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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.

Replace coordinate inequalities by the Loewner order and coordinate sums by matrix trace. The maximizing semidefinite program isIts feasible set is the fantope. In an orthonormal eigenbasis of $X$, its objective is ${\sum }_i{\lambda }_i(X)Y_{ii}$, and the diagonal entries of $Y$ belong to the capped simplex. Choosing $Y$ as the orthogonal projection matrix onto the largest $k$ eigenvectors attains the sum of the largest eigenvalues, as in the Ky Fan maximum principle.

The minimizing semidefinite program isFor example, its Lagrangian arises by assigning $S⪰0$ to $I-Y⪰0$, keeping $Y⪰0$ in the domain, and using free $t$ for $\mathrm{tr}Y=k$. Finiteness of the supremum over $Y$ requires $S+tI-X⪰0$.

Equality follows directly by choosing $S$ to have eigenvalues $({\lambda }_i(X)-t{)}_{+}$ in the same orthonormal eigenbasis, with $t$ between the $k$th and $(k+1)$st eigenvalues, or below the smallest one for $k=n$. This is the threshold semidefinite program for the largest eigenvalues, and it also verifies the endpoint case $k=n$.