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.)


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 $\kappa_2(\Lambda) = \lvert\lambda_\max(B)\rvert \lvert\lambda_\max(B^{-1})\rvert \leq \sigma_\max(B)\sigma_\max(B^{-1}) = \kappa_2(B)$.

| cite | improve this answer | |
  • $\begingroup$ ...hmm, but why is the largest singular value greater than the largest eigenvalue? $\endgroup$ – Jason S Feb 28 '16 at 21:16
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.