Exercise about eigenvalue

Question

Consider $A: n \times n$, Hermitian matrix, and the ordering of eigenvalues: $\lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_n$.

Prove that:

$\lambda_1 = \max (x^{H}Ax)$ and $\lambda_n = \min (x^{H}Ax)$, for $x \in \mathbb{C}^n$; $|| x ||_{2} = 1$.

I could not solve this problem. I have a feeling that solving for $\lambda_1$ the other result (for $\lambda_n$) is analogous.

My attempt was that since $A$ is Hermitian, then $A$ is unitarily similar to a diagonal matrix $D$, that is, $A = UDU^{H}$ where $U$ is unitary and $D = diag(\lambda_1, ..., \lambda_n)$. I have this $$ x^{H}Ax = x^{H}UDU^{H}x= \left( U^{H}x\right)^{H}DU^{H}x = y^{H}Dy = \sum_{i=1}^{n}\lambda_{i}\overline{y_i}y_i = \sum_{i=1}^{n}\lambda_{i}|y_i|^{2}.$$

where $y = [y_1 \;\;y_2\;\;\ldots y_n]^{T}$.

But now I don´t know what I can do it.

Answer 1

Hint: since $U$ is invertible, then certainly $\min_{\|x\| = 1}(x^HAx) = \min_{\|Ux\| = 1}\min((Ux)^HA(Ux))$. Since $U$ is unitary, then certainly $\|Ux\| = \|x\|$.

Answer 2

$\textbf{Hint:}$ Take $x_i=Ue_i$, where $e_i$ is the $i$-th vector of the canonical basis, and see what you get. The $x_i$'s also form a basis so you can compute other vectors in terms of these.