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 aii.
(ii) In a skew symmetric matrix, any aij must be equal to −aji.
•
So any aii must be equal to −aii
•
That means:
aii = −aii
⇒ 2aii = 0
⇒ aii = 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.
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