## Problem

Suppose $$A$$ is $$n \times n$$ Hermitian matrix, and unitarily diagonalized as follows

$$A = U\Sigma U^\ast$$

where $$\Sigma$$ is the diagonal matrix with eigenvalues and $$U$$ is the matrix of orthonormal eigenvectors.

Show, for $$\forall y$$ such that $$\Vert y \Vert_2 = 1$$,

$$\min_{\lambda \in \sigma(A)} |\lambda| \le \vert y^\ast \Sigma y \vert \le \max_{\lambda \in \sigma(A)} |\lambda|$$

where $$\sigma(A)$$ denotes the set of all eigenvalues of $$A$$.

## Try

Let $$u_1, \cdots, u_n$$ be the columns of $$U$$. Then

$$y = \sum_{i=1}^n a_i u_i$$

with $$\sum_i a_i^2 = 1$$. Thus

$$\vert y^\ast \Sigma y \vert = \vert \sum_{i=1}^n a_i^2 \lambda_i \vert$$

but I’m not sure we can say

$$\min_{\lambda \in \sigma(A)} |\lambda| \le \vert \sum_{i=1}^n a_i^2 \lambda_i \vert \le \max_{\lambda \in \sigma(A)} |\lambda|$$

Any help will be appreciated.