# 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:

 $\displaystyle 1$ $\displaystyle=$ $\displaystyle\|\tilde{u}\|_{p}$ $\displaystyle=$ $\displaystyle\|X(D-\tilde{\lambda}I)^{-1}X^{-1}r\|_{p}\leq\|X\|_{p}\|(D-\tilde% {\lambda}I)^{-1}\|_{p}\|X^{-1}\|_{p}\|r\|_{p}$ $\displaystyle=$ $\displaystyle\kappa_{p}(X)\|(D-\tilde{\lambda}I)^{-1}\|_{p}\|r\|_{p}.$

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

 $\|(D-\tilde{\lambda}I)^{-1}\|_{p}=\max\limits_{\|x\|_{p}\neq 0}\frac{\|(D-% \tilde{\lambda}I)^{-1}x\|_{p}}{\|x\|_{p}}=\max\limits_{\lambda\in\sigma(A)}% \frac{1}{|\lambda-\tilde{\lambda}|}=\frac{1}{\min\limits_{\lambda\in\sigma(A)}% |\lambda-\tilde{\lambda}|}$

whence:

 $\min\limits_{\lambda\in\sigma(A)}|\lambda-\tilde{\lambda}|\leq\kappa_{p}(X)||r% ||_{p}.$
Title proof of Bauer-Fike theorem ProofOfBauerFikeTheorem 2013-03-22 15:33:08 2013-03-22 15:33:08 Andrea Ambrosio (7332) Andrea Ambrosio (7332) 9 Andrea Ambrosio (7332) Proof msc 15A42