Math 431  Assignment 7
Solutions
Section 1.3
28
.
Clearly, the
n
×
n
zero matrix
O
is in
W
1
.
Let
A, B
be in
W
1
and
let
c
be in
F
. Then,
A
+
B
and
cA
are in
W
1
because
(
A
+
B
)
t
=
A
t
+
B
t
=

A

B
=

(
A
+
B
)
,
and
(
cA
)
t
=
cA
t
=
c
(

A
) =

(
cA
)
.
Therefore,
W
1
is a subspace of
M
n
×
n
(
F
). Now assume that
char
(
F
)
6
= 2. If
A
is in
W
1
∩
W
2
, then
A
=
A
t
=

A,
and so 2
A
=
O
. Since
char
(
F
)
6
= 2, we conclude that
A
=
O
, i.e.
W
1
∩
W
2
=
{
O
}
. To complete the proof of
M
n
×
n
(
F
) =
W
1
⊕
W
2
, we need only show that
every matrix
M
in
M
n
×
n
(
F
) can be written as a sum of symmetric matrix
A
and a skewsymmetric
B
. That is,
M
=
A
+
B,
with
A
t
=
A
and
B
t
=

B
. Assuming that we have such a sum, taking the
transpose of both parts gives us
M
t
=
A
t
+
B
t
=
A

B.
Adding and subtracting the above two equations gives us
M
+
M
t
= 2
A
and
M

M
t
= 2
B,
respectively. Therefore, our candidates are
A
=
M
+
M
t
2
and
B
=
M

M
t
2
.
Finally, it is easy to verify that
A
t
=
A
and
B
t
=

B
to complete the proof.
30
.
Assume that
V
=
W
1
⊕
W
2
.
Then each
v
in
V
can be written as the
sum
x
1
+
y
2
with
x
1
∈
W
1
and
y
2
∈
W
2
, because
V
=
W
1
+
W
2
. Assume that
v
=
x
1
+
x
2
=
y
1
+
y
2
,
for some
x
i
, y
i
∈
W
i
(i=1,2). Then
x
1

y
1
=
y
2

x
2
,
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and so
x
1

y
1
and
y
2

x
2
are in
W
1
∩
W
2
=
{
0
}
. Therefore,
x
1
=
y
1
and
x
2
=
y
2
,
that is, each
v
in
V
can be written
uniquely
as the sum
x
+
y
with
x
∈
W
1
and
y
∈
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 Spring '10
 FUCKHEAD
 Matrices, Skewsymmetric matrix, Det, Transpose, Symmetry in mathematics

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