Symmetric Matrix

A symmetric matrix is a square matrix that equals its own transpose. In this page we present the definition, key properties, worked examples, and practice questions with detailed solutions.

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 \( A \).

In other words, \( A_{ij} = A_{ji} \) for every pair of indices \( (i, j) \) — the entries mirrored across the main diagonal are equal.

Key idea. A symmetric matrix is completely determined by the entries on and above its main diagonal. The diagonal entries \( A_{ii} \) are unconstrained — they can be any real numbers.

Example of a \( 4 \times 4 \) Symmetric Matrix

\[ \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 5 & 6 & 7 \\ 3 & 6 & 8 & 9 \\ 4 & 7 & 9 & 0 \end{bmatrix} \]

(a) A \( 2 \times 2 \) Symmetric Matrix

\[ A = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} \]

The entries mirrored across the main diagonal are equal (both are \( -2 \)), so \(A\) is symmetric. Its transpose is obtained by swapping rows and columns:

\[ A^T = \begin{bmatrix} 1 & -2 \\ -2 & 5 \end{bmatrix} = A. \]

(b) A \( 3 \times 3 \) Symmetric Matrix

\[ B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & -5 \\ 3 & -5 & 6 \end{bmatrix}. \]

\[ B^T = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & -5 \\ 3 & -5 & 6 \end{bmatrix} = B. \]

Since \(B^T = B\), matrix \(B\) is symmetric.

The Product of Any Matrix and Its Transpose Is Symmetric

For any \( m \times n \) matrix \( A \), the products \( A A^T \) and \( A^T A \) are symmetric, with sizes \( m \times m \) and \( n \times n \) respectively.

Let

\[ A = \begin{bmatrix} -3 & -3 & 9 \\ 2 & -7 & 1 \end{bmatrix}. \]

Its transpose is

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

Compute \(A A^T\):

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

The result is a \( 2 \times 2 \) symmetric matrix (note the mirrored off-diagonal entries \(24\)).

Compute \(A^T A\):

\[ A^T A = \begin{bmatrix} -3 & 2 \\ -3 & -7 \\ 9 & 1 \end{bmatrix} \begin{bmatrix} -3 & -3 & 9 \\ 2 & -7 & 1 \end{bmatrix} = \begin{bmatrix} 13 & -5 & -25 \\ -5 & 58 & -34 \\ -25 & -34 & 82 \end{bmatrix}. \]

This is a \(3 \times 3\) symmetric matrix.

Properties of Symmetric Matrices

Some of the most important properties are listed below.

  1. If \(A\) is symmetric, then \(A^T = A\).
  2. If \(A\) and \(B\) are symmetric, then \(A \pm B\) is symmetric.
  3. If \(A\) and \(B\) are symmetric matrices of the same size, then \(AB + BA\) is symmetric. (Note: \(AB\) alone is not necessarily symmetric.)
  4. If \(A\) is symmetric and invertible, then \(A^{-1}\) is also symmetric.
  5. If \(A\) is symmetric and \(k\) is any real scalar, then \(kA\) is symmetric.
  6. If \(A\) is symmetric, then \(A^n\) is symmetric for every positive integer \(n\).
  7. For any matrix \(B\), the products \(B B^T\) and \(B^T B\) are symmetric.
  8. Every real symmetric matrix has only real eigenvalues, and its eigenvectors can be chosen to form an orthonormal basis.
  9. Every real symmetric matrix is orthogonally diagonalizable: \(A = Q D Q^T\), where \(Q\) is orthogonal and \(D\) is diagonal (the Spectral Theorem).
Complementary concept — skew-symmetric matrices. A square matrix \( A \) is skew-symmetric if \( A^T = -A \). Every square matrix decomposes uniquely as \[ A = \tfrac{1}{2}(A + A^T) + \tfrac{1}{2}(A - A^T), \] where the first term is symmetric and the second is skew-symmetric.

Examples with Solutions

Example 1 — Identify Symmetric Matrices

Which of 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

  • A: not symmetric — it is not even square (it is \(2 \times 3\)), and symmetry is defined only for square matrices.
  • B: symmetric — every entry \(B_{ij} = B_{ji}\): \(B_{12}=0=B_{21}\), \(B_{13}=1=B_{31}\), \(B_{23}=1=B_{32}\).
  • C: not symmetric — for example \(C_{12} = -7\) but \(C_{21} = 5\), so \(C \neq C^T\).
  • D: symmetric — every mirrored pair matches: \(D_{12}=4=D_{21}\), \(D_{13}=0=D_{31}\), \(D_{14}=9=D_{41}\), \(D_{23}=-4=D_{32}\), \(D_{24}=0=D_{42}\), \(D_{34}=0=D_{43}\).

Therefore, matrices \(B\) and \(D\) are symmetric.

Example 2 — Sum and Difference of Symmetric Matrices

Given the symmetric matrices

\[ A = \begin{bmatrix} -1 & 1 \\ 1 & 2 \end{bmatrix}, \quad B = \begin{bmatrix} 4 & 2 \\ 2 & -3 \end{bmatrix}, \]

show that \(A + B\) and \(A - B\) are symmetric (Property 2).

Solution

\[ A + B = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}, \qquad A - B = \begin{bmatrix} -5 & -1 \\ -1 & 5 \end{bmatrix}. \]

Both results have mirrored off-diagonal entries, so both are symmetric.

Example 3 — Inverse of a Symmetric Matrix

Find \(A^{-1}\) for \(A = \begin{bmatrix} -1 & -2 \\ -2 & 0 \end{bmatrix}\) and verify that it is symmetric (Property 4).

Solution

Use the \(2 \times 2\) inverse formula:

\[ \begin{bmatrix} x & y \\ z & w \end{bmatrix}^{-1} = \frac{1}{xw - yz}\begin{bmatrix} w & -y \\ -z & x \end{bmatrix}. \]

\[ A^{-1} = \frac{1}{(-1)(0) - (-2)(-2)} \begin{bmatrix} 0 & 2 \\ 2 & -1 \end{bmatrix} = -\frac{1}{4}\begin{bmatrix} 0 & 2 \\ 2 & -1 \end{bmatrix} = \begin{bmatrix} 0 & -\tfrac{1}{2} \\ -\tfrac{1}{2} & \tfrac{1}{4} \end{bmatrix}. \]

The result is symmetric, as expected.

Example 4 — Power of a Symmetric Matrix

Let \(A = \begin{bmatrix} -3 & 1 \\ 1 & 7 \end{bmatrix}\). Compute \(A^3\) and confirm that it is symmetric (Property 6).

Solution

\[ A^2 = \begin{bmatrix} -3 & 1 \\ 1 & 7 \end{bmatrix} \begin{bmatrix} -3 & 1 \\ 1 & 7 \end{bmatrix} = \begin{bmatrix} 10 & 4 \\ 4 & 50 \end{bmatrix}. \]

\[ A^3 = A^2 \cdot A = \begin{bmatrix} 10 & 4 \\ 4 & 50 \end{bmatrix} \begin{bmatrix} -3 & 1 \\ 1 & 7 \end{bmatrix} = \begin{bmatrix} -26 & 38 \\ 38 & 354 \end{bmatrix}. \]

The result is symmetric.

Example 5 — Find Constants That Make a Matrix Symmetric

Find real \(a, b, c\) such that \(A = \begin{bmatrix} 0 & a+b & c+2 \\ a & 2 & c \\ 4 & a+b & 4 \end{bmatrix}\) is symmetric.

Solution

Set mirrored entries equal:

  • \(A_{12} = A_{21}\): \(a + b = a \Rightarrow b = 0\).
  • \(A_{13} = A_{31}\): \(c + 2 = 4 \Rightarrow c = 2\).
  • \(A_{23} = A_{32}\): \(c = a + b \Rightarrow 2 = a + 0 \Rightarrow a = 2\).

Therefore \(a = 2\), \(b = 0\), \(c = 2\).

Example 6 — Verify That \(AB + BA\) Is Symmetric

Given 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 symmetric (Property 3).

Solution

\[ AB = \begin{bmatrix} 20 & 19 \\ 44 & 60 \end{bmatrix}, \qquad BA = \begin{bmatrix} 20 & 44 \\ 19 & 60 \end{bmatrix}. \]

\[ AB + BA = \begin{bmatrix} 40 & 63 \\ 63 & 120 \end{bmatrix}. \]

The result has mirrored off-diagonal entries, so \(AB + BA\) is symmetric.

Questions with Solutions

Question 1 — Inverses of Symmetric Matrices

Matrices \(A\) and \(B\) satisfy \(AB = BA = I\). Show that if one of them is symmetric, then so is the other.

Solution

Since \(AB = BA = I\), the matrices are inverses: \(B = A^{-1}\) and \(A = B^{-1}\). By Property 4, the inverse of a symmetric matrix is symmetric. Hence if \(A\) is symmetric, so is \(B\), and vice versa.

Question 2 — Transpose of a Matrix Sum

Given \(A = \begin{bmatrix} -5 & -2 \\ -1 & 2 \end{bmatrix}\) and \(B = \begin{bmatrix} 3 & 1 \\ 0 & 4 \end{bmatrix}\), compute \(A + B\) and explain why \((A+B)^T = A + B\).

Solution

\[ A + B = \begin{bmatrix} -2 & -1 \\ -1 & 6 \end{bmatrix}. \]

The result has equal mirrored off-diagonal entries (\(-1\) and \(-1\)), so it is symmetric, hence \((A+B)^T = A + B\).

Question 3 — Finding Unknowns in a Matrix

Let \(A = \begin{bmatrix} -1 & -b \\ c & c^2 \end{bmatrix}\). Find real \(b, c\) such that \(A\) is symmetric and \(\det(A) = -1\).

Solution

For symmetry: \(-b = c\), so \(b = -c\).

Determinant: \(\det(A) = (-1)(c^2) - (-b)(c) = -c^2 + bc\).

Substitute \(b = -c\):

\[ -c^2 - c^2 = -2c^2 = -1 \;\Rightarrow\; c^2 = \tfrac{1}{2}. \]

Two solution pairs:

  • \(c = \tfrac{\sqrt{2}}{2}\) and \(b = -\tfrac{\sqrt{2}}{2}\).
  • \(c = -\tfrac{\sqrt{2}}{2}\) and \(b = \tfrac{\sqrt{2}}{2}\).
Question 4 — Matrix Inverses and Powers

Show that for \(A = \begin{bmatrix} a & \sqrt{1-a^2} \\ \sqrt{1-a^2} & -a \end{bmatrix}\) with \(|a| < 1\), we have \(A = A^T = A^{-1}\), and find \(A^n\).

Solution

By inspection, \(A^T = A\), so \(A\) is symmetric.

Determinant: \(\det(A) = (a)(-a) - (\sqrt{1-a^2})^2 = -a^2 - (1-a^2) = -1\).

\[ A^{-1} = \frac{1}{-1}\begin{bmatrix} -a & -\sqrt{1-a^2} \\ -\sqrt{1-a^2} & a \end{bmatrix} = \begin{bmatrix} a & \sqrt{1-a^2} \\ \sqrt{1-a^2} & -a \end{bmatrix} = A. \]

So \(A = A^T = A^{-1}\), and \(A^2 = I\). Therefore:

  • \(A^n = I\) if \(n\) is even.
  • \(A^n = A\) if \(n\) is odd.

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