In the previous section, we saw the theorem 1 related to symmetric and skew symmetric matrices. In this section, we will see theorem 2.
Theorem 2
Any square matrix can be expressed as the sum of a symmetric matrix and a skew symmetric matrix.
Proof:
Let A be a square matrix. First, we will prove that,
$A = \frac{1}{2} \left(A + A' \right)~+~\frac{1}{2} \left(A - A' \right)$
$\begin{array}{ll}
{~\color{magenta} 1 } &{{}} &{{}} &{{}}
&{\frac{1}{2} \left(A + A' \right)~+~\frac{1}{2} \left(A - A'
\right)} \\
{~\color{magenta} 2 } &{{}}
&{{}} & {~=~} &{\frac{1}{2} A ~+~\frac{1}{2} A'
~+~\frac{1}{2} A ~-~\frac{1}{2} A'} \\
{~\color{magenta} 3
} &{{}} &{{}} & {~=~} &{\frac{1}{2} A
~+~\frac{1}{2} A ~+~\frac{1}{2} A' ~-~\frac{1}{2} A'} \\
{~\color{magenta} 4 } &{{}} &{{}} & {~=~} &{\frac{1}{2} A ~+~\frac{1}{2} A} \\
{~\color{magenta} 5 } &{{}} &{{}} & {~=~} &{A} \\
\end{array}$
•
Now we can write the proof for theorem 2 in steps:
1.
We just proved that, A can be written as the sum of two terms.
$\frac{1}{2} \left(A + A' \right)$ and $\frac{1}{2} \left(A - A'
\right)$
2. Consider the first term.
•
In this term, we know (from theorem 1) that, (A+A') is a symmetric matrix.
• We need to prove that, $\frac{1}{2} \left(A + A' \right)$ is also a symmetric matrix.
•
For that, we need to prove:
$\frac{1}{2} \left(A + A' \right)~=~\left[\frac{1}{2} \left(A + A' \right) \right]'$
•
In other words, we need to prove:
$\frac{1}{2} B~=~\left[\frac{1}{2} B \right]'$
♦ Where B = A + A'
• This can be proved in 4 steps:
(i) B is symmetric.
♦ That means, B = B'
♦ That means, $\frac{1}{2} B ~=~ \frac{1}{2} B'$
(ii) For any matrix X, we have: kX' = (kX)'
(iii) So the result in (i) becomes:
$\frac{1}{2} B ~=~ \frac{1}{2} B' ~=~\left[\frac{1}{2} B \right]'$
(iv) By picking the first and last terms in (iii), we get:
$\frac{1}{2} B ~=~\left[\frac{1}{2} B \right]'$.
•
This is same as:
$\frac{1}{2} \left(A + A' \right)~=~\left[\frac{1}{2} \left(A + A' \right) \right]'$
•
So $\frac{1}{2} \left(A + A' \right)$ is a symmetric matrix.
3. Consider the second term.
•
In this term, we know (from theorem 1) that, (A−A') is a skew symmetric matrix.
•
We need to prove that, $\frac{1}{2} \left(A - A' \right)$ is also a skew symmetric matrix.
•
For that, we need to prove:
$\frac{1}{2} \left(A - A' \right)~=~- \left[\frac{1}{2} \left(A - A' \right) \right]'$
•
In other words, we need to prove:
$\frac{1}{2} C~=~- \left[\frac{1}{2} C \right]'$
♦ Where C = A − A'
•
This can be proved in 4 steps:
(i) C is skew symmetric.
♦ That means, C = -C'
♦ That means, $\frac{1}{2} C ~=~ -\frac{1}{2} C'$
(ii) For any matrix X, we have: kX' = (kX)'
(iii) So the result in (i) becomes:
$\frac{1}{2} C ~=~ -\frac{1}{2} C' ~=~- \left[\frac{1}{2} C \right]'$
(iv) By picking the first and last terms in (iii), we get:
$\frac{1}{2} C ~=~-\left[\frac{1}{2} C \right]'$.
•
This is same as:
$\frac{1}{2} \left(A - A' \right)~=~- \left[\frac{1}{2} \left(A - A' \right) \right]'$
•
So $\frac{1}{2} \left(A - A' \right)$ is a skew symmetric matrix.
4. So the first term is a symmetric matrix. Also, the second term is a skew symmetric matrix.
•
Thus we effectively wrote matrix A as the sum of a symmetric matrix and a skew symmetric matrix.
•
Theorem 2 is proved.
Solved example 19.15
Express the matrix B = $\left[\begin{array}{r}
3 &{ 11 } &{ 2 } \\
-10 &{ -4 } &{ 12 } \\
8 &{ -9 } &{ 13 } \\
\end{array}\right]
$ as the sum of a symmetric matrix and a skew symmetric matrix.
Solution:
1. Write the transpose of B:
B' = $\left[\begin{array}{r}
3 &{ -10 } &{ 8 } \\
11 &{ -4 } &{ -9 } \\
2 &{ 12 } &{ 13 } \\
\end{array}\right]
$
2. Let $P = \frac{1}{2} (B+B')$ and $Q = \frac{1}{2} (B-B')$
3. Finding P and proving that, it is symmetric:
• If we try to write P', we will get the same P. So P is a symmetric matrix.
4. Finding Q and proving that, it is skew symmetric:
• We see that, Q' = -Q. So it is a skew symmetric matrix.
5. Checking the sum:
The link below gives a few more solved examples:
Exercise 19.3
In the next section, we will see elementary operation of a matrix and invertible matrices.
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