proof of Bauer-Fike theorem

We can assume $\tilde{\lambda }\notin \sigma (A)$ (otherwise, we can choose $\lambda =\tilde{\lambda }$ and theorem is proven, since ${\kappa }_p(X)>1$).Then ${(A-\tilde{\lambda }I)}^{-1}$ exists, so we can write:

$$\tilde{u}={(A-\tilde{\lambda }I)}^{-1}r=X{(D-\tilde{\lambda }I)}^{-1}{X}^{-1}r$$

since $A$ is diagonalizable; taking the p-norm (http://planetmath.org/VectorPnorm) of both sides, we obtain:

$1$$=$${\parallel \tilde{u}\parallel }_p$

$=$${\parallel X{(D-\tilde{\lambda }I)}^{-1}{X}^{-1}r\parallel }_p\le {\parallel X\parallel }_p{\parallel {(D-\tilde{\lambda }I)}^{-1}\parallel }_p{\parallel X^{-1}\parallel }_p{\parallel r\parallel }_p$

$=$${\kappa }_p(X){\parallel {(D-\tilde{\lambda }I)}^{-1}\parallel }_p{\parallel r\parallel }_p.$

But, since ${(D-\tilde{\lambda }I)}^{-1}$ is a diagonal matrix, the p-norm is easily computed, and yields:

$${\parallel {(D-\tilde{\lambda }I)}^{-1}\parallel }_p=\underset{{\parallel x\parallel }_p\ne 0}{\max }\frac{{\parallel {(D-\tilde{\lambda }I)}^{-1}x\parallel }_p}{{\parallel x\parallel }_p}=\underset{\lambda \in \sigma (A)}{\max }\frac{1}{|\lambda -\tilde{\lambda }|}=\frac{1}{\underset{\lambda \in \sigma (A)}{\min }|\lambda -\tilde{\lambda }|}$$

whence:

$$\underset{\lambda \in \sigma (A)}{\min }|\lambda -\tilde{\lambda }|\le {\kappa }_p(X){||r||}_p.$$