= Trace orthogonality of symmetric and antisymmetric matrices
{title2=$\operatorname{tr}(SA)=0$}
Let $S$ be a real <symmetric matrix> and $A$ a real <antisymmetric matrix> of the same size. Invariance of the <trace> under <transpose> and the cyclic <trace> identity give
$$
\operatorname{tr}(SA)=\operatorname{tr}((SA)^T)=\operatorname{tr}(A^TS^T)=-\operatorname{tr}(AS)=-\operatorname{tr}(SA).
$$
Hence $\operatorname{tr}(SA)=0$. Equivalently these two <matrices> are <orthogonal> for the <Frobenius inner product>, since $\langle S,A\rangle_F=\operatorname{tr}(S^TA)=\operatorname{tr}(SA)$. Together with the unique <symmetric and antisymmetric parts of a matrix>, this is an <orthogonal decomposition> of the real square <matrices>.
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