Express matrix A = [1 2 2-1] as the sum of a symmetric and skew-symmetric matrix.
Answers
Answer:
Let A=[
3
1
5
−1
], then A
′
=[
3
5
1
−1
]
Now, A+A
′
=[
3
1
5
−1
]+[
3
5
1
−1
]=[
6
6
6
−2
]
Let P=
2
1
(A+A
′
)=
2
1
[
6
6
6
−2
]=[
3
3
3
−1
]
Now, P
′
=[
3
3
3
−1
]=P
Thus, P=
2
1
(A+A
′
) is a symmetric matrix.
Now, A−A
′
=[
3
1
5
−1
]−[
3
5
1
−1
]=[
0
−4
4
0
]
Let Q=
2
1
(A−A
′
)=
2
1
[
0
−4
4
0
]=[
0
−2
2
0
]
Now, Q
′
=[
0
−2
2
0
]=Q
Thus, Q=
2
1
(A−A
′
) is a skew-symmetric matrix.
Representing A as sum of P and Q:
P+Q=[
3
3
3
−1
]+[
0
−2
2
0
]=[
3
1
5
−1
]=A
(ii) Let =
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
, then A
′
=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
Now, A+A
′
=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
+
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
12
−4
4
−4
6
−2
4
−2
6
⎦
⎥
⎥
⎤
Let P=
2
1
(A+A
′
)=
2
1
⎣
⎢
⎢
⎡
12
−4
4
−4
6
−2
4
−2
6
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
Now, P
′
=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
=P
Thus, P=
2
1
(A+A
′
) is a symmetric matrix.
Now, A−A
′
=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
−
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
0
0
0
0
0
0
0
0
0
⎦
⎥
⎥
⎤
Let Q=
2
1
(A−A
′
)=
⎣
⎢
⎢
⎡
0
0
0
0
0
0
0
0
0
⎦
⎥
⎥
⎤
Now, Q
′
=
⎣
⎢
⎢
⎡
0
0
0
0
0
0
0
0
0
⎦
⎥
⎥
⎤
=−Q
Thus, Q=
2
1
(A−A
′
) is a skew-symmetric matrix.
Representing A as sum of P and Q:
P+Q=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
+
⎣
⎢
⎢
⎡
0
0
0
0
0
0
0
0
0
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
6
−2
2
−2
3
−1
2
−1
3
⎦
⎥
⎥
⎤
=A
(iii) Let A=
⎣
⎢
⎢
⎡
3
−2
−4
3
−2
−5
−1
1
2
⎦
⎥
⎥
⎤
, then A
′
=
⎣
⎢
⎢
⎡
3
3
−1
−2
−2
1
−4
−5
2
⎦
⎥
⎥
⎤
Now, A+A
′
=
⎣
⎢
⎢
⎡
3
−2
−4
3
−2
−5
−1
1
2
⎦
⎥
⎥
⎤
+
⎣
⎢
⎢
⎡
3
3
−1
−2
−2
1
−4
−5
2
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
6
1
−5
1
−4
−4
−5
−4
4
⎦
⎥
⎥
⎤
Let P=
2
1
(A+A
′
)=
2
1
⎣
⎢
⎢
⎡
6
1
−5
1
−4
−4
−5
−4
4
⎦
⎥
⎥
⎤
=
⎣
⎢
⎢
⎡
3
2
1
−
2
5
2
1
−2
−2
−
2
5
−2
2
⎦
⎥
⎥
⎤
Now, P
′
=
⎣
⎢
⎢
⎡
3
2
1
−
2
5
2
1
−2
−2
−
2
5
−2
2
⎦
⎥
⎥
⎤
=P
Thus, P=
2
1
(A+A
′
) is a symmetric matrix.