary vector in null Ρ Π range Ρ, this implies that null Ρ Π range Ρ = {0}.
Now suppose υ € V. Then obviously
υ = (υ - Ρυ) + Ρυ.
Note that Ρ{υ - Ρυ) = Ρυ - Ρ2υ = 0, so (υ - Ρυ) € null P. Clearly Ρυ €
range P. Thus the equation above shows that υ € null Ρ + range P. Because
υ was an arbitrary vector in V, this implies that V = null Ρ + range P.
We have shown that null Ρ Π range Ρ = {0} and V = null Ρ + range P.
Thus V = null Ρ θ range Ρ (by 1.9).

47 Chapter 5. Eigenvalues and Eigenvectors
22. Suppose V = U θ W, where U and W are nonzero subspaces of V. Find all
eigenvalues and eigenvectors of F\j,w·
Solution: Because V = U θ W, each vector υ G V can be written
uniquely in the form
υ = и + w,
where и G U and w G W. Recall that if υ is represented as above, then
Рцу/υ = u.
Suppose λ G F is an eigenvalue of Рцлу/. Then there exists a nonzero
vector ν € V such that Pu,wv = λυ. Writing this equation using the
representation of υ given above, we have и = X(u + w). Thus
(1 - X)u-Xw = 0.
Because V = U θ W, if 0 is written as the sum of a vector in U and a vector
in W, then both vectors must be 0. Thus the equation above implies that
(1 - X)u = Xw = 0. Because и and w are not both 0 (because υ φ 0), this
implies that λ = 1 or λ = 0.
For ν G V with representation as above, the equation Puy/v = 0 is
equivalent to the equation и = 0, which is equivalent to the equation υ = w,
which is equivalent to the statement that υ G W. This means that 0 is an
eigenvalue of P(j,w (because W is a nonzero subspace of V) and that W
equals the set of eigenvectors corresponding to the eigenvalue 0.
For ν G V with representation as above, the equation P(j,wv — ν is
equivalent to the equation υ = и, which is equivalent to the statement that
ν G U. This means that 1 is an eigenvalue of P(j,w (because U is a nonzero
subspace of V) and that U equals the set of eigenvectors corresponding to
the eigenvalue 1.
23. Give an example of an operator Τ € £(R4) such that Τ has no (real)
eigenvalues.
Solution-. Define Τ e £(R4) by
Suppose λ G R. For this particular operator, the eigenvalue-eigenvector
equation Т(х\,Х2,хз,хл) = λ(χι,Χ2>^3ι^4) becomes the system of
equations

48 Chapter 5. Eigenvalues and Eigenvectors
—£2 = λχι
Xl = XX2
—χ\ = Ахз
хз = λχ4.
Multiplying together the first two equations and also multiplying together
the last two equations gives -X1X2 = λ2χΐΧ2 and —X3X4 = А2хзХ4. If either
xi or X2 does not equal 0, then the first two equations show that neither of
xi, X2 equals 0. Similarly, if either X3 or X4 does not equal 0, then the last
two equations show that neither of X3, X4 equals 0. Thus if λ is an eigenvalue
of T, then there is a solution to the system of equations above with X1X2 φ 0
or X3X4 φ 0. Either way, we conclude that -1 = λ2, which is impossible for
any real number λ. Thus Τ has no real eigenvalues.
24. Suppose V is a real vector space and Τ € C(V) has no eigenvalues. Prove
that every subspace of V invariant under Τ has even dimension.
Solution: Suppose U is a subspace of V that is invariant under T.
Thus T\u € C{U). If dim C/ were odd, then T\u would have an eigenvalue
λ € R (by 5.26), so there would exist a nonzero vector и € U such that
Т\ии = ku.
Obviously this would imply that Tu = Xu, which would imply that λ is
an eigenvalue of T. But Τ has no eigenvalues, so dirnC/ must be even, as
desired.

Chapter 6
Inner-Product Spaces
1. Prove that if x, у are nonzero vectors in R , then
(x,y) = ||z||||]/||cos0,
where θ is the angle between χ and у (thinking of χ and у as arrows with
initial point at the origin). Hint: draw the triangle formed by x, y, and x-y;
then use the law of cosines.
Solution: Suppose that x,y are nonzero vectors in R2 and θ is the
angle between χ and y. Consider the triangle formed by x, y, and χ — у:
The law of cosines states that
iix-y|i2 = iNi2 + iiyii2-2iixiniyiicos0.
As usual, we compute the norm of a vector squared by taking the inner
product of the vector with itself:
l|s-y||2 = (x-y,x-y)
= (x, x) - (x, y) - (y, x) + (y, y)
= \\x\\2 + \\y\\2-2(x,y).
49

50
Chapter 6. Inner-Product Spaces
Substitute the last expression for ||r — y\\2 into the left side of the law of
cosines, obtaining
||x||2 + ||y||2 - 2(x,y) = \\xf + \\yf - 2\\x\\\\y\\ cos*.
Now subtract ||x||2 + ||y||2 from both sides of the equation above, and then
divide both sides by —2, obtaining
<*,!/> = IMIIIl/ll a**·
2. Suppose u, ν € V. Prove that (u, v) = 0 if and only if
Nl<||u + a«||
for all a € F.
SOLUTION: First suppose that (u,v) = 0. Let α € F. Then u,av are
orthogonal. The Pythagorean theorem thus implies that
||u + H|2 = |H2 + ||H|2
> INI2·
Taking square roots gives ||u|| < ||ιι + αυ||, as desired.
To prove the implication in the other direction, now suppose that ||u|| <
||ιι + αυ|| for all α € F. Squaring this inequality, we get
H|2<||u + H|2
= (u + av, и + αν)
= (и, и) + (и, αν) + (αν, и) + (αν, αν)
= \\u\\2 + a(u,v)+afcv) + \a\2\\v\\2
= \\u\\2 +2Bea(u,v) + \a\2\\v\\2
for all α € F. Thus
-2Rea(u,v) < \a\2\\v\\2
for all α € F. In particular, we can let α equal — t(u, v) for t > 0.
Substituting this value for α into the inequality above gives
2i|(u,u>|2<i2|<u,t;>|2H2
for all t > 0. Divide both sides of the inequality above by t, getting

51 Chapter 6. Inner-Product Spaces
2|<u,u>|2<i|<u,t;>|2H|2
for all t > 0. If υ = 0, then (u,v) = 0, as desired. If ν φ 0, set t equal to
1/||υ||2 in the inequality above, getting
2\{u,v)\2 <\{u,v)\\
which implies that (u, v) = 0, as desired.
3. Prove that
(£<*)'*(έν)(έ¥)
.7=1 i=i i=i J
for all real numbers αϊ,..., α„ and bi,..., bn.
SOLUTION: Suppose αϊ,... ,On,bi,... ,bn € R. Using the usual inner
product on Rn, we have
(EaA-)2 = (E(^i)(Vv7))s
3=1 .7 = 1
= ((αϊ, y/2a2,..., Vnan), (bi, &2/V2,..., bn/y/n))
< ||(a,, V2a2,..., v^rOflKbb&a/v^,..., bn/Vn)\\2
= (Σ^-2)(Σ^·),
i=i J=i
where the inequality above comes from the Cauchy-Schwarz inequality.
Suppose u, ν € V are such that
||u||=3, ||и + «|=4, ||u-«||=6.
What number must ||υ|| equal?
Solution: Prom the parallelogram equality, we have
|M|2 = \\u + v\\2 + \\u-v\\2-2\\u\\2
_ 16 + 36- 18
2
= 17.

52
Chapter 6. Inner-Product Spaces
Thus ||v|| = ν/Ϊ7.
5. Prove or disprove: there is an inner product on R2 such that the associated
norm is given by
||(xi,x2)|| = |xi| + |x2|
for all (xi,X2) € R2·
SOLUTION: We will show that there does not exist an inner product on
R2 such that the associated norm is given by the formula above by showing
that the parallelogram equality is violated. Let
и = (3,2) and υ =(1,3).
Then
u + v = (4., 5) and u — υ = (2, — 1).
Using the formula above, we then have
||u + u||2 + ||u-t;||2 = 81+9
= 90
and
2(N|2 + H2) = 2(25 + 16)
= 82.
Thus the parallelogram equality fails, as desired.
6. Prove that if V is a real inner-product space, then
Цц + г,Ц2-1|и-г;112
{и, ν) =
for all и, υ € V.
Solution: Suppose V is a real inner-product space and u, ν € V. Then
\\u + v\\2 — ||u - v\\2 _ (u + v, и + ν) — (и — ν, и — ν)
4 = 4
[[и»2 + 2(ti, υ) + HI2 - ([[и»2 - 2(ц, υ) + IN2)
4
_ 4(u, ν)
= 4~
= (u,v),

53 Chapter 6. Inner-Product Spaces
as desired.
7. Prove that if V is a complex inner-product space, then
. . ||u + t;||2- ||u - v||2 + ||u + tt;||2i - ||u - гг;||2г
{и, ν) =
for all и, υ € V.
Solution: Suppose V is a complex inner-product space and u,v € V.
Then
Ци + υΙΙ2 = (u + v,u + v)
= \\u\\2 + (u,v) + (v,u) + \\v\\2
and
— \\u — v\\2 = —(u — v,u — v)
= -\\u\\2 + (u,v) + (v,u)-\\v\\2
and
i\\u + iv\\2 = i(u + iv, и + iv)
= i\\u\\2 + (u,v)-(v,u)+i\\v\\2
and
—i\\u — iv\\2 = —i(u — iv,u — iv)
= -г||и||2 + (и, ν) - (υ,и) - i\\vf.
Adding the four equations, we have
||и + г;||2 - ||ιι-υ||2 -(-гЦи + гг;^ - i||u - iv\\2 = 4.(u,v),
as desired.
8. A norm on a vector space U is a function || ||: U —* [0, oo) such that ||u|| = 0
if and only if и = 0, \\au\\ = |a|||u|| for all a € F and all u € 17, and
II" + υ|| < ||u|| + ||υ|| for all u,v € U. Prove that a norm satisfying the
parallelogram equality comes from an inner product (in other words, show
that if || || is a norm on U satisfying the parallelogram equality, then there
is an inner product ( , ) on U such that ||u|| = (u,u)1/2 for all и € 17).

54 Chapter 6. Inner-Product Spaces
Comment: This is among the hardest exercises in the book. Instructors
may want to simplify this exercise slightly by allowing students to consider
only the case where F = R.
Solution·. Suppose that U is a vector space and || || is a norm on U
satisfying the parallelogram equality. We want to find an inner product (, )
on U such that ||u|| = (u,u)1/2 for all и € U.
First consider the case where F = R. For u, υ € U, define (u, v) by
\\u+v\\2-\\u-v\\2
{u,v) = .
This definition is motivated by Exercise 6 of this chapter, which gives a
formula for the inner product in terms of norms.
For и € U we have
\\u + u\\2-\\u-u\\2
(u, u) = -
Pull2 - 11Q112
4
= INI2-
Thus ||u|| = (u, u)1/2, as desired. However, we still must show that ( , )
satisfies the properties required of an inner product.
Because (u, u) = \\u\\2 (as shown above), we have (u, u) > 0 for all u € 17,
with equality if and only if и = 0; these properties follow from the properties
of a norm. Thus ( , ) satisfies the positivity and definiteness properties
required of an inner product.
To prove that ( , ) is additive in the first slot, let u, v,w € U. Then
4((u + v, w) — (u, w) - (v, to))
= \\u + v + io||2 -\\u + v- io||2 - ||u + w||2
+ ||u - wf - \\v + w\\2 + \\v - wf
= \\u + v + w\\2 + (||u - w\\2 + \\v - HI2)
-\\u + v- w\\2 - (||t* + wf + \\v + wf),
where the first equality comes from the definition of ( , ). In the last
equality above, the parentheses indicate groupings to which we will apply

55
Chapter 6. Inner-Product Spaces
the parallelogram equality, which asserts that the sum of the norms squared
of two vectors x,y € U can be computed from the formula
Μ|.+μ. = Ιξ+ϊΙ! + 1-ιϊ1!.
Applying the parallelogram equality to the two terms in parentheses above
(take χ = и — го, у = ν — го for the first sum in parentheses, then χ =
u + w,y = υ + w for the second term in parentheses) gives
4((u + v, w) — (u, го) — {υ, го))
2 ||u-Ы> - 2го||2 ||u-i>||2
\u + v — го
2 2
2 ||и + г; + 2го||2 ||и-г;||2
/ιι н2 и н2ч llu + υ — 2го||2
= (||u + v + wf + HI2) + ^ ϋ-
/н n2 ιι н2\ ||u + г; +2го||2
-(||и + г;-го||2 + ||го||2)-^^ U-.
Applying the parallelogram equality to the two terms in parentheses above
(take χ = u + г; + го, у = го for the first sum in parentheses, then χ =
u + v — w,y = w for the second term in parentheses) gives
4((u + v, w) - (u, w) — (v, го))
||u + v + 2го||2 ||и + г;||2 ||u + υ - 2го||2
2 + ~^_ + 2
||и + г;||2 ||ц + ν - 2гоЦ2 ||u + v + 2го||2
= 0.
Thus
(u + v, w) = (и, го) + (ν, го),
completing the proof that ( , ) is additive in the first slot.
To prove that ( , ) is homogeneous in the first slot, let u,v € U. If η is
a positive integer, then

56 Chapter 6. Inner-Product Spaces
(nu,v) = (u + · · · + u,v)
η times
= (u,v) -\ \-(u,v)
N ν '
η times
= n(u,v),
where the second equality comes from additivity in the first slot, which we
have already verified. Replacing it with u/n in the equality above gives
(u,v) = n(u/n,v), which implies that
(-») = ~(u,v).
η η
Let 77i be another positive integer, and replace и with mu in the equality
above, getting
τη . 1
(—u, v) = —lmu,v)
η η
771. .
= — {u,v},
η
where the second equality holds because we have already shown that ( , )
is homogeneous in the first slot with respect to positive integers. We have
now shown that (, ) is homogeneous in the first slot with respect to positive
rational numbers.
Prom the definition of ( , ), we have
(-u,v) =
\\-u + v\\2
\\u + v\\2
-(u,v).
—
4
4
||-u
ll«-
-«II2
«II2
Combining this with the result from the previous paragraph, we can now
conclude that (, ) is homogeneous in the first slot with respect to all rational
numbers.
Now suppose that λ € R. There exists a sequence η, гг,... of rational
numbers such that Ηπίπ-,α, r„ = λ. Thus

57 Chapter 6. Inner-Product Spaces
X(u, v) = lim rn(u,v)
n—»oo
= lim (rnu,v)
= Шпд-- '2 2
|r„u + t;r- \\rnu-v\
In the next paragraph we will show that limn_*oo |jr„it + г>|| = ||Au + v|| and
ϋπΐη-,οο ||rnu —υ|| = ||Au —υ||. Combining this with the last equation above
we can conclude that
\\Xu+v\\2-\\ku-v\\2
X(u, v) =
= (Au,«),
which will complete the proof that ( , ) is homogeneous in the first slot.
If x, у € U, then
Ы = \\у + (х-у)\\
<Ы\ + \\х-у\\
and thus
INI-||y||<l|x-y||-
Interchanging the roles of χ and y, we get
IMI-||*|| < II*-y||.
Because |||x|| - \\y\\\ equals ||x|| - \\y\\ or ||y|| - ||x||, we can now conclude
that
|IMI-llvll|<ll*-vl|.
With χ = rnu + ν and у = ku + υ, this inequality gives
|||r„u + v||-||Au + v||| < ||r„u - Au||
= |r„-A|H|.
Because Ηπίπ-,α, r„ = λ, this shows that
lim ||r„u + v|| = ||λ« + υ||.
n—*oo
Replacing ν with —v, we have

58 Chapter 6. Inner-Product Spaces
lim ||rnu - υ|| = ||Au —υ||.
η—»oo
The last two equations are the promised ingredients that were needed for
the proof that ( , ) is homogeneous in the first slot.
Finally, we must show that (u, v) = (v,u) (recall that we are considering
the case where F = R). This last step is easy:
(u, v) =
||u + t;||2-||u-
4
ll« + u||2-||t;-
-Hl2
-«||2
4
= (v,u).
This completes the proof that ( , ) is an inner product when F = R. Whew!
Now consider the case where F = C. For u,υ € U, define (u,v) by
(u, v) =
llu + υ||2 - llu - v||2 + llu + iv\\2i - llu - iv\\2i
4
This definition is motivated by Exercise 7 of this chapter, which gives a
formula for the inner product in terms of norms.
For и € U we have
(u, u) =
\\u + u||2 - ||u - u||2 + ||u + ttt||2z - ||u - iu||2i
4
||2и||2 + |1 + г|2Н|2г-|1-г-|2Н2г
4
4||и||2+2|Ы|2г-2|Ы|2г
= IMI2.
Thus ||u|| = (u, u)1/2, as desired. However, we still must show that ( , )
satisfies the properties required of an inner product.
Because (u, u) = \\u\\2 (as shown above), we have (u, u) > 0 for all и € U,
with equality if and only if и = 0; these properties follow from the properties
of a norm. Thus ( , ) satisfies the positivity and definiteness properties
required of an inner product.
For convenience, let's define ( , )r by
(«,f>R =
||и + г;||2-||и-г;||2

59 Chapter 6. Inner-Product Spaces
Here the subscript R reminds us that (, ) r was the inner product we defined
when considering the case F = R. Now we are assuming that F = C, but
( , )r is still well defined. Note that
(u,v) = (u,i>)r + (u,iv)Ri.
We have already proved that ( , )r is additive in the first slot, and now
we use that information. Let u, v, w € U. Then
(u + v, w) = (u + v, w)n + (u + v, iw)ni
= (u,w)R + (v,w)K + (u,iw)Ki + (v,iw)Ki
= {(u, w)R + (u, iw)Ki) + ((v, w)R + (v, iw}ni)
= (u,w) + (υ, го).
Thus ( , ) is additive in the first slot.
To prove that (, ) is homogeneous in the first slot, let u, υ € U. If λ € R,
then
(\u,v) = (λιι, v)r + (Au, £i>)rz
= A(u,v)R + X(u,iv)Ki
= X(u,v),
where we have used the homogeneity of ( , )r in the first slot. The last
equation above shows that ( , ) is homogeneous in the first slot with respect
to all real numbers. We must still extend this result to complex numbers.
Note that
.. . ||t'u + i>||2 — \\iu — i>||2 + Цш + гирг — \\iu — iv\\2i
(ги, υ) =
_ \\г(и + г;)||2г - \\г(и - v)\\2i - \\г(и + iv)\\2 + |К(ц - гг;)||2
_ ||ц + г> ||2г - ||ц - г> ||2г - ||ц + tf||2 + ||u - if||2
4
= i(u,v).
Combining this result with additivity and homogeneity with respect to real
numbers, we get that
((a + bi)u, v) = (a + bi)(u, v)

60 Chapter 6. Inner-Product Spaces
for all a, b € R. In other words, ( , ) is homogeneous in the first slot with
respect to all complex numbers.
Finally, we must show that (u,v) = (v,u). This last step is easy:
llu + v\\2 - llu - v\\2 + llu + iv\\2i - llu - iv\\2i
(u, υ) =
4
\\υ + u\\2 - \\υ - u\\2 + \\i{-iu +.«)||2t - ||(-i)(w + v)\\2i
4
llv + u\\2 - \\v - u\\2 + \\v + iu\\4 - llv - iu\\2i
= (υ, it).
This completes the proof that ( , ) is an inner product when F = C.
Suppose η is a positive integer. Prove that
/1 sinx sin2x sinnx cosx cos2x cosnx\
is an orthonormal list of vectors in C[—π, π], the vector space of continuous
real-valued functions on [—ж, ж] with inner product
(f,9) = / f{x)g(x)cb:.
J -τ
Comment: This orthonormal list is often used for modeling periodic
phenomena such as tides.
Solution: First we need to show that each element of the list above
has norm 1. This follows easily from the following formulas:
/
/
4, , 2jt — s\n2jt
(sin jt)2dt= Aj J
(cos jtfdt
2 2jt + sin 2jt
4j
Next we need to show that any two distinct elements of the list above
are orthogonal. This follows easily from the following formulas, valid when
ЗФк:

61
Chapter 6. Inner-Product Spaces
/
/
/
(sin jf) (sin kt)dt =
js'm(j — k)t + ks'm(J — k)t — jsin(j + k)t + ks\n(j + k)t
20' - k)U + k)
(sin j t) (cos kt) dt =
j cos(j — k)t + kcos(j — k)t + j cos(J + k)t — kcos(j + k)t
2(k-j)(j + k)
(cos j t) (cos kt) dt =
js\n(J — k)t + fcsin(j — k)t + js'm(j + k)t — fesin^" +k)t
(sin jf) (cos jt) dt =
W-k){j + k)
(cos jt)2
2i
10. On V2(R), consider the inner product given by
(P> Я) = / P(x)q(x) dx-
Jo
Apply the Gram-Schmidt procedure to the basis (Ι,χ,χ2) to produce an
orthonormal basis of ^(R-)·
Solution: Applying the Gram-Schmidt procedure to (1, x, x2) produces
(using elementary calculus and some arithmetic) the following orthonormal
basis of V2(R):
(1, V3(-l + 2x),\/5(l - 6x + 6x2)).
11. What happens if the Gram-Schmidt procedure is applied to a list of vectors
that is not linearly independent?
Solution: Suppose (v\,..., vm) is a linearly dependent list of vectors
in V.
livi = 0, then at the first step of applying the Gram-Schmidt procedure
to (v\,..., vm) we will be dividing by 0 when trying to set ei = υι/||υι||.
If v\ Φ 0, then by the linear dependence lemma (2.4), some Vj is in
span(ui,.. ,,Vj-i); here we choose j to be the smallest positive integer
with this property. If we apply the Gram-Schmidt procedure to produce
(ei,..., ej-i) at the end of step j, then
span(ui,... ,Vj-X) = span(eb ... ,ej-i).

62
Chapter 6. Inner-Product Spaces
Thus Vj € span(ei,..., e_,_i). By 6.17, this implies that
Vj = (vj,ei)ei + · · · + (vj, ej_i)ej_i.
Thus the Gram-Schmidt formula 6.23 for ej includes a division by 0, which
is not allowed.
12. Suppose V is a real inner-product space and (v\,..., vm) is a linearly
independent list of vectors in V. Prove that there exist exactly 2m orthonormal
lists (ei,..., em) of vectors in V such that
span(ui,..., Vj) = span(ei,..., e_,·)
for all j € {1,.. .,m}.
Solution: For j = 1, the condition above states that
span(ui) = span(ei).
Because there are only two vectors in span(ui) with norm 1 (these two
vectors are υι/||υι|| and —vi/|jvi ||), we have only these two choices for e\.
Now suppose that j > 1 and that an orthonormal list (ei,..., e_j_i) has
been chosen such that
span(ui ,...,Vj-i) = span(ei,..., e,-_i).
The Gram-Schmidt procedure produces e_, € V such that (ei,...,ej) is an
orthonormal list and
span(ui,..., Vj) = span(ei,..., ej).
Suppose ej' € V is another vector with these properties, meaning that
(ei,..., ej-i, ej) is an orthonormal list and
span(ui,... ,Vj) = span(ei,... ,ej_i,e/).
The last two equations show that span(ei,..., ej-i,ej') = span(ei) · · · ι ey)-
In particular, ej' € span(ei,... ,ej), which implies that
€/ = (β/ι ei>ei + · · · + (ej', ej)ej
= {ej , ej)ej,
where the first equality comes from 6.17 (with span(ei,..., ej) replacing V
and j replacing n) and the second equality holds because (ei,... ,e_j_i,e/)

63
Chapter 6. Inner-Product Spaces
is an orthonormal list. Taking norms of both sides of the last equation, and
recalling that ej and e/ both have norm 1, we see that |(e/,e_/)| = 1. Thus
(ej',ej) = 1 or (е/,е/} = —1. Hence the last equation above implies that
e/ = ej or e/ = —ej.
We have shown that there are exactly two possible choices for each ej.
As j ranges from 1 to m, this gives us exactly 2Ш possible choices for
(βι,.-.,βπ,).
13. Suppose (ei,...,em) is an orthonormal list of vectors in V. Let υ € V.
Prove that
IMI2 = K«,ei>|a + ... + |<«iem>|2
if and only if υ € span(ei,..., em).
Solution: Extend (βι,.,.,βπ,) to an orthonormal basis (βι,.,.,βη)
of V. Then
ν = (v, ei)ei -i \-(v, en)e„
and
N2 = |(t;1e1>|2 + -.. + |(t;)en>|2;
see 6.17. Prom the last equation, we see that
1Н12 = |<»,е1>|а + - + |<»1ет>|а
if and only if (υ,β,η+ι) = · · - = (υ, e„) = 0. From the first equation above,
this happens if and only if
υ = (v, ei)ei + - - · + (v, eTO>em,
which happens if and only if υ € span(ei,..., em).
14. Find an orthonormal basis of 7*2(R) (with inner product as in Exercise 10)
such that the differentiation operator (the operator that takes ρ to p7)
on 7b(R) has an upper-triangular matrix with respect to this basis.
Solution: Because l' = 0, x' = 1, and (x2)' = 2x, the differentiation
operator on 7*2(R) has an upper-triangular matrix with respect to the basis
(Ι,χ,χ2). However, (Ι,χ,χ2) is not an orthonormal basis. But, as can be
seen from the proof of 6.27, if the Gram-Schmidt procedure is applied to this

64 Chapter 6. Inner-Product Spaces
basis, we will get an orthonormal basis with respect to which the
differentiation operator has an upper-triangular matrix. As we saw in Exercise 10 of
this chapter, the Gram-Schmidt procedure applied to (Ι,χ,χ2) gives
(1, V3(-l + 2x),\/5(l - 6x + 6x2)).
which is our desired orthonormal basis of ^(R)·
15. Suppose U is a subspace of V. Prove that
dimJ7-L = dimVr-dimJ7.
Solution.- Prom 6.29, we know that
V = U@UL.
Thus by Exercise 17 in Chapter 2, we have
dim V = dim U + dim J7X,
which implies that dim J7X = dim V — dim U.
16. Suppose U is a subspace of V. Prove that UL = {0} if and only if U = V.
Solution: From 6.29, we know that
v = ueu-L.
This clearly implies that UL = {0} if and only if U = V.
17. Prove that if Ρ € C(V) is such that P2 = Ρ and every vector in null Ρ is
orthogonal to every vector in range P, then Ρ is an orthogonal projection.
Solution: Suppose Ρ € C{V) is such that Ρ2 = Ρ and every vector
in null Ρ is orthogonal to every vector in range P. Let U = range P. We
will show that Ρ equals the orthogonal projection Рц. То do this, suppose
υ € V. Then
υ = Pv + (v- Pv).
Clearly Pv € range Ρ = U. Also, P(v — Pv) = Pv — P2v = 0, which means
that ν — Pv € null P. Thus ν — Pv is orthogonal to every vector in U. In
other words, ν — Pv € J7X. Thus the equation above writes ν as the sum of
a vector in U and a vector in UL. In this decomposition, the vector in U
equals, by definition, Pyv. Hence Pv = Puv, as desired.

65 Chapter 6. Inner-Product Spaces
18. Prove that if Ρ G C(V) is such that P2 = Ρ and
\\Pv\\ < imi
for every ν G V, then Ρ is an orthogonal projection.
Solution: Suppose и G range Ρ and w g null P. If we can show that
(u,w) = 0, then by the previous exercise we can conclude that Ρ is an
orthogonal projection.
Because и G range P, there exists u' G V such that
и = Pu'.
Applying Ρ to both sides of this equation, we have
Pu = PV
= Pu'
= u.
Because w G null P, this implies that
P(u + aw) = и
for every α G F. Thus
\\u\\2 = \\P(u + aw)\\2
< ||u + ato||2
for every a G F, where the second line follows from our hypothesis that
||Ρυ|| < ||υ|| for every υ € V. The inequality above implies (see Exercise 2
of this chapter) that (u, w) = 0, as desired.
19. Suppose Τ G £>{V) and U is a subspace'of V. Prove that U is invariant
under Τ if and only if РцТРц = ТРи.
Solution: First suppose that U is invariant under T. Let υ G V.
Then P(jv G U and hence T(Puv) € U (because U is invariant under T).
Thus Pu(T(Puv)) = T(P(jv). Because υ was an arbitrary vector in V, this
implies that РцТРи = ТРу, as desired.
To prove the implication in the other direction, now suppose that
PuTPu = ТРи.

66 Chapter 6. Inner-Product Spaces
Suppose и € U. Then P(ju = u, so applying both sides of the equation
above to и gives P(j(Tu) = Tu, which implies that Tu € U. Because и was
an arbitrary vector in U, this implies that Τ is invariant under U, as desired.
20. Suppose Τ € C{V) and U is a subspace of V. Prove that U and UL are
both invariant under Τ if and only if РцТ — ТРц.
Solution: First suppose that U and J7X are both invariant under T.
By the previous exercise, this implies that
PuTPu = TPu
and
Ρί/χΓΡί/χ=ΓΡί/χ.
But Рц± = I — P(j, so the last equation becomes
(/ - Pu)T(I - Pu) = T(I - Pv).
Expanding both sides of the equation above and rearranging terms, we get
ΡυΤΡυ = PuT.
Combining this with the first equation above, we get РцТ = TPu, as desired.
To prove the implication in the other direction, suppose now that
ΡυΤ = ΤΡυ.
Then
PuTPu = (ΡυΤ)Ρυ
= (TPu)Pu
= TPu2
= TPu,
which implies (by the previous exercise) that U is invariant under T, as
desired. Also,
Ρί/χΓΡί/χ = ((/-Ρί/)Γ)Ρί/χ
= {Τ-ΡυΤ)Ρυ±
= (Τ-ΤΡυ)Ρυ±
= T(l - Ρυ)Ρυ±
= TPu J
= ΤΡυ±,

67
Chapter 6. Inner-Product Spaces
which implies (by the previous exercise) that £/ is invariant under T, as
desired.
21. In R4, let
tf = span((l, 1,0,0), (1,1,1,2)).
Find и &U such that \\u - (1,2,3,4)|| is as small as possible.
Solution: First we find an orthonormal basis of U by applying the
Gram-Schmidt procedure to ((1,1,0,0), (1,1,1,2)), getting
«'(js-js·0·0)
e2 = (0·0·^'^)-
Thus with ei,62 as above, (еьбг) is an orthonormal basis of U. By 6.36
and 6.35, the closest point и € U to (1,2,3,4) is
((l,2,3,4),ei)ei + ((l,2,3,4),e2)e2,
which equals
/3 3 11 22\
V2'2' 5' δ/'
22. Find ρ e P3(R.) such that p(0) = 0, ?/(0) = 0, and
f |2 + 3r-p(x)|2iix
Jo
is as small as possible.
Solution: Define an inner product on ^(R) by
(/. 9)= I f(x)9(x) <&·
Jo
Let q(x) = 2 + 3x, and let
U = {ρ € V3(K) : p(0) = 0,p'(0) = 0}.
With this notation, our problem is to find the closest point ρ € U to q. To
do this, first we find an orthonormal basis of U.

68
Chapter 6. Inner-Product Spaces
A polynomial ρ satisfying p(0) = 0, p'(O) = 0 has constant term 0 and
first degree term also equal to 0. Thus a basis of U is
(x2,x3).
Apply the Gram-Schmidt procedure to this basis, getting
ei = VEx2
e2 = V7(-5x2 + 6x3).
Thus with е\,в2 as above, (еьвг) is an orthonormal basis of U. By 6.36
and 6.35, the closest point ρ € U to q is given by the formula
p= (q,ei)ei +(q,e2)e2.
A short computation now shows that
p(x) = 24x2 - —x3.
23. Find ρ € Vs(R) that makes
/| sin χ - p(x)|2dx
-7Γ
as small as possible. (The polynomial 6.40 is an excellent approximation
to the answer to this exercise, but here you are asked to find the exact
solution, which involves powers of ж. A computer that can perform symbolic
integration will be useful.)
Solution: Let C[-ir, π] denote the real vector space of continuous real-
valued functions on [—π, π] with inner product
(f,g)= Г f(x)g(x)dx.
J—π
Let υ € C[—π, π] be the function defined by v(x) = sinx. Let U denote the
subspace of C[-n, π] consisting of the polynomials with real coefficients and
degree at most 5. We need to find ρ € U such that ||υ — p\\ is as small as
possible.
First find an orthonormal basis of U by applying the Gram-Schmidt
procedure (using the inner product above) to the basis (l,x,x2,x3,x4,x5)
of U, producing the orthonormal basis (61,62,63,64,65,ее), where

69 Chapter 6. Inner-Product Spaces
ei
v/2^'
62 ~ π·3/2 '
ез
e4 =
e5 =
3/2 '
y/Ifr2 - 3x2)
2тг5/2
^/f (Зтг2* - 5x3)
2^72 '
3(3тг4 - 30тг2х2 + 35x4)
8V2tt9/2
у/Ц{\ЪжАх - 70тг2х3 + 63х5)
66 8тг»/2
Now compute Р{/*> using 6.35 (with m = 6), getting
105(1485 - 1537Г2 + тг4) 315(1155 - 125тг2 + тг4) 3
PuV= 8^ x ^ x
693(945 - 1057Γ2 + π4) 5
+ 8тг>° Х ■
Finally, 6.36 and the discussion following 6.42 show that the function above
is the one we seek.
24. Find a polynomial q € V2 (R) such that
1 fl
for every ρ G ^(R).
Solution: We will need an orthonormal basis of ^(R), where the inner
product of two polynomials in ^(R) is defined to be the integral from 0 to
1 of the product of the two polynomials. An orthonormal basis of ^(R)
was already computed in Exercise 10 of this chapter. Specifically, let
ei(x) = 1
e2(x) =V3(-l+2x)
e3(x) = \/5(l-6x + 6x2).

70
Chapter 6. Inner-Product Spaces
Then (61,62,63) is an orthonormal basis of ^(R).
Define a linear functional φ on T-^R) by
φ(ρ) = ρ{^)·
We seek q e 7MR) such that φ(ρ) = (p,q) for every ρ € Яг (R)· By the
formula given in the proof of 6.45, we have
q = <p(ei)ei + v?(e2)e2 + <£>(e3)e3.
Evaluate the right side of the equation above to get
3
q(x) = -- + 15x - 15x2.
25. Find a polynomial q € ^(R) such that
/ p(x)(coswx)dx = J p(x)q(x)dx
Jo Jo
for every ρ € ЯгСН.)-
Solution: Define a linear functional φ on ^(R) by
φ{ρ) — J p(x)(costtx)<Ix.
Jo
We seek q 6 7MR) such that φ{ρ) = (p,q) for every ρ G ^(R), where
the inner product on ^(R) is defined as in the previous exercise. Letting
βι,β2,β3 be as in the previous exercise, but using our new definition of φ,
we again have
q = <p(ei)ei + <p(e2)<a + ¥>(ез)ез-
Evaluate the right side of the equation above to get
12-24x
q(x)
7Г2
26. Fix a vector υ e V and define Τ e C(V, F) by Tu = (u, v). For α 6 F, find
a formula for T*a.
Solution: Because Τ e C(V,F), we know that T* G £(F,f). Fix
α 6 F. Then T*a is the unique vector in V such that

71 Chapter 6. Inner-Product Spaces
(a) (Tu, a) = (u, T*a)
for all ύ € U. The inner product on the right is the inner product in V, but
the inner product on the left is the usual inner product on F: the product
of the entry in the first slot with the complex conjugate of the entry in the
second slot. Thus
(Tu,a) = {Tu)a
= (u,v)a
(b) = (u,av).
Comparing (a) with (b) gives
(u, T*a) = (u, av)
for all и € U. Thus T*a = av.
27. Suppose η is a positive integer. Define Τ € £(Fn) by
T{zi, . . . , Z„) = (0,2l,...,2n-l).
Find a formula for T*{z\,..., Zn).
Solution: Fix (z\,..., z„) € Fn. Then for every {w\,..., wn) € Fn, we
have
((u>i,..., ton), Τ*{ζχ,..., Zn)) = (T{wi,..., wn), (zi,..., Zn))
= ((0,Wi, . . . ,Wn^),(zi, . . . , Zn))
= WiZl-\ l-lUn-iZn"
= ({wi,... ,wn),{z2,... ,zn,0)).
Thus
T*(zi,...,zn) = (z2,...,zn,0).
28. Suppose Τ € C(V) and λ € F. Prove that λ is an eigenvalue of Γ if and
only if λ is an eigenvalue of T*.
Solution: We have

72
Chapter 6. Inner-Product Spaces
λ is an not eigenvalue of Τ <=ϊ Τ — XI is invertible
<=ϊ S(T - XI) = (T- XI)S = /
for some S € C(V)
<=ϊ(Τ- XI)*S* = S*(T - XI)* = I
for some S € C(V)
<=ϊ (Τ - XI)* is invertible
<=ϊ Τ* - XI is invertible
·<=>· λ is not an eigenvalue of T*.
Thus λ is an eigenvalue of Τ if and only if λ is an eigenvalue of Γ*.
29. Suppose Τ € C(V) and U is a subspace of V. Prove that U is invariant
under Τ if and only if J7X is invariant under T*.
Solution: First suppose that U is invariant under T. To prove that
UL is invariant under T*, let ν € UL. We need to show that T*v € J7X.
But
(u,T*v) = (Tu,v)
-0
for every и € U (because if и € U, then Tu € U and hence Tu is orthogonal
to v, an element of UL). Thus T*v € J7X, and hence J7X is invariant under
T*, as desired.
To prove the other direction, now suppose that J7X is invariant under T*.
Then by the first direction, we know that (U-1)-1 is invariant under (T1*)*.
But (U^ = U (by 6.33) and (T*)* = T, so U is invariant under T,
completing the proof.
30. Suppose Τ € C(V, W). Prove that
(a) T is injective if and only if T* is surjective;
(b) Τ is surjective if and only if T* is injective.
Solution: First we prove (a):
Τ is injective <=*► null Τ = {0}
<=ϊ (rangeΓ*)χ = {0}
<=ϊ rangeT* = W
<=$■ Τ* is surjective ,

73 Chapter 6. Inner-Product Spaces
where the second line comes from 6.46(c).
Now that (a) has been proved, (b) follows immediately by replacing Τ
with T* in (a).
31. Prove that
dim null T* = dim null Τ + dim W - dim V
and
dim range T* = dim range Τ
for every Τ € C(V, W).
Solution: Let Γ € C(V, W). Then
dim null Γ* = dim(rangeT)-1-
= dim W — dim range Τ
= dim null Τ + axmW - dim V,
where the first equality comes from 6.46(a), the second equality comes from
Exercise 15 of this chapter, and the third equality comes from 3.4. This
proves the first equality that we seek.
To prove the second equality, note that
dim range T* = dim W - dim null T*
= dim V - dim null Τ
= dim range T,
where the first and third equalities come from 3.4 and the second equality
comes from the first part of this exercise. This proves the second equality
that we seek.
32. Suppose Λ is an m-by-n matrix of real numbers. Prove that the dimension
of the span of the columns of A (in Rm) equals the dimension of the span
of the rows of A (in Rn).
Solution: Let Γ € £(Rn,Rm) be such that the matrix of Τ (with
respect to the standard bases) equals A. Then range Γ equals the span of
the columns of A. Thus

74
Chapter 6. Inner-Product Spaces
dimension of the span of the columns of A
= dim range Τ
— dim range Τ1*
= dimension of the span of the columns of λΛ{Τ*)
= dimension of the span of the columns of the transpose of A
= dimension of the span of the rows of A,
where the second equality comes from the previous exercise and the fourth
equality comes from 6.47.

Chapter 7
Operators on
Inner-Product Spaces
1.
Make ^(R·) into an inner-product space by defining
(Р,Я)= / p{x)q(x)dx.
Jo
Define Τ € £(-p2(R)) by Т(<ю + ац -f- a2x2) = ац.
(a) Show that Τ is not self-adjoint.
(b) The matrix of Τ with respect to the basis (ΐ,χ,χ2) is
" 0
0
0
0
1
0
0 "
0
0
This matrix equals its conjugate transpose, even though Τ is not self-
adjoint. Explain why this is not a contradiction.
Solution: (a): Note that
but
(Γ1,χ> = (0,χ>
= 0
(1,Гх) = (1,х>
_ 1
~ 2*
75

76
Chapter 7. Operators on Inner-Product Spaces
Thus (ΤΊ,χ) φ (Ι,Τχ), which shows that Τ is not self-adjoint.
(b): The result stating that the matrix of T* is the conjugate transpose
of the matrix of Τ has as a hypothesis that we are working with orthonormal
bases (see 6.47). Because (ΐ,χ,χ2) is not an orthonormal basis of ^(R), we
cannot compute the matrix of T* with respect to this basis by taking the
conjugate transpose of the matrix of T.
Prove or give a counterexample: the product of any two self-adjoint
operators on a finite-dimensional inner-product space is self-adjoint.
Solution: Let S,T € £(F2) be the operators whose matrices (with
respect to the standard basis) are given by
M(S)
1 0
0 2
and M{T)
0 1
1 0
Each of these matrices obviously equals its conjugate transpose, and hence
S, Τ are self-adjoint. Now
M{ST) = M(S)M(T)
0 1
2 0
Because M{ST) does not equal its conjugate transpose, ST is not self-
adjoint. Thus we have an example of two self-adjoint operators whose
product is not self-adjoint.
Of course there are also many other examples.
Comment: Suppose S,T € C{V) are self-adjoint. Then ST is self-
adjoint if and only if ST = TS (as is easy to see).
(a) Show that if V is a real inner-product space, then the set of self-adjoint
operators on V is a subspace of C(V).
(b) Show that if V is a complex inner-product space, then the set of self-
adjoint operators on V is not a subspace of C(V).
Solution: (a): Suppose V is a real inner-product space. Obviously the
zero operator is self-adjoint. Furthermore, if S, Τ € C(V) are self-adjoint,
then
(S + T)* = S* + T*
= S + T,

77 Chapter 7. Operators on Inner-Product Spaces
ала thus S+T is self-adjoint. Finally, if Г € C(V) is self-adjoint and a € R,
then
(aT)* = aT*
= aT,
and thus aT is self-adjoint. We have shown that the set of self-adjoint
operators on V contains the zero operator and that it is closed under addition
and scalar multiplication. Thus the set of self-adjoint operators on V is a
subspace of C(V).
(b): Suppose now that V is a complex vector space. The identity operator
/ is self-adjoint, but (il)* = -il so il is not self-adjoint. Thus the set of
self-adjoint operators on V is not closed under scalar multiplication and
hence it is not a subspace of C(V).
4. Suppose Ρ € C(V) is such that P2 = P. Prove that Ρ is an orthogonal
projection if and only if Ρ is self-adjoint.
Solution: First suppose that Ρ is an orthogonal projection. Thus there
is a subspace U of V such that Ρ = Рц. Suppose v\, щ. € V. Write
V\ = U\+W\, V2 = U2 + U>2,
where u\,U2 € U and w\,W2 € J7X (see 6.29). Now
(Pvl,v2) = (ui,u2 + w2)
= (ltl,U2>+ (U1,W2)
= (ltl,U2>
= (ttl,U2> + (lui,U2>
= (щ +WX,U2)
= (vi,Pvi).
Thus Ρ = P*, and hence Ρ is self-adjoint.
To prove the implication in the other direction, now suppose that Ρ is
self-adjoint. Let υ € V. Because P(v — Pv) = Pv — P2v = 0, we have
ν - Pv € null Ρ = (range P*)x = (range P)x,
where the first equality comes from 6.46(c). Writing
v = Pv + (v- Pv),

78
Chapter 7. Operators on Inner-Product Spaces
we have Pv € range Ρ and (v — Pv) € (range P)x. Thus Pv = Prangcpv.
Because this holds for all ν € V, we have Ρ = PI&ngcp, which shows that Ρ
is an orthogonal projection.
5. Show that if dim V > 2, then the set of normal operators on V is not a
subspace of C(V).
Solution: Suppose dimV > 2. Let (ei,...,e„) be an orthonormal
basis of V. Define S, Τ € C(V) by
5(αιβι + 1- Οηβη) = α.2&\ — α\&ι
and
T{aiei Η 1- a„en) — α^&\ + α\&ι.
A simple calculation verifies that
5*(αιβι + 1- a„en) = —α.2&ι + а\&ч.·
From this formula, another simple calculation shows that SS* = S*S. Yet
another simple calculation shows that Τ is self-adjoint. Thus both S and Τ
are normal. However, S + Τ is given by the formula
(5 + Γ)(αιβι+ ··· + αηβη) = 2a2ei.
A simple calculation verifies that
{S + T)* (aid + ··· + a„e„) = 2axe2.
A final simple calculation shows that (S + T)(S + Τ)* φ (S + T)*(S + T).
In other words, S + Τ is not normal. Thus the set of normal operators on
V is not closed under addition and hence is not a subspace of C(V).
6. Prove that if Γ € C{V) is normal, then
range Τ = range T*.
Solution: Suppose Τ is normal. Then
range Γ = (null Γ*)x
= (nullT)x
= rangeT*,

79
Chapter 7. Operators on Inner-Product Spaces
where the first equality comes from 6.46(d), the second equality comes from
7.6 (see especially the marginal comment at 7.6), and the third equality
comes from 6.46(b).
7. Prove that if Γ € C(V) is normal, then
null Tk = null Τ and range Tk = range Τ
for every positive integer k.
Solution: Suppose Τ € C(V) is normal and that к is a positive integer.
Obviously we can assume that к > 2.
First we will prove that null 7* = nullT. If υ € null Γ, then
Tkv = Tk-l(Tv)
= Tk'l0
= 0,
and so υ € nullTfc. Thus null Г С nullT*.
To prove an inclusion in the other direction, suppose now that ν € null Tk.
Then
(T*Tk-lv,T*Tk-lv) = (TT*Tk-lv,Tk-lv)
= (TTkv,Tk-lv)
= (0,Tk-lv)
= 0,
where the second equality holds because T*T = TT*. The last equality
above implies that Т*Тк~1у = О. Thus
0 = (Γ*3*~Ιν>Γ*~2«>
= (Tk-lv,Tk-lv).
Hence Tk~lv = 0. In other words, υ € null Τ*-1. The same argument, with
к replaced with к — I, shows that υ € null Τ*-2. Repeat this process until
reaching the conclusion that υ € null Г. This shows that null Т* С null Г,
completing that proof that nullT* = nullT.
Now we will show that range Tk = range T. If υ € range Tk, then there
exists и € V such that υ = Tku = T(Tk~l)u, which implies that υ € range T.
Thus range Tk С range T. Note that

80 Chapter 7. Operators on Inner-Product Spaces
dim range Tk = dim V - dim null Tk
= dim V- dim null Τ
= dim range Τ,
where the first and third equalities come from 3.4 and the second equality
comes from the first part of this exercise. Because range Tk and range Γ
have the same dimension and one of them is contained in the other, these
two subspaces of V must be equal, completing the proof.
8. Prove that there does not exist a self-adjoint operator Τ € £(R.3) such that
r(l,2,3) = (0,0,0)andT(2,5,7) = (2,5,7).
Solution: Suppose Τ € £(R3) is such that T(l,2,3) = (0,0,0) and
T(2,5,7) = (2,5,7). Obviously (1,2,3) is an eigenvector of Γ with
eigenvalue 0 and (2,5,7) is an eigenvector of Τ with eigenvalue 1. If Γ were
self-adjoint, then eigenvectors corresponding to distinct eigenvalues would
be orthogonal (see 7.8). Because (1,2,3) and (2,5,7) are not orthogonal,
Τ cannot be self-adjoint.
9. Prove that a normal operator on a complex inner-product space is self-
adjoint if and only if all its eigenvalues are real.
Comment: This exercise strengthens the analogy (for normal operators)
between self-adjoint operators and real numbers.
Solution: Suppose V is a complex inner product space and Τ € C(V)
is normal.
If Γ is self-adjoint, then by 7.1 all its eigenvalues are real.
Conversely, suppose that all the eigenvalues of Τ are real. By the
complex spectral theorem (7.9), there is an orthonormal basis (ei,..., e„) of V
consisting of eigenvectors of T. Thus there exist real numbers λι,..., λη
such that Tej = Xjej for j = 1,..., n. .The matrix of Τ with respect to
the basis (ei,..., e„) is the diagonal matrix with λι,..., λη on the diagonal.
This matrix equals its conjugate transpose. Thus Τ = T*. In other words,
Τ is self-adjoint, as desired.
10. Suppose V is a complex inner-product space and Τ € C(V) is a normal
operator such that T9 = T8. Prove that Τ is self-adjoint and T2 = T.
Solution: By the complex spectral theorem (7.9), there is an
orthonormal basis (ei,..., e„) of V consisting of eigenvectors of T. Let λι,..., A„ be
the corresponding eigenvalues. Thus

81 Chapter 7. Operators on Inner-Product Spaces
Tej = Xjej
for j = 1,..., n. Applying Τ repeatedly to both sides of the equation above,
we get l^ej = Xj9ej and T^ej = Xj8ej. Thus λ,·9 = А_Д which implies that
Xj equals 0 or 1. In particular, all the eigenvalues of Τ are real. This implies
(by the previous exercise) that Τ is self-adjoint.
Applying Τ to both sides of the equation above, we get
I ej = Xj ej
= Xjtj
= Tej,
where the second equality holds because Xj equals 0 or 1. Because T2 and
Τ agree on a basis, they must be equal.
11. Suppose V is a complex inner-product space. Prove that every normal
operator on V has a square root. (An operator S € C(V) is called a square
root of Γ € C(V) if S2 = T.)
Solution: Suppose Τ € C(V) is normal. By the complex spectral
theorem (7.9), there is an orthonormal basis (ei,... ,βη) of V consisting of
eigenvectors of T. Thus there exist complex numbers Ai,..., A„ such that
Tej = Xjej for j = l,...,n. Define S to be the operator on V such that
Sej = Xj1'2ej for j = l,...,n; here Xj1'2 denotes a complex square root
of Xj (every nonzero complex number has two square roots—it does not
matter which one is chosen). Then, as is easy to verify, S2 = T. Thus S is
a square root of T.
12. Give an example of a real inner-product space V and Τ € C(V) and real
numbers α, β with a2 < 4/3 such that T2 + aT + βΐ is not invertible.
Comment: This exercise shows that the hypothesis that Τ is self-adjoint
is needed in 7.11, even for real vector spaces.
Solution: Let Τ e C(R2) be the counterclockwise rotation on R2; so
T{x,y) = (~У,х) for (г.У) € R2· Thus T2 = -/. Taking α = 0 and β = 1,
we have a2 < 4/3 and
Τ2 + αΤ + βΙ = T2 + I
= 0.
In particular, T2 + aT + βΐ is not invertible.

82
Chapter 7. Operators on Inner-Product Spaces
13. Prove or give a counterexample: every self-adjoint operator on V has a cube
root. (An operator S € C(V) is called a cube root of Γ € C(V) if S3 = T.)
Solution: Suppose Τ G C(V) is self-adjoint. By the spectral theorem
(7.13), there is an orthonormal basis (ei,...,en) of V consisting of
eigenvectors of T. The corresponding eigenvalues must be real (by 7.1). Thus
there exist real numbers Ai,..., An such that Tej = Xjej for j = 1,..., n.
Define S to be the operator on V such that Se3· = Aj1'3e;· for j = 1,..., n.
Then, as is easy to verify, 53 = T. Thus S is a cube root of T, completing
the proof that every self-adjoint operator on V has a cube root.
14. Suppose Τ € C(V) is self-adjoint, λ € F, and e > 0. Prove that if there
exists υ € V such that ||w|| = 1 and
||Γυ-λι/||<6,
then Τ has an eigenvalue A' such that |A — A'| < e.
Solution: By the spectral theorem (7.13), there is an orthonormal
basis (ei,..., en) of V consisting of eigenvectors of T. Let Ai,..., An be the
corresponding eigenvalues.
Suppose ν G V is such that ||w|| = 1 and ||Γυ — λυ|| < e. From 6.17 we
have
ν = (υ, ei)ei 4- · · · + (υ, e,i)en,
and so
Τυ = Αι (υ, ei)ei + ■ ■ ■ + λ„(υ, en)e„.
Thus
£2>||Γι/-λι/||2
= ΙΙ(λι - λ)(υ,ei)ei + · · · + (A„ - AXt/.e^e»!!2
= |A1-A|2|(l;1e1>|2 + .-- + |An-A|2Kl;,en>|2
>(min{|A1-A|2,...,|An-A|2})(|(l;,e1>|2 + ... + |(l;,eIl>|2)
= min{|A,-A|2,...,|An-A|2}
Thus 6 > |Aj — A| for some j. In other words, there is an eigenvalue whose
distance from A is less than e, as desired.

83
Chapter 7. Operators on Inner-Product Spaces
15. Suppose U is a finite-dimensional real vector space and Τ € C(U). Prove
that U. has a basis consisting of eigenvectors of Τ if and only if there is an
inner product on U that makes Τ into a self-adjoint operator.
Solution: First suppose that U has a basis (ei,..., en) of eigenvectors
of T. Because (ei,..., e„) is a basis of J7, every element of U can be uniquely
written as a linear combination of (ei,...,en). Thus we can define an inner
product on U by
(aiei -I 1- a„en, biei -I 1- bne„) = a\bi -\ 1- anbn.
It is easy to verify that this is indeed an inner product on U and that
(ei,..., en) is on orthonormal basis of U with respect to this inner product.
Because each ej is an eigenvector of Г, the operator Τ has a diagonal matrix
with respect to the orthonormal basis (ei,..., en). Thus Τ is self-adjoint.
Conversely, now suppose that there is an inner product on U that makes
Τ into a self-adjoint operator. Then by the spectral theorem (7.13), U has
a basis consisting of eigenvectors of T.
16. Give an example of an operator Τ on an inner product space such that Τ
has an invariant subspace whose orthogonal complement is not invariant
under T.
Comment: This exercise shows that 7.18 can fail without the hypothesis
that Τ is normal.
Solution: Define Τ G £(F2) by T(to, ζ) = (ζ, 0). Then T(w, 0) = (0,0)
for all to € F. Thus the subspace U defined by U = {(го,0) : w € F} is
invariant under T. However, UL = {(0, z) : G F}, which is not invariant
under Τ because (0,1) e Ux but Γ(0,1) = (1,0) φ UL.
Of course there are also many other examples.
17. Prove that the sum of any two positive operators on V is positive.
SOLUTION: Suppose S and Τ are positive operators on V. Because S
and Τ are self-adjoint, so is S + T. Furthermore,
((S + 7>, υ) = (St/, υ) + (Τυ, υ)
>0.
Thus S + Τ is a positive operator, as desired.
18. Prove that if Γ € £{V) is positive, then so is Tk for every positive integer k.

84
Chapter 7. Operators on Inner-Product Spaces
Solution: Suppose Τ € C(V) is positive and к is a positive integer.
Then Tk is self-adjoint (because Τ is self-adjoint).
First consider the case where к is an even integer. Then we can write
к = 2m for some positive integer m. Now
(Tkv,v) = (T2mv,v)
= (Tmv,Tmv)
>0
for every υ e V, where the second equality holds because Τ is self-adjoint.
The inequality above shows that Tk is positive, as desired.
Now consider the case where к is an odd integer. Then we can write
к = 2m + 1 for some nonnegative integer m. Now
(Tkv,v) = (T2m+lv,v)
= (T(Tmv),Tmv)
>0
for every ν € V, where the second equality holds because Τ is self-adjoint
and the inequality holds because Τ is positive. The inequality above shows
that Tk is positive, as desired.
19. Suppose that Γ is a positive operator on V. Prove that Τ is invertible if
and only if
(Tv, v)>0
for every иб^\ {О}.
Solution: First suppose that Τ is invertible. By 7.27, there exists an
operator S € C(V) such that Τ = S*S. Suppose ν € V \ {0}. Then Sv φ 0
because otherwise we would have Tv = S*Sv = 0, which would contradict
the invertibility of T. Now
(Tv, v) = (S'Sv, v)
= (Sv, Sv)
>0,
as desired.
Now suppose that (Tv, v) > 0 for every ν € V \ {0}. In particular, this
means that Tv φ 0 for every ν € V \ {0}. Thus Τ is injective, and hence Τ
is invertible (see 3.21), as desired.

85
Chapter 7. Operators on Inner-Product Spaces
20. Prove or disprove: the identity operator on F2 has infinitely many self-
adjoint square roots.
SOLUTION: For each t € [—1,1], the operator whose matrix (with respect
to the standard basis) equals
" t ν/Γ^ί5 ■
ч/ПЙ? -t
is self-adjoint and a square root of the identity operator, as can be verified by
squaring the matrix above. Thus the identity operator has infinitely many
self-adjoint square roots.
21. Prove or give a counterexample: if S € C(V) and there exists an orthonormal
basis (ei,..., en) of V such that ||5е;|| = 1 for each ej, then S is an isometry.
Solution: Define S e £(F2) by
S(w,z) = (to+ 2,0).
With the usual inner product on F2, the standard basis ((1,0), (0,1)) is an
orthonormal basis of F2. Note that ||5(1,0)|| = ||5(0,1)|| = 1. However, S
is not an isometry because ||5(1, —1)|| = 0.
Of course there are also many other examples.
22. Prove that if S € £(R3) is an isometry, then there exists a nonzero vector
χ e R3 such that S2! = x.
Solution: Suppose S € £(R3) is an isometry. Then there is a basis of
R3 with respect to which S has a block diagonal matrix, where each block
on the diagonal is a 1-by-l matrix containing 1 or —1 or is a 2-by-2 matrix
(see 7.38). Because R3 has odd dimension, at least one of these blocks must
be a 1-by-l matrix. In other words, either 1 or —1 must be an eigenvalue
of S. Thus there is a nonzero vector χ € R3 such that Sx = Xx, where
λ = ±1. Hence
S2x = S(Sx) = S(Xx) = XSx = \2x = x.
23. Define Τ € £(F3) by
T{zi,z2,z3) = (23,2.21,3.22).
Find (explicitly) an isometry S € £(F3) such that Τ = Sy/T'T.

86
Chapter 7. Operators on Inner-Product Spaces
24.
0
2
0
0
0
1
0
0
3
2
0
0
1
0
0 _
0
3
0
, which e<
' 4
0
0
0
9
0
0
0
1
Solution: With respect to the standard basis of F3, we have
M(T) =
Thus
M{T) =
Computing the product M(T*)M(T), which equals M{T*T), we get
M(TT) =
Prom the matrix above, we see that {T*T){z\, z2l zz) = (4.21,9.22,23). Thus
Vf^T(zi1z2, 23) = (2.2i, З22,23). Hence if we define S € £(F3) by
5(21,22,23) = (23.21.22),
then S is an isometry and Τ = SVT*T.
Suppose Τ € C(V), S € C(V) is an isometry, and R G C(V) is a positive
operator such that Τ = SR. Prove that R = \/T*T.
Comment: This exercise shows that if we write Τ as the product of an
isometry and a positive operator (as in the polar decomposition), then the
positive operator must equal y/T*T.
Solution: Taking adjoints of both sides of the equation Τ = SR, we
have
T* = R*S*
= RS\
where the last equation holds because R is positive (and hence self-adjoint).
Multiplying together our formulas for T* and T, we get
T*T = RS'SR
= R\

87 Chapter 7. Operators on Inner-Product Spaces
where the last equation holds because S is an isometry (and hence S*S = I
by 7.36). The equation above asserts that R is a square root of T*T; because
R is positive, this implies that R = VT*T.
25. Suppose Τ G C{V). Prove that Τ is invertible if and only if there exists a
unique isometry S G C(V) such that Τ = SVT'T.
Solution: First suppose that Τ is invertible. The polar decomposition
(7.41) states that there exists an isometry S G C(V) such that
Τ = SVT*T.
Because Τ is invertible, this implies that y/T*T is invertible (see Exercise 22
in Chapter 3). Thus the equation above implies that S = T(y/T*T) .
Because S must be given by this formula, we see that there is a unique
operator S G C(V) such that Τ = Sy/T'T, as desired.
Now suppose that there exists a unique isometry S € C(V) such that
T = SVT*T. This means that the linear map 5г in the proof of the polar
decomposition (7.41) must be 0 because otherwise we could replace 5г with
—S2 and get another choice for S. But range S2 equals (range T)x, and
hence (range T)x = {0}. This implies that range Г = V, which implies that
Τ is invertible (by 3.21), as desired.
26. Prove that if Τ G C(V) is self-adjoint, then the singular values of Τ equal
the absolute values of the eigenvalues of Τ (repeated appropriately).
Solution: Suppose Τ G C{V) is self-adjoint. There exists an orthonor-
mal basis (ei,..., e„) of V consisting of eigenvectors of T. Thus
Tej = Xjej
for each j, where Ai,..., λη 6 R are the eigenvalues of T. Thus
TTej = T2ej
= (Α,)4·
for each j. The equation above implies that y/T*Tej = |Aj|e_j for each j.
Thus the singular values of Τ are |Ai|,..., |λη|, as desired.
27. Prove or give a counterexample: if Τ G C(V), then the singular values of T2
equal the squares of the singular values of T.
Solution: DeEne Τ e £(F2) by

88 Chapter 7. Operators on Inner-Product Spaces
T(zuz2) = (z2,0).
Then T*T{zi,z2) = (0,z2) and hence VT*T(z1,z2) = {0,z2). Thus the
eigenvalues of \/T*T are 0,1. Hence the singular values of Τ are 0,1.
However, T2 = 0, so the singular values of T2 are 0,0. Thus for this
operator T, the singular values of T2 do not equal the squares of the singular
values of T.
Of course there are also many other examples.
28. Suppose Τ € C(V). Prove that Τ is invertible if and only if 0 is not a
singular value of T.
Solution: If S e C{V) and ST=TS = /, then taking adjoints we get
T*S* = S'T* = I. Thus if Γ is invertible, then so is T\
Now
Τ is invertible <=> Τ and T* are invertible
<=> T*T is invertible
<=> y/T*TVT*T is invertible
<=> VT'T is invertible
<=> 0 is not an eigenvalue of VT*T
<=> 0 is not a singular value of T,
where the second and fourth equivalences follow from Exercise 22 in
Chapter 2.
29. Suppose Τ € C(V). Prove that dim range Γ equals the number of nonzero
singular values of T.
Solution: By the singular value decomposition (7.46), there exist or-
thonormal bases (u\,..., Un) and (w\,..., wn) of V such that
Tv = si (υ, ui)wi -I 1- sn(i>, un)wn
for every υ € V, where s\,... ,sn are the singular values of T. For each j,
we have Tu3· = SjWj. Thus each wj corresponding to a nonzero Sj is in
rangeT. The equation above also shows that the toy's corresponding to
nonzero Sj's span range T. Thus dim range Τ equals the number of nonzero
singular values of T.
30. Suppose S Ε C(V). Prove that S is an isometry if and only if all the singular
values of S equal 1.

89 Chapter 7. Operators on Inner-Product Spaces
Solution: We have
S is an isometry <=> S*S = I
<=> all the eigenvalues of VS*S equal 1
<=> all the singular values of S equal 1,
where the first equivalence comes from 7.36 and the third equivalence comes
from the spectral theorem (7.9 or 7.13) applied to the self-adjoint operator
31. Suppose Γι, T2 € C(V). Prove that Γι and T2 have the same singular values
if and only if there exist isometries Si,S2 € C(V) such that Γι = SiT2S2.
Solution: First suppose that Γι and Γ2 have the same singular values
s\,..., sn. By the singular-value decomposition (7.46), there exist orthonor-
mal bases (eb ... ,e„), (/1,... ,/n), {е\,... ,e'n), (/{,..- ,f'n) of V such that
Txv = si(v,ei)fi + --- + sn(v,en)fn,
T2v = si(v, e\)f[ + ■■■ + sn(v,e'n)f'n
for every υ 6 V. Define Si,S2€ C(V) by
Si (αι/ί + · · · + anf'n) = 01/1+··- + anfn,
S2(aiei +■■■ + a„e„) = сце\ + ■■■ + One'n.
Then
||51(α1/ί + ··· + αη/;)||2 = ||α1/1+..-+αη/η||2
= |&1|2 + ..· + |αη|2
= ||α1/ί+.-·+αη/;||2,
and thus Si is an isometry. Similarly, S2 is an isometry. This implies that
52* = S2-1 (see 7.36). In particular, Sfe'j = ej. Now for υ € V we have
T2(S2v) = si^t/.e'jj/i + · · · + sn(S2v,e'n)f'n
= si(v, S2*e\)f[ +■■■ + sn(v, S2*e'n)f'n
= si(v,ei)f[ + ■■■ +sn(v,en)f'n.

90
Chapter 7. Operators on Inner-Product Spaces
Thus
Si{T2S2v) = si(v,ei)Sif[ +·■· + SnfaejSifn
= si(v,ei)fi +--- + sn(v,en)fn
= Τιυ
for every υ eV. Hence S1T2S2 = ΤΊ, as desired.
To prove the implication in the other direction, now suppose that there
exist isometries 5ь5г € С-(У) such that ΤΊ = S1T2S2· Using 7.36, we have
1\ 1\ — 02 -12 &1 011202
= 1S2 72*72ιί?2·
This implies that T\*Ti and ТУТг have the same eigenvalues (and that the
corresponding spaces of eigenvectors have the same dimensions). Thus T\
and T2 have the same singular values.
32. Suppose Τ € £(^) has singular-value decomposition given by
Τυ = 5ι(υ,βι)/ι + ··· + sn(v1en)fn
for every ν € V, where s\,... ,sn are the singular values of Τ and (ei,..., en)
and (/1,..., /„) are orthonormal bases of V.
(a) Prove that
T'v = si (υ, /i)ei + · · · + sn(v, fn)en
for every υ € V.
(b) Prove that if Τ is invertible, then
T-i„ <",/i)ei <«,/η)βη
1 υ = + · · · Η
Sl Sn
for every υ € V.
Solution: (a): Fix υ e V. Then
(to, T*v) = (Tw, υ)
= (si (w, ei)/i + · · · + sn(w, e„)/n, v)
= si(w,ei)(fi,v) + --- + sn(w,en)(fn,v)
= (w,si(v,fi)ei +--- + sn(v,fn)en)

91 Chapter 7. Operators on Inner-Product Spaces
for all w G V. This implies that
T*v = si (υ, /i)ei + --- + sn(v, /n)e„,
as desired.
(b): Suppose Τ is invertible. Let υ €V and let
("■/i)ei , , (v,fn)en
w — 1 Λ ;
si sn
none of the singular values si,...,sn equals 0 (see Exercise 28 of this
chapter), so this makes sense. Now
T (v,fi)Tei , , («,/η)Γβ»
lw = 1 -i
Si 8n
(υ, fi)sifi , , (v,fn)snfn
— 1- ·· · Η
Si Sn
= <«,/l>/l +··· + <«,/n>/n
= V.
Thus ш = Γ-1υ, as desired.
33. Suppose Τ G C(V). Let s denote the smallest singular value of T, and let s
denote the largest singular value of T. Prove that
sibll < \\Tv\\ < s\\v\\
for every υ €V.
Solution: Let ν G V. By the singular value decomposition (7.46),
there exist orthonormal bases (ui,..., Un\ and {w\,..., wn) of V such that
Tv = si(v, ui)wi -1 1- sn(v,un)wn,
where si,..., sn are the singular values of T. Because (ui,..., un) and
(toi,..., wn) are both orthonormal bases of V, we have
ё2\М2 = з2(\(ь,щ)\2 + ... + \(ь,ип)\2)
<si2\(v,u1)\2 + -- + sn2\(v,un)\2
= \\Tvf,

92 Chapter 7. Operators on Inner-Product Spaces
giving the first desired inequality. Also,
\\Tv\\2 = s12\(v,u1)\2 + --- + sn2\(v,un)\2
<*2(|<υ,"ι>|2 + ··· + Κ^η>|2)
= s2\\v\\2,
giving the second desired inequality.
34. Suppose T", T" 6 C(V). Let s' denote the largest singular value of T", let s"
denote the largest singular value of T", and let s denote the largest singular
value of T" + T". Prove that s < s' 4- s".
Solution: Let Τ = Τ" + Τ". Because s is a singular value of T, we
know that s is an eigenvalue of VT*T. Thus there exists a vector υ Ε V
such that ||υ|| = 1 and y/T*Tv = sv. Now
s = \\sv\\
= Hv^ft/Ц
= \\Tv\\
= \\rv + r'v\\
< \\rv\\ + \\Г'ь\\
<s'\\v\\+s"\H
= s' + s",
where the third line above comes from 7.42 and the sixth line above comes
from the previous exercise.

Chapter 8
Operators on
Complex Vector Spaces
1. Define Τ € £(C2) by
T(w,z) = (z,0).
Find all generalized eigenvectors of T.
Solution: Suppose λ is an eigenvalue of T. For this particular operator,
the eigenvalue-eigenvector equation T{w, z) = A(tu, z) becomes the system
of equations
ζ = Xw
0 = Xz.
If λ φ 0, then the second equation implies that ζ = 0, and the first equation
then implies that w = 0. Because an eigenvalue must have a nonzero
eigenvector, this shows that 0 is the only possible eigenvalue of T. For λ = 0, the
equations above show that ζ must equal 0, but w can be arbitrary. Thus 0
is indeed an eigenvalue of T, and the set of eigenvectors corresponding to
this eigenvalue is
{(w)0):weF}.
Note that T2 = 0. Thus every vector in C2 is a generalized eigenvector
of Τ (corresponding to the eigenvalue 0).
93

94
Chapter 8. Operators on Complex Vector Spaces
2. Define Τ G £(C2) by
T(w,z) = (-z,w).
Find all generalized eigenvectors of T.
Solution: On page 78 of the textbook we saw that the eigenvalues of
Τ are i and -i. Note that T2 = -I.
The set of generalized eigenvectors of Τ corresponding to the eigenvalue г
equals ηιι11(!Γ — il)2 (by 8.7). To compute this, note that
(T - il)2 = T2-2iT-I
= -21 - 2iT
= -2i(T - il).
Thus пи11(!Г — il)2 equals the set of eigenvectors of Τ corresponding to
the eigenvalue i. On page 78 of the textbook we noted that this equals
{(a,-ia) :a€ C}.
The set of generalized eigenvectors of Τ corresponding to the
eigenvalue —i equals n\i\\(T + H)2 (by 8.7). To compute this, note that
(T + il)2 = T2 + 2iT - I
= -21 + 2iT
= 2i{T + iI).
Thus пи11(!Г + il)2 equals the set of eigenvectors of Τ corresponding to
the eigenvalue —i. On page 78 of the textbook we noted that this equals
{(a,ia) : a G CK
3. Suppose Τ G C{V), m is apositive integer, and υ G V is such that Tm~1v φ Ο
but T^-v = 0. Prove that
(υ,Γυ,Γν..,"!™-1!/)
is linearly independent.
Solution: Suppose ao, a.\, аг,..., am_i G F are such that
oqv + aiTv + a2T2v + ■ ■ · + am-1Tn-lv = 0.
Because Τ^υ = 0, if we apply Τ™-1 to both sides of the equation above, we
get oqI^-^v = 0. Because Γ171-^ φ 0, this implies that Go = 0. Thus the
equation above can be rewritten as

95
Chapter 8. Operators on Complex Vector Spaces
aiTv + a2T2v + ■■■+ am-iT"1-1!/ = 0.
Applying Τ™-2 to both sides of this equation, we get αιΤ™-^ = 0. Thus
a\ = 0. Continuing in this fashion, we have clq = αϊ = a2 = · · · = Om-i = 0,
which means that (υ, Tv, T2v,..., Τ"1- ιυ) is linearly independent.
4. Suppose Τ € £(C3) is denned by T(zi, z2, zz) = (z2, z3,0). Prove that Τ has
no square root. More precisely, prove that there does not exist S € £(C3)
such that S2 = T.
Solution: Note that Γ3 = 0. Suppose there exists S € £(C3) such
that S2 = T. Then S6 = Γ3 = 0, so S is nilpotent. By 8.8, this implies that
S3 = 0. Thus
T2 = S4
= SS3
= 0.
But Tl2(,zi,Z2,Z3) = (23,0,0), so T2 is not the 0 operator, contradicting the
equation above. This contradiction shows that our supposition that there
exists S € £(C3) such that S2 = Τ must have been false.
5. Suppose S,T € £(V). Prove that if ST is nilpotent, then TS is nilpotent.
Solution: Suppose ST is nilpotent. Thus there exists a positive integer
η such that (ST)n = 0. Now
(TS)n+l = (TS)(TS)... (TS)
= T(ST)(ST)...(ST)S
= T(ST)nS
= CO(o)(S)
= 0,
and thus TS is nilpotent.
6. Suppose ./V G £(V) is nilpotent. Prove (without using 8.26) that 0 is the
only eigenvalue of N.
Solution: There is a positive integer m such that Nm = 0. This implies
that ./V is not injective, so 0 is an eigenvalue of N.
Conversely, suppose A is an eigenvalue of N. Then there exists a nonzero
vector υ G V such that

96
Chapter 8. Operators on Complex Vector Spaces
λυ = Nv.
Repeatedly applying N to both sides of this equation shows that
Xmv = Ν™υ
= 0.
Thus λ = 0, as desired.
7. Suppose V is an inner-product space. Prove that if TV € C(V) is self-adjoint
and nilpotent, then N = 0.
Solution: Suppose ./V g £(V) is self-adjoint and nilpotent. Because ./V
is self-adjoint, there is an orthonormal basis (ei,... , en) of V consisting of
eigenvectors of ./V (by the spectral theorem). Because N is nilpotent, 0 is the
only eigenvalue of ./V (see Exercise б of this chapter). Thus the eigenvalue
corresponding to each ey must equal 0. In other words, Ne3- = 0 for each j.
Because (ei,..., en) is a basis of V, this implies that N = 0.
8. Suppose N G C(V) is such that nullWdimV/-1 φ nullNdimV. Prove that N
is nilpotent and that
dim null N3 = j
for every integer j with 0 < j < dim V.
Solution: Because null Ndim v~l φ null Ndim v, we know (by 8.5) that
null Ν3~λ φ null Nj whenever 0 < j < dim V. Thus
{0} = nullJV°c шШЛ^С ...С шШЛГВт1'-1с null7VdimV/.
1 J + + + +
At each of the strict inclusions in the chain above, the dimension must
increase by at least 1. However, if the dimension increases by more than 1
at any step, we would end up with dim null Ndim v > dimV, a
contradiction because a subspace of V cannot have dimension larger than dim V.
Thus the dimension increases by exactly one at each step. In other words,
dim null N] = j for every integer j with 0 < j < dimV". In particular,
taking j = dimV, we have dim null 7Vdim v = dimK. This means that
nullNdimV = у Thus jy-dimV = 0) ша SQ N is nilpotent.
9. Suppose Τ G C(V) and m is a nonnegative integer such that
range Τ"1 = range T"*l.
Prove that range Tk = range T™ for all к > τη.

97 Chapter 8. Operators on Complex Vector Spaces
Solution: Suppose и G rangeTm+l. Thus there exists a vector υ in V
(the domain of T) such that и = T"1*1^ Now Ί^υ is in range Ί™, which
by our hypothesis equals range Tm+1. Thus there exists w G V such that
Ί^υ = Гт+1ш. Putting all this together, we have
u = г"И-1,,
= Г(Т,т+1го)
= t^+V
Thus и G rangeT"171"*"2. Because и was an arbitrary vector in rangeT'm+1) we
have shown that range Т4™"1 С rangeT"n+2. We also have an easy inclusion
in the other direction, so we conclude that range Tm+l = range T"171"*"2.
In the paragraph above, we showed that rangeT™ = rangeT,m+1 implies
rangeT"171"*"1 = range Τ™*2. Apply that result, with m replaced with m + 1,
to conclude that range T"1*2 = range J™*3. Continuing in this fashion, we
see that
range Τ"1 = range Τ""-*-1 = range Τ"η+2
as desired.
10. Prove or give a counterexample: if Τ G £(У), then
V = null Τ Θ range Т.
Solution: Define Τ G £(F2) by T(w,z) = (z,0). Thus
null Τ = rangeT = {(to,0) : to G F},
which clearly implies that F2 is not the direct sum of null Τ and range T.
Of course there are also many other examples.
11. Prove that if Τ G C(V), then
Vr = nullT,r*©rangeTn,
where η = dim V.
Solution: Let Τ G £(V). First we show that
V = null 7™ + range T".

98 Chapter 8. Operators on Complex Vector Spaces
To do this, let υ € V. Then
v = (v-Tnu)+Tnu
for any vector и € V. Obviously Г"и € range Τ". Thus looking at the
equation above, we see that we need to show that there exists и € V such
that υ — T^u € null Τ". In other words, we want a vector и € V such
that T^iv - T^u) = 0, which is equivalent to Τ"υ = T2nu. But T^v €
range?"1, and ranger" = rangeT2" (by 8.9), so Τ"υ € rangeT2". Thus
there indeed exists и € V such that T^v = T2nu, completing our proof
υ € null T" + range T™. Because υ was an arbitrary vector in V, this implies
that V = nullT" + range Γ".
For any linear map (and in particular for I"1), the dimension of the
domain equals the sum of the dimensions of the null space and range (by 3.4).
In other words,
dim V = dim nullT" + dim range T".
This equation, along with the equation V = null IT1 + range T™, implies that
V = nullT" Φ range Τ" (by 2.19).
12. Suppose V is a complex vector space, N € C(V), and 0 is the only eigenvalue
of N. Prove that N is nilpotent. Give an example to show that this is not
necessarily true on a real vector space.
Solution: Because 0 is the only eigenvalue of N, 8.23(a) implies that
every vector in V is a generalized eigenvector of Τ corresponding to the
eigenvalue 0. This implies that N is nilpotent.
Define Τ € £(R3) by
T(x,y,z) = (-j/,x,0).
Then 0 is an eigenvalue of Τ because Γ(0,0,1) = (0,0,0). As can be
verified from the definition of eigenvalue, Τ has no other eigenvalues (which
must be in R, because Γ is an operator on a real vector space). However,
T^x, y, z) = (y, —x, 0). In particular, Τ3 φ 0. Thus Τ is not nilpotent.
Of course there are also many other examples.
13. Suppose that V is a complex vector space with dim V = η and Τ € C(V) is
such that
nullTn-2^nullTn-1.
Prove that Τ has at most two distinct eigenvalues.

99
Chapter 8. Operators on Complex Vector Spaces
Solution: Because nullT"-2 Φ nullT"-1, we see that dimnullTJ' is
at least 1 more than dim null !P'"l for j = I,...,n - 1 (by 8.5). Thus
dim null T"-1 > n- 1. In particular, 0 is an eigenvalue of Γ with multiplicity
at least η - I. Because the sum of the multiplicities of all the eigenvalues
of Τ equals η (by 8.18), this implies that Τ can have at most one additional
eigenvalue.
14. Give an example of an operator on C4 whose characteristic polynomial
equals (z - 7)2(z - 8)2.
Solution: Define Τ G £(C4) by
T(zu Z2,z3,Zi) = (7ζι,7ζ2,8ζ3,8ζ4).
Then ηιι11(!Γ - 71) is the two-dimensional subspace
{(21,22,0,0): 21,22 GC}
and пи11(!Г - 81) is the two-dimensional subspace
{(0,0,23,24): 23,24 GC}.
Thus 7 is an eigenvalue of Τ with multiplicity at least 2 and 8 is an eigenvalue
of Γ with multiplicity at least 2. Because 2+2 = 4 = dim C4, there can be no
other eigenvalues of Τ and the eigenvalues 7 and 8 must have multiplicity 2
(by 8.18). Thus the characteristic polynomial of Γ equals (2 - 7)2(2 — 8)2.
Of course there are also many other examples.
15. Suppose V is a complex vector space. Suppose Τ G C(V) is such that 5 and
6 are eigenvalues of Τ and that Τ has no other eigenvalues. Prove that
(T-5I)n-l(T-6I)n-1 =0,
where η = dim V.
Solution: Because 5 and 6 are eigenvalues of Τ and Τ has no other
eigenvalues, the characteristic polynomial of Τ must be of the form
(z-5)d>(z-6)d\
where 1 < di and 1 < d^. Because di + d2 = n, we must also have di < n- 1
and d2 < η — 1. The Cayley-Hamilton theorem implies that
(T-5I)dl(T-6l)d2 =0.

100 Chapter 8. Operators on Complex Vector Spaces
Because di < η - 1 and ά-ι < η - 1, we can multiply the equation above by
appropriate powers of Τ -51 and T- 61 to get (Γ-5/)η_1(Γ-6/)η_1 = 0.
16. Suppose V is a complex vector space and Τ € C{V). Prove that V has a basis
consisting of eigenvectors of Τ if and only if every generalized eigenvector
of Τ is an eigenvector of T.
Comment: For complex vector spaces, this exercise adds another
equivalence to the list given by 5.21.
Solution: First suppose that V has a basis consisting of eigenvectors
of T. Thus there exists a basis («i,..., vn) of V and Ai,..., An € С such
that Tvj = XjVj for each j. Suppose υ € V is a generalized eigenvector of Τ
corresponding to an eigenvalue A. Because («i,..., υη) is a basis of V, there
exist αϊ,..., an € С such that
ν = αινι Η 1- anvn.
Thus
(Γ - λ/)υ = (Αι - λ)αιυι + ■ ■ ■ + (λη - λ)αηυη.
Applying Τ — XI repeatedly to both sides of this equation, we get
(Γ - λ/)ηυ = (λ! - Χ)ηαιυι + .. - + (λ„ - A)na„t;n.
The left side of the equation above equals 0 (because υ is a generalized
eigenvector of Τ corresponding to the eigenvalue A). Thus the right side
of the equation above equals 0. Thus (Ay — A)nay = 0 for each j. This
implies that the indices j such that a.j φ 0 must all have all satisfy Ay = A,
which means that υ is a linear combination of the Vj's that correspond to
eigenvalue A, which means that υ is itself an eigenvector corresponding to
eigenvalue A, as desired.
To prove the other direction, now suppose that every generalized
eigenvector of Γ is an eigenvector of T. By 8.25, there exists a basis of V consisting
of generalized eigenvectors of T. Because every generalized eigenvector of Τ
is an eigenvector of T, this gives a basis of V consisting of eigenvectors of T,
as desired.
17. Suppose V is an inner-product space and N € C(V) is nilpotent. Prove
that there exists an orthonormal basis of V with respect to which N has an
upper-triangular matrix.

101 Chapter 8. Operators on Complex Vector Spaces
Solution: By 8.26, there is a basis of V with respect to which N
has an upper-triangular matrix. By 6.27, this basis can be chosen to be
orthonormal. (If V is a complex inner-product space, then we can just use
6.28 directly, so the hypothesis that N is nilpotent is not needed.)
18. Define N € £(F5) by
N(xi,x2,X3,X4,Xb) = (2х2,Зхз,-Х4,4х5,0).
Find a square root oi I + N.
Solution: Note that
N2(xi,X2,X3,X4,Xb) = (6X3,-3X4,-4X5,0,0)
iV3(xi,X2,X3,X4,X5) = (-6X4,-12X5,0,0,0)
iV4(xi, X2, X3, хл, хь) = (-24x5,0,0,0,0)
Ν5(χ1,χ2,Χ3,χ4,Χ5) = (0,0,0,0,0).
Because TV5 = 0, the proof of 8.30 shows that
is a square root oi I + N. Using the formulas above, we calculate that the
operator S € £(F5) defined by
5(X1,X2,X3,X4,X5) =
/ Зхз 3X4 15X5 Зхз 3X4 ЗХ5
(x1+x2- —- —+—, x2 + —+ —-—,
Хз _ ~2~ + ~2~' Xa + 2X5' Xb)
is a square root of / + N.
19. Prove that if V is a complex vector space, then every invertible operator
on V has a cube root.
Solution: First suppose that ./V € £(V) is nilpotent. We will show
that I + N has a cube root by imitating the proof of 8.30. Specifically, we
guess that there is a cube root of / + N of the form
/ + aiN + a2N2 + ■··+ (hn-iN

102
Chapter 8. Operators on Complex Vector Spaces
where m is such that Nm = 0. Having made this guess, we can try to choose
αϊ, α2,..., Om-ι so that the operator above has its cube equal to I+N. Now
(I + alN + a2N2 + a3N3 + ---+am-iNm-1)3
= I + 3<nN + (За2 + 3ai 2)W2 + (3a3 + 6ai a2 + ai 3)W3 + · · ·
+ (За™-! + terms involving ai,... ,Οπ,^ίΓ1"1.
We want the right side of the equation above to equal I + N. Hence choose
αϊ so that 3αι = 1 (thus ai = 1/3). Next, choose a2 so that За2 + 3ai2 = 0
(thus a2 = —1/9). Then choose аз so that the coefficient of N3 on the
right side of the equation above equals 0 (thus аз = 5/81). Continue in this
fashion for j = 4,..., m— 1, at each step solving for ay so that the coefficient
of TV·7 on the right side of the equation above equals 0. Actually we don't
care about the explicit formula for the ay's. We need only know that some
choice of the ay's gives a cube root of / + N.
Having shown that the identity plus a nilpotent always has a cube root,
we now look at the proof that every invertible operator on a complex vector
space has a square root (see 8.32). In that proof, if we replace the words
"square root" with "cube root", we get a proof that every invertible operator
on a complex vector space has a cube root.
20. Suppose Τ € C(V) is invertible. Prove that there exists a polynomial
ρ € V(F) such that T~l = p(T).
Solution: Let ao + a\z + · ■ · + Om-iz"1-1 + zm denote the minimal
polynomial of T. This is the monic polynomial of smallest degree such that
aQI + aiT + --+am-iTm-l+Tm = 0.
If clq were equal to 0, then we could multiply both sides of the equation
above by T~l to get
a1I + a2T + --+am-iTm-2 + Tm-1 =0,
which would give a monic polynomial q of smaller degree such that q(T) = 0,
contradicting the definition of the minimal polynomial. Thus clq φ 0.
Because αο φ 0, we can solve the first equation above for the identity
operator /, getting
r _ _aJ.<Ti _ . . . _ a"1~1 rpm-l Lt™
ao ao ao

103 Chapter 8. Operators on Complex Vector Spaces
Now multiply both sides of the equation above by Γ ι, getting
-1 _ _alr_a2T атп-1ггт-2 1
Setting
rp-l _ _ZLj _ ^_ψ _ ... _ am-lrpm-2 _rpm-l
CLQ CtQ Oq CLQ
we thus have T~l = p(T).
21. Give an example of an operator on C3 whose minimal polynomial equals z2.
Solution: DeHne Τ € £(C3) by
T(wuW2yW3) = (ш3)0,0).
Clearly T2 = 0. In other words, the polynomial z2 when applied to Τ gives 0.
Thus the minimal polynomial of Γ is a divisor of z2 (by 8.34). But the only
monic polynomials that divide z2 are 1, z, and z2. The polynomial 1 applied
to Τ gives the identity operator, which is not 0, and the polynomial ζ applied
to Τ gives T, which is also not 0. Thus the minimal polynomial of Τ must
hez2.
Of course there are also many other examples.
22. Give an example of an operator on C4 whose minimal polynomial equals
Φ-i)2·
SOLUTION: Define Τ € £(C4) by
T(wi, io2, юз, wA) = (0, w-i + to4, to3, wA).
Then
(T - I)(wi,W2,W3iW4) = (-tub 104,0,0),
which implies that
(Τ - Ι)2{υ>ι.,υ)2,υ)3,ιυΑ) = {wu 0,0,0),
which implies that
T{T - I)2 = 0.

104
Chapter 8. Operators on Complex Vector Spaces
In other words, the polynomial z(z - l)2 when applied to Τ gives 0. Thus
the minimal polynomial of Γ is a divisor of z(z - l)2 (by 8.34).
Note that 0 is an eigenvalue of Γ because T(l, 0,0,0) = (0,0,0,0) and 1
is an eigenvalue of Τ because T(0,1,0,0) = (0,1,0,0). Thus 0 and 1 must
both be roots of the minimal polynomial of Τ (by 8.36).
The only monic polynomials that divide z(z — l)2 and have 0,1 as roots
are z{z - 1) and z(z — l)2. Because T(T — Ι) φ 0, as is easy to check, this
implies that z{z — l)2 is the minimal polynomial of T.
Of course -there are also many other examples.
23. Suppose V is a complex vector space and Τ € C[V). Prove that V has a
basis consisting of eigenvectors of Τ if and only if the minimal polynomial
of Τ has no repeated roots.
Comment: For complex vector spaces, this exercise adds another
equivalence to the list given by 5.21.
Solution: First suppose that there is a basis («i,..., vn) of V consisting
of eigenvectors of T. Let Ai,..., Am be the distinct eigenvalues of T. Then
for each vj, there exists At with (T - \kl)vj = 0. Thus
{T-\iI)...{T-^1)4 = 0
for each j (because all the operators in sight commute, for each j the
appropriate Τ — \kl can be moved to the last position in the product above). An
operator that sends each vector in a basis to the 0 vector is the 0 operator,
so
(T-\lI)...(T-\mI) = 0.
Thus the polynomial (z — Ai)... (z - Am) when applied to Τ gives 0. Thus
the minimal polynomial of Γ is a divisor of {z — \\)... (z — Am) (by 8.34).
Because (z — Ai)... (z - Am) has no repeated roots, this implies that the
minimal polynomial of Τ has no repeated roots, as desired. (Because each
eigenvalue of Τ must be a root of the minimal polynomial of T, the minimal
polynomial of Τ actually equals (z - Ai)... (z - Am).)
To prove the implication in the other direction, now suppose that the
minimal polynomial of Τ has no repeated roots. Letting Ai,..., Am denote
the distinct eigenvalues of T, this means that the minimal polynomial of Τ
equals
(ζ- Χι)...(ζ- Am).

105
Chapter 8. Operators on Complex Vector Spaces
Thus
(T-XlI)...(T-XmI) = 0.
Let Um be the subspace of generalized eigenvectors corresponding to the
eigenvalue Am. Recall that Um is invariant under Τ (see 8.23(b)). Suppose
υ € Um- Let и = (Τ - Xml)v, so и € Um. The equation above implies that
(T\Um - λι J)... {T\Um - Xm-JJu ={T- λι/)... (Г - XmI)v
= 0.
Because (T — XmI)\(jm is nilpotent (see see 8.23(c)), 0 is the only eigenvalue
of (Г - XmI)\Um (this follows from 8.26 and 5.18). Thus T\Um - Ay/ is
invertible (as an operator on Um) for j = 1,..., m — 1. The equation above
thus implies that и = 0. In other words, υ is an eigenvector of T.
We have shown that every generalized eigenvector of Τ corresponding to
the eigenvalue Am is an eigenvector of T. There is nothing special about the
eigenvalue Am—we could have relabeled the eigenvalues so that any of them
was called Am. Thus we can conclude that every generalized eigenvector of
Τ is actually an eigenvector of T. Because there is a basis of V consisting
of generalized eigenvectors of Τ (see 8.25), this means that there is a basis
of V consisting of eigenvectors of T, as desired.
24. Suppose V is an inner-product space. Prove that if Γ € £>{V) is normal,
then the minimal polynomial of Τ has no repeated roots.
Solution: Suppose Τ € £(V) is normal. Let ρ denote the minimal
polynomial of T. Suppose A € F is an eigenvalue of T. We can write
ρ(ζ) = (ζ-ΧΓς(ζ)
where m is a positive integer and q is a mojaic polynomial such that д(А) φ 0.
Our goal is to prove that m = 1, which implies that ρ has no repeated roots.
Because p[T) = 0, we know that (T - XI)mq(T) = 0. This is equivalent
to the statement that
range q(T) С null(T - A/)m.
Because Τ is normal, so is Τ - XI, and thus null(T - A/)m = null(T - XI)
(by Exercise 7 in Chapter 7). Hence the set inclusion above becomes
range9(T)CnuU(T-A/),

106
Chapter 8. Operators on Complex Vector Spaces
which implies that (T—\I)q(T) = 0. Letting pi(z) = (z—\)q(z), this means
that pi-is a monic polynomial with the property that Pi(T) =0. If m > 1,
then the degree of pi would be less than the degree of p, contradicting the
definition of minimal polynomial. Thus we can conclude that m = 1, as
desired.
Comment: The proof given above works on both real and complex
vector spaces. If V is a complex vector space, then this exercise can be done
by using the complex spectral theorem (7.9) and Exercise 23 of this chapter.
25. Suppose Τ € C(V) and ν € V. Let ρ be the monic polynomial of smallest
degree such that
p(T)v = 0.
Prove that ρ divides the minimal polynomial of T.
Solution: Let q denote the minimal polynomial of T. By the division
algorithm (4.5), there exist polynomials s,r € V(F) such that
q = sp + r
and deg г < deg p. Thus
q(T)v = s(T)p(T)v + r(7>.
Because q(T) = 0 and p(T)v = 0, the equation above shows that r(T)v =
0. This implies that г = 0 (otherwise we could multiply г by a scalar to
get a monic polynomial with degree smaller than deg ρ that when applied
to Τ gives an operator having υ in its null space, which would contract
the definition of ρ as the monic polynomial of smallest degree with this
property).
Using the information that г = 0, rewrite the formula above for q as
q = sp.
Thus ρ divides q, the minimal polynomial of T.
26. Give an example of an operator on C4 whose characteristic and minimal
polynomials both equal z(z — l)2(z — 3).
Solution: Define Τ € £(C4) by
T(wi, W2,W3,Wi) = (Q,W2 + W3,W3,3W4).

107 Chapter 8. Operators on Complex Vector Spaces
An easy computation shows that T(T — I)2{T — 31) = 0. Thus the
minimal polynomial of Γ is a divisor of z(z — l)2(z — 3) (by 8.34).
Note that 0 is an eigenvalue of Τ because T(l,0,0,0) = (0,0,0,0) and 1
is an eigenvalue of Τ because T(0,1,0,0) = (0,1,0,0) and 3 is an eigenvalue
of Τ because T(0,0,0,1) = (0,0,0,3). Thus 0, 1, and 3 must be roots of the
minimal polynomial of Τ (by 8.36).
The only monic polynomials that divide z(z — l)2(z—3) and have 0,1,3 as
roots are z(z-l)(z-3) and z(z-l)2(z-Z). BecauseT(T- 1)(Τ-3) φ 0, as
is easy to check, this implies that z(z — l)2(z — 3) is the minimal polynomial
of T, as desired.
Because Τ is an operator on a four-dimensional complex vector space and
the minimal polynomial of Τ has degree 4, the characteristic polynomial of
Τ (which is a monic polynomial of degree 4 that is divisible by the minimal
polynomial of T) must equal the minimal polynomial of T.
Of course there are also many other examples.
27. Give an example of an operator on C4 whose characteristic polynomial
equals z(z — l)2(z - 3) and whose minimal polynomial equals z(z - l)(z - 3).
Solution: Define Τ € £(C4) by
T{wi,W2,WZ,Wi) = (0,tu2,tu3,3tu4).
An easy computation shows that T(T—I)(T—31) = 0. Thus the minimal
polynomial of Τ is a divisor of z{z — l)(z — 3) (by 8.34).
Note that 0 is an eigenvalue of Τ because T(l,0,0,0) = (0,0,0,0) and 1
is an eigenvalue of Τ because Γ(0,1,0,0) = (0,1,0,0) and 3 is an eigenvalue
of Г because T(0,0,0,1) = (0,0,0,3). Thus 0, l,and3 must be roots of the
minimal polynomial of Τ (by 8.36).
The only monic polynomial that is a divisor of z(z — l)(z — 3) and has
0,1,3 as roots is z(z — \)(z — 3). Thus z(z — l)(z — 3) is the minimal
polynomial of T, as desired.
Note that every vector in {(0,102,103,0) : 102,103 € C) is an eigenvector
of Τ corresponding to the eigenvalue 1. Thus the eigenvalue 1 of Γ has
multiplicity at least 2. The eigenvalues 0 and 3 of Γ have multiplicity at
least 1.
Because Γ is an operator on a four-dimensional complex vector space, the
sum of the multiplicities of all the eigenvalues equals 4 (by 8.18). Thus the
each use of the phrase "at least" in the previous paragraph can be replaced
by "equal" because if any of the eigenvalues had larger multiplicity, the sum
of the multiplicities of all the eigenvalues would exceed 4.

108
Chapter 8. Operators on Complex Vector Spaces
Because 0 and 3 are eigenvalues of Τ with multiplicity 1 and 1 is an
eigenvalue of Τ with multiplicity 2 and Τ has no other eigenvalues (the
multiplicities of the eigenvalues mentioned already sum to 4), the characteristic
polynomial of Τ equals z(z - l)2(z - 3), as desired.
Of course there are also many other examples.
28. Suppose Go,... ,αη-ι € C. Find the minimal and characteristic polynomials
of the operator on Cn whose matrix (with respect to the standard basis) is
0 —ao
1 0 -ai
1 '" · -a2
0 -a„_2
1 -an_!
Comment: This exercise shows that every monic polynomial is the
characteristic polynomial of some operator.
Solution: Suppose that Τ € £(Cn) has matrix as above with respect
to the standard basis (ei,..., e„) of C". Thus
Tex = e2
T2ei = Te2 = e3
r'-'ei =Te„_i = e„
T^ei = Ten = -aoei — a\ei Οη-ιβη.
Thus
(еьГеьГ2еь...,Гп-1е1) = (еье2,ез,...,еп).
In particular, (ei,Te\,T2ei,... ,Tn~le\) is linearly independent. Thus if ρ
is a monic polynomial with degree less than n, then p{T)e\ φ 0. Thus the
minimal polynomial of Τ must have degree n.
Writing l^ei as a linear combination of (ei,T'ei,T2ei,...,Tn_1ei) is
possible in only one way:
"icy = — clqBi — a\Te\ — · · · — On-iT" e\.
Thus setting

109 Chapter 8. Operators on Complex Vector Spaces
p(z) =go +a\z-\ |-ап_1.гп-1 + zn,
we see that ρ is the only monic polynomial of degree η such that p(T)e\ = 0.
Hence ρ must equal the minimal polynomial of T.
Because Τ is an operator on an η-dimensional complex vector space and
the minimal polynomial of Τ has degree n, the characteristic polynomial
of Τ (which is a monic polynomial of degree η that is divisible by the
minimal polynomial of T) must equal the minimal polynomial of T. Thus the
characteristic polynomial of Τ also equals p.
29. Suppose N € C(V) is nilpotent. Prove that the minimal polynomial of N
is zm+l, where m is the length of the longest consecutive string of l's that
appears on the line directly above the diagonal in the matrix of N with
respect to any Jordan basis for N.
Solution: Suppose (vi,...,vn) is a Jordan basis for N and that m
equals the length of the longest consecutive string of l's that appears on
the line directly above the diagonal of M(N, (v\,...,υη)). The diagonal of
Λ4(Ν, (v\, ... ,vn)) contains only 0's (by Exercise 6 of this chapter). Thus
λ4(Ν, (υι,... ,vn)) is a block diagonal matrix whose blocks have the form
"01 0 *
"·. 1
0 0
and the largest such block in an (τη + l)-by-(m + 1) matrix. For each Vj,
we see that Nm+lVj = 0. Because Nm+l equals 0 on a basis of V, we
conclude that Nm+l = 0. Thus the minimal polynomial of N most divide
zm+i ^y 8.34) and hence must be of the form zk for some к < m + 1. But
there is a basis vector Vj such that N^-Vj ч= Vj-m φ 0. Thus Nm φ 0, which
implies that the minimal polynomial of ./V equals zm+l.
30. Suppose V is a complex vector space and Τ € C[V). Prove that there
does not exist a direct sum decomposition of V into two proper subspaces
invariant under Τ if and only if the minimal polynomial of Τ is of the form
(z-\)d'mV for some λ € С.
Solution: First suppose that there does not exist a direct sum
decomposition of V into two proper subspaces invariant under T. Thus Τ has only
one eigenvalue (by 8.23), which we will call λ.

110
Chapter 8. Operators on Complex Vector Spaces
There is a Jordan basis of Τ (by 8.47), meaning that with respect to this
basis Τ has a block diagonal matrix
Ai 0
0 Am
where each Aj is an upper-triangular matrix of the form
λ 1 0
Aj =
0
1
λ
If m > 1, then we could let U denote the span of the basis vectors
corresponding to A\ and let W denote the span of the basis vectors corresponding
to U2, ■ ■., Um; we would have V = U® W, where U and W would be proper
subspaces of V invariant under T. This contradiction shows that m = 1.
Thus the matrix of Τ with respect to our Jordan basis is just A\. In other
words, there is a basis (yi,..., v<am y) of V such that the matrix of Τ - XI
with respect to this basis is
0
0
0
1
0
The previous problem now implies that the minimal polynomial of Τ - XI
equals zdim v. This clearly implies that the minimal polynomial of Τ equals
(z -X)dimV, as desired.
To prove the implication in the other direction, suppose that the minimal
polynomial of Τ equals (z - X)dimV. Suppose that there exist two proper
subspaces J7i, U2 of V, each invariant under T, such that V = U\ φ С/г-
Let pi denote the minimal polynomial of T\u1 and рг denote the minimal
polynomial of T\u2. If ui € U\, then
(ΡΙΡ2)(Ϊ>1=Ρ2(Ϊ>1(2>1
= 0.
Similarly, if ич € U2, then

Ill
Chapter 8. Operators on Complex Vector Spaces
31.
{piP2)(T)u2 = Pi(T)p2(T)u2
= 0.
Because every vector in V can be written in the form щ +U2, where щ € U\
and U2 € U2, the equations above imply that (p\P2)(T) = 0. Thus the
minimal polynomial of T, which equals (z - X)dmxV, is a divisor of рда
(by 8.34).
The degree of the monic polynomial p\p2 equals the degree of pi plus the
degree of P2, which is less than or equal to dimJ7i + dim[/2, which equals
dim V. Because (z - X)dhaV is a divisor of P1P2, this implies that
pl(z)P2(z) = (z-\f™v.
This implies that
Pl(z) = (z - X)dlmUi and p2(z) = (z - X)dimU\
Let m = max{dimJ7i,dimJ72}. Let p(z) = (z - X)m. The equations above
show that ρ(Τ)\υι = 0 and p(T)\u2 = 0. Thus p(T) = 0. Because the degree
of ρ is less than dim V, this contradicts our hypothesis that (z-X)dlmV is the
minimal polynomial of T. This contradiction means that our assumption
that there exists a direct sum decomposition of V into two proper subspaces
invariant under Τ must have been false, as desired.
Suppose Τ € C{V) and (υι,...,υη) is a basis of V that is a Jordan
basis for T. Describe the matrix of Τ with respect to the basis (υη,..,υι)
obtained by reversing the order of the v's.
Solution: The matrix of Τ with respect to (υι,,
agonal matrix
Ai 0
0 Am
where each Aj is an upper-triangular matrix of the form
Xj 1 0
. ,vn) is a block di-
Aj =
1
Xj

112
Chapter 8. Operators on Complex Vector Spaces
Temporarily fix j, and let (ui,..., Uk) be the part of (v\,..., vn)
corresponding to the block Aj. Thus
Tu\ = XjUi
TU2 =U\ + \jU2
Tuz = u2 + Aju3
Тик = ujt_i +XjUk.
Write the equations above in the form
Тик = \jUk +Uk-i
Tuk-l = XjUk-l +Uk-2
TU2 = AjU2 + Ul
Tu\ = \jU\.
Thus the matrix of Т^ра^щ,...,Ufc) ^vit-h respect to (ujt,... ,щ) is
0
1 Xj
In other words, Bj is obtained from Aj by reflection across the diagonal, so
in Bj the 1 's he below the diagonal instead of above it. Now we see that the
matrix of Τ with respect to (vn,... ,v\) is the block diagonal matrix
0 Bi
Bj =

Chapter 9
Operators on
Real Vector Spaces
Prove that 1 is an eigenvalue of every square matrix with the property that
the sum of the entries in each row equals 1.
Solution: Suppose that A is an n-by-n matrix such that the sum of
the entries in each row of A equals 1. Let χ be the n-by-1 matrix all of
whose entries equal 1. Then from the definition of matrix multiplication we
see that the entry in row j, column 1 of Ax equals that sum of the entries
in row j of A, which equals 1. Thus Ax = x, which implies that 1 is an
eigenvalue of A.
Consider a 2-by-2 matrix of real numbers
A =
а с
b d
Prove that A has an eigenvalue (in R) if and only if
(a - d)2 + Abe > 0.
Solution: A number λ € R is an eigenvalue of A if and only if there
exist numbers x, у € R, not both 0, such that
α с
b d
X
. У .
= λ
χ
. У .
Q££ -U (ΥΠ
The left side of this equation equals [ , ], so the equation above is
equivalent to the system of equations
113

114
Chapter 9. Operators on Real Vector Spaces
3.
(a - λ)χ + су = О
Ъх + (d - Х)у = 0.
It is easy to see that this system of equations has a solution other than
x = У = 0 if and only if
(a - X)(d - λ) = be,
which is equivalent to the equation
X2-(a + d)X + (ad-bc) = 0.
There is a real number λ satisfying the equation above if and only if
(a + d)2 - 4(ad - be) > 0.
The left side of the inequality above equals (a - d)2 + 4bc, and thus we
conclude that A has a real eigenvalue if and only if
(a - d)2 + Abe > 0.
Suppose A is a block diagonal matrix
Ai 0
A =
0
where each Aj is a square matrix. Prove that the set of eigenvalues of A
equals the union of the eigenvalues of A\,..., Am.
Solution: Suppose that Aj has size rij-by-rij. Let
x\
χ =
If
where each Xj is an rij-by-l matrix. Then
Aixi
Ax =
sLmXri
Thus the equation Ax = Xx is equivalent to the system of equations

115
Chapter 9. Operators on Real Vector Spaces
A\X\ = Xxi
A-mXxn — AXm.
Suppose that λ is an eigenvalue of A. Then there is a nonzero vector χ
satisfying the system of equations above. Because χ is nonzero, there exists
к such that xjt is nonzero. Because A^Xk = X%k, this implies that λ is an
eigenvalue of Ajt- Thus λ is in the union of the eigenvalues of A\,..., Am,
as desired.
To prove the implication in the other direction, suppose now that λ is in
the union of the eigenvalues of Ai,..., Am. Then there exists к such that λ
is an eigenvalue of A^. Thus there exists a nonzero n^-by-l vector Xk such
that AfcXfc = Axjfc. For j φ к, define Xj to be the rij-by-1 matrix all of whose
entries equal 0, and define χ to be the matrix determined by xi,... ,xm as
above. Then χ is nonzero (because xjt is nonzero) and Ax = λχ, which
shows that λ is an eigenvalue of A, as desired.
4. Suppose A is a block upper-triangular matrix
Ai *
A =
0 Am
where each Aj is a square matrix. Prove that the set of eigenvalues of A
equals the union of the eigenvalues of A\,..., Am.
Comment: Clearly Exercise 4 is a stronger statement than Exercise 3.
Even so, students may want to do Exercise 3 first because it is easier than
Exercise 4.
Solution: We will prove that 0 is an eigenvalue of A if and only if 0 is
an eigenvalue of at least one of the Ak's. This is all we need to do, because
for arbitrary λ € F, we can replace A with A - XI and each Ak with Ak - XI,
concluding that λ is an eigenvalue of A if and only if λ is an eigenvalue of at
least one of the A^'s. This last statement implies that the set of eigenvalues
of A equals the union of the eigenvalues of A\,..., Am.
Suppose that A has size n-by-n and each Aj has size ny-by-ny. We can
write a typical n-by-1 matrix χ in the form
Xl
χ =
Xt,

116 Chapter 9. Operators on Real Vector Spaces
where each Xj is an ny-by-l matrix. The product Ax can be computed by
multiplying together the block matrices of A and χ given above, using the
same formula as one would use when multiplying matrices of numbers (this
follows from the definition of matrix multiplication).
First suppose that 0 is an eigenvalue of A. Thus there exists a nonzero
n-by-1 matrix χ such that Ax = 0. Write χ in the form above, and let к be
the largest index with nonzero x^; thus we can write
' x\
0
(If к = τη, then the 0's shown above at the tail of χ do not appear.) If we
break Ax into blocks of the same size as was done for x, then the fcth block
of Ax will equal A^Xk', this follows from the block upper-triangular form of
A and the 0's that appear in χ after the fcth block. But Ax = 0, so the fcth
block of Ax equals 0, so A^Xk = 0. Because Xk φ 0, this implies that 0 is
an eigenvalue of Ak, as desired.
To prove the implication in the other direction, suppose now that 0 is an
eigenvalue of some Ak- This means that the operator on Mat(n, 1,F) (the
vector space of n-by-1 matrices) that sends Xk G Mat(n, 1, F) to AkXk is not
injective. Thus this operator is not surjective (by 3.21). Thus the operator
on Mat(ni Η l· Пк, 1, F) that sends
X\
to
A\ *
_ 0 Ak
is not surjective (because the last block in the product above will be AkXk,
which cannot be an arbitrary n^-by-l matrix). Again using 3.21, this means
that the last operator is not injective. In other words, there exists a nonzero
vector
χι
Xk

117
Chapter 9. Operators on Real Vector Spaces
x\
Xk
€Mat(m +--- + пь 1,F)
such that
Αχ
0
A*
x\
Xk
= 0.
Adjoining an appropriate number of 0's, this implies that
Ak
Ajt+i
X\
Xk
0
= 0.
In other words, 0 is an eigenvalue of A, as desired.
Suppose V is a real vector space and Τ € C(V). Suppose α, β € R are
such that T2 + aT + βΐ = 0. Prove that Τ has an eigenvalue if and only if
a2 > 4/3.
Solution: First suppose that Τ has an eigenvalue λ € R. Thus there
exists a nonzero vector υ € V such that Tv = Xv. Applying Τ to both sides
of the last equation, we get T^v — X2v. Thus
0 = (Γ2 + aT + βΙ)υ
= X2v + aXv + βν
= (Χ2 + αΧ + β)ν.
Because ν φ 0, the last equation implies that
λ2 + αλ + β = 0,
which implies (recall that λ, α, and β are all real) that a2 > 4/3, as desired.
To prove the implication in the opposite direction, suppose now that
a2 > 4/3. Then the polynomial χ2 + αχ+β has two real roots, which means
that we can write

118
Chapter 9. Operators on Real Vector Spaces
x2 + ax + β = (χ - г\) (χ - гг)
for some π, гг € R. Thus
0 = Τ2 + αΤ+βΙ
= (Τ - ΓιΙ)(Τ - r2I).
In particular, (Τ - riI)(T - Γ2Ι) is not injectivc, which implies that at least
one of Τ — r\I and Τ — ril is not injective. In other words, at least one of
r\, гг must be an eigenvalue of T. Thus Τ has an eigenvalue, as desired.
6. Suppose V is a real inner-product space and Τ € C{V). Prove that there
is an orthonormal basis of V with respect to which Τ has a block upper-
triangular matrix
A\ *
where each Aj is a 1-by-l matrix or a 2-by-2 matrix with no eigenvalues.
Solution: We know that there is a basis (υι,..., vn) of V with respect
to which the matrix of Τ has the block upper-triangular form above, where
each Aj is a 1-by-l matrix or a 2-by-2 matrix with no eigenvalues (see 9.4).
Apply the Gram-Schmidt procedure to (v\,..., vn), getting an
orthonormal basis (ei,..., en) of V such that
span(ui,... ,vj) = span(ei,...,ej)
for j = Ι,.,.,η (see 6.20). This condition on the spans implies that the
matrix of Τ with respect to (ei,...,en) is also a block upper-triangular of
the form above, where each Aj is a 1-by-l matrix or a 2-by-2 matrix (these
Aj's may differ from the Aj's corresponding to the matrix of Τ with respect
to (υι,..., vn)). All that remains is to show that, if necessary, we can modify
our orthonormal basis so that none of the 2-by-2 blocks on the diagonal have
eigenvalues.
Suppose that Aj is a 2-by-2 block on the diagonal of the matrix of Τ with
respect to (ei,...,en) and that Aj has an eigenvalue λ. Thus there exist
x,y € R, not both 0, such that
X
У
=
X
У

119 Chapter 9. Operators on Real Vector Spaces
Let ejt, ejt+i denote the basis vectors corresponding to Aj. Then, as is easy
to verify,
T(xek + yek+i) = u + X(xek + yek+i)
for some и € span(ei,..., ejt-i). Let
= xek + yek+i
Jk ||xefc + yefc+ill
and choose Д+ι € V such that (fk, fk+i) is orthonormal and
span(/fc, fk+i) = span(efc,efc+i).
Then the matrix of Τ with respect to the orthonormal basis
(ei,. ..,ejt-i,/fc,/fc+iiefc+2i··· a)
will be a block upper-triangular matrix with the same entries on the diagonal
as previously, except that Aj will be replaced by the 2-by-2 matrix
λ * '
0 * "
In other words, where we had Ay on the diagonal, we can now think of
having two 1-by-l matrices on the diagonal (and we still have a block upper-
triangular matrix because of the 0 in the lower-left entry of the matrix
above).
Repeating, when necessary, the procedure described above, we obtain an
orthonormal basis of V with respect to which Τ has a block upper-triangular
matrix
A\ *
where each Aj is a 1-by-1 matrix or a 2-by-2 matrix with no eigenvalues.
7. Prove that if Γ € C(V) and j is a positive integer such that j < dim V, then
Τ has an invariant subspace whose dimension equals j — 1 or j.
SOLUTION: Suppose Τ € C(V) and j is a positive integer such that
j < dim V. If V is a complex vector space, then Τ has an invariant subspace

120 Chapter 9. Operators on Real Vector Spaces
whose dimension equals j (by Exercise 17 in Chapter 5). So we can assume
that V is a real vector space.
By 9.4, there is a basis (υι,..., vn) of V with respect to which Τ has a
block upper-triangular matrix
A\ *
where each A^ is a 1-by-l matrix or a 2-by-2 matrix. Either Vj-\ or Vj must
be the last vector in a block of vectors corresponding to some A^. Thus
either span(ui,..., i>j-i) or span(ui,..., Vj) must be invariant under T.
8. Prove that there does not exist an operator Τ € £(R7) such that Τ2 +Τ + Ι
is nilpotent.
Solution: Suppose Τ € £(R7). Prom part (b) of 9.9 (combined with
9.4, which insures that Τ has a matrix of the form 9.10 with respect to some
basis of R7), we see that
dimnuU(r2 + T + /)7
must be an even integer. In particular, dimnuu(!r2 + Τ + Ι)7 φ 7, which
implies that null(T2 + Τ +1)7 φ R7, which implies that (Τ2 + Τ +1)7 φ 0,
which implies that Τ is not nilpotent (by 8.8).
9. Give an example of an operator Τ € £(C7) such that T2 + T+I is nilpotent.
Solution: Let λ = (-1 + 3i)/2. Define Τ € £(C7) by Τ = XI. Then
T2 + Τ + / = (λ2 + λ + 1)/
= 0.
In particular, Τ2 + Τ + I is nilpotent.
10. Suppose V is a real vector space and Τ € C(V). Suppose α, β € R are such
that a2 < 40. Prove that
null(r2 + aT + /3/)fc
has even dimension for every positive integer k.
Solution: Let к be a positive integer, and let U = mill(T2+aT+fiI)k.
We need to prove that dim U is even.

121
Chapter 9. Operators on Real Vector Spaces
Because U is invariant under Τ (by 8.22, with p(x) = (x2 + ax + /3)fc), we
can define S € C(U) by S = Τ\υ· Clearly (S2 + aS + pi)k is the 0 operator
on U. Thus S2 + aS + βΐ is a. nilpotent operator on U, which implies that
nuU(52 +aS +pi)dlmU = U (by 8.8). Now part (b) of 9.9, applied to S and
U instead of Τ and V, shows that dirnC/ is an even integer, as desired.
11. Suppose V is a real vector space and Τ € C(V). Suppose α, β € R are such
that a2 < 4/3 and T2 + aT + βΐ is nilpotent. Prove that dim V is even and
(T2+aT+/3/)dimV//2 = 0.
Solution: Let S = T2 + aT + βΐ. Because S is nilpotent, there is a
smallest positive integer m such that S™1 = 0. Thus
{0} = nullS0 С nullS С nullS2 С ...С nuUSm = V,
1 J + + + +
where the proper inclusions come from 8.5. The previous exercise states
that each nullSfc has even dimension (in particular V, which equals nullS"1,
has even dimension). Hence the dimension must increase by at least 2 in all
the proper inclusions above. Thus dim S"1 > 2m, which clearly implies that
m < (dim V)/2. Because S™ = 0, this implies that S6™ vl2 = 0, as desired.
12. Prove that if Г € £(R3) and 5,7 are eigenvalues of T, then Τ has no
eigenpairs.
Solution: Suppose that Τ € £(R3) and 5,7 are eigenvalues of T. Of
course each of these two eigenvalues must have multiplicity at least 1. By
9.17, the sum of the multiplicities of all the eigenvalues of Τ plus twice the
sum of the multiplicities of all the eigenpairs of Τ equals 3. Because the sum
of the multiplicities of all the eigenvalues of Τ is at least 2, there is no room
for an eigenpair, which would add at least 2 more to the sum (because we
take twice the sum of the multiplicities of all the eigenpairs). Thus Τ has
no eigenpairs.
13. Suppose V is a real vector space with dim V = η and Τ € C{V) is such that
nulir^VnullT"-1.
Prove that Τ has at most two distinct eigenvalues and that Τ has no
eigenpairs.
Solution: Because пи]1Тп~2 ф null Τ"-1, we see that dim null Tj is
at least 1 more than dim null T·7-1 for j = 1,... ,n - 1 (by 8.5). Thus

122 Chapter 9. Operators on Real Vector Spaces
dim null T"1-1 > η - 1. In particular, 0 is an eigenvalue of Τ with
multiplicity at least η — I. Because the sum of the multiplicities of all the
eigenvalues of Τ plus the sum of twice the multiplicities of all the eigenpairs
of Τ equals η (by 9.17), this implies that Τ can have no eigenpairs and at
most one additional eigenvalue.
14. Suppose V is a vector space with dimension 2 and Τ € C(V). Prove that if
а с
b d
is the matrix of Τ with respect to some basis of V, then the characteristic
polynomial of Τ equals (z - a)(z - d) - be.
Comment: As usual unless otherwise specified, here V may be a real
or complex vector space.
Solution: Let q(z) = (z - a)(z - d) - be. If {νγ,υ-ΐ) is the basis of V
with respect to which Τ has the matrix above, then
Tv\ = av\ + bi)2 and Тщ = evi + dvi.
Prom these equations you can easily verify that g(T)vi = 0 and q(T)t)2 = 0.
Because q(T) is 0 on a basis of V, we conclude that q{T) = 0.
Because q is a monic polynomial of degree 2 and q(T) = 0, we conclude
that the minimal polynomial of Τ has degree 1 or 2.
Suppose first that the minimal polynomial of Τ has degree 2. Because
the minimal polynomial of Γ is a divisor of q (by 8.34), and because a monic
polynomial of degree 2 can be a divisor of another monic polynomial of
degree 2 only if the two polynomials are equal, we conclude that q is the
minimal polynomial of T. The Cayley-Hamilton theorem now implies that
q is a divisor of the characteristic polynomial of T, which is also a monic
polynomial of degree 2. This implies that q is the characteristic polynomial
of 7\ as desired.
Now consider the only remaining possible case, which is that the minimal
polynomial of Τ has degree 1, meaning that it equals ζ — λ for some λ € F.
This implies that Τ = λ/, which implies that the characteristic polynomial
of Τ equals (ζ—λ)2. Because Τ = XI, we must have α = d = X and b = с = 0.
Thus q(z) = (z - λ)2. In particular, q is the characteristic polynomial of T,
as desired.
Comment: Note that we did not need to find the eigenvalues of Γ to
do this exercise.

123
Chapter 9. Operators on Real Vector Spaces
15. Suppose V is a real inner-product space and S € C(V) is an isometry. Prove
that if {α, β) is an eigenpair of S, then /3=1.
Solution: There is a basis of V with respect to which S has a block
diagonal matrix, where each block on the diagonal is a 1-by-l matrix or a
2-by-2 matrix of the form
cos θ -sin θ
sin θ cos θ '
with θ € (0, π) (see 7.38). The characteristic polynomial of the matrix above
is (x - cos Θ)2 + sin2 Θ, which equals x2 — 2(cos θ)χ + 1.
If (α, β) is an eigenpair of S, then
dim null(52 + aS + /3/)<Um v > 0,
which implies that x2 + ax + β is the characteristic polynomial of a 2-by-2
matrix of the form displayed above (see 9.9). Thus /3=1.

Chapter 10
Trace and Determinant
1. Suppose that Τ € C{V) and (vi,...,vn) is a basis of V. Prove that
Л4(Т, (v\,... ,vn)) is invertible if and only if Τ is invertible.
Solution: First suppose that M(T) is an invertible matrix (because the
only basis is sight is (v\,..., vn), we can leave the basis out of the notation).
Thus there exists an n-by-n matrix В such that
M(T)B = BM{T) = I.
There exists an operator S € C(V) such that M(S) = В (see 3.19). Thus
the equation above becomes
M(T)M(S) =M(S)M(T) = I,
which we can rewrite as
M{TS) = M{ST) = M(I),
which implies that
TS = ST = I.
Thus Τ is invertible, as desired, with inverse S.
To prove the implication in the other direction, suppose now that Τ is
invertible. Thus there exists S € C(V) such that
TS = ST = I.
This implies that
124

125
Chapter 10. Trace and Determinant
M{TS) = M{ST) = M(I),
which implies that
M(T)M(S) = M(S)M(T) = I.
Thus M(T) is invertible, as desired, with inverse M(S).
2. Prove that if A and В are square matrices of the same size and AB = I,
then В A = I.
Solution: Suppose that A and В are n-by-n matrices and AB = I.
There exist S, Τ € £(Fn) such that
M{S) = A and M(T) = B;
here we are using the standard basis of Fn (the existence of S, Τ € £(Fn)
satisfying the equations above follows from 3.19). Because AB = I, we have
M(S)M(T) = I, which implies that M(ST) = M(I), which implies that
ST = I, which implies that TS = I (by Exercise 23 in Chapter 3). Thus
В А = M(T)M(S)
= M(TS)
= M(I)
= 1.
3. Suppose Τ € C(V) has the same matrix with respect to every basis of V.
Prove that Γ is a scalar multiple of the identity operator.
Solution: We begin by proving that (υ, Τν) is linearly dependent for
every υ € V. To do this, fix υ € V, and suppose that (υ, Τυ) is linearly
independent. Then (v, Tv) can be extended to a basis (v, Tv, щ,..., u„.)
of V. The first column of the matrix of Τ with respect to this basis is
' 0 "
1
0
_ 0 _
Clearly (2v, Tv, щ,..., Un) is also a basis of V. The first column of the
matrix of Τ with respect to this basis is

126
Chapter 10. Trace and Determinant
0
2
0
0 _
Thus Τ has different matrices with respect to the two bases we have
considered. This contradiction shows that (υ, Τυ) is linearly dependent for every
ν € V. This implies that for every vector in V is an eigenvector of T. This
implies that Γ is a scalar multiple of the identity operator (by Exercise 12
in Chapter 5).
4. Suppose that (щ,..., un) and (υι,..., vn) are bases of V. Let Τ € £(V) be
the operator such that Tvk = Uk for к = 1,..., η. Prove that
Μ (Τ, (υι,..., vn)) = Μ (I, (щ,..., tin), (v\,. · ·, vn)).
Solution: Fix к. Write
uk = αιυι Η hanvn,
where a\,...,On € F. Because Tv^ = щ, the fcth column of the matrix
M. (T, (v\ ,..., vn)) consists of the numbers a\,..., an. Because 1щ = «jt>
the fcth column of Μ (I, (щ,..., un), (υι,..., vn)) also consists of the
numbers αϊ,... ,αη.
Because M(T, (νι,...,νη)) and M(I, {щ,... ,un),{vi, ...,vn)) have the
same columns, these two matrices must be equal.
5. Prove that if β is a square matrix with complex entries, then there exists
an invertible square matrix A with complex entries such that A~lBA is an
upper-triangular matrix.
SOLUTION: Suppose β is an n-by-n matrix with complex entries. Let
(ei,... ,en) denote the standard basis of Cn. There exists Τ € £(Cn) such
that M(T, (ei,..., e„)) = В (see 3.19).
There is a basis (υι,..., vn) of V such that M. (T, (v\,..., vn)) is an upper-
triangular matrix (see 5.13). Let A = Λ4((ν\,. ..,vn), (e\,... ,en))- Then
A is invertible (by 10.2) and
A-lBA = A-lM{T,{ei,...,en))A
= M(T,(vu...,vn)),

127 Chapter 10. Trace and Determinant
where the second equality comes from 10.3. Thus A~XBA is an upper-
triangular matrix.
6. Give an example of a real vector space V and Τ € C(V) such that
trace(T2) < 0.
Solution: Define Τ € £(R2) by
T(x,y) = (-y,x).
Then Γ2 = -/, so traceiT2) = -2.
7. Suppose V is a real vector space, Τ 6 C(V), and V has a basis consisting of
eigenvectors of T. Prove that trace(T2) > 0.
Solution: Let (v\,..., vn) be a basis of V consisting of eigenvectors
of T. Thus there exist λι,...,λη 6 R such that Tv3- = XjVj for each j.
Clearly the matrix of T2 with respect to the basis (v\,..., vn) is the diagonal
matrix
"λ,2 0 "
_ 0 λη2
Thus trace Τ2 = λι2 + · · · + λ„2 > 0.
8. Suppose V is an inner-product space and v, w € C{V). Define Τ € C(V) by
Tu = (u, v)w. Find a formula for trace T.
SOLUTION: First suppose that υ φ 0. Extend (тглт) to an orthonormal
basis (A,ei,... ,en) of V. Note that for each j, we have Tej = 0 (because
(ej,v) = 0). The trace of Τ equals the sum of the diagonal entries in the
matrix of Τ with respect to the basis (Л, ei,..., en). Thus
tracer = (Пщ), щ) + (Τβχ,βχ) + ■■■ + (Te^e»)
= (w,v).
If ν = 0, then Τ = 0 and so trace Γ = 0 = (w,v). Thus we have the
formula

128
Chapter 10. Trace and Determinant
traceT = (го, υ)
regardless of whether or not ν = 0.
9. Prove that if Ρ 6 C(V) satisfies P2 = P, then trace Ρ is a nonnegative
integer.
Solution: Suppose that Τ € C{V) satisfies P2 = P. Let (щ,... ,ιι™)
be a basis of range Ρ and let {v\,..., vn) be a basis of null P. Then
(Ul,...,lim,l'i,...,l'n)
is a basis of V (this holds because V = range Γ θ null Γ; see Exercise 21
in Chapter 5). For each Uj we have Pu3- = щ and for each Vk we have
Pi>k = 0. Thus the matrix of Ρ with respect to the basis above of V is
a diagonal matrix whose diagonal contains m l's followed by η 0's. Thus
trace Ρ = m, which is a nonnegative integer, as desired. In fact, we have
shown that
trace Ρ = dim range P.
10. Prove that if V is an inner-product space and Τ € C(V), then
traceT* = traceT.
Solution: Suppose that V is an inner-product space and Τ € £{V).
Let (ei,..., en) be an orthonormal basis of V. The trace of any operator on
V equals the sum of the diagonal entries on the matrix of the operator with
respect to this basis. Thus
trace Γ* = (Γ·βι,βι> + ··· + (Γ*βη, en)
= (ei,Tei>+--- + (en,ren>
= (Tei,ei)+·*·· +(Ten, e„)
= (rei,ei>+--- + (Ten,en>
= traceT1.
11. Suppose V is an inner-product space. Prove that if Τ € C(V) is a positive
operator and trace Τ = 0, then Τ = 0.
Solution: Suppose Τ € C{V) is a positive operator and traceT = 0.
There exists an operator S € C{V) such that Τ = S*S (by 7.27). Let
(ei,... ,en) be an orthonormal basis of V. Then

129
Chapter 10. Trace and Determinant
0 = traceT
= (Tel,el)+--- + (Ten,en)
= (S*Sel,el) + --- + (S*Sen,en)
= \\Sel\\2 + --- + \\Sen\\2.
The equation above implies that Sej = 0 for each j. Because S is 0 on a
basis of V, we have 5 = 0. Because Τ = S*S, this implies that Τ = 0.
12. Suppose Τ € £(C3) is the operator whose matrix is
" 51 -12 -21 '
60 -40 -28 .
57-68 1
Someone tells you (accurately) that -48 and 24 are eigenvalues of T.
Without using a computer or writing anything down, find the third eigenvalue
of T.
Solution: The sum of the eigenvalues of Τ equals the sum of the
diagonal terms of the matrix above (both quantities equal trace T). The
sum of the diagonal terms of the matrix above equals 12. The sum of two of
the eigenvalues of T, -48 and 24, equals -24. Because the sum of all three
eigenvalues of Τ must equal 12, the third eigenvalue of Τ must be 36.
13. Prove or give a counterexample: if Τ € C(V) and с € F, then trace(cT) =
с trace T.
Solution: Suppose Τ € C{V) and с € F. To prove that trace(cT) =
с trace T, consider a basis of V. Then trace Τ equals the sum of the diagonal
terms of the matrix of Τ with respect to this basis. The matrix of cT, with
respect to the same basis, equals с times the matrix of T. Thus the sum
of the diagonal terms of the matrix of cT equals с times the sum of the
diagonal terms of the matrix of T. In other words, trace(cT') = с trace T.
14. Prove or give a counterexample: if S,Τ € C(V), then
trace(5T) = (trace S) (trace T).
Solution: Define S, Τ € £(F2) by S(x,y) = T(x,y) = (-y,x). Then
with respect to the standard bases the matrix of S (which of course equals
the matrix of T) is

130
Chapter 10. Trace and Determinant
0 -1
1 0 '
Thus traceS = trace Γ = 0. However, ST = -I, so trace ST = -2. Thus
for this choice of S and T, we have trace(ST) φ (trace S) (trace T).
Of course there are also many other examples.
15. Suppose Τ € C(V). Prove that if trace(5TJ = 0 for all S € C(V), then
T = 0.
Solution: Suppose that trace(5T') = 0 for all S € C(V). Then
trace(TS) = 0 for all S € C(V) (by 10.12). Suppose that there exists υ € V
such that Tv φ 0. Then (Tv) can be extended to a basis (Tv, щ,..., Un)
of V. Define S € C(V) by
S(aTv + b\u\ Η 1- bnun) = av.
Thus S(Tv) = υ and Suj = 0 for eachj. Hence (TS)(Tv) = T(S{Tv)) = Tv
and (TS)(uj) = 0 for each j. This implies that with respect to the basis
(Tv,u\,... ,un), the matrix of TS consists of all 0's except for a 1 in the
upper-left corner. Thus trace(r5) = 1. This contradiction shows that our
assumption that Tv φ 0 must have been false. Thus Tv = 0 for every ν € V,
which means that Τ = 0.
16. Suppose V is an inner-product space and Τ € C(V). Prove that if (ei,..., e„)
is an orthonormal basis of V, then
trace(r*r) = ||rei||2 + --- + ||ren||2.
Conclude that the right side of the equation above is independent of which
orthonormal basis (ei,..., en) is chosen for V.
Solution: Suppose that (ei,..., en) is an orthonormal basis of V. Then
trace T*T = (T*Tei,ei) +■■■ + (T'Ten,^)
= (Te1,Te1) + --- + (Ten,Ten)
= \\Te1\\2 + --- + \\Ten\\2.
Because trace T*T does not depend upon the choice of a basis of V,
the formula above shows that ||Tei||2 + ■ ■ ■ + ||Ге„||2 is independent of the
orthonormal basis (ei,...,en).

131
Chapter 10. Trace and Determinant
17. Suppose V is a complex inner-product space and Τ € C(V). Let Ai,..., λη
be the eigenvalues of T, repeated according to multiplicity. Suppose
αι,ι
<4,n
On,l ··· an,n
is the matrix of Τ with respect to some orthonormal basis of V. Prove that
\Μ\2 + - + \\η\2<ΥΣ\α^\2·
Solution: Suppose that (ei,...,en) is the orthonormal basis with
respect to which Τ has the matrix above. Thus for each k, we have
which implies that
Thus
Тек = alike\ -i 1- anifce„,
||Tefc||2 = |a1,^ + ..- + |anifc|^.
||Γβι||2 + ·.· + ||Γ^||2 = 53Σ|αΜ|2.
fc=l j=l
By the previous exercise, the left side of this equation equals trace(T'*T').
This reduces the exercise at hand to proving that
|А1|2 + --- + |А„|2<1гасе(ГГ).
There is an orthonormal basis (/i,..., /n) with respect to which Τ has
an upper-triangular matrix (by 6.28). The diagonal entries of the matrix of
Τ with respect to (/i,. ·., /n) axe precisely Ai,..., λη (by 8.10), where we
can relabel the eigenvalues of Τ so that they appear in the order λι,..., A„
along the diagonal. In other words,
M{T,(fu- .·,/„)) =
λι
0
A„
From the matrix above, we see that

132
Chapter 10. Trace and Determinant
μ*|2 < ЦГЛЦ2
for each k. Thus
|λ,|2 + ··· + |λη|2<||Τ/,||2 + ··· + ||Γ/η||2
= trace(r*T),
as desired (here the last equality comes from the previous exercise).
18. Suppose V is an inner-product space. Prove that
(S,T) =trace(5T*)
defines an inner product on C(V).
Solution: Suppose that (·,·) is defined as above and R,S,T € C(V).
Then (T,T) = trace(TT*), and thus (T,T) > 0 (by the formula given in
Exercise 16 of this chapter, with Τ replaced with T*). Because TT* is a
positive operator (see 7.27), we also see that (T, T) = 0 if and only if Τ = 0
(by Exercise 11 of this chapter).
Now
(R + S,T) = trace((i? + S)T*)
= trace(flT* + ST)
= trace(RT*) + trace(5T*)
= (R,T) + (S,T),
where the third equality comes from 10.12.
For с € F, we have
(cS, T) = trace(c5T*)
= с trace(ST*)
= c(S,T),
where the second equality comes from Exercise 13 of this chapter.
Finally,
(S,T) = trace(Sr*)
= trace((Sr*)*)
= trace(T5*)

133
Chapter 10. Trace and Determinant
where the second equality comes from Exercise 10 of this chapter.
We have shown that (·, ·) satisfies all the properties required of an inner
product.
Comment: Suppose (ei,... ,βη) is an orthonormal basis of V and
0.1,1 ■■■ ai,n
αη,ι ... αΠιη
is the matrix of Τ with respect to this basis. Then
(T,T> = trace(3T*)
= trace(T*T)
k=l j=\
where the second equality comes from 10.12 and the third equality comes
■from Exercise 16 of this chapter. Thus the norm on C(V) induced by (-,-)
is the same as the standard norm on Fn (here we are identifying each
operator with its matrix, which has n2 entries). Because norms determine
the inner product (see Exercises 6 and 7 in Chapter 6), this means that the
inner product (·, ■) is the same as the standard inner product of Fn (again
using the identification via matrices with respect to the orthonormal basis
(ei,...,e„)).
19. Suppose V is an inner-product space and Τ € C(V). Prove that if
llT'till < ΙΙΓ11ΙΙ
for every υ € V, then Τ is normal.
Solution: Suppose that ||Γ*ν|| < ||Γυ|| for every υ € V. Suppose
и € V with ||u|| = 1. Extend (u) to an orthonormal basis (u,ei,...,en)
of V. Then
traced) = ||7"u||2 + ||Г'в11|2 + · · · + ||T*en||2
<||Ти||2 + ||Ге1||2 + --- + ||Геп||2
= trace(T*T)
= trace(7T*),

134
Chapter 10. Trace and Determinant
where the first and third lines come from Exercise 16 of this chapter and
the last line comes from 10.12. Because the first and last lines above are
equal, we must have equality throughout. Thus ||T*u|| = \\Tu\\. This clearly
implies that
\\T'(au)\\ = \\T(au)\\
for every a G F. Because every vector in V can be written in the form au for
some α G F and some и G V with ||u|| = 1, this implies that \\T*v\\ = \\Tv\\
for every υ G V. This implies that Τ is normal (by 7.6).
Comment: This exercise fails on infinite-dimensional inner-product
spaces, leading to what are called hyponormal operators, which have a well-
developed theory.
20. Prove or give a counterexample: if Τ G C[V) and с G F, then det(cT) =
cdimUdetT.
Solution: Let η = dim V. If A is an n-by-n matrix, then
det(cA) = cndetA
for every с G F (this follows immediately from the definition 10.25). Now
suppose that Τ G C(V). Because det Τ equals the determinant of the matrix
of Τ with respect to any basis (see 10.33), the equation above implies that
for every с G F we have
det(cT) = detM(cT)
= dei{cM[T))
= cn det M(T)
= c"detT,
as desired.
21. Prove or give a counterexample: if S, Τ G £(V), then
det(5 + T) = det S + det T.
Solution: Define S, Τ G £(F2) by
S(x,y) = (x,0) and T(x,y) = (0,y).
Then S and Τ are both not invertible. Thus det S = det Γ = 0 (by 10.14).
However, S + Τ = I and det/ = 1, so det(5 + Τ) φ detS + detT.

135
Chapter 10. Trace and Determinant
Of course there are also many other examples
22. Suppose A is a block upper-triangular matrix
A\ *
A =
0
where each Aj along the diagonal is a square matrix. Prove that
detA = (detAi)...(detAm).
Solution: First consider the case where m = 2, so we can write A in
the form
A =
В
0
*
С
where В is an n-by-n matrix
В =
and С is a p-by-p matrix
C =
bi.i
bn,l
cl,l
°РЛ
Ί,η
bn,n
C1,P
°P,P
Let Qjjt denote the entry in row j, column к of A. Note that a3·^ = 0 if j > η
and k<n (this follows from the block upper-triangular form of A). Because
A is an (n +p)-by-(n +p) matrix, to compute det A we need to consider a
typical permutation (mi,..., nin+p) £ perm(n + p). If any of mi,..., пц is
greater than n, then
Thus in computing det A we need only consider permutations in which the
first η coordinates all come from {Ι,.,.,η}, which means that the last ρ
coordinates all come from {n + 1,..., η + ρ}. We can break any such
permutation (mi,... ,тпп+р) into two pieces: a permutation of {1,... ,n} and
(relabeling η + 1,..., η + ρ as 1,... ,p to correspond to the labeling of the

136
Chapter 10. Trace and Determinant
entries of C) a permutation of {1,..., p}. Clearly the sign of (mi,..., mn+P)
will equal the product of the signs of these two permutations. Putting all
this together, we have
detA= 22 (sign(mi,...,m„+p))amiil ...amn+pin+p
(mi ,...,тпп+р)ерстт(п+р)
Σ [(signO"i,...,yn))(sign(fci1...,fcP))·
(Λ·—Jn)€permn
(fci,...,fcp)€permp
bjul ■ ■ -bjn.nCki.l ■ ■ -Ck^p]
Σ (sign(ii,...,in))bj,,i--.bjnin·
C7b--Jn)€pcrmn
^ (signCX:! *p))cfclli - - - cj^
(fcl,...,fcp)6permp
= (det5)(detC),
completing the proof when m = 2.
Suppose now that m > 2 and
P.P
Writing
we have
Thus
A =
В =
Αι
0
Αι
0
*
"m
*
Άτη— 1
A =
В *
0 Ат
detA = (det#)(detAm)
= (detAi)...(detAm_i)(detAm),

137
Chapter 10. Trace and Determinant
where the first equality holds by the m = 2 case proved above and the
second equality comes from induction on m (meaning that we can assume
the desired result holds when m is replaced with m — 1).
23. Suppose Λ is an n-by-n matrix with real entries. Let S € £(Cn) denote
the operator on Cn whose matrix equals A, and let Τ € £(Rn) denote the
operator on Rn whose matrix equals A. Prove that trace S = trace Τ and
det S = detT.
Solution: The formulas defining the trace and determinant of a matrix
do not depend upon whether we think of the matrix entries as real or complex
numbers. We have traceA = traceS and traceA = tracer (by 10.11).
Thus trace S = trace T. Similarly, we have det A = det S and det A = det Τ
(by 10.33). Thus det S = detT.
24. Suppose V is an inner-product space and Τ € C(V). Prove that
det Γ* = detT.
Use this to prove that |det T\ = det y/T*T, giving a different proof than was
given in 10.37.
Solution: Let η = dim V. If λ € F, then ((T - XI)n)* = {T* - XI)n,
which implies that
dim nuU(T - λ/)η = dim null(T* - AJ)n,
where we have used Exercise 31 in Chapter 7. The equation above shows that
the eigenvalues of T* are precisely the complex conjugates of the eigenvalues
of T, with the same multiplicities. If F = C, the determinant equals the
product of the eigenvalues, counting multiplicity, so we have detT* = detT
(so far just on complex vector spaces).
Now suppose F = R, so we must consider eigenpairs. If α, β € R, then
dim null(T2 + aT + βΙ)η = dim null((T*)2 + aT* + βΙ)η,
again by Exercise 31 in Chapter 7. Thus Τ and T* have the same eigenpairs
with the same multiplicities. From the paragraph above, Τ and T* have
the same eigenvalues with the same multiplicities. Because the determinant
equals the product of the eigenvalues (counting multiplicity) times the
product of the second coordinates of the eigenpairs (counting multiplicity), this
implies that det T* = det Τ = detT.

138
Chapter 10. Trace and Determinant
At this point, we know that det T* = det Τ regardless of whether F = С
or F = R. Thus
(det yfFff = (det VT*T) (det y/T*T)
= det(T*T)
= (det T) (det T)
= (ditT)(detT)
= |det7f.
Taking square roots (and recalling that the positive operator y/T*T has a
nonnegative determinant), we have det VT*T = |det!T|, as desired.
25. Let a,b, с be positive numbers. Find the volume of the ellipsoid
2 2 2
{(s,y,s)€R3:^ + ^ + ^<l}
by finding a set Ω с R3 whose volume you know and an operator Τ € £(R3)
such that Γ(Ω) equals the ellipsoid above.
Solution: Let Ε denote the ellipsoid defined above and let Ω be the
ball with radius 1 defined by
Ω = {(x,y, z) € R3 : x2 + y2 + z? < 1}.
Of course volume Ω = |тг. Define Τ € £(R3) by
T{x,y,z) = {ax,by,cz).
If (x, y, ζ) € Ω, then
Μ2 , (by)2 , (c*)2 · 2 о г
<1,
which shows that T(x, y, z) € E. Thus Γ(Ω) С Е.
Conversely, if (χ, у, ζ) € Ε, then
/Χ\2 /y\2 fZ\2 Χ2 у2 Ζ2
(α) +(!) +(с) =^ + Ь + 1*
<1,

139
Chapter 10. Trace and Determinant
which shows that (|, £, f) € Ω. Because T(§, £, f) = (х.у.г), this implies
that Ε с Τ(Ω).
The last two paragraphs show that Ε = Τ(Ω). Thus
volume Ε = volume Τ(Ω)
= |detT| (volume Ω)
4жаЬс
where the second equality comes from 10.38.