Frobenius inner product ($⟨A,B{⟩}_F=\mathrm{tr}(A^{\ast }B)$)

For complex matrices, the inner product $⟨A,B{⟩}_F=\mathrm{tr}(A^{\ast }B)$ is conjugate-linear in its first argument. On real symmetric matrices it becomes $\mathrm{tr}(AB)$. In particular $⟨A,x{x}^T{⟩}_F=x^{T}Ax$, identifying quadratic forms with linear tests on matrices.

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 giveHence $\mathrm{tr}(SA)=0$. Equivalently these two matrices are orthogonal for the Frobenius inner product, since $⟨S,A{⟩}_F=\mathrm{tr}(S^{T}A)=\mathrm{tr}(SA)$. Together with the unique symmetric and antisymmetric parts of a matrix, this is an orthogonal decomposition of the real square matrices.