Density operators have an orthogonal spectral ensemble

By [F1]–[F2], choose an orthonormal eigenbasis $e_1^{\mathrm{\prime }},\dots ,e_d^{\mathrm{\prime }}$ with $\rho e_k^{\mathrm{\prime }}={\mu }_k{e}_k^{\mathrm{\prime }}$ and ${\mu }_k\ge 0$. In this basis the representing matrix is diagonal, so [F1] and [F3] give ${\sum }_{k=1}^d{\mu }_k=\mathrm{tr}(\rho )=1$. Thus the index set $J=\{k:{\mu }_k>0\}$ is nonempty, and omitting its complement removes exactly the zero eigenvalues.

List the finite set $J$ in increasing order as $k_1,\dots ,k_m$ and put ${\lambda }_j={\mu }_{k_j}$ and $e_j=e_{k_j}^{\mathrm{\prime }}$. By [F4], each unit eigenvector has a rank-one projector $\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$; the eigenbasis gives $\rho ={\sum }_{j=1}^m{\lambda }_j\mathrm{\mid }{e}_j⟩⟨e_j\mathrm{\mid }$, with positive weights summing to one. The operator maps each $e_{k_j}^{\mathrm{\prime }}$ to the nonzero vector ${\mu }_{k_j}{e}_{k_j}^{\mathrm{\prime }}$ and maps every other eigenbasis vector to zero, so its image is exactly the span of the $e_{k_j}^{\mathrm{\prime }}$. Thus [F5] gives $\mathrm{rank}(\rho )=m$. Distinct basis vectors give orthogonal projectors.