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If I have a diagonalizable matrix $A = V\Lambda V^{-1}$, is there a way to show that for any similar $B$ such that $B = T\Lambda T^{-1}$, the Euclidean condition number $\kappa_2(B) \geq \kappa_2(\Lambda)$? Or is this statement false?

(and if it's a well-known result, is there a good linear algebra reference that mentions this? I can't seem to find one.)

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It is a consequence of the fact that the largest singular value is greater than the largest eigenvalue (in modulus): $\lvert\lambda_\max(B)\rvert \leq \sigma_\max(B)$, and $\lvert\lambda_\max(B^{-1})\rvert \leq \sigma_\max(B^{-1})$, so $\kappa_2(\Lambda) = \lvert\lambda_\max(B)\rvert \lvert\lambda_\max(B^{-1})\rvert \leq \sigma_\max(B)\sigma_\max(B^{-1}) = \kappa_2(B)$.

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  • $\begingroup$ ...hmm, but why is the largest singular value greater than the largest eigenvalue? $\endgroup$ – Jason S Feb 28 '16 at 21:16
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    $\begingroup$ The largest singular value is the Euclidean norm of a matrix, and that is greater than the largest eigenvalue because $|\lambda|\|v\| = \|Av\| \leq \|A\| \|v\|$. $\endgroup$ – Federico Poloni Feb 28 '16 at 21:37

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