Set of square matrices = symmetric matrices + antisymmetric matrices

Question

Let $R^{n×n}$ denote the set of $n×n$ real matrices. Let $S ⊂ R ^{n×n}$ denote the subspace of $n×n$ symmetric matrices and $A ⊂ R^{n×n}$ denote the subspace of $n × n$ antisymmetric matrices. Show that $R^{n×n} = S + A$.

Answer 1

HINT

Given $M\in\textbf{M}_{n}$, we have

\begin{align*} M = \frac{M + M^{T}}{2} + \frac{M - M^{T}}{2} \end{align*}

EDIT

Notice that $\displaystyle \frac{M + M^{T}}{2}\in S$. Indeed, we have \begin{align*} \left[\frac{M + M^{T}}{2}\right]^{T} = \frac{M^{T} + (M^{T})^{T}}{2} = \frac{M^{T} + M}{2} = \frac{M + M^{T}}{2} \end{align*}

Analogously, observe that $\displaystyle\frac{M - M^{T}}{2}\in A$. Indeed, we have \begin{align*} \left[\frac{M - M^{T}}{2}\right]^{T} = \frac{M^{T} - (M^{T})^{T}}{2} = \frac{M^{T} - M}{2} = -\left[\frac{M - M^{T}}{2}\right] \end{align*}

Hence the claim has been proved.