Symmetric Matrix

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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 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.
\[ A = \begin{bmatrix} -9 & \color{red}2 & \color{blue}{-3} & \color{red}{7}\\ \\ \color{red}2 & 2 & \color{green}{-4} & \color{blue}{0}\\ \\ \color{blue}{-3} & \color{green}{-4} & 4 & \color{red}5\\ \\ \color{red}{7} & \color{blue}{0} & \color{red}5 & -5 \end{bmatrix} \]
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

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