1). This is impossible, because
T
*
(
v
) =

T
(
v
) =
v
.
3.
Since
T

λI
is not injective, it follows that (
T

λI
)
0
=
T
0

λI
is not surjective; since
V
is finitedimensional, this means that
T
0

λI
is not injective, so
λ
is an eigenvalue of
T
0
.
4.
Since
v
and
w
are eigenvectors of a normal operator for distinct eigenvalues, they are
orthogonal; by Pythagoras’ theorem,
k
Tv
+
Tw
k
2
=
k
3
v
+ 4
w
k
2
=
k
3
v
k
2
+
k
4
w
k
2
= 9
k
v
k
2
+ 16
k
w
k
2
= 100
,
so
k
Tv
+
Tw
k
= 10
.
5.
(i) There are many examples; for example, on
C
2
, we can take
S
=
0
1

1
0
and
T
=
1
0
0

1
.
(ii) Since
S
and
T
are normal, it follows that
range(
S
*
) = range(
S
) and range(
T
*
) = range(
T
);
in particular, range(
S
*
) and range(
T
*
) are also orthogonal. By the Pythagorean theorem,
since
Sv
and
Tv
, and also
S
*
v
and
T
*
v
, are orthogonal for any
v
∈
V,
k
(
S
+
T
)
v
k
2
=
k
Sv
k
2
+
k
Tv
k
2
=
k
S
*
v
k
2
+
k
T
*
v
k
2
=
k
(
S
+
T
)
*
v
k
2
for all
v
∈
V
. Therefore,
S
+
T
is normal.
6.
(i) Assume that
P
=
P
U
.
Then
P
is selfadjoint, because:
for any
v, w
∈
V
, since
v

Pv, w

Pw
∈
range(
P
)
⊥
, it follows that
h
Pv, w
i  h
v, Pw
i
=
h
Pv, w

Pw
i
+
h
Pv

v, Pw
i
= 0 + 0 = 0
.
(ii) Assume that
P
is selfadjoint, and define
U
:= range(
P
). Let
v
∈
V.
For any
w
∈
U
,
h
v

Pv, w
i
=
P
(
v, w
)

P
(
Tv, w
i
) =
P
(
v, w
)

P
(
v, Tw
) =
P
(
v, w
)

P
(
v, w
) = 0
,
so
v

Pv
∈
U
⊥
. This implies that
Pv
=
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 Fall '08
 GUREVITCH
 Math, Linear Algebra, Algebra