Matrix Symmetries

Certain terminology is used to describe matrices depending on how the transpose operation modifies them.

Symmetry

A square matrix $A$ is said to be symmetric when the transpose operation has no effect. In other words, a matrix is symmetric if and only if

A^T = A.

You can also say that a matrix $A_{n \times n}$ is symmetric if and only if,

[A]_{ij} = [A]_{ji}, \quad \text{for each} \; 1 \le i,j \le n.

Example

Take the matrix

A = \begin{bmatrix} 1 & 9 & 3 \\ 9 & 8 & 6 \\ 3 & 6 & 4 \end{bmatrix}.

This matrix is symmetric, since

\begin{bmatrix} 1 & 9 & 3 \\ 9 & 8 & 6 \\ 3 & 6 & 4 \end{bmatrix}^T = \begin{bmatrix} 1 & 9 & 3 \\ 9 & 8 & 6 \\ 3 & 6 & 4 \end{bmatrix}.

Now take the matrix

B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}.

This matrix is not symmetric, since

\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}^T = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \end{bmatrix} \ne \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}

Skew-Symmetry

A square matrix $A$ is said to be skew-symmetric when the transpose of a matrix equals its additive inverse. That is, a matrix is skew-symmetric if and only if

A^T = -A.

As we did above, the definition can also be expressed element-wise:

A matrix $A_{n \times n}$ is said to be skew-symmetric if and only if

[A]_{ij} = -[A]_{ji}, \quad \text{for each} \; 1 \le i,j \le n.

Example

Take the matrix

A = \begin{bmatrix} 0 & 7 & -6 \\ -7 & 0 & -4 \\ 6 & 4 & 0 \end{bmatrix}.

This matrix is skew-symmetric, since

\begin{align*} \begin{bmatrix} 0 & 7 & -6 \\ -7 & 0 & -4 \\ 6 & 4 & 0 \end{bmatrix}^T &= \begin{bmatrix} 0 & -7 & 6 \\ 7 & 0 & 4 \\ -6 & -4 & 0 \end{bmatrix} \\ &= -\begin{bmatrix} 0 & 7 & -6 \\ -7 & 0 & -4 \\ 6 & 4 & 0 \end{bmatrix} = -A. \end{align*}

Notice how, for a matrix to be skew-symmetric, the elements along its diagonal must all be $0$ . That is because for the diagonal entries we have that

a_{ii} = -a_{ii},

and therefore

\begin{align*} 2 a_{ii} &= 0 \\ a_{ii} &= 0 \end{align*}

Diagonal Matrices

In this context I would like to introduce the concept of diagonal matrices. A diagonal matrix - which is always a square matrix - of the size $n \times n$ , is a matrix of the form

D = \begin{bmatrix} \lambda_1 & 0 & \cdots & 0 \\ 0 & \lambda_2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & \lambda_n \end{bmatrix}.

Notice how each time I referred to symmetry in the context of matrices, what I meant was symmetric along the diagonal axis which goes from the top-left to the bottom-right. That is because the transpose operation “mirrors” the matrix along this axis.

Therefore, a diagonal matrix as the one shown above is always symmetric, since it only has potential non-zero elements along this axis and all the other elements are $0$ . Keep in mind that (out of pure coincidence), it could be the case that $\lambda_i = 0$ for some or all $i$ .

Therefore, the square zero matrix ( $0_{n \times n}$ ) is also a diagonal matrix. In fact, the square zero matrix is special because not only is it symmetric, it is also skew-symmetric. In fact, the zero matrix of size $n \times n$ is the only kind of matrix which is both symmetric and skew-symmetric at the same time.

Final Remark

Notice how only square matrices can be symmetric or skew-symmetric. That is because obviously, when you transpose a rectangular matrix of shape $m \times n$ , where $m \ne n$ , you end up with a matrix of shape $n \times m$ , so the matrices cannot be the same.

On the other hand, when you transpose a square matrix of shape $n \times n$ , you still get a matrix of shape $n \times n$ , so there’s at least a chance for them to be equal if the elements also match up.