MATH 108B HW 2 SOLUTIONS
RAHUL SHAH
Problem 1.
[
§
6.27]
Solution.
Given
z
= (
z
1
, . . . z
n
)
,
z
0
= (
z
0
1
, . . . z
0
n
)
∈
F
n
, notice that
z
,
z
0
=
*
n
X
i
=1
z
i
e
i
,
n
X
i
=1
z
0
i
e
i
+
=
n
X
i
=1
*
z
i
e
i
,
n
X
j
=1
z
0
j
e
j
+
=
n
X
i,j
=1
z
i
z
0
j
h
e
i
,
e
j
i
=
n
X
i
=1
z
i
z
0
i
Now define
T
*
(
z
1
, . . . z
n
) = (
z
2
, . . . z
n
,
0) and notice that
T
(
z
)
,
z
0
=
n
X
i
=2
z
i

1
z
0
i
and
z
, T
*
(
z
0
)
=
n

1
X
i
=1
z
i
z
0
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.
Tv
=
λv
for some
v
6
= 0. Thus
h
Tv, v
i
=
h
λv, v
i
=
λ
h
v, v
i
=
v,
λv
Thus
h
v, T
*
v
i
=
h
Tv, v
i
=
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
.
h
u, T
*
w
i
=
h
Tu, w
i
= 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.
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 Winter '09
 Math, Linear Algebra, Diagonal matrix, Conjugate transpose, RAHUL SHAH

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