Symmetric and Skew Symmetric Orthogonal Subspaces

Question

If $R$ is the subspace of n x n symmetric matrices, and $S$ is the subspace of n x n skew symmetric matrices, how do we show that $S^{\perp} \subseteq R$ with respect to the inner product $<x,y> = Tr(x^{T}y)$?

I can show that $R \subseteq S^{\perp}$, which relies on the fact that one can show $Tr(rs) = 0$ for any $r\in R$ and $s\in S$. I dont really see, formally, how this implies the direction above (that $S^{\perp} \subseteq R$, i.e. that $x\in S^{\perp} \implies x \in R$)? How does one prove $S^{\perp} \subseteq R$ formally?

Answer 1

You want to show that if $\operatorname{Tr}(A^TB)=0$ for all $B\in S$, then $A=A^T$. For this, they key is to write $A=A_1+A_2$, with $A_1$ symmetric and $A_2$ skew-symmetric. This is simply $$ A=\frac{A+A^T}2+\frac{A-A^T}2. $$ Now $$ 0=\operatorname{Tr}(A^TB)=\operatorname{Tr}(A_1^TB)+\operatorname{Tr}(A_2^TB)=\operatorname{Tr}(A_2^TB). $$ And we are free to choose $B$, so in particular we may use $B=A_2$. Thus $$ 0=\operatorname{Tr}(A_2^TA_2), $$ implying that $A_2=0$. Then $A=A^T$.