Sure Questions on Symmetric and Skew-Symmetric Matrices

This lesson works through the "sure questions" on symmetric and skew-symmetric matrices: a short proof that a skew-symmetric matrix has zeros all down its diagonal, the result that every square matrix splits into a symmetric and a skew-symmetric part, and a worked example carrying out that split.

Diagonal entries of a skew-symmetric matrix are zero

Let $A = [a_{ij}]$ be a square matrix that is skew-symmetric, which means $A^{\mathsf T} = -A$. Entry by entry this says: $a_{ij} = -a_{ji} \quad \text{for all } i, j.$

Now look at a diagonal entry by setting $i = j$: $a_{ii} = -a_{ii} \implies a_{ii} + a_{ii} = 0 \implies 2a_{ii} = 0 \implies a_{ii} = 0.$

Since this holds for every $i$, all of the diagonal entries $a_{11}, a_{22}, a_{33}, \dots$ are zero. So a skew-symmetric matrix always has a zero diagonal.

Any square matrix as a symmetric plus skew-symmetric sum

Let $A$ be any square matrix. We claim it can be written as $A = \tfrac{1}{2}\left(A + A^{\mathsf T}\right) + \tfrac{1}{2}\left(A - A^{\mathsf T}\right).$

Expanding the right side, the $\tfrac{1}{2}A^{\mathsf T}$ terms cancel and the two $\tfrac{1}{2}A$ terms add, leaving $A$, so the identity holds.

The first part is symmetric. Because $\left(A + A^{\mathsf T}\right)^{\mathsf T} = A^{\mathsf T} + A = A + A^{\mathsf T}$, the matrix $A + A^{\mathsf T}$ equals its own transpose, and so does $\tfrac{1}{2}(A + A^{\mathsf T})$.

The second part is skew-symmetric. Because $\left(A - A^{\mathsf T}\right)^{\mathsf T} = A^{\mathsf T} - A = -\left(A - A^{\mathsf T}\right)$, the matrix $A - A^{\mathsf T}$ is the negative of its transpose, and so is $\tfrac{1}{2}(A - A^{\mathsf T})$.

Writing $P = \tfrac{1}{2}(A + A^{\mathsf T})$ and $Q = \tfrac{1}{2}(A - A^{\mathsf T})$, we have $A = P + Q$ with $P$ symmetric and $Q$ skew-symmetric.

Worked example: setting up A and its transpose

Express $A = \begin{bmatrix} 3 & 5 \\ 1 & 1 \end{bmatrix}$ as the sum of a symmetric and a skew-symmetric matrix.

First write the transpose by interchanging rows and columns: $A^{\mathsf T} = \begin{bmatrix} 3 & 1 \\ 5 & 1 \end{bmatrix}.$

Building the symmetric part P

Add $A$ and $A^{\mathsf T}$, then halve: $A + A^{\mathsf T} = \begin{bmatrix} 6 & 6 \\ 6 & 2 \end{bmatrix}, \qquad P = \tfrac{1}{2}\begin{bmatrix} 6 & 6 \\ 6 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 3 \\ 3 & 1 \end{bmatrix}.$

Checking, $P^{\mathsf T} = \begin{bmatrix} 3 & 3 \\ 3 & 1 \end{bmatrix} = P$, so $P$ is symmetric.

Building the skew-symmetric part Q

Subtract $A^{\mathsf T}$ from $A$, then halve: $A - A^{\mathsf T} = \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}, \qquad Q = \tfrac{1}{2}\begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}.$

Checking, $Q^{\mathsf T} = \begin{bmatrix} 0 & -2 \\ 2 & 0 \end{bmatrix} = -Q$, so $Q$ is skew-symmetric.

Checking that P plus Q recovers A

Adding the two parts back together: $P + Q = \begin{bmatrix} 3 & 3 \\ 3 & 1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix} = \begin{bmatrix} 3 & 5 \\ 1 & 1 \end{bmatrix} = A.$

So $A = P + Q = \tfrac{1}{2}(A + A^{\mathsf T}) + \tfrac{1}{2}(A - A^{\mathsf T})$, as required.

Key takeaways

• A skew-symmetric matrix satisfies $A^{\mathsf T} = -A$, which forces every diagonal entry to be zero.
• For any square matrix, $\tfrac{1}{2}(A + A^{\mathsf T})$ is symmetric and $\tfrac{1}{2}(A - A^{\mathsf T})$ is skew-symmetric.
• Every square matrix is the sum of these two parts, $A = \tfrac{1}{2}(A + A^{\mathsf T}) + \tfrac{1}{2}(A - A^{\mathsf T})$, and adding them back recovers the original matrix.