Download MATH 108B HW 2 SOLUTIONS Problem 1. [§6.27] Solution. Given z

MATH 108B HW 2 SOLUTIONS
RAHUL SHAH
Problem 1. [§6.27]
Solution. Given z = (z1 , . . . zn ), z0 = (z10 , . . . zn0 ) ∈ Fn , notice that
* n
+
n
X
X
0
0
z, z
=
zi ei ,
zi e i
i=1
=
n
X
i=1
*
zi ei ,
i=1
+
zj0 ej
j=1
n
X
=
n
X
zi zj0 hei , ej i
i,j=1
=
n
X
zi zi0
i=1
Now define T ∗ (z1 , . . . zn ) = (z2 , . . . zn , 0) and notice that
n
X
T (z), z0 =
zi−1 zi0
i=2
and
X
n−1
0
zi zi+1
.
z, T ∗ (z0 ) =
i=1
Both these quantities are equal and thus T ∗ is the adjoint of T .
Problem 2. [§6.28]
Solution. Let λ ∈ F s.t. T v = λv for some v 6= 0. Thus
hT v, vi
=
hλv, vi
=
λ hv, vi
=
v, λv
Thus hv, T ∗ vi = hT v, vi = v, λv . Thus v, T ∗ v − λv = 0. Since v 6= 0, we find that T ∗ v − λv = 0 and thus λ is an
eigenvalue of T ∗ . Now notice that since (T ∗ )∗ = T , and λ = λ, we find that λ an eigenvalue of T ∗ ⇒ λ an eigenvalue
of T .
Problem 3. [§6.29]
Solution. Assume that T (U ) ⊂ U , for U a subspace of V . Let w ∈ U ⊥ . Let u ∈ U . hu, T ∗ wi = hT u, wi = 0, since
T (U ) ⊂ U and w ∈ U ⊥ . Thus T ∗ w ∈ U ⊥ and thus T (U ⊥ ) ⊂ U ⊥ . Hence U ⊥ is invariant under T ∗ . Since (T ∗ )∗ = T
and (U ⊥ )⊥ = U for U a subspace, we find that the reverse implication also holds.
1
2
RAHUL SHAH
Problem 4. [§7.1]
Solution.
R1
R1
R1
R1
a. Notice that 0 T (1)x2 dx = 0 0 dx = 0. However, 0 1T (x2 ) dx = 0 x2 dx =
1, T (x2 ) . Thus T is not self-adjoint.
1
.
3
Thus T (1), x2 6=
b. The matrix is given with respect to the basis {1, x, x2 }, which is not orthonormal. The Real Spectral Theorem
says that if and only if we have a basis of orthonormal eigenvectors, do we find that our operator is self-adjoint.
Since the given basis is not orthonormal, it is not a contradiction that in spite of having a diagonal matrix
of eigenvalues with a basis of eigenvectors, our operator is not self-adjoint; thus it is not a contradiction that
our matrix is its conjugate transpose but is not self-adjoint.
Problem 5. [§7.2]
Solution. Let V = R2 , and with respect to the orthonormal basis {e1 , e2 }, let T : (a, b) 7→ (a + b, a + b) and let
S : (a, b) 7→ (−a + b, a + b). Both T and S are self-adjoint (the conjugate transpose is itself), however, T S : (a, b) 7→
(2b, 2b), which is not self-adjoint. Each of these facts require proof ; hint: look at each of their matrices.
Problem 6. [§7.4]
Solution.
‘⇒’ Assume that P is an orthogonal projection. Thus V = range P ⊕ null P . Let u, v ∈ V . u = u1 + u2 and
v = v1 + v2 , where u1 , v1 ∈ range P and u2 , v2 ∈ null P . Notice that
hP u, vi
Thus P is self-adjoint.
=
hP u1 + P u2 , v1 + v2 i
=
hP u2 , v1 + v2 i
=
hP u1 , v1 i + hP u1 , v2 i
=
hP u1 , v1 i ; since P u1 ∈ range P ⇒ hP u1 , v2 i = 0
=
hu1 , v1 i
=
hu1 , P v1 i
=
hu1 + u2 , P v1 + P v2 i
=
hu, P vi
MATH 108B HW 2 SOLUTIONS
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‘⇐’ Assume that for each u, w ∈ V , hP u, wi = hu, P wi. Let u ∈ null P and let w ∈ range P . Thus w = P v for
some v ∈ V . We find that
hu, wi
=
hu, P vi
=
hP u, vi
=
h0, vi
=
0
Thus V = null P ⊕ range P and range P ⊥ = null P . Since we also know that P 2 = P , P is an orthogonal
projection.
Problem 7. [§7.6]
Solution. Notice that null T ∗ = null T by Proposition 7.6 (kT vk = 0 ⇔ kT ∗ vk = 0). Thus range T = (null T ∗ )⊥ =
(null T )⊥ = range T ∗ .
Problem 8. [§7.8]
Solution. Notice that (1, 2, 3) is an eigenvector of eigenvalue 0 and (2, 5, 7) is an eigenvector of eigenvalue 1. If T was
self-adjoint, then by Corollary 7.14, h(1, 2, 3), (2, 5, 7)i = 0, which is a contradiction since h(1, 2, 3), (2, 5, 7)i = 33 6=
0.
Problem 9. [§7.10]
Solution. We will prove the first part of [§7.7], that for T a normal operator, null T k = null T . That null T ⊂ null T k
is clear. Let v ∈ null T k . Then
D
T ∗ T k−1 v, T ∗ T k−1 v
E
D
=
T T ∗ T k−1 v, T k−1 v
D
E
T ∗ T k v, T k−1 v
D
E
T ∗ 0, T k−1 v
D
E
0, T k−1 v
=
0
=
=
=
E
Thus T ∗ T k−1 v = 0. Now notice that
D
T k−1 v, T k−1 v
E
D
=
T ∗ T k−1 v, T k−2 v
D
E
0, T k−2 v
=
0
=
E
Thus T k−1 v = 0 and thus v ∈ null T k−1 . By induction on k, we find that null T k ⊂ null T , and thus they are equal.
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RAHUL SHAH
Assume that T is a normal operator such that T 9 = T 8 . Then for all v, T 9 v − T 8 v = 0. Thus T 8 (T v − v) = 0.
Thus T v − vnull T 8 = null T .Thus T (T v − v) = 0. We immediately find that T 2 v = T v for all v and thus T 2 = T . By
the Complex Spectral Theorem, T has a diagonal matrix with respect to some orthonormal basis. Let the diagonal
entries be λ1 , . . . λn . Since T 2 = T , we find that λ2i = λi (since T 2 is the diagonal matrix with the diagonal entries
given by λ21 , . . . λ2n . Thus λi = 0 or λi = 1. In either case, the diagonal entry is real, and all off-diagonal entries are
0. Thus T equals its conjugate transpose. Thus T is self-adjoint.