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.

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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.

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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.

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

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19.10 - Transpose of a Matrix

In the previous section, we completed a discussion on multiplication of matrices. In this section, we will see transpose of a matrix.

Transpose of a Matrix

This can be written in 9 steps:
1. Suppose that, we are given a matrix A.
• Based on A, we can write a new matrix A' in such a way that:
    ♦ Rows in A, are the columns in A'.
    ♦ Columns in A, are the rows in A'.
2. In other words,
    ♦ First row of A, becomes the first column in A'.   
    ♦ Second row of A, becomes the second column in A'.   
    ♦ Third row of A, becomes the third column in A'.   
    ♦ so on . . .
3. This is same as:
    ♦ First column of A, becomes the first row in A'.   
    ♦ Second column of A, becomes the second row in A'.   
    ♦ Third column of A, becomes the third row in A'.
    ♦ so on . . .
4. Let us see an example:
If A = $\left[\begin{array}{r}               
-7    &{    0    }    \\
8    &{    \sqrt{2}    }    \\
-{\frac{1}{3}}    &{    3    }    \\
\end{array}\right]               
$, then A' = $\left[\begin{array}{r}                                       
-7    &{    8    }    &{    -{\frac{1}{3}}    }       \\
0    &{    \sqrt{2}    }    &{    3    }        \\
\end{array}\right]                                       
$
5. Consider any element, say $-{\frac{1}{3}}$ in A. It’s position is: row 3, column 1.
• But in A', the position of $-{\frac{1}{3}}$ is: row 1, column 3.
    ♦ That is., the row and column are interchanged.
6. This interchanging of rows and columns is applicable to all elements in general.
• We can write:
    ♦ In the original matrix, an element is at position i,j.
    ♦ Then in the new matrix, that element will be at the position j,i.
7. Also note that:
If the order of the original matrix is m×n, then the order of the new matrix will be n×m.
8. Based on the above steps, we can write the definition:
If A = [a_ij]_m×n, then A' = [a_ji]_n×m .
9. The new matrix A' is called the transpose of A. It can be denoted either as A' or A^T.

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Properties of transpose of matrices

• We have to be familiar with four properties:
For any matrices A and B of suitable orders, we have
I. (A')' = A                II. (kA)' = kA' (where k is any constant)
III. (A+B)' = A' + B'        IV. (AB)' = B' A'.

[We mention ‘suitable orders’ because, any two given matrices cannot be added or multiplied. If we want (A+B), then A and B must be of the same order. Similarly, if we want (AB), then number of columns in A must be equal to the number of rows in B]

• We will see the proofs of the four properties in higher classes. At present, we will verify them using examples.
Let A = $\left[\begin{array}{r}                       
9    &{    6    }    &{    -4    }\\
0    &{    \sqrt{2}    }    &{    1    }\\
\end{array}\right]                       
$ and B = $\left[\begin{array}{r}                       
-1    &{    2    }    &{    5    }\\
11    &{    7    }    &{    3    }\\
\end{array}\right] $.

1. Verifying property I:

2. Verifying property II:

• From (4) and (6), we get:
(kB)' = kB' (where k is any constant) 

3. Verifying property III:

• From (5) and (7), we get:
(A+B)' = A' + B'

4. Verifying property IV:
Let X = $\left[\begin{array}{r}                       
9    &{    6    }    &{    -4    }\\
0    &{    {2}    }    &{    1    }\\
\end{array}\right]                       
$ and Y = $\left[\begin{array}{r}                       
-1    &{    2    }    &{    5    }\\
11    &{    7    }    &{    3    }\\
7    &{    1    }    &{    4    }\\
\end{array}\right] $.

• From (3) and (7), we get:
(XY)' = Y' X'
(note that, positions of X and Y are interchanged)

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In the next section, we will see symmetric and skew symmetric matrices.

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19.9 - Multiplicative Identity

In the previous section, we saw two properties of multiplication of matrices. In this section, we will see the third property.

Property III: The existence of multiplicative identity
• Consider any square matrix A of the order (m×m).
• We can write an identity matrix I of the same order.
That I will satisfy the equation: AI = IA = A
• Let us see an example. It can be written in 4 steps:

1. Let A = $\left[\begin{array}{r}                           
-8    &{    2    }    &{    5    }    \\
0    &{    -7    }    &{    -3    }    \\
3    &{    2    }    &{    4    }    \\
\end{array}\right]$

• Then I = $\left[\begin{array}{r}                           
1    &{    0    }    &{    0    }    \\
0    &{    1    }    &{    0    }    \\
0    &{    0    }    &{    1    }    \\
\end{array}\right]                           
$               

2. First we find AI:

3. Next we find IA:

4. Based on (2) and (3), we can write:
AI = IA = A

◼ We will see the actual proof in higher classes.

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Now we will see a solved example.

Solved example 19.14
If A = $\left[\begin{array}{r}                           
1    &{    3    }    &{    2    }    \\
2    &{    0    }    &{    -1    }    \\
1    &{    2    }    &{    3    }    \\
\end{array}\right]                           
$, then show that A³ - 4A² - 3A + 11I = O
Solution:
1. First we will write A²:

2. Next we will write A³:

3. Next we will write 4A²:

3. Next we will write 3A:

4. Finally we will write 11I:

5. Substituting the values, we get:

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The link below gives a few more solved examples:

Exercise 19.2

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In the next section, we will see transpose of a matrix.

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