Theorem 6.1, proof — $\mathrm{trace}(A)\,\kappa_f(A) \ge \mathrm{rank}(A)\,\lambda_n$

Let $A \in \mathbb{R}^{n\times n}$ be a nonzero symmetric positive semi-definite matrix, let $\lambda_n$ be its largest eigenvalue, and let $\kappa_f(A)$ be the ratio between its largest and smallest nonzero eigenvalue. Then

This linear-algebra inequality converts the largest eigenvalue, which bounds a single sample $z^TAz$, into a multiple of the trace, so that the range of each sample is measured relative to the quantity being estimated.

Formalization Note The page derives the inequality from an indexing $0 = \lambda_1 = \cdots = \lambda_k \le \cdots \le \lambda_n$ with $k = n - \mathrm{rank}(A) + 1$ and writes $\kappa_f(A) = \lambda_n/\lambda_k$; with that $k$, $\lambda_k = 0$, whereas $\kappa_f$ needs $\lambda_k$ to be the smallest nonzero eigenvalue (an off-by-one slip). Only the inequality is stated, with $\kappa_f$ as defined in kappaF. The hypothesis $A \ne 0$ excludes the placeholder value of $\kappa_f(0)$.