Chapter 7: Operators on Inner Product Spaces

Linear Algebra Done Right, by Sheldon Axler

A: Self-Adjoint and Normal Operators

Problem 1
Suppose n is a positive integer. Define T ∈ L(Fn ) by

T (z1 , . . . , zn ) = (0, z1 , . . . , zn−1 ).

Find a formula for T ∗ (z1 , . . . , zn ).

Proof. Fix (y1 , . . . , yn ) ∈ Fn . Then for all (z1 , . . . , zn ) ∈ Fn , we have

h(z1 , . . . , zn ), T ∗ (y1 , . . . , yn )i = hT (z1 , . . . , zn ), (y1 , . . . , yn )i
= h(0, z1 , . . . , zn−1 ), (y1 , . . . , yn )i
= z1 y2 + z2 y3 + · · · + zn−1 yn
= h(z1 , . . . , zn−1 , zn ), (y2 , . . . , yn , 0)i.
∗
Thus T is the left-shift operator. That is, for all (z1 , . . . , zn ) ∈ Fn , we have
T ∗ (z1 , . . . , zn ) = (z2 , . . . , zn , 0),
as desired.

Problem 2
Suppose T ∈ L(V ) and λ ∈ F. Prove that λ is an eigenvalue of T if and
only if λ̄ is an eigenvalue of T ∗ .

Proof. Suppose λ is an eigenvalue of T . Then there exists v ∈ V such that

T v = λv. It follows
λ is not an eigenvalue of T ⇐⇒ T − λI is invertible
⇐⇒ S(T − λI) = (T − λI)S = I
for some S ∈ L(V )
⇐⇒ S (T ∗ − λI)∗ = (T − λI)∗ S ∗ = I ∗
∗

for some S ∗ ∈ L(V )

⇐⇒ (T − λI)∗ is invertible
⇐⇒ T ∗ − λ̄I is invertible
⇐⇒ λ̄ is not an eigenvalue of T ∗ .

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Since the first statement and the last statement are equivalent, so too are their
contrapositives. Hence λ is an eigenvalue of T if and only if λ̄ is an eigenvalue
of T ∗ , as was to be shown.

Problem 3
Suppose T ∈ L(V ) and U is a subspace of V . Prove that U is invariant
under T if and only if U ⊥ is invariant under T ∗ .

Proof. (⇒) First suppose U is invariant under T , and let x ∈ U ⊥ . For any
u ∈ U , it follows

hT ∗ x, ui = hx, T ui
= 0,

where the second equality follows since T u ∈ U (by hypothesis). Thus T ∗ x ∈ U ⊥

for all x ∈ U ⊥ . That is, U ⊥ is invariant under T ∗ .
(⇐) Now suppose U ⊥ is invariant under T ∗ , and let y ∈ U . For any u0 ∈ U ⊥ ,
it follows

hT y, u0 i = hy, T ∗ u0 i
= 0,

where the second equality follows since T ∗ u0 ∈ U ⊥ (by hypothesis). Thus T y ∈ U

for all y ∈ U . That is, U is invariant under T , completing the proof.

Problem 5
Prove that

dim null T ∗ = dim null T + dim W − dim V

and

dim range T ∗ = dim range T

for every T ∈ L(V, W ).

Proof. Let T ∈ L(V, W ). Notice

⊥
dim null T ∗ = dim (range T )
= dim W − dim range T
= dim W + dim null T − dim V,

2
where the first equality follows by 7.7(a), the second equality follows by 6.50,
and the third equality follows by the Fundamental Theorem of Linear Maps.
Next notice
⊥
dim range T ∗ = dim (null T )
= dim V − dim null T
= dim range T,

where the first equality follows by 7.7(b), and the second and third equalities
follow again by the same theorems above.

Problem 7
Suppose S, T ∈ L(V ) are self-adjoint. Prove that ST is self-adjoint if and
only if ST = T S.

Proof. (⇒) Suppose ST is self-adjoint. We have

ST = (ST )∗
= T ∗S∗
= T S,

where the second equality follows by 7.6(e).

(⇐) Conversely, suppose ST = T S. It follows

(ST )∗ = (T S)∗
= S∗T ∗,

where the second equality again follows by 7.6(e), completing the proof.

Problem 9
Suppose V is a complex inner product space with V 6= {0}. Show that
the set of self-adjoint operators on V is not a subspace of L(V ).

Proof. Let A denote the set of self-adjoint operators on V , and suppose T ∈

A. By 7.6(b), notice (iT )∗ = −iT ∗ , so that A is not closed under scalar
multiplication. Thus A is not a subspace of L(V ).

Problem 11
Suppose P ∈ L(V ) is such that P 2 = P . Prove that there is a subspace
U of V such that P = PU if and only if P is self-adjoint.

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Proof. (⇒) First suppose there is a subspace U ⊆ V such that P = PU , and let
v1 , v2 ∈ V . It follows

hP v1 , v2 i = hu1 , u2 + w2 i
= hu1 , u2 i + hu1 , w2 i
= hu1 , u2 i
= hu1 , u2 i + hw1 , u2 i
= hu1 + w1 , u2 i
= hv1 , P v2 i,

and thus P = P ∗ .
(⇐) Conversely, suppose P = P ∗ . Let v ∈ V . Notice P (v−P v) = P v−P 2 v =
⊥
0, and hence v − P v ∈ null P . By 7.7(c), we know null P = (range T ∗ ) . By
⊥
hypothesis, P is self-adjoint, and hence we have v − P v ∈ (range T ) . Notice
we may write

v = P v + (v − P v),
⊥
where P v ∈ range P and v − P v ∈ (range T ) . Let U = range P . Since the
above holds for all v ∈ V , we conclude P = PU , and the proof is complete.

Problem 13
Give an example of an operator T ∈ L(C4 ) such that T is normal but
not self-adjoint.

Proof. Let T be the operator on C4 whose matrix with respect to the standard
basis is
 
2 −3 0 0
3 2 0 0
 .
0 0 0 0
0 0 0 0

We claim T is normal and not self-adjoint. To see that T is not self-adjoint,

notice that the entry in row 2, column 1 does not equal the complex conjugate
of the entry in row 1 column 2.
Next, notice
    
2 −3 0 0 2 3 0 0 13 0 0 0
3 2 0 0 −3 2 0 0  0 13 0 0
M(T T ∗ ) = 
    
0 0 0 0  0 0 0 0 =  0
 
0 0 0
0 0 0 0 0 0 0 0 0 0 0 0

4
and
    
2 3 0 0 2 −3 0 0 13 0 0 0
−3  3
2 0 0 2 0 0 0 13 0 0
M(T ∗ T ) = 
 
= ,
0 0 0 0 0 0 0 0  0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0

and hence T T ∗ and T ∗ T have the same matrix. Thus T T ∗ = T ∗ T , and T is

normal.

Problem 15
Fix u, x ∈ V . Define T ∈ L(V ) by

T v = hv, uix

for every v ∈ V .
(a) Suppose F = R. Prove that T is self-adjoint if and only if u, x is
linearly dependent.
(a) Prove that T is normal if and only if u, x is linearly dependent.

Proof. We first derive a useful formula for T ∗ which we’ll use in both (a) and
(b). Let w1 , w2 ∈ V and notice

hw1 , T ∗ w2 i = hT w1 , w2 i
= hhw1 , uix, w2 i
= hw1 , uihx, w2 i
= hw1 , hx, w2 i ui
= hw1 , hw2 , xiui,

and thus T ∗ w2 = hw2 , xiu. Since w2 was arbitrary, we may rewrite this as
T ∗ v = hv, xiu for all v ∈ V .
(a) (⇒) Suppose T is self-adjoint. Then we have

hv, uix − hv, xiu = T v − T ∗ v = 0

for all v ∈ V . In particular, we have

hu, uix − hu, xiu = 0.

We may assume both u and x are nonzero, for otherwise there is nothing
to prove. Hence hu, ui =
6 0, which forces hu, xi to be nonzero as well, and
thus the equation above shows u, x is linearly dependent.
(⇐) Conversely, suppose u, x is linearly dependent. We may again

5
 assume both u and x are nonzero, for otherwise T = 0, which is self-adjoint.
Thus there exists a nonzero α ∈ R such that u = αx. It follows

T v = hv, uix
1
= hv, αxi u
α
= hv, xiu
= T ∗,

and thus T is self-adjoint, completing the proof.

(b) (⇒) Suppose T is normal and let v ∈ V . It follows

hhv, uix, xiu = T ∗ (hv, uix)

= T ∗T v
= T T ∗v
= T (hv, xiu)
= hhv, xiu, uix.

We may assume both u and x are nonzero, for otherwise there is nothing to
prove. Since the above holds for v = u, we may conclude hhv, uix, xi 6= 0,
which also forces hhv, xiu, ui =
6 0. Thus u, x is linearly dependent.
(⇐) Conversely, suppose u, x is linearly dependent. We may again
assume both u and x are nonzero, for otherwise T = 0, which is normal.
Thus there exists a nonzero α ∈ R such that u = αx. It follows

T ∗ T v = T ∗ (hv, uix)
= hhv, uix, xiu
 
1 1
= hv, αxi u, u αx
α α
= hhv, xiu, uix
= T (hv, xiu)
= T T ∗ v,

and thus T is normal, completing the proof.

Problem 16
Suppose T ∈ L(V ) is normal. Prove that

range T = range T ∗ .

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Proof. Suppose T ∈ L(V ) is normal. We first prove null T = null T ∗ . It follows
v ∈ null T ⇐⇒ T v = 0
⇐⇒ kT vk = 0
⇐⇒ kT ∗ vk = 0
⇐⇒ T ∗ v = 0
⇐⇒ v ∈ null T ∗ ,
where the third equivalence follows by 7.20, and indeed we have null T = null T ∗ .
This implies (null T )⊥ = (null T ∗ )⊥ , and by 7.7(b) and 7.7(c), this is equivalent
to range T ∗ = range T , as desired.

Problem 17
Suppose T ∈ L(V ) is normal. Prove that

null T k = null T and range T k = range T

for every positive integer k.

Proof. To show null T k = null T , we first prove null T k = null T k+1 for all k ∈ Z+ .
Let m ∈ Z+ . If m = 1, there’s nothing to prove, so we may assume m > 1.
Clearly, if v ∈ null T m , then v ∈ null T m+1 , and hence null T m ⊆ null T m+1 .
Next, suppose v ∈ null T m+1 . Then T (T m v) = 0, and hence T m v ∈ null T . By
Problem 16, this implies T m v ∈ null T ∗ , and by 7.7(a) we have T m ∈ (range T )⊥ .
Since of course T m v ∈ range T as well, we must have T m v = 0. Thus v ∈ null T m ,
and therefore null T m+1 ⊆ null T m . Thus null T m = null T m+1 . Since m was
arbitrary, this implies null T k = null T for all k ∈ Z+ , as desired.
Now we will show range T k = range T for all k ∈ Z+ . Let n ∈ Z+ . If n = 1,
there’s nothing to prove, so we may assume n > 1. Suppose w ∈ range T n .
Then there exists v ∈ V such that T n v = w, and hence T (T n−1 v) = w, so that
w ∈ range T as well and we have range T n ⊆ range T . Next, notice
dim range T n = dim V − dim null T n
= dim V − dim null T
= dim range T,
where the second equality follows from the previous paragraph. Since range T n
is a subspace of range T of the same dimension, it must equal range T . And since
n was arbitrary, we conclude range T k = range T for all k ∈ Z+ , completing the
proof.

Problem 19
Suppose T ∈ L(C3 ) is normal and T (1, 1, 1) = (2, 2, 2). Suppose
(z1 , z2 , z3 ) ∈ null T . Prove that z1 + z2 + z3 = 0.

7
Proof. By Problem 16, null T = null T ∗ , hence T ∗ (z1 , z2 , z3 ) = 0. Therefore, we
have

2(z1 + z2 + z3 ) = h(2, 2, 2), (z1 , z2 , z3 )i

= hT (1, 1, 1), (z1 , z2 , z3 )i
= h(1, 1, 1), T ∗ (z1 , z2 , z3 )i
= h(1, 1, 1), (0, 0, 0)i
= 0,

and so z1 + z2 + z3 = 0, as was to be shown.

Problem 21
Fix a positive integer n. In the inner product space of continuous real-
valued functions on [−π, π] with inner product
Z π
hf, gi = f (x)g(x)dx,
−π

let

V = span(1, cos x, cos 2x, . . . , cos nx, sin x, sin 2x, . . . , sin nx).

(a) Define D ∈ L(V ) by Df = f 0 . Show that D∗ = −D. Conclude

that D is normal but not self-adjoint.
(b) Define T ∈ L(V ) by T f = f 00 . Show that T is self-adjoint.

Proof. From Problem 4 of 6B, recall that

1 cos x cos 2x cos nx sin x sin 2x sin nx
√ , √ , √ ,..., √ , √ , √ ,..., √
2π π π π π π π

is an orthonormal list, and hence it is an orthonormal basis of V .

(a) For k = 1, . . . , n, define

cos(kx) sin(kx)
ek = √ and fk = √ .
π π

Notice
k sin(kx) k cos(kx)
Dek = − √ = −kfk and Dfk = √ = kek ,
π π

8
 and thus, for any v, w ∈ V , it follows

hv, D∗ wi = hDv, wi
*   n
 +
1 1 X

= D  v, √ √ + hv, ek iek + hv, fk ifk  , w
2π 2π k=1
* n +
X

= −khv, ek ifk + khv, fk iek , w
k=1
Xn n
X
=− khv, ek ihfk , wi + khv, fk ihek , wi
k=1 k=1
Xn Xn
=− khw, fk ihv, ek i + khw, ek ihv, fk i
k=1 k=1
n
X n
X
= khw, ek ihv, fk i − khw, fk ihv, ek i
k=1 k=1
* n
+ * n
+
X X
= v, khw, ek ifk − v, khw, fk iek
k=1 k=1
* n
+
X

= v, khw, ek ifk − khw, fk iek
k=1
*  +
  n
1 1 X

= v, −D  w, √ √ + hw, ek iek + hw, fk ifk 
2π 2π k=1
= hv, −Dwi,

and thus D∗ = −D, showing that D is not self-adjoint. Moreover, notice

that this implies

DD∗ = D(−D) = −DD = (D∗ )D = D∗ D,

so that D is normal, completing the proof.

(b) Notice T = D2 , and hence

T ∗ = (DD)∗ = D∗ D∗ = (−D)(−D) = D2 = T.

Thus T is self-adjoint.

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B: The Spectral Theorem
Problem 1
True or false (and give a proof of your answer): There exists T ∈ L(R3 )
such that T is not self-adjoint (with respect to the usual inner product)
and such that there is a basis of R3 consisting of eigenvectors of T .

Proof. The statement is true. To see this, consider the linear operator T defined
by its action on the basis (1, 0, 0), (0, 1, 0), (0, 1, 1):

T (1, 0, 0) = (0, 0, 0)
T (0, 1, 0) = (0, 0, 0)
T (0, 1, 1) = (0, 1, 1).

Notice T (1, 0, 0) = 0 · (1, 0, 0) and T (0, 1, 0) = 0 · (0, 1, 0), so that (1, 0, 0) and
(0, 1, 0) are eigenvectors with eigenvalue 0. Also, (0, 1, 1) is an eigenvector
with eigenvalue 1. Thus (1, 0, 0), (0, 1, 0), (0, 1, 1) is a basis of R3 consisting of
eigenvectors of T . That T is not self-adjoint follows from the contrapositive of
7.22, since (0, 1, 0) and (0, 1, 1) correspond to distinct eigenvalues yet they are
not orthogonal.

Problem 3
Give an example of an operator T ∈ L(C3 ) such that 2 and 3 are the
only eigenvalues of T and T 2 − 5T + 6I 6= 0.

Proof. Define T ∈ L(C3 ) by its action on the standard basis:

T e1 = 2e2
T e2 = e1 + 2e2
T e3 = 3e3 .

Then
 
2 1 0
M(T ) = 0 2 0 .
0 0 3

By 5.32, the only eigenvalues of T are the entries on the diagonal: 2 and 3. Now

10
notice
(T 2 − 5T + 6I)e2 = (T − 3I)(T − 2I)e2
= (T − 3I)(T e2 − 2e2 )
= (T − 3I)(e1 + 2e2 − 2e2 )
= (T − 3I)e1
= T e1 − 3e1
= −e1 ,
so that T 2 − 5T + 6I 6= 0. Thus T is an operator of the desired form.

Problem 5
Suppose F = R and T ∈ L(V ). Prove that T is self-adjoint if and only
if all pairs of eigenvectors corresponding to distinct eigenvalues of T are
orthogonal and

V = E(λ1 , T ) ⊕ · · · ⊕ E(λm , T ),

where λ1 , . . . , λm denote the distinct eigenvalues of T .

Proof. (⇐) Suppose all pairs of eigenvectors corresponding to distinct eigenvalues

of T are orthogonal and
V = E(λ1 , T ) ⊕ · · · ⊕ E(λm , T ),
where λ1 , . . . , λm denote the distinct eigenvalues of T . By 5.41, V has a basis
consisting of eigenvectors of T . Dividing each element of the basis by its norm
produces an orthonormal basis consisting of eigenvectors of T . By the Real
Spectral Theorem, T is self-adjoint, as desired.
(⇒) Conversely, suppose T is self-adjoint as suppose v1 , v2 ∈ V are eigen-
vectors of T corresponding to eigenvalues λ1 , λ2 ∈ R such that λ1 6= λ2 . It
follows
0 = hT v1 , v2 i − hv1 , T v2 i
= hλ1 v1 , v2 i − hv1 , λ2 v2 i
= λ1 hv1 , v2 i − λ2 hv1 , v2 i
= λ1 hv1 , v2 i − λ2 hv1 , v2 i
= (λ1 − λ2 )hv1 , v2 i.
Since λ1 6= λ2 , it must be that hv1 , v2 i = 0. Thus all pairs of eigenvectors
corresponding to distinct eigenvalues of T are orthogonal. By the Real Spectral
Theorem, since T is self-adjoint, T is diagonalizable. And by 5.34, this implies
V = E(λ1 , T ) ⊕ · · · ⊕ E(λm , T ),
where λ1 , . . . , λm denote the distinct eigenvalues of T , completing the proof.

11
 Problem 6
Prove that a normal operator on a complex vector space is self-adjoint if
and only if all its eigenvalues are real.

Proof. Let T be a normal operator on a complex vector space, V .

(⇒) Suppose T is self-adjoint. Then by 7.13, all eigenvalues of T are real.
(⇐) Conversely, suppose all eigenvalues of T are real. By the Complex
Spectral Theorem, there exists an orthonormal basis v1 , . . . , vn of V consisting
of eigenvectors of T . Thus there exist λ1 , . . . , λn ∈ R such that T vk = λk vk for
k = 1, . . . , n. Thus M(T ) is diagonal, and all entries along the diagonal are real.
Therefore M(T ) equals the conjugate transpose of M(T ). By 7.10, this implies
M(T ) = M(T ∗ ), and we conclude T = T ∗ , so that T is indeed self-adjoint.

Problem 7
Suppose V is a complex inner product space and T ∈ L(V ) is a normal
operator such that T 9 = T 8 . Prove that T is self-adjoint and T 2 = T .

Proof. By the Complex Spectral Theorem, since T is normal, V has an orthonor-

mal basis v1 , . . . , vn consisting of eigenvectors of T . Let λ1 , . . . , λn ∈ C be the
corresponding eigenvalues, so that

T vk = λk vk

for k = 1, . . . , n. Repeatedly applying T to both sides of the equation above 8

times yields

T 9 vk = (λk )9 vk and T 8 vk = (λk )8 vk .

Since T 9 = T 8 , we conclude (λk )9 = (λk )8 and thus λk ∈ {0, 1}. In particular,

all eigenvalues of T are real, hence by Problem 6 we have that T is self-adjoint.
To see that T 2 = T , notice

T 2 vk = (λk )2 vk
= λ k vk
= T vk ,

where the second equality follows from the fact that λk ∈ {0, 1}, and the proof
is complete.

Problem 9
Suppose V is a complex inner product space. Prove that every normal
operator on V has a square root. (An operator S ∈ L(V ) is called a
square root of T ∈ L(V ) if S 2 = T .)

12
Proof. Suppose T ∈ L(V ) is normal. By the Complex Spectral Theorem,
V has an orthonormal basis v1 , . . . , vn consisting of eigenvectors of T . Let
λ1 , . . . , λn ∈ C be the corresponding eigenvalues, so that

T vk = λk vk

for k = 1, . . . , n. Define S ∈ L(V ) by its action on this basis:

p
Svk = λk vk ,
√
choosing the complex square root λk by some definite rule. Let v ∈ V . Then
there exist α1 , . . . , αn ∈ C such that v = α1 v1 + · · · + αn vn . It follows

S 2 v = S 2 (α1 v1 + · · · + αn vn )
 p p 
= S α1 λ1 v1 + · · · + αn λn vn
= α1 λ1 v1 + · · · + αn λn vn
= α1 T v1 + · · · + αn T vn
= T (α1 v1 + · · · + αn vn )
= T v.

Thus S 2 = T , and indeed T has a square root, as was to be shown.

Problem 11
Prove or give a counterexample: every self-adjoint operator on V has a
cube root. (An operator T ∈ L(V ) is called a cube root of T ∈ L(V ) if
S 3 = T .)

Proof. Suppose T ∈ L(V ) is self-adjoint. Regardless of whether F = R or F = R,

both Spectral Theorems imply that V has an orthonormal basis v1 , . . . , vn
consisting of eigenvectors of T . By 7.13, all eigenvalues of T are real. So let
λ1 , . . . , λn ∈ R be the eigenvalues corresponding to v1 , . . . , vn , so that

T vk = λk vk

for k = 1, . . . , n. Define S ∈ L(V ) by its action on this basis:

1
Svk = (λk ) 3 vk ,

Let v ∈ V . Then there exist α1 , . . . , αn ∈ C such that v = α1 v1 + · · · + αn vn . It

13
follows

S 3 v = S 3 (α1 v1 + · · · + αn vn )
 1 1

= S 2 α1 (λ1 ) 3 v1 + · · · + αn (λn ) 3 vn
 2 2

= S α1 (λ1 ) 3 v1 + · · · + αn (λn ) 3 vn
= α1 λ1 v1 + · · · + αn λn vn
= α1 T v1 + · · · + αn T vn
= T (α1 v1 + · · · + αn vn )
= T v.

Thus S 3 = T , and indeed T has a cube root. Thus, all self-adjoint operators on
a finite-dimensional inner product space have a cube root.

Problem 13
Give an alternative proof of the Complex Spectral Theorem that avoids
Schur’s Theorem and instead follows the pattern of the proof of the Real
Spectral Theorem.

Proof. Suppose (c) holds, so that T has a diagonal matrix with respect to some
orthonormal basis of V . The matrix of T ∗ (with respect to the same basis) is
obtained by taking the conjugate transpose of the matrix of T ; hence T ∗ also
has a diagonal matrix. Any two diagonal matrices commute; thus T commutes
with T ∗ , which means that T is normal. That is, (a) holds.
We will prove that (a) implies (b) by induction on dim V . For our base case,
suppose dim V = 1. Since 5.21 guarantees the existence of an eigenvector of T ,
clearly (b) is true in this case. Next assume that dim V > 1 and that (a) implies
(b) for all complex inner product spaces of smaller dimension.
Suppose (a) holds, so that T is normal. Let u be an eigenvector of T with
kuk = 1, and set U = span(u). Clearly U is invariant under T . By Problem
3 of 7A, this implies that U ⊥ is invariant under T ∗ as well. But of course
T ∗ is also normal, and since dim U ⊥ = dim V − 1, our inductive hypothesis
implies that there exists an orthonormal basis of U ⊥ consisting of eigenvectors
of T |U ⊥ . Adjoining u to this basis gives an orthonormal basis of V consisting of
eigenvectors of T , completing the proof that (a) implies (b).
We have proved that (c) implies (a) and that (a) implies (b). Clearly (b)
implies (c), and the proof is complete.

14
 Problem 15
Find the value of x such that the matrix
 
1 1 0
0 1 1
1 0 x

is normal.

Proof. Let M be the above matrix. We wish to find x ∈ F such that M M ∗ =

M ∗ M . Notice
  
1 1 0 1 0 1
M M ∗ = 0 1 1  1 1 0
1 0 x 0 1 x
 
2 1 1
= 1 2 x 
1 x 1 + x2

and
  
1 0 1 1 1 0
M ∗ M = 1 1 0  0 1 1
0 1 x 1 0 x
 
2 1 x
= 1 2 1 .
x 1 1 + x2

Thus it must be that x = 1.

C: Positive Operators and Isometries

Problem 1
Prove or give a counterexample: If T ∈ L(V ) is self-adjoint and there
exists an orthonormal basis e1 , . . . , en of V such that hT ej , ej i ≥ 0 for
each j, then T is a positive operator.

Proof. The statement is false. To see this, let e1 , e2 ∈ R2 be the standard basis
and consider T ∈ L(R2 ) defined by

T e1 = e1
T e2 = −e2 .

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Then
 
1 0
M(T ) = ,
0 −1
and since M(T ) is diagonal, T must be self-adjoint by the Real Spectral Theorem.
But notice that the basis
1
v1 = √ (e1 + e2 )
2
1
v2 = √ (e1 − e2 )
2
is orthonormal and that
hT v1 , v1 i = hv2 , v1 i = 0
and
hT v2 , v2 i = hv1 , v2 i = 0.
Thus T is of the desired form, but T is not a positive operator, since
hT e2 , e2 i = h−e2 , e2 i = −1,
completing the proof.

Problem 3
Suppose T is a positive operator on V and U is a subspace of V invariant
under T . Prove that T |U ∈ L(U ) is a positive operator on U .

Proof. That T |U is self-adjoint follows by 7.28. Let u ∈ U . Then, since

hT |U (u), ui = hT u, ui > 0,
T |U is a positive operator on U , as was to be shown.

Problem 5
Prove that the sum of two positive operators on V is positive.

Proof. Let S, T ∈ L(V ) be positive operators. Notice

(S + T )∗ = S ∗ + T ∗ = S + T,
hence S + T is self-adjoint. Next, let v ∈ V . It follows
h(S + T )v, vi = hSv + T v, vi
= hSv, vi + hT v, vi
≥ 0,
and thus S + T is a positive operator as well.

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 Problem 7
Suppose T is a positive operator on V . Prove that T is invertible if and
only if

hT v, vi > 0

for every v ∈ V with v 6= 0.

Proof. Let T be a positive operator on V .

(⇒) Suppose T is invertible and let v ∈ V \{0}. Since T is a positive operator,
by 7.35(e) there exists R ∈ L(V ) such that T = R2 . Since T is invertible, so is
R. In particular, R is injective, and thus Rv 6= 0. It follows

hT v, vi = hR2 , vi
= hRv, R∗ vi
= hRv, Rvi
2
= kRvk
> 0,

completing the proof in one direction.

(⇐) Now suppose hT v, vi > 0 for every v ∈ V \ {0}. Assume by way of
contraction that T is not invertible, so that there exists w ∈ V \ {0} such that
T w = 0. But then hT w, wi = h0, wi = 0, a contradiction. Thus T must be
invertible, completing the proof.

Problem 9
Prove or disprove: the identity operator on F2 has infinitely many self-
adjoint square roots.

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