 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 $B=(A+A^{†})/2$ and $C=(A-A^{†})/(2i)$ are Hermitian. Expand $B,C$ in the real orthonormal basis and combine their coefficients. The resulting complex expansion is unique because $\mathrm{Tr}({\sigma }_m{\sigma }_k)={\delta }_{mk}$. Taking the Hilbert-Schmidt inner product with ${\sigma }_m$ givesIf $A$ is Hermitian, then $a_k^{\ast }=\mathrm{Tr}(({\sigma }_{k}A{)}^{†})=\mathrm{Tr}(A{\sigma }_k)=a_k$ 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.