MATH 110  SOLUTIONS TO PRACTICE FINAL
LECTURE 1, SUMMER 2009
GSI: SANTIAGO CA
˜
NEZ
1.
Axler, 6.18 (Hint: By 6.17 it is enough to show that every vector in null
P
is orthogonal to
every vector in range
P
— use 6.2 to show this)
Proof.
First, we know that
P
is the projection operator corresponding to the decomposition
V
= range
P
⊕
null
P.
Let
u
∈
range
P
and
w
∈
null
P
. We must show that
h
u, w
i
= 0. Now, for any
a
∈
F
. we have
P
(
u
+
aw
) =
u
since
aw
∈
null
P
. Thus
k
u
k ≤ k
u
+
aw
k
for any scalar
a
by the assumption that
k
Pv
k ≤ k
v
k
for any
v
. By problem 6.2, this implies that
h
u, w
i
= 0 as required. We conclude that
P
is an
orthogonal projection by problem 6.17.
2.
Axler, 6.20
Proof.
Suppose that
U
and
U
⊥
are
T
invariant. To show that
P
U
T
=
TP
U
, we must show
that both sides give the same result when applied to an arbitrary
v
∈
V
.
To this end, let
v
∈
V
and write it as
v
=
u
+
w
according to the decomposition
V
=
U
⊕
U
⊥
. Then
Tu
∈
U
and
Tw
∈
U
⊥
since
U
and
U
⊥
are
T
invariant. Thus
P
U
T
(
v
) =
P
U
(
Tu
+
Tw
) =
Tu
and
TP
U
(
v
) =
T
(
P
U
(
u
+
w
)) =
Tu,
so
P
U
T
=
TP
U
as required.
Conversely, suppose that
P
U
T
=
TP
U
. Let
u
∈
U
and
w
∈
U
⊥
. We must show that
Tu
∈
U
and
Tw
∈
U
⊥
. We have
P
U
T
(
u
+
w
) =
P
U
(
Tu
+
Tw
)
and
TP
U
(
u
+
w
) =
T
(
P
U
(
u
+
w
)) =
Tu.
For these to be equal means that
Tu
itself is the piece of
Tu
+
Tw
which is in
U
and that
Tw
must then be in
U
⊥
. Hence
U
and
U
⊥
are
T
invariant as claimed.
3.
Suppose that
V
is an innerproduct space and that
U
and
W
are subspaces of
V
. Show
that
U
⊆
W
if and only if
W
⊥
⊆
U
⊥
.
Date
: August 6, 2009.
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GSI: SANTIAGO CA
˜
NEZ
Proof.
Suppose that
U
⊆
W
and let
v
∈
W
⊥
. Then
h
v, w
i
= 0 for any
w
∈
W
. In particular
then, since
U
⊆
W
,
h
v, u
i
= 0 for any
u
∈
U
. Thus
w
∈
U
⊥
so
W
⊥
⊆
U
⊥
.
Conversely, suppose that
W
⊥
⊆
U
⊥
. By what we just shown, we have
(
U
⊥
)
⊥
⊆
(
W
⊥
)
⊥
,
and hence
U
⊆
W
since taking two orthogonal complements gives back the original space.
4.
Suppose that
S
and
T
are selfadjoint operators on an innerproduct space
V
. Show that
ST
is selfadjoint if and only if
S
and
T
commute.
Proof.
We have (
ST
)
*
=
T
*
S
*
=
TS
. Thus (
ST
)
*
=
ST
if and only if
ST
=
TS
, which is the
desired claim. (Easiest proof ever!)
5.
Axler, 7.6
Proof.
Recall that if
T
is normal, then null
T
= null
T
*
.
This follows from the fact that
k
Tv
k
=
k
T
*
v
k
for all
v
. Thus
range
T
= (null
T
*
)
⊥
= (null
T
)
⊥
= range
T
*
as required.
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 Spring '08
 GUREVITCH
 Math, Linear Algebra, Hilbert space, complex vector space, Axler

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