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 the Symmetric Matrices
Some of the most important properties of the symmetric matrices are given below.
- If \( A \) is a symmetric matrix, then \( A^T = A \), where\( A^T \) is the transpose of matrix \( A \).
- If \( A \) and \( B \) are symmetric matrices, then \( A \pm B \) are also symmetric.
- If \( A \) and \( B \) are symmetric matrices, then \( AB + BA \) is also symmetric. (\( A \) and \( B \) must be of the same size)
- If \( A \) is a symmetric matrix, then its inverse matrix \( A^{-1} \) ,if it exists ,is also symmetric.
- If \( A \) is a symmetric matrix, then \( k A \) is also symmetric; where \( k \) is any real number.
- If \( A \) is a symmetric matrix, then \( A^n \) is also symmetric; \( n \) is any positive integer.
- 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
More References and links
-
Matrices with Examples and Questions with Solutions.
-
Transpose of a Matrix.
-
Determinant of a Square Matrix.
-
Inverse Matrix Questions with Solutions.
- Elementary Linear Algebra - 7 th Edition - Howard Anton and Chris Rorres
- Introduction to Linear Algebra - Fifth Edition (2016) - Gilbert Strang
- Linear Algebra Done Right - third edition, 2015 - Sheldon Axler
- Linear Algebra with Applications - 2012 - Gareth Williams