Inner product space and symmetric matrix

Question

I can't prove this claim:

Let $A \in M^\mathbb{R} _{n\times n}$ be a symmetric matrix, and $\lambda_1 \ge \lambda_2 \ge...\ge \lambda_n$ all the eigenvalues of $A$.

Prove that for every linear subspace $U \neq0 \subseteq \mathbb{R}^n$ such that $\dim U=k$ there is $v \in U$ that $\dfrac{(Av,v)}{(v,v)} \le \lambda_k$ where $(\cdot ,\cdot )$ is the standard inner product of $\mathbb{R}^n$.

I'm sure that I need to use somehow in the orthonormal basis of eigenvectors of $A$ but I don't know how.

Answer 1

Let $u_1,...,u_n$ be an orth. basis of $\mathbb R^n$ such that $Au_j = \lambda u_j$

Then $\dfrac{(Au_j,u_j)}{(u_j,u_j)}= \lambda_j$

Your turn !