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For a feasible vector, set $X=x{x}^T$. It is a positive semidefinite matrix, andThus every original feasible point supplies an equally valuable SDP feasible point. Dropping the rank-one restriction is the semidefinite relaxation of slab-constrained quadratic maximization, provingThe original PDF has $|a_i^{T}x|\le 1$; the absolute values are missing from the extracted TeX and are essential for this relaxation.

For the subsequent finite rounding argument, the constraint vectors must span ${\mathbb{R}}^n$. Otherwise a nonzero common-kernel vector $d$ makes both $x=td$ and $X=td{d}^T$ unbounded feasible families, so both objective suprema are infinite. Under spanning, $H={\sum }_i{a}_i{a}_i^T$ is a positive-definite matrix andThe first inequality uses positive semidefinite trace nonnegativity. The SDP feasible set is closed and bounded, hence compact, so an optimum exists. Its trace is positive: a sufficiently small positive multiple of the identity is feasible when $n\ge 1$. These facts justify the optimum and nonzero denominator used below. Assume $m,n\ge 1$.

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.