$\underset{\dim(S) = k}{\max} \underset{y \in S}{\min} \frac{y^T (X^T A X) y}{y^Ty} \ge \underset{y \in S_0}{\min} \frac{y^T (X^T A X) y}{y^Ty}$?

Question

Let $A \in \mathbb{R}^{n \times n}$ be symmetric and let $X \in \mathbb{R}^{n \times n}$ be nonsingular. Let $\lambda_k$ be the $k^{th}$ largest eigenvalue of $A$. Suppose for some $k$ we have that $\lambda_k > 0$ and define the subspace $S_0 \subseteq \mathbb{R}^n$ by

$$ S_0 = \{X^{-1}q_1, \dots, X^{-1}q_k\}, \quad q_i \neq 0$$

where $Aq_i = \lambda_iq_i$ and $i = 1, \dots, k$.

Then we have

$$\underset{\dim(S) = k}{\max} \underset{y \in S}{\min} \frac{y^T (X^T A X) y}{y^Ty} \ge \underset{y \in S_0}{\min} \frac{y^T (X^T A X) y}{y^Ty}.$$

I cannot see how this is obvious. I understand that $S_0$ has been constructed in a certain so that the inequality will be satisfied, but I can't see why it works? What makes $S_0$ so special?

Answer 1

This is not about $S_0$. You just need the general inequality $$\max_{a\in A}\min_{b\in a} f(b)\geq \min_{b\in a_0} f(b)$$ for some fixed $a_0\in A$. But that's trivial, because on the left hand side, the maximum is over all $a's$ (including $a_0$).