Symmetric and Skew Symmetric Matrices

• In matrix algebra, many problems become easier when we recognize special types of matrices.

• Two important special types are symmetric matrices and skew-symmetric matrices, which are defined using the transpose of a matrix.

• Understanding these helps in simplifying matrices, solving systems, and in higher topics like quadratic forms and eigenvalues.

A square matrix \[A = [a_{ij}]_{n \times n}\] is called symmetric if

i.e., \[a_{ij} = a_{ji}\] for all i and j.

A square matrix \[A = [a_{ij}]_{n \times n}\] is called skew-symmetric if \[A^T = -A\]

i.e.,\[a_{ij} = -a_{ji}\] for all i and j.

Theorem 1: For any square matrix A with real number entries, A + A′ is a symmetric matrix and A − A′ is a skew-symmetric matrix.

Proof:

Part 1: Symmetric Matrix
Let B = A + A′, then

Take transpose on both sides:
B′ = (A + A′)′
= A′ + (A′)′ (as (A + B)′ = A′ + B′)
= A′ + A (as (A′)′ = A)
= A + A′ (as A + B = B + A)
= B

Therefore, B = A + A′ is a symmetric matrix

Part 2: Skew-Symmetric Matrix

Now let
C = A − A′

C′ = (A − A′)′ = A′ − (A′)′ (Why?)
= A′ − A (Why?)
= −(A − A′) = −C

Therefore
C = A − A′ is a skew-symmetric matrix.

Theorem 2: Any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix.

Proof: Let A be a square matrix, then we can write

\[\mathrm{A=\frac{1}{2}(A+A^{\prime})+\frac{1}{2}(A-A^{\prime})}\]

From Theorem 1, we know that (A + A′) is a symmetric matrix and (A − A′) is a skew-symmetric matrix.

Multiplying by \[\frac{1}{2}\] does not change these properties.

Since for any matrix A, (kA)′ = kA′, it follows that \[\frac{1}{2}(\mathrm{A}+\mathrm{A}^{\prime})\] is symmetric matrix and \[\frac{1}{2}(\mathrm{A}-\mathrm{A}^{\prime})\] is skew symmetric matrix.

Thus, any square matrix can be expressed as the sum of a symmetric and a skew-symmetric matrix. 

Express the matrix \[\mathbf{B}= \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}\] as the sum of a symmetric and a skew-symmetric matrix.

Solution: Here \[\mathrm{B}^{\prime}= \begin{bmatrix} 2 & -1 & 1 \\ -2 & 3 & -2 \\ -4 & 4 & -3 \end{bmatrix}\]

Let \[\begin{aligned} & \mathrm{P}=\frac{1}{2}\left(\mathrm{B}+\mathrm{B}^{\prime}\right)=\frac{1}{2} \begin{bmatrix} 4 & -3 & -3 \\ -3 & 6 & 2 \\ -3 & 2 & -6 \end{bmatrix}= \begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ \\ \frac{-3}{2} & 3 & 1 \\ \\ \frac{-3}{2} & 1 & -3 \end{bmatrix}, \end{aligned}\]

Now \[\mathrm{P^{\prime}}= \begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ \frac{-3}{2} & 3 & 1 \\ \\ \frac{-3}{2} & 1 & -3 \end{bmatrix}=\mathrm{P}\]

Thus \[\mathrm{P}=\frac{1}{2}(\mathrm{B}+\mathrm{B}^{\prime})\] is a symmetric matrix.

Also, let \[\mathrm{Q}=\frac{1}{2}\left(\mathrm{B}-\mathrm{B}^{\prime}\right)=\frac{1}{2} \begin{bmatrix} 0 & -1 & -5 \\ 1 & 0 & 6 \\ 5 & -6 & 0 \end{bmatrix}= \begin{bmatrix} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix}\]

Then \[\mathrm{Q^{\prime}}= \begin{bmatrix} 0 & \frac{1}{2} & \frac{5}{3} \\ \frac{-1}{2} & 0 & -3 \\ \\ \frac{-5}{2} & 3 & 0 \end{bmatrix}=-\mathrm{Q}\]

Thus \[\mathrm{Q}=\frac{1}{2}(\mathrm{B}-\mathrm{B}^{\prime})\] is a skew symmetric matrix.

Now \[\mathbf{P}+\mathbf{Q}= \begin{bmatrix} 2 & \frac{-3}{2} & \frac{-3}{2} \\ \frac{-3}{2} & 3 & 1 \\ \frac{-3}{2} & 1 & -3 \end{bmatrix}+ \begin{bmatrix} 0 & \frac{-1}{2} & \frac{-5}{2} \\ \frac{1}{2} & 0 & 3 \\ \frac{5}{2} & -3 & 0 \end{bmatrix}= \begin{bmatrix} 2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3 \end{bmatrix}=\mathbf{B}\]

Thus, B is represented as the sum of a symmetric and a skew-symmetric matrix.

• A square matrix is symmetric if \[A^T = A\].

• A square matrix is skew-symmetric if \[A^T = -A\].

• In a skew-symmetric matrix, all diagonal elements are zero.

• For any square matrix A:

  • \[A + A^T\] is symmetric.

  • \[A - A^T\] is skew-symmetric.

• Any square matrix A can be written as

• The decomposition into symmetric and skew-symmetric parts is unique.