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Purdue University
MA 353: Linear Algebra II with Applications
Homework 10, due Apr. 2, Some Solutions
(I)
Sec. 1.6#34:
(a) Prove that if
W
1
is any subspace of a ﬁnitedimensional vector space
V
, then there exists a subspace
W
2
of
V
such that
V
=
W
1
⊕
W
2
.
Proof.
Let
γ
1
=
{
v
1
,v
2
,...,v
k
}
be a basis of
W
1
. Then extend
it to a basis
γ
=
{
v
1
,v
2
,...,v
k
,v
k
+1
,...,v
n
}
of
V
.
Let
γ
2
=
{
v
k
+1
,...,v
n
}
and
W
2
= span
γ
. Then
V
=
W
1
⊕
W
2
. This can be
shown as follows. First, one can see that
V
=
W
1
+
W
2
because
W
1
+
W
2
= span
γ
1
+ span
γ
2
= span
γ
1
∪
γ
2
= span
γ
=
V
. Next
we have to show that
W
1
∩
W
2
=
{
0
}
. Let
x
∈
W
1
∩
W
2
. Then
since
x
∈
W
1
,
x
=
a
1
v
1
+
a
2
v
2
+
···
+
a
k
v
k
,
and since
x
∈
W
2
,
x
=
a
k
+1
v
k
+1
+
a
k
+2
v
k
+1
+
···
+
a
n
v
n
.
Hence
a
1
v
1
+
a
2
v
2
+
···
+
a
k
v
k
=
a
k
+1
v
k
+1
+
a
k
+2
v
k
+1
+
···
+
a
n
v
n
which gives
a
1
v
1
+
a
2
v
2
+
···
+
a
k
v
k

a
k
+1
v
k
+1

a
k
+2
v
k
+1
 ··· 
a
n
v
n
= 0
.
Now since
{
v
1
,v
2
,...,v
k
,v
k
+1
,...,v
n
}
is linearly independent, we
have
a
1
=
a
2
=
···
=
a
k
=
a
k
+1
=
···
=
a
n
= 0
,
namely
x
= 0. This completes the proof.
±
(b) Let
V
=
R
2
and
W
1
=
{
(
a
1
,
0) :
a
1
∈
R
}
. Give examples of
two diﬀeren subspaces
W
2
and
W
0
2
such that
V
=
W
1
⊕
W
2
and
V
=
W
1
⊕
W
0
2
.
Answer.
Let
W
2
=
{
(0
,a
2
) :
a
2
∈
R
}
and
W
0
2
=
{
(
a,a
) :
a
∈
R
}
.
Then one can verify that those are desired examples.
±
1
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View Full DocumentSec. 2.3#16: Let
V
be a ﬁnitedimensional vector space, and let
T
:
V
→
V
be linear.
(a) If rank(
T
) = rank(
T
2
), prove that
R
(
T
)
∩
N
(
T
) =
{
0
}
and deduce
that
V
=
R
(
T
)
⊕
N
(
T
).
Proof.
Assume rank(
T
) = rank(
T
2
). Then by RankNullity Theo
rem, this implies dim
N
(
T
) = dim
N
(
T
2
). But notice that
N
(
T
)
⊆
N
(
T
2
) because if
x
∈
N
(
T
), then
T
(
x
) = 0 and hence
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 Spring '08
 Staff
 Linear Algebra, Algebra, Vector Space

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