Higher Secondary Mathematics: skew symmetric matrix

Thursday, March 7, 2024

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.

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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Wednesday, March 6, 2024

19.11 - Symmetric and Skew Symmetric Matrices

In the previous section, we saw transpose of a matrix. In this section, we will see symmetric and skew symmetric matrices.

Symmetric Matrix

This can be written in 5 steps:
1. Consider any square matrix A. We know how to write it's transpose A'.
2. If A' is equal to A, then we say that, A is a symmetric matrix.
3. An example is given below:
Let A = $\left[\begin{array}{r}
3    &{    \sqrt2    }    &{    11    }\\
\sqrt2    &{    5    }    &{    -8    }\\
11    &{    -8    }    &{    9    }\\
\end{array}\right]
$.
• If we write A', we will get A itself. So A is a symmetric matrix.
4. Consider any element say the $\sqrt2$ at second row, first column. Let us interchange the row and column. So we want first row, second column. What is the element at the interchanged position? It is also $\sqrt2$
5. This is applicable to any element of a symmetric matrix.
• We can write:
   ♦ Let a be any element of a symmetric matrix. Let it be at the row i, column j.
   ♦ The element at row j, column i will also be the same element a.

Skew Symmetric Matrix

This can be written in 5 steps:
1. Consider any square matrix A. We know how to write it's transpose A'.
2. If A' is equal to −A, then we say that, A is a skew symmetric matrix.
3. An example is given below:
Let A = $\left[\begin{array}{r}
0    &{    -3    }    &{    11    }\\
3    &{    0    }    &{    -7    }\\
-11    &{    7    }    &{    0    }\\
\end{array}\right]
$.
• If we write A', we will get:
A' = $\left[\begin{array}{r}
0    &{    3    }    &{    -11    }\\
-3    &{    0    }    &{    7    }\\
11    &{    -7    }    &{    0    }\\
\end{array}\right]
$.

• We see that, A' = −A. So A is a skew symmetric matrix.
4. Consider any non-diagonal element in A. Say the 7 at third row, second column. Let us interchange the row and column. So we want second row, third column. What is the element at the interchanged position? It is -7.
5. This is applicable to any non-diagonal element of a skew symmetric matrix.
• We can write:
   ♦ Let a be any non-diagonal element of a skew symmetric matrix. Let it be at the row i, column j.
   ♦ The element at row j, column i will be the -ve of element a.
6.The property written in (5) is applicable to diagonal elements also.This can be explained in 3 steps:
(i) For any diagonal element, the row number and column number will be the same.
• So any diagonal element, can be represented as a_ii.
(ii) In a skew symmetric matrix, any a_ij must be equal to −a_ji.
• So any a_ii must be equal to −a_ii
• That means:
a_ii = −a_ii
⇒ 2a_ii = 0
⇒ a_ii = 0
(iii) We can write:
If A is a skew symmetric matrix, then all diagonal elements of A will be zero.

Now we will see two theorems related to symmetric and skew symmetric matrices.

Theorem 1
For any square matrix A,
(i) (A + A') is a symmetric matrix
(ii) (A – A') is a skew symmetric matrix.

Proof for part (i):
$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{Let}}    &{B}    & {~=~}    &{A+A'}    \\
{~\color{magenta}    2    }    &{\implies}    &{B'}    & {~=~}    &{(A+A')'}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{A' + (A')'}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{A' + A}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{A+A'}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{B}    \\
\end{array}$

◼ Remarks:
(3) (A+B)' = A' + B'
(4) (A')' = A
(5) A+B = B+A

• We see that, B' = B. That means, B is a symmetric matrix. That means, (A+A') is a symmetric matrix.

Proof for part (ii):
$\begin{array}{ll} {~\color{magenta}    1    }    &{\text{Let}}    &{C}    & {~=~}    &{A-A'}    \\
{~\color{magenta}    2    }    &{\implies}    &{C'}    & {~=~}    &{(A-A')'}    \\
{~\color{magenta}    3    }    &{{}}    &{{}}    & {~=~}    &{A' - (A')'}    \\
{~\color{magenta}    4    }    &{{}}    &{{}}    & {~=~}    &{A' - A}    \\
{~\color{magenta}    5    }    &{{}}    &{{}}    & {~=~}    &{-(A-A')}    \\
{~\color{magenta}    6    }    &{{}}    &{{}}    & {~=~}    &{-C}    \\
\end{array}$

◼ Remarks:
(3) (A−B)' = [A+(−B)]'= [A+(−1B)]'= [A'+(−1B)']
= [A'+(−1)(B)'] = A' − B'
(4) (A')' = A
(5) A−B = [A + (−B)] = [(−B) + A] = −[B − A]

• We see that, C' = −C. That means, C is a skew symmetric matrix. That means, (A−A') is a skew symmetric matrix.

In the next section, we will see the second theorem.