Eigenvalues of symmetric matrix in real inner product space I got the following exercise to solve:
Let $A\in\mathbb{R}^{n\times n}$
  be a symmetric matrix and let $\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{n}$
  be its eigenvalues sorted in a desecnding order. Show that: $$\lambda_{k}=\max_{V,\,\dim\left(V\right)=k}\left\{ \min\frac{\left<x,Ax\right>}{\left<x,x\right>}\,:\, x\in V\,,\, x\neq0\right\} $$
 There is no mention of some specific inner product being taken here so I assume it needs to be proven for a general inner product on an $n$
  dimensional real inner-product space.
I'm completely stumped with this, help would be greatly appreciated! 
 A: Fix an orthonormal basis $e_1,\ldots,e_n$ of $V$. As $A$ is symmetric, we can choose the basis so that $A$ is diagonal, i.e. $Ae_j=\lambda_je_j$.
Note that
$$
 \min\left\{\frac{\left<x,Ax\right>}{\left<x,x\right>}\,:\, x\in V\,,\, x\neq0\right\}
= \min\left\{{\left<x,Ax\right>}\,:\, x\in V\,,\, \|x\|=1\right\}
$$
For any $x=\sum_{j=1}^nc_je_j$ with $\sum_jc_j^2=1$, we have
$$
\langle x,Ax\rangle = \left\langle \sum_{j=1}^nc_je_j,\sum_{j=1}^nc_j\lambda_je_j\right\rangle=\sum_{j=1}^n\lambda_j\,c_j^2.
$$
Fix a subspace $V$, where $\dim V=k$. Consider the subspace $W=\text{span}\{e_k,e_{k+1},\ldots,e_{n}\}$. As $\dim W=n-k+1$, $V\cap W$ cannot be zero (otherwise, we would find $n+1$ linearly independent elements of $V$). Any $x\in V\cap W$ with norm one is of the form $x=\sum_{j=k}^nc_je_j$ with $\sum c_j^2=1$, so
$$
\langle x,Ax\rangle = \sum_{j=k}^n\lambda_jc_j^2\leq\lambda_k.
$$
This shows that
$$
\min\left\{{\left<x,Ax\right>}\,:\, x\in V\,,\, \|x\|=1\right\}\leq\lambda_k.
$$
As $V$ was arbitrary, we have shown that
$$
\max_{\dim V=k}\left\{\min\left\{{\left<x,Ax\right>}\,:\, x\in V\,,\, \|x\|=1\right\}\right\}\leq\lambda_k.
$$
Since the equality is achieved for $V=\text{span}\{e_1,\ldots,e_k\}$, we are done.
