# Symmetric Matrix

## Definition of a Symmetric Matrix

A square matrix A is symmetric if and only if A = AT where AT is the transpose of matrix A .
A symmetric matrix may be recognized 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.

These are examples of symmetric matrices.
a)

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
Note that AT = A and therefore matrix A is symmetric.

b)

The transpose of matrix B is obtained by interchanging the column of the matrix into a row. The transpose of matrix B is
Note that BT = 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 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 of the same size, then $$AB + BA$$ is also symmetric.
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