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3 Math 115B, Winter 2007: Homework 3

Exercise 3.1. (FIS, 6.5 Exercise 10) LetA be annn real symmetric [complexnormal] matrix. Then

tr(A) =

ni=1

i, tr(AA) =

ni=1

|i|2,

where the is are the (not necessarily distinct) eigenvalues of A.

Proof. By Theorems 6.19 and 6.20 from FIS, A is orthogonally [unitarily] equiv-alent to a diagonal matrix D with the eigenvalues ofA on the diagonal, i.e. thereexists an orthogonal [unitary] matrix P such that A = PDP. Since similarmatrices have the same trace, tr(A) = tr(PDP). Moreover, for any squarematrices M and N, tr(M N) = tr(N M), so in particular,

tr(PDP) = tr((PD)P) = tr(P(PD)) = tr((P P)D)

= tr(ID) = tr(D) =n

i=1

i.

For the second claim, note that A = (PDP) = PD(P) = PDP,

and that the diagonal entries of D are the is. This implies that DD is a

diagonal matrix with entries |i|2, so that we have

tr(AA) = tr(PDP PDP) = tr(PDDP)

= tr(DP PD) = tr(DD) =n

i=1

|i|2

Exercise 3.2. (FIS, 6.5 Exercise 31) Let V be a finite dimensional complexinner product space, and let u be a unit vector in V. Define the Householderoperator Hu : V V by Hu(x) = x 2 x|u u for all x V.(a) Hu is linear.(b) Hu(x) = x if and only if x|u = 0.(c) Hu(u) = u.(d) Hu = Hu and H

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u = IV, and hence Hu is a unitary operator on V.Note that if V is a real inner product space, Hu is a reflection.

Proof. (a) For any a C, and x, y V,

Hu(ax + y) = ax + y 2 ax + y|u u = ax + y 2a x|u u 2 y|u u= a(x 2 x|u u) + (y 2 y|u u) = aHu(x) + Hu(y).

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(b) For any x V,

Hu(x) = x x 2 x|u u = x 2 x|u u = 0 x|u = 0.

(c) Hu(u) = u 2 u|u u = u 2u = u.(d) For any x and y V, we have

x|Hu(y) = x|y 2 y|u u = x|y + x| y|u u= x|y 2y|u x|u = x|y 2 u|y x|u= x|y + 2 x|u u|y = x|y + 2 x|u u|y= x 2 x|u u|y = Hu(x)|y ,

so the uniqueness of the adjoint guarantees that Hu = Hu. To see that H2

u = IV,simply compute

H2u(x) = Hu(x 2 x|u u) = Hu(x) 2 x|u Hu(u)= (x 2 x|u u) + 2 x|u u = x = IV(x).

Exercise 3.3. (FIS, 6.6 Exercise 6) Let T be a normal operator on a finite di-mensional inner product space. IfT is a projection, then T is also an orthogonalprojection.

Proof. An orthogonal projection is simply a self-adjoint projection, so it sufficesto prove that T is self-adjoint. As a corollary to the spectral theorem, we knowthat a normal operator is self-adjoint if and only if its eigenvalues are real. Theeigenvalues of a projection are real. Indeed, if is an eigenvalue and x aneigenvector for , then

x = T(x) = T2

(x) = 2

(x),

therefore (1) = 0, i.e. = 0 or 1. Hence T is an orthogonal projection.

Exercise 3.4. (FIS, 6.6 Exercise 7) Let T be a normal operator on a finite-dimensional complex inner product space V. Using the spectral decomposition

T = 1T1 + ... + kTk,

we prove that(a) If g is a polynomial, then

g(T) =

ki=1

g(i)Ti.

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(b) If Tn = 0 for some n, then T = 0.(c) A linear operator U : V V commutes with T if and only if U commuteswith each Ti.(d) There exists a normal operator U : V V such that U2 = T.(e) T is invertible if and only if i

= 0 for 1

i

k.

(f) T is a projection if and only if every eigenvalue of T is 1 or 0.

Proof. (a) We proceed by induction on deg(g). For constant polynomials, theresult is trivial. Let g = anX

n + ... + a0 and suppose that the result holds forany polynomial of degree n 1. Then

g(T) = anTn + (an1T

n1 + ... + a0)

= anT Tn1 + (an1(

n11

T1 + ... + n1k Tk) + ... + a0)

= an

k

r=1

rTr

k

s=1

n1s Ts

+

n1i=0

ai k

j=1

ijTj

= an k,k

r=1,s=1r

n1

s TrT

n1

s

+

n1,ki=0,j=1

ai

i

j

Tj

=

k

r=1

(annr )T

nr

+

n1,ki=0,j=1

ai

ij

Tj

=

n,ki=0,j=1

ai

ij

Tj =

kj=1

g(j)Tj

(b) This follows from the fact that T1,...,Tk are linearly independent inhomF(V, V). Indeed, since Ti maps V to Wi, every T(v) can be written uniquelyas a linear combination of the (finitely many) basis vectors {vi,j} for Wi. Thenwe have

Tn(v) = n1

T1(v) + ... + nkTk(v) =

ii=1

ni

j

ai,jvi,j

= 0

i = 1, 2, . .., n ni ai,j = 0.

Since this holds for every v V, we may select v so that for each i, at least oneai,j = 0. Then we must have ni = 0 for all i = 1,...,n. In particular, i = 0 foreach i so that the eigenvalues of T are all zero. Hence T = 0.

(c) We have, by linear independence of the Tis,

U T T U = 1(U T1 T1U) + ... + k(U Tk TkU) = 0 U Ti = TiU i = 1, 2,...,k

consider

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(d) Every complex number has a square root, so setting U =k

i=1

iTi,

one sees immediately by (a) that

U2 =k

i=1

(i)2Ti =

k

i=1

iTi = T.

As a corollary to the spectral theorem, we know that U = g(U) =k

i=1 g(

i)Ti,which obviously commutes with each Ti. By (c), U

U = U U, therefore U isnormal.

(e) This one is almost tautological, since an eigenvector is by definitionnonzero.

T is invertible ker(T) = {0} 0 is not an eigenvalue of T

(f) As we saw in (b), the Tis are linearly independent, so that

(T2

T) =

k

i=1

(2i

i)Ti = 0

2i i = 0 i = 1, 2, ...,n.

In other words, T2 = T if and only if every i is a root ofX(X1), i.e. T2 = Tiff i = 0 or 1.

Exercise 3.5. (FIS, 6.6 Exercise 8) If T is a normal operator on a complexfinite dimensional inner product space andU is a linear operator that commuteswith T, then U commutes with T.

Proof. As a corollary to the Spectral Theorem, we know that T = g(T) forsome polynomial g. If g(X) = anX

n + ... + a0, then

U T = U g(T) = U(anTn + ... + a0) = anU T

n + ... + a0U

= an(T U)Tn1 + ... + a0U = anT

nU + ... + a0U

= (anTn + ... + a0)U = g(T)U = T

U.

Exercise 3.6. (FIS, 6.6 Exercise 10) Simultaneous Diagonalization. LetU and T be normal operators on a finite-dimensional complex inner productspace V such that T U = U T. Prove that there exists an orthonormal basis forV consisting of vectors that are eigenvectors of both T and U.

Proof. Let 1,...,k be the eigenvalues of T, and set Wi = ker (T iI). EachWi is trivially T- and T

-invariant. That Wi is U-invariant follows directly

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from the fact that U and (T iIV) commute. Indeed, by Exercise 3.5, Ucommutes with T, therefore

(T iIV)U = (T U iU) = (UT iU) = U(T iIV).

Whenever x W, (T iIV)(U

(x)) = U

(T iIV)(x) = U

(0) = 0, i.e.U(x) W. Replacing U by U in the above two lines, one sees that Wi is alsoU-invariant. Since W is both T and U-invariant, we know that W is bothT and W-invariant. This allows us to consider T and W as operators on eachWi and W

i .

From this point, the proof proceeds by induction on dim(V). Ifdim(V) = 1there is nothing to prove - every operator is diagonalizable. Suppose that theresult holds for any space of dimension strictly less than dim(V). In particular,dim(Wi) < dim(V). By the induction hypothesis, W1 and W

1

have bases{ui}iI and {wj}jJ consisting of simultaneous eigenvectors for T and U. Itsuffices to prove that the union of these two bases is a basis for V. Every vectorv V has a unique expression v = u + w for some u W1 and w W1 (seeTheorem 6.6, FIS), so that v = i aiui +j bjwj is a unique. In particular,the zero vector is a unique linera combination so that = {ui}iI {wj}jJis linearly independent and spans V. Hence is a basis for V consisting ofsimultaneous eigenvectors for T and U.

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