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For an integer $0\le k\le n$, the fantope is the convex hull of rank-$k$ orthogonal projection matrices in ${\mathbb{R}}^n$. Equivalently, it consists of real symmetric matrices with eigenvalues in $[0,1]$ and matrix trace $k$. To prove the equivalence, apply the spectral theorem for real symmetric matrices and express the eigenvalue vector as a convex combination of the zero-one extreme points of the capped simplex. The sum of the largest eigenvalues is its support function.

For a real symmetric matrix $X$ and $1\le k\le n$,Indeed $Y=Q{Q}^T$ is a rank-$k$ orthogonal projection matrix in the fantope, and projecting onto the eigenvectors of the largest eigenvalues attains the support function maximum. Equivalently, the largest matrix trace of a compression to a $k$-dimensional vector subspace is the sum of the largest eigenvalues.

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 $k=n$.

The sum of the largest eigenvalues has the semidefinite program representationFor every feasible $Y$ in the fantope, positive semidefinite trace nonnegativity gives $\mathrm{tr}(XY)\le kt+\mathrm{tr}S$. To attain equality, use an orthonormal eigenbasis of $X$ and choose $S=\mathrm{diag}(({\lambda }_i-t{)}_{+})$ in that basis, with the same threshold choice as in the threshold formula for the sum of the largest components. This argument also covers $k=n$, without requiring strict feasibility of the maximization program.