Homework 9 Solutions

Homework 9 Solutions
1. Let V , W be nite dimensional inner product spaces with dim(V ) ≤ dim(W ). Prove there is a linear
map T : V → W such that hT (v), T (v 0 )iW = hv, v 0 iV for all v, v 0 ∈ V .
Proof. Let {v1 , . . . , vm }, {w1 , . . . , wn } be orthonormal bases for V, W respectively. These exist by the
Gram-Schmidt process. By the dimension condition, m ≤ n. Dene
Pm T as the unique
Pm linear map with
T (vk ) = wk for each 1 ≤ k ≤ m. Then, for v, v 0 ∈ V , we write v = k=1 ak vk , v 0 = k=1 bk vk . We have
hT (v), T (v 0 )iW = h
m
X
m
X
ak T (vk ),
k=1
bk T (vk )i = h
k=1
m
X
ak wk ,
k=1
m
X
bk wk i
k=1
As hvj , vk i = hwj , wk i for each j, k, we have
h
m
X
ak wk ,
k=1
m
X
bk wk i = h
k=1
m
X
ak vk ,
k=1
m
X
bk vk i = hv, v 0 iV
k=1
2. Let T : V → V be an orthogonal projection on a subspace of an inner product space V . Prove that
kT (v)k ≤ kvk for all v ∈ V .
Proof.
For v ∈ V , write v = w + z where w ∈ R(T ), z ∈ R(T )⊥ , so T (v) = w. Then,
2
2
2
2
2
kT (v)k = kwk ≤ kwk + kzk = kw + zk = kvk
as hw, zi = 0.
3. Let V be a nite dimensional inner product space and T : V → V be a projection such that
kT (v)k ≤ kvk for each v ∈ V . Prove T is an orthogonal projection.
Proof. Suppose T be projection along Z ; that is, N (T ) = Z . Suppose T is not an orthogonal projection.
Thus, if W = R(T ), we have Z 6= W ⊥ . Hence, Z ⊥ 6= W (if Z ⊥ = W , then (Z ⊥ )⊥ = Z = W ⊥ ). As
dim Z ⊥ = dim W , we have Z ⊥ \W 6= ∅. Let x ∈ Z ⊥ \W . We show kT (x)k > kxk. Write T (x) = z + z 0 where
z ∈ Z and z 0 ∈ Z ⊥ . This is possible as V = Z ⊕ Z ⊥ . Write x = wx + zx where wx ∈ W , zx ∈ Z . Then,
T (x) = wx = z + z 0 , so wx − z 0 = z ∈ Z . Thus, x − z 0 = (wx − z 0 ) + zx ∈ Z and is also in Z ⊥ , so x = z 0 .
Hence, kT (x)k2 = kz + z 0 k2 = kzk2 + kz 0 k2 = kzk2 + kxk2 . As x ∈/ W , T (x) 6= x, so z 6= 0. Thus, kzk2 > 0,
so kT (x)k2 > kxk2 .
4. Let T be a normal operator on a nite dimensional inner product space. Suppose T is also a projection. Prove T is an orthogonal projection.
Proof. As T is normal, there is an orthonormal basis β = {v1 , . . . , vn } for V such that [T ]β is diagonal.
As T is a projection, the eigenvalues of T allPbelong to {0, 1}. Thus, if {v1 , . . . , vm } have eigenvalue 1 and
the rest have eigenvalue 0, we have, for v = nk=1 ck vk ∈ V
hT (v), T (v)i = h
n
X
k=1
ck T (vk ),
n
X
k=1
ck (vk )i = h
m
X
c k vk ,
k=1
m
X
k=1
ck vk i =
m
X
k=1
|ck |2 ≤
n
X
k=1
|ck |2 = h
n
X
k=1
ck v k ,
n
X
ck vk i = hv, vi
k=1
so by problem 3, T is an orthogonal projection.
Let T be a normal operator on a nite dimensional complex inner product space such that T n = 0 for
some n > 0. Prove T = 0.
5.
1
Proof. As T is normal, it has a basis of eigenvectors, say {v1 , . . . , vm } with eigenvalues λ1 , . . . , λm .
Then, T n (vk ) = λnk vk = 0 for all n, so λk = 0 for all k. Hence, T = 0 on a basis, so T = 0.
6. Let T be a normal operator on a nite dimensional complex inner product space. Prove for any integer
n > 1 there is a linear operator S such that T = S n .
Proof. As T is normal, there is an orthonormal basis β = {v1 , . . . , vm } of eigenvectors, say with eigenvalues λ1 , . . . , λm . For each λk , let µk be such that µnk = λk . These exist by the fundamental theorem
of algebra. Dene S by S(vk ) = µk vk . Then, S n (vk ) = µnk vk = T (vk ) for each k, so S n = T as these
transformations are equal on a basis.
7. Let T be a unitary operator on a nite dimensional inner product space and W ⊆ V a nite dimensional T -invariant subspace. Prove that W ⊥ is also T -invariant.
Let x ∈ W ⊥ and w ∈ W . Then, hw, T (x)i = hT ∗ (w), xi = hT −1 (w), xi as T ∗ = T −1 . As
T is unitary, it is an isomorphism, so T −1 is also an isomorphism. Hence, dim T −1 (X) = dim(X) for
any subspace X ⊆ V . In particular, dim T −1 (W ) = dim(W ), but W ⊆ T −1 (W ) as W is T -invariant, so
W = T −1 (W ). Thus, T −1 (w) ∈ W , so hT −1 (w), xi = 0 as x ∈ W ⊥ . As this holds for all w ∈ W , T (x) ∈ W ⊥ .
Proof.
8.
Find an orthogonal matrix whose rst row is ( 31 , 32 , 23 ).
Take any basis where ( 13 , 32 , 23 ) is the rst element and perform Gram-Schmidt. Then, use the
results as the rows of the orthogonal matrix. For example, we could take {( 13 , 23 , 23 ), (−4, 1, 1), (0, 1, −1)} (to
−1
)} giving the matrix
start o orthogonally) and normalize to {( 31 , 32 , 23 ), ( √−418 , √118 , √118 ), (0, √12 , √
2
Solution.
1
3
−4
 √18

0
2
3
√1
18
√1
2
2 
3
√1 
18
−1
√
2
9. Let T be a unitary operator on a complex nite dimensional inner product space V . Prove there
exists a unitary operator S such that T = S 2 .
Proof. Let S be as constructed in problem 6. We show S is unitary. Let β = {v1 , . . . , vm } be as in
problem 6, so S(vk ) = µk vk , T (vk ) = λk vk , µ2k = λk . Then, as β is an orthonormal basis, S ∗ (vk ) = µk vk .
Hence, SS ∗ (vk ) = |µk |2 vk = |λk |vk . Now, T is unitary, so T ∗ (vk ) = λk vk and, as T ∗ = T −1 , we have
T ∗ (vk ) = T −1 (vk ) = λ1k vk . Thus, λk λk vk = vk , which is |λk |2 = 1. Hence, SS ∗ (vk ) = vk for all k , so
SS ∗ = Id.
10. Let T be a self-adjoint positive denite operator on a nite dimensional inner product space V .
Prove there is an operator S such that S ∗ S = T .
Proof. Let β = {v1 , . . . , vn } be an orthonormal basis for V such that T (vk ) = λk vk for each vk ∈ β . This
√
exists as T is self-adjoint. As T is positive denite, λk > 0 for each k. Let S : V → V by S(vk ) = λk vk
√
√
(the positive square root). Then, as β is an orthonormal basis, S ∗ (vk ) = λk vk = λk vk for each vk . Thus,
S ∗ S(vk ) = λk vk = T (vk ) for all k , so S ∗ S = T .
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