= 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 decomposition
$$
A=B+iC,\qquad B=\frac{A+A^\dagger}{2},\qquad
C=\frac{A-A^\dagger}{2i},
$$
with $B,C$ 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:
$$
\operatorname{Tr}(\sigma_jA)=\sum_k a_k\operatorname{Tr}(\sigma_j\sigma_k)=a_j.
$$
Consequently
$$
\boxed{A=\sum_{k=1}^{N^2}\operatorname{Tr}(\sigma_kA)\sigma_k.}
$$
If $A$ is Hermitian, cyclicity of the <trace> gives
$$
a_j^*=\operatorname{Tr}[(\sigma_jA)^\dagger]
=\operatorname{Tr}(A\sigma_j)=\operatorname{Tr}(\sigma_jA)=a_j.
$$
Thus its coordinate vector is real. The product $\sigma_jA$ need not itself be Hermitian; <trace> cyclicity is what justifies the conclusion.
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