Problem Set 4  Solutions
J. Scholtz
October 20, 2006
Question 1: 2.1/26
Let
T
:
V
→
V
be a projection along
W
1
, where
V
=
W
1
⊕
W
2
.
part a)
T
is linear. Proof:
T
(
x
) =
T
(
x
1
+
x
2
) =
x
1
, where
x
1
∈
W
1
. Then
T
(
a
+
b
) =
T
(
a
1
+
b
1
+
a
2
+
b
2
) =
a
1
+
b
1
=
T
(
a
)+
T
(
b
). The first stap is possible because each
a
can be split into
a
1
∈
W
1
and
a
2
∈
W
2
, and clearly then if
a
+
b
=
w
1
+
w
2
, then
w
1
=
a
1
+
b
1
∈
W
1
and
w
2
=
a
2
+
b
2
∈
W
2
. Claim:
W
1
=
{
x
∈
V
:
T
(
x
) =
x
}
. Proof:
If
w
∈
W
1
, then
T
(
w
) =
w
by definition. Hence
W
1
⊂ {
x
∈
V
:
T
(
x
) =
x
}
.
On the other hand, suppose that
T
(
x
) =
x
, but
x /
∈
W
1
.
Then this means
T
(
x
) =
T
(
x
1
+
x
2
) =
x
1
, which implies that
x
2
= 0.
But this means that
x
=
x
1
∈
W
1
 contradiction. Therefore:
W
1
=
{
x
∈
V
:
T
(
x
) =
x
}
.
part b)
By the definition of the map
R
(
T
)
⊂
W
1
. However, since
T
(
W
1
) =
W
1
, then
W
1
⊂
R
(
T
). Together this means
R
(
T
) =
W
1
.
Suppose
v
∈
W
2
, then
T
(
w
) = 0, hence
W
2
⊂
N
(
T
). Yet at the same time, if
T
(
x
) = 0, then
T
(
x
1
+
x
2
) =
x
1
= 0, which means
x
1
= 0,
x
=
x
2
∈
W
2
. We
can conclude
N
(
T
) =
W
2
.
part c)
If
W
1
=
V
, then
W
2
= 0 and also
T
(
x
) =
x
for any
x
, hence
T
=
I
.
part d)
If
W
1
= 0, then
N
(
T
) =
W
2
=
V
, therefore
T
(
x
) = 0 for all
x
.
Question 2: 2.2/12
Let
T
still be a projection on
W
along
W
, where
V
=
W
⊕
W
. Then we can
pick a basis for the subspace
W
,
β
W
=
{
w
1
, . . . , w
k
}
and extend it onto a basis
1
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for
V
,
β
V
=
{
w
1
, . . . , w
k
, v
k
+1
, . . . , v
n
}
. Then we know that the basis for
W
is the set
{
w
k
+1
, . . . , w
n
}
(it is certainly linearly independent as it is a part of
another basis, it also must span since
V
=
W
⊕
W
).
Then, by the previous
exercise we know that:
T
(
w
i
) =
w
i
and
T
(
w
j
) = 0 for all
i
or
j
. Therefore the
matrix with repsect to this basis will look like:
[
T
]
β
=
1
.
.
.
1
0
.
.
.
0
Where the rest of entries is zeros.
Question 3: 2.3/12
Let
V, W, Z
be vectorspaces,
T
:
V
→
W
and
U
:
W
→
Z
.
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