How to prove the upper and lower bound of $ |\lambda_i(A)| $ for non-symmetry matrices?

Question

Suppose $A\in \mathbb{R}^{n\times n}$ (not necessary symmetry). $\lambda_i(A)$ is the $i$th eigenvalue for $A$. How to prove the bound for $\lambda_i(A)$ is: $$\min_{x\in\mathbb{C}^n,\|x\|_2=1}\|x^HAx\|\le |\lambda_i(A)|\le \max_{x\in\mathbb{C}^n,\|x\|_2=1}\|x^HAx\|,\forall i\in\{1,\cdots,n\}$$

My thinking:

This is very similiar to Rayleigh-Ritz Theorem (http://math.oregonstate.edu/~restrepo/475A/Notes/sourcea-/node79.html ) But here, we calculate the abs value of the eigenvalues and we do not suppose that the matrix $A$ is symmetry.

If we use $x^TAx=x^T\frac{A+A^T}{2}x$, the eigenvalue of $A$ cannot be handdled.

Answer 1

Fix $i$ and suppose $x$ is an eigen vector of norm 1 corresponding to eigen value $\lambda_i(A)$. Then $x^{H}Ax=\lambda_i (A)x^{H}x=\lambda_i (A)$. Hence $|\lambda_i (A)| =|x^{H}Ax|$. Can you complete the proof from this?