I am trying to understand this theorem better and show simple application of it.
For $A$ symmetric $n \times n$ matrix with Eigenvalues $\lambda_1 \leq \lambda_2 \leq \dots \leq \lambda_n$ then the $k$th largest eigenvalue can be written in the form:
$$\lambda_k = \min_{v_1,\dots,v_{k-1}} \max_{f \perp v_1,\dots,v_{k-1}} \frac{f^TAf}{ff}$$
So, what I am understanding is that if we fix $v_i$ to something specific say the standard basis then we have: $$\lambda_k = \min_{v_1,\dots,v_{k-1}} \max_{f \perp v_1,\dots,v_{k-1}} \frac{f^TAf}{ff} \leq \max_{f \perp e_1,\dots e_{k-1}} \frac{f^TAf}{ff}$$
My question/ application is that I read that the minimum is achieved when we fix $v_i = u_i$ where $u_i$ are orthonormal eigenvectors of $A$ i.e.
$$\lambda_k = \min_{v_1,\dots,v_{k-1}} \max_{f \perp v_1,\dots,v_{k-1}} \frac{f^TAf}{ff}=\max_{f \perp u_1,\dots u_{k-1}} \frac{f^TAf}{ff}$$
This is not so obvious to me and would really appreciate if someone explains it to me. The source where I got this is https://en.wikipedia.org/wiki/Min-max_theorem more specifically on the proof for compact operators.