Questions about a proof for the Min-Max-Theorem for compact operators

Question

I'm looking at a proof that is written here: https://en.wikipedia.org/wiki/Min-max_theorem#Compact_operators

The questions I have are about the second part where we show that $$min_{S_{k-1}}max_{x\in S_{k-1}^\perp,||x||=1} (Ax,x)=\lambda_k$$

$1)$ Why does $S'\cap S_{k-1}^\perp\neq\{0\}$ imply that there is an $x\in S_{k-1}^\perp$ with $||x||=1$ and $(Ax,x)\geq\lambda_k$? I understand the part with $||x||=1$, but not the inequality.

$2)$ How do we get the last equation by setting $S_{k-1}=span\{u_1,...,u_{k-1}\}$?

Thanks in advance.

Answer 1

1. Any $x\in S'$ satisfies the inequation: $x\in S'\implies x=\sum_{n=1} ^k {a_n u_n}$ and $<Ax,x>=\sum_{n=1} ^k {(a_n)^2 <Au_n,u_n>}=\sum_{n=1} ^k {(a_n)^2 \lambda_n}\geq\sum_{n=1} ^k {(a_n)^2 \lambda_k}=\lambda_k$ and $\sum_{n=1} ^k {(a_n)^2}=||x||$.

2. You already have one inequation, you need to show the other one. Picking a specific $S_{k-1}$ can only make the expression bigger, as you are taking the infimum of all $S_{k-1}$'s. Now you can see the $u_k\in S_{k-1} ^\perp$ and it's easy to see that he gives the maximum - $\lambda_k$, thus we got the other direction.