Proving $\left\|(A-\lambda I)^{-1}\right\|_{2}$ with Inverse of A

[jschmid2]

Could someone please help me show that if A is Hermitian
\left\|(A-\lambda I)^{-1}\right\|_{2}=\frac{1}{min_{\lambda_{i}\in\sigma(A)}|\lambda-\lambda_{i}|}
where \sigma(A) denotes the eigenvalues of A.

I have figured out how to solve the norm without an inverse, but the inverse confuses me a bit.
Recall, that \left\|\cdot\right\|=\sqrt{r_{\sigma}(A^{*}A)}, which is to say the square root of the largest eigenvalue of A^{*}A.

 

[Dick]

If A is hermitian you can diagonalize it with the eigenvalues lying along the diagonal. Think about what your expressions look like with A in that form.