19.12 - Second Theorem

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.

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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:

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The link below gives a few more solved examples:

Exercise 19.3

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In the next section, we will see elementary operation of a matrix and invertible matrices.

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