Why Every Square Matrix Splits into Symmetric and Skew-Symmetric Parts

Let $M_n(F)$ be the vector space of all $n\times n$ matrices over a field $F$. Assume that $2\ne 0$ in $F$—in particular, this holds when $F=\mathbb{R}$ or $\mathbb{C}$. Then every matrix is the sum of a symmetric matrix and a skew-symmetric matrix, and that representation is unique. This is exactly what it means to say that the two subspaces form a direct sum.

The two subspaces

A matrix $S$ is symmetric if $S^T=S$. A matrix $K$ is skew-symmetric if $K^T=-K$, where $T$ denotes transpose. Both collections are subspaces of $M_n(F)$: they contain the zero matrix and are closed under addition and scalar multiplication.

The claim can therefore be written

$$M_n(F)=\operatorname{Sym}_n(F)\oplus\operatorname{Skew}_n(F).$$

To prove it, show two things: every matrix is a sum of matrices from the stated subspaces, and the only matrix in both subspaces is zero. The second fact ensures the sum is direct and the representation is unique.

Construct the decomposition

Take any $A\in M_n(F)$ and define

$$S=\frac{A+A^T}{2},\qquad K=\frac{A-A^T}{2}.$$

These definitions make sense because $2$ has a multiplicative inverse in $F$. First, their sum is the original matrix:

$$S+K=\frac{A+A^T+A-A^T}{2}=A.$$

Next, check the transpose of each part:

$$S^T=\left(\frac{A+A^T}{2}\right)^T =\frac{A^T+A}{2}=S,$$

so $S$ is symmetric. Similarly,

$$K^T=\left(\frac{A-A^T}{2}\right)^T =\frac{A^T-A}{2}=-K,$$

so $K$ is skew-symmetric. Thus every $A$ is a sum of one matrix from each subspace.

Prove the sum is direct

Suppose a matrix $X$ is both symmetric and skew-symmetric. Then $X^T=X$ and $X^T=-X$, so

$$X=-X\quad\Longrightarrow\quad 2X=0.$$

Since $2\ne 0$ in the field, it follows that $X=0$. Therefore

$$\operatorname{Sym}_n(F)\cap\operatorname{Skew}_n(F)=\{0\}.$$

A sum of two subspaces with zero intersection is a direct sum. This also proves uniqueness: if $A=S_1+K_1=S_2+K_2$, then $S_1-S_2=K_2-K_1$ lies in both subspaces, so it must be zero. Hence $S_1=S_2$ and $K_1=K_2$.

Worked example

Let

$$A=\begin{pmatrix}1&4\\-2&3\end{pmatrix},\qquad A^T=\begin{pmatrix}1&-2\\4&3\end{pmatrix}.$$

The symmetric part is

$$S=\frac{A+A^T}{2} =\frac{1}{2}\begin{pmatrix}2&2\\2&6\end{pmatrix} =\begin{pmatrix}1&1\\1&3\end{pmatrix},$$

and the skew-symmetric part is

$$K=\frac{A-A^T}{2} =\frac{1}{2}\begin{pmatrix}0&6\\-6&0\end{pmatrix} =\begin{pmatrix}0&3\\-3&0\end{pmatrix}.$$

Indeed, $S^T=S$, $K^T=-K$, and

$$S+K=\begin{pmatrix}1&4\\-2&3\end{pmatrix}=A.$$

A dimension check

There are $n(n+1)/2$ independent entries in a symmetric matrix and $n(n-1)/2$ in a skew-symmetric matrix. In the skew-symmetric case, the diagonal entries must be zero, and entries on one side of the diagonal determine the opposite side. The dimensions add to

$$\frac{n(n+1)}{2}+\frac{n(n-1)}{2}=n^2,$$

the dimension of $M_n(F)$. This agrees with the decomposition, though the construction and zero-intersection proof establish it directly.

Common mistakes and caveats

• Forgetting the factor of $1/2$. The formulas require that 2 be invertible in the field. Over the real numbers this is automatic.
• Checking only that the parts add up. A decomposition proof must also verify that one part is symmetric and the other skew-symmetric.
• Assuming the claim holds in every field. In characteristic 2, $-X=X$, so symmetric and skew-symmetric matrices are not generally disjoint. For example, over a field of characteristic 2, every symmetric matrix is also skew-symmetric, and nonsymmetric matrices cannot be sums of symmetric matrices.
• Confusing transpose with conjugate transpose. Over complex numbers, this result uses ordinary transpose. If instead you want a decomposition into Hermitian and skew-Hermitian matrices, use the conjugate transpose $A^*$ and the parts $(A+A^*)/2$ and $(A-A^*)/2$.
• Using rectangular matrices without changing the claim. Symmetric and skew-symmetric are properties of square matrices, since the transpose must have the same dimensions as the matrix.