MATH 108B HW 3 SOLUTIONS
RAHUL SHAH
Problem 1. [§7.11]
Solution. Assume T is a normal operator on a complex vector space. Then, by the complex spectral theorem, we find
that T has an orthonormal basis consisting of eigenvectors. Thus M(T ) is a diagonal matrix (with respect to the
given orthonormal basis of eigenvectors). Let exp ıθj be the j-th diagonal entry of M(T ). Let λj = exp ı
θj
2
(choose
one of the square roots - exists since z 2 − c always has a root over the complex numbers for any c, a complex number).
Let M(T ) be a diagonal matrix with the j-th diagonal entry given by λj and let S be the operator that has matrix
M(S) with respect to the given orthonormal basis of eigenvectors of T . Notice that S 2 = T .
Problem 2. [§7.12]
Solution. Let V = R2 , and let T be the operator given by a 90◦ rotation. Let α = 0 and let β = 1. Then α2 < 4β.
Notice that T 2 = −I and thus T 2 + αT + βI = 0, which is clearly not invertible.
Problem 3. [§7.13]
Solution. Notice that for V , either a real or a complex vector space, and T a self-adjoint operator on T , T has an
orthonormal basis consisting of eigenvectors (since every self-adjoint operator is normal, this also works for complex
spaces). Thus M(T ) is a diagonal matrix respect to the given orthonormal basis of eigenvectors. Since every complex
number (strictly real or not) has a cube-root, we can find a diagonal matrix M(S) such that S 3 = T by picking the
diagonal entries as in [§7.11]. If we define S to be the operator that has the matrix M(S) with respect to the given
orthonormal basis of eigenvectors of T , we find that S 3 = T .
Problem 4. [§7.15]
Solution.
‘⇒’ Given an operator T on a finite dimensional real vector space V such that T has a basis of eigenvectors,
let {v1 , . . . vk } be the basis of eigenvectors and let λi be the eigenvalue associated to vi . For each v ∈ V ,
v = a1 v1 + . . . ak vk , and this representation is unique. Thus define
*
X
i≤k
+
ai vi ,
X
bj vj
=
X
j≤k
ai bi .
i≤k
Notice that hv, vi ≥ 0, since a2i ≥ 0 for all ai ∈ R. If
P
a2
i
= 0, then each ai is 0. Thus the given function is
positive-definite. Similarly, we can (and you need to) check the rest of the conditions to show that the given
1
2
RAHUL SHAH
function is an inner product on V . Notice that {v1 , . . . vk } is an orthonormal basis with respect to the given
inner product. Now notice that
hT (a1 v1 + . . . ak vk ), b1 v1 + . . . bk vk i
=
=
ha1 λ1 v1 + . . . ak λk vk , b1 v1 + . . . bk vk i
X
hai λi vi , b1 v1 + . . . bk vk i
i≤k
=
X
λi hai vi , bi vi i
i≤k
=
X
hai vi , λi vi i
i≤k
We similarly find that ha1 v1 + . . . ak vk , T (b1 v1 + . . . bk vk )i =
P
i≤k
hai vi , λi vi i and thus T is self-adjoint.
‘⇐’ Suppose there is an inner-product on V for which T is self-adjoint. Then the result follows from the real
spectral theorem.
Problem 5. [§7.16]
Solution. Let V = R2 and let T be the operator defined by the matrix:


1 1
.

0 1
Notice that T (1, 0) = (1, 0) and thus the linear subspace, U , spanned by (1, 0) is invariant under T . Notice that T ⊥
is the linear subspace spanned by (0, 1). However, T (0, 1) = (1, 1) ∈
/ U ⊥ . Thus U ⊥ is not invariant under T .
Problem 6. [§7.18]
Solution. Suppose that T is a positive operator. Then T is self-adjoint by Theorem 7.27. Let k be even. Then
E
E
k
D k
D k+1
k−1
k
T v, v = T 2 v, T 2 v ≥ 0 by positive-definiteness. Now let k > 1 be odd. Then T k v, v = T 2 +1 v, T 2 v =
hT (w), wi ≥ 0 for w = T
k−1
2
v. For k = 1, the result is trivial.
Problem 7. [§7.22]
Solution. Since R3 is odd-dimensional, S has an eigenvalue, λ and an associated eigenvector v. Since S is also an
isometry, kSvk = |λv| = kvk. Thus λ = ±1. Then S 2 v = v.
Problem 8. [§7.23]
√
Solution. You can check that T ∗ T (a, b, c) = (4a, 9b, c). Thus T ∗ T (a, b, c) = (2a, 3b, c). Since T (a, b, c) = (c, 2a, 3b),
√
we know that if S(a, b, c) = (c, a, b) then S T ∗ T = T . It is also clear that S is an isometry, since it permutes the
orthonormal basis.
Problem 9. [§7.25]
Solution.
MATH 108B HW 3 SOLUTIONS
3
√
√
‘⇒’ Assume that T is invertible. Since T = S T ∗ T , we see that T ∗ T must also be invertible (since it is equal
√
−1
and thus S is unique.
to S −1 T ). Thus S = T T ∗ T
√
‘⇐’ Assume T is not invertible. Thus T ∗ T is not invertible. Since T ∗ T is self-adjoint, construct a basis
√
{v1 , . . . vn } of orthonormal eigenvectors. Without loss of generality, we can assume that v1 ∈ null T ∗ T .
Define U on the basis as U (v1 ) = −S(v1 ) and U (vj ) = S(vj ) for j 6= i. Since S is an isometry, so is U , and
U 6= S, since S(v1 ) 6= U (v1 ) unless S(v1 ) = 0, which is impossible, since S is an isometry. It is easy to see
√
that T = U T ∗ T .