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


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.