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Use an orthonormal eigenbasis of the Hermitian operator $A$ and set $b_j=⟨{\alpha }_j|B|{\alpha }_j⟩$. Since $B$ is a positive contraction,For $0<r<d$, the missing weight among the first $r$ entries equals the weight in the remaining entries:The ordered eigenvalues satisfy ${\lambda }_j\ge {\lambda }_{r-1}$ for $j<r$ and ${\lambda }_j\le {\lambda }_{r-1}$ for $j\ge r$. ThereforeThis argument does not require the eigenvalues of $A$ to be positive. For $r=0$, positivity and zero trace give $B=0$; for $r=d$, the same argument applied to $I-B$ gives $B=I$. Thus in every caseThe orthogonal projection onto the leading $r$ eigenvectors attains equality. This is the Hermitian effect variational principle, extending the Ky Fan maximum principle to all positive contractions with the prescribed trace.

If $\rho$ is a density operator with decreasing eigenvalues ${\lambda }_j$ and $T$ has matrix rank at most $k$, let $\Pi$ project onto the image of $T$. The Cauchy-Schwarz inequality for the Hilbert-Schmidt inner product givesThe Hermitian effect variational principle bounds $\mathrm{Tr}(\rho \Pi )$ by ${\sum }_{j<k}{\lambda }_j$, proving the claimed estimate without assuming $\rho$ is invertible.