 Solution

A basis of Hermitian matrices is a real basis for the Hermitian subspace, but it also supplies a complex basis for all matrices. Indeed every operator has the decompositionwith $B,C$ Hermitian. Expand each in the given real basis, then combine their coefficients to expand $A$. Orthogonality for the Hilbert-Schmidt inner product determines those complex coefficients uniquely:ConsequentlyIf $A$ is Hermitian, cyclicity of the trace givesThus its coordinate vector is real. The product ${\sigma }_{j}A$ need not itself be Hermitian; trace cyclicity is what justifies the conclusion.