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.)



Similarity transform of a diagonalizable matrix that minimizes the Euclidean condition number
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 $\kappa\_2(\Lambda) = \lvert\lambda\_\max(B)\rvert \lvert\lambda\_\max(B^{-1})\rvert \leq \sigma\_\max(B)\sigma\_\max(B^{-1}) = \kappa\_2(B)$.


