 Solution

Use the real vector space of Hermitian matrices and its Hilbert-Schmidt inner product $⟨A,B⟩=\mathrm{Tr}(AB)$. A Hermitian $N$-by-$N$ matrix has $N$ real diagonal entries and two real parameters for each off-diagonal pair, hence real dimension $N+2\left(\genfrac{}{}{0ex}{}{N}{2}\right)=N^2$. Imposing zero trace removes one real dimension. Each supplied generalized Pauli matrix is Hermitian and traceless, and there are $N(N-1)$ off-diagonal matrices and $N-1$ diagonal matrices, totaling $N^2-1$.

Their assumed orthonormality makes them linearly independent. Since this equals the dimension of the traceless space, they are an orthonormal basis of the traceless Hermitian matrices. To extend the family, the required additional generator is ${\sigma }_0=I/\sqrt{N}$, which is implicit in the indexing of the question. It has unit Hilbert-Schmidt norm and is orthogonal to every traceless generator, so the extended family is an orthonormal basis of all Hermitian matrices.

For a density operator $\rho$, cyclicity of the trace givesThus $s\in {\mathbb{R}}^{N^2-1}$. Since $\mathrm{Tr}\rho =1$, the resulting Generalized Bloch representation is