= Solution
The <Hermitian operators> form a <real vector space> of dimension $N^2$. To see that the same basis spans all complex operators, decompose an arbitrary $A$ as $A=B+iC$, where $B=(A+A^\dagger)/2$ and $C=(A-A^\dagger)/(2i)$ are Hermitian. Expand $B,C$ in the real <orthonormal basis> and combine their coefficients. The resulting complex expansion is unique because $\operatorname{Tr}(\sigma_m\sigma_k)=\delta_{mk}$. Taking the <Hilbert-Schmidt inner product> with $\sigma_m$ gives
$$
\boxed{A=\sum_{k=1}^{N^2}a_k\sigma_k,\qquad a_k=\operatorname{Tr}(\sigma_kA)}.
$$
If $A$ is Hermitian, then $a_k^*=\operatorname{Tr}((\sigma_kA)^\dagger)=\operatorname{Tr}(A\sigma_k)=a_k$ by cyclicity. Hence $\boxed{\vec a\in\mathbb R^{N^2}\text{ for Hermitian }A}$. The basis is a real basis of <Hermitian operators> and its complexification is a complex basis of all operators.
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