Symmetric and Skew-Symmetric Matrices

Definitions

Both symmetric and skew-symmetric matrices are defined only for square matrices, since the definitions compare a matrix to its own transpose, and $A$ and $A^T$ can only be equal (or negatives of each other) when they have the same order -- which forces $A$ to be square.

Definition (symmetric matrix). A square matrix $A$ is symmetric if $A^T=A$, i.e. $a_{ji}=a_{ij}$ for every $i,j$ -- the matrix is a mirror image of itself across the main diagonal.

Definition (skew-symmetric matrix). A square matrix $A$ is skew-symmetric if $A^T=-A$, i.e. $a_{ji}=-a_{ij}$ for every $i,j$.

Consequence -- the diagonal of a skew-symmetric matrix is always zero. Setting $i=j$ in the skew-symmetric condition gives $a_{ii}=-a_{ii}$, i.e. $2a_{ii}=0$, so $a_{ii}=0$ for every $i$. (Exercise: Transpose, Symmetric and Skew-Symmetric Matrices, Q4 asks for this proof in full, and Q2 uses it as a quick first check when testing whether a given matrix is skew-symmetric: if even one diagonal entry is non-zero, the matrix cannot be skew-symmetric, without checking anything else.)

Every Square Matrix Splits Into a Symmetric Part and a Skew-Symmetric Part

Theorem. Every square matrix $A$ can be written as $A=P+Q$, where $P=\dfrac{1}{2}(A+A^T)$ is symmetric and $Q=\dfrac{1}{2}(A-A^T)$ is skew-symmetric, and this decomposition is unique.

Proof (that $P$ is symmetric and $Q$ is skew-symmetric). Using $(A^T)^T=A$ and $(A+B)^T=A^T+B^T$ from Section 6:

so $P^T=P$, i.e. $P$ is symmetric. Similarly,

so $Q^T=-Q$, i.e. $Q$ is skew-symmetric. Finally $P+Q=\tfrac{1}{2}(A+A^T)+\tfrac{1}{2}(A-A^T)=A$, confirming the decomposition. $\blacksquare$ (The general form of this argument, without specific numbers, is proved again as Miscellaneous, Q1; Example 8 and Exercise: Transpose, Symmetric and Skew-Symmetric Matrices, Q3 apply it to specific $2\times2$ matrices.)

Worked illustration. For $A=\begin{bmatrix}2&3\\5&7\end{bmatrix}$, $A^T=\begin{bmatrix}2&5\\3&7\end{bmatrix}$, so

$P=\tfrac12(A+A^T)=\tfrac12\begin{bmatrix}4&8\\8&14\end{bmatrix}=\begin{bmatrix}2&4\\4&7\end{bmatrix},\qquad Q=\tfrac12(A-A^T)=\tfrac12\begin{bmatrix}0&-2\\2&0\end{bmatrix}=\begin{bmatrix}0&-1\\1&0\end{bmatrix},$ …