# Symmetric Matrix

  

## Definition of a Symmetric Matrix

A square matrix $A$ is symmetric if and only if $A = A^T$ where $A^T$ is the transpose of matrix $M$.
A symmetric matrix may be reconized visually: The entries that are symmetrically positioned with respect to the main diagonal are equal as shown in the example below of a symmetric matrix.
$A = \begin{bmatrix} -9 & \color{red}2 & \color{blue}{-3} & \color{red}{7}\\ \\ \color{red}2 & 2 & \color{green}{-4} & \color{blue}{0}\\ \\ \color{blue}{-3} & \color{green}{-4} & 4 & \color{red}5\\ \\ \color{red}{7} & \color{blue}{0} & \color{red}5 & -5 \end{bmatrix}$
These are examples of symmetric matrices.
a)
$A = \begin{bmatrix} 1 & -2 \\ -2 & 9 \end{bmatrix}$
By simple inspection, the entries that are symmetrically positioned with respect to the main diagonal are both equal to $-2$ and therefore matrix $A$ is symmetric.
Also, the transpose of matrix $A$ is obtained by interchanging the row of the matrix into a column. The transpose of matrix $A$ is $A^T = \begin{bmatrix} 1 & -2 \\ -2 & 9 \end{bmatrix}$.
Note that $A^T = A$ and therefore matrix $A$ is symmetric.

b)
$B = \begin{bmatrix} 9 & 8 & -1\\ 8 & 7 & 0\\ -1 & 0 & -6 \end{bmatrix}$
The transpose of matrix $B$ is obtained by interchanging the column of the matrix into a row. The transpose of matrix $B$ is $B^T = \begin{bmatrix} 9 & 8 & -1\\ 8 & 7 & 0\\ -1 & 0 & -6 \end{bmatrix}$
Note that $B^T = B$ and therefore matrix $B$ is symmetric.

## The Product of Any Matrix and it Transpose is Symmetric

For any matrix $A$ of size $m \times n$ , the matrices $A A^T$ and $A^T A$ are symmetric and have sizes $m \times m$ and $n \times n$ respectively.
Let $A = \begin{bmatrix} -3 & -3 & 9\\ 2 & -7 & 1 \end{bmatrix}$ is a $2 \times 3$ matrix

$A^T = \begin{bmatrix} -3 & 2\\ -3 & -7 \\ 9 & 1 \end{bmatrix}$

$A A^T = \begin{bmatrix} -3 & -3 & 9\\ 2 & -7 & 1 \end{bmatrix} \begin{bmatrix} -3 & 2\\ -3 & -7 \\ 9 & 1 \end{bmatrix}$
$\quad \quad = \begin{bmatrix} 99 & 24\\ 24 & 54 \end{bmatrix}$

Hence $A A^T$ is a $2 \times 2$ symmetric matrix.

$A^T A = \begin{bmatrix} -3 & 2\\ -3 & -7 \\ 9 & 1 \end{bmatrix} \begin{bmatrix} -3 & -3 & 9\\ 2 & -7 & 1 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} 13 & -5 & -25\\ -5 & 58 & -34\\ -25 & -34 & 82 \end{bmatrix}$ is a $3 \times 3$

Hence $A^T A$ is a $3 \times 3$ symmetric matrix.

## Properties of the Symmetric Matrices

Some of the most important properties of the symmetric matrices are given below.

1.    If $A$ is a symmetric matrix, then $A^T = A$, where$A^T$ is the transpose of matrix $A$.
2.    If $A$ and $B$ are symmetric matrices, then $A \pm B$ are also symmetric.
3.    If $A$ and $B$ are symmetric matrices, then $AB + BA$ is also symmetric. ($A$ and $B$ must be of the same size)
4.    If $A$ is a symmetric matrix, then its inverse matrix $A^{-1}$ ,if it exists ,is also symmetric.
5.    If $A$ is a symmetric matrix, then $k A$ is also symmetric; where $k$ is any real number.
6.    If $A$ is a symmetric matrix, then $A^n$ is also symmetric; $n$ is any positive integer.
7.    For any matrix $B$, the matrices $B B^T$ and $B^T B$ are symmetric.

## Examples with Solutions

Example 1
Which if the following matrices are symmetric?
a) $A = \begin{bmatrix} 0 & -1 & 1 \\ -1 & 0 & 0 \end{bmatrix}$      b) $B = \begin{bmatrix} -2 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{bmatrix}$      c) $C = \begin{bmatrix} 5 & -7 & 0 & 0\\ 5 & -7 & 0 & 0\\ 5 & -7 & 0 & 0\\ 5 & -7 & 0 & 0 \end{bmatrix}$      d) $D = \begin{bmatrix} 1 & 4 & 0 & 9\\ 4 & 2 & -4 & 0\\ 0 & -4 & 4 & 0\\ 9 & 0 & 0 & -5 \end{bmatrix}$

Solution
Matrices $B$ and $D$ are symmetric.

Example 2
Symmetric matrices $A$ and $B$ are given by $A = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}$.
Show that $A+B$ and $A-B$ are symmetric. (Verify property 2)

Solution
Calculate $A + B$
$A + B = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix} + \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}$
$\quad = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$

Calculate $A - B$
$A - B = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix} - \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}$
$\quad = \begin{bmatrix} -5 & -1 \\ -1 & 5 \end{bmatrix}$

Hence $A + B$ and $A - B$ are also symmetric matrices.

Example 3
Find the inverse of the symmetric matrix $A = \begin{bmatrix} -1 & -2\\ -2 & 0 \end{bmatrix}$ and show that this inverse matrix is symmetric.(Verify properties 4)

Solution
Use the formula of the inverse of a $2 \times 2$ matrix $\begin{bmatrix} x & y \\ z & w \\ \end{bmatrix}^{-1} = \dfrac{1}{xw - yz} \begin{bmatrix} w & - y \\ - z & x \\ \end{bmatrix}$ to find $A^{-1}$
$A^{-1} = \dfrac{1}{-4} \begin{bmatrix} 0 & 2\\ 2 & -1 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} 0 & -\dfrac{1}{2}\\ -\dfrac{1}{2} & \dfrac{1}{4} \end{bmatrix}$
Hence $A^{-1}$ is also symmetric.

Example 4
Let matrix $A = \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix}$
Calculate $A^3$ and verify that it is symmetric. (Verify properties 6)

Solution
$A^3 = \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix} \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix} \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} 10 & 4\\ 4 & 50 \end{bmatrix} \begin{bmatrix} -3 & 1\\ 1 & 7 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} -26 & 38\\ 38 & 354 \end{bmatrix}$
Hence $A^3$ is a symmetric matrix.

Example 5
Find the real numbers $a$, $b$ and $c$ so that the matrix $A = \begin{bmatrix} 0 & a+b & c+2 \\ a & 2 & c \\ 4 & a+b & 4 \end{bmatrix}$ is symmetric.

Solution
For the given matrix to be symmetric, we have to have the following equations satisfied simultaneously:
$a = a + b$ , $c + 2 = 4$ , $c = a + b$
Solve the system of equations in $a,\; b,\; c$ above to obtain
$b = 0 , \; c = 2 , \; a = 2$ which are the values so that the given matrix is symmetric.

Example 6
Given the symmetric matrices $A = \begin{bmatrix} -1 & 3 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 8 \\ 8 & 9 \end{bmatrix}$, verify that $AB + BA$ is also symmetric (Property 3 above)

Solution
Calculate $AB$
$AB = \begin{bmatrix} -1 & 3 \\ 3 & 4 \end{bmatrix} \begin{bmatrix} 4 & 8 \\ 8 & 9 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} 20 & 19 \\ 44 & 60 \end{bmatrix}$

Calculate $BA$
$BA = \begin{bmatrix} 4 & 8 \\ 8 & 9 \end{bmatrix} \begin{bmatrix} -1 & 3 \\ 3 & 4 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} 20 & 44\\ 19 & 60 \end{bmatrix}$

Calculate $AB + BA$
$AB + BA = \begin{bmatrix} 20 & 19 \\ 44 & 60 \end{bmatrix} + \begin{bmatrix} 20 & 44\\ 19 & 60 \end{bmatrix}$

$\quad \quad = \begin{bmatrix} 40 & 63\\ 63 & 120 \end{bmatrix}$
Hence $AB + BA$ is a symmetric matrix.

## Questions (with solutions given below)

• Part 1
Matrices $A$ and $B$ are such that $A B = B A = I$ where $I$ is the identity matrix. Use any of the properties above to explain that if one of the matrices is symmetric, then the other one is also symmetric.
• Part 2
Given the matrices $A = \begin{bmatrix} -5 & -2 \\ - 1 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix}$.
Calculate $A + B$ and explain why $(A + B)^T = A + B$ without any further computations.
• Part 3
Let matrix $A = \begin{bmatrix} - 1 & - b\\ c & c^2 \end{bmatrix}$. Find the real numbers $b$ and $c$ so that matrix $A$ is symmetric and $Det (A) = - 1$.
• Part 4
Show that any symmetric matrix of the form $A = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$ , with $|a| \lt 1$, is such that $A = A^T = A^{-1}$ and calculate $A^n$ where $n$ is a positive integer.

### Solutions to the Above Questions

• Part 1
Given that $A B = B A = I$ where $I$ is the identity matrix, and according to the definition of the inverse of a matrix $A$ and $B$ are inverses of each other. Hence according to property 4, if $A$ is symmetric, then $B$ its inverse is also symmetric and if $B$ is symmetric, then its inverse $A$ is also symmetric.

• Part 2
$A + B = \begin{bmatrix} -5 & -2 \\ - 1 & 2 \end{bmatrix} + \begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} -2 & -1 \\ -1 & 6 \end{bmatrix}$
Note that $A + B$ is symmetric and therefore according to the definition (or property 1)
$(A + B)^T = A + B$

• Part 3
For matrix $A$ to be symmetric, we have to have $- b = c$ or $b = - c$.
$Det(A) = - c^2 + b c = - 1$
Hence the equation
$- c^2 + b c = - 1$
Substitute $b$ by $- c$ in the equation above
$-c^2 - c^2 = - 1$
Solve for $C$ to obtain two solutions
$c_1 = \dfrac{\sqrt 2}{2}$ and $c_1 = - \dfrac{\sqrt 2}{2}$
Hence two pairs of solutions
$c = \dfrac{\sqrt 2}{2}$ and $b = - \dfrac{\sqrt 2}{2}$
or
$c = - \dfrac{\sqrt 2}{2}$ and $b = \dfrac{\sqrt 2}{2}$

• Part 4
Given $A = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$

Determine $A^T$

$A^T = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$

Use the formula of the inverse of a $2 \times 2$ matrix $\begin{bmatrix} x & y \\ z & w \\ \end{bmatrix}^{-1} = \dfrac{1}{xw - yz} \begin{bmatrix} w & - y \\ - z & x \\ \end{bmatrix}$ to find $A^{-1}$

$A^{-1} = \dfrac{1}{- a^2 - (\sqrt{1-a^2})^2} \begin{bmatrix} - a & - \sqrt{1-a^2} \\ \\ - \sqrt{1-a^2} & a \\ \end{bmatrix}$

Simplify
$A^{-1} = (-1) \begin{bmatrix} - a & - \sqrt{1-a^2} \\ \\ - \sqrt{1-a^2} & a \\ \end{bmatrix}$

$\quad \quad = \begin{bmatrix} a & \sqrt{1-a^2} \\ \\ \sqrt{1-a^2} & - a \\ \end{bmatrix}$

and we note that $A = A^T = A^{-1}$
$A^2 = A A = A A^{-1} = I$
$A^3 = A^2 A = I A = A$
$A^4 = A^3 A = A A = I$
$A^5 = A^4 A = I A = A$
We may easily conclude that
$A^n = I$ if $n$ is even
and
$A^n = A$ if $n$ is odd