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Let $\|\cdot\|$ be any norm on $\mathbb{C}^n$. Let $A\in \mathbb{C}^{n\times n}$ We define the matrix norm by $||A||=\max_{||x||=1}||Ax||$. If $A=diag(\lambda_1,...,\lambda_n)$ and it is invertible, then do we always have $$||A||\cdot||A^{-1}||=\frac{\max_i|\lambda_i|}{\min_i|\lambda_i|}?$$

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    $\begingroup$ Yes, that is correct. $\endgroup$ – Paul Oct 24 '17 at 2:11
  • $\begingroup$ @Paul Thanks. Could you give some hints for the proof? $\endgroup$ – Right Oct 24 '17 at 2:19
  • $\begingroup$ Well, how do you take the norm of a diagonal matrix, first of all? $\endgroup$ – J. M. is a poor mathematician Oct 24 '17 at 2:32
  • $\begingroup$ @J.M.isnotamathematician It is given by the subordinate norm $||A||=\max_{||x||=1}||Ax||$ $\endgroup$ – Right Oct 24 '17 at 2:35
  • $\begingroup$ That's the general definition. I'm asking you how you'd evaluate it for a diagonal matrix. If you figure that out, then the expression for the condition number is not too hard to derive. $\endgroup$ – J. M. is a poor mathematician Oct 24 '17 at 2:36
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I mistakenly thought $\|\cdot\|$ was the $\ell^2$-norm. See below the line for the general situation.

Hint: $$\|Ax\|^2=\left\|\begin{bmatrix}\lambda_1 x_1 \\ \vdots \\ \lambda_n x_n\end{bmatrix}\right\|^2 = \lambda_1^2 x_1^2 + \cdots + \lambda_n^2 x_n^2 \le (\max_i \lambda_i^2) (x_1^2 + \cdots + x_n^2) = (\max_i \lambda_i^2) \|x\|^2.$$ By looking at the definition of $\|A\|$, can you now compute $\|A\|$? Computing $\|A^{-1}\|$ is similar, since it is also a diagonal matrix.


General situation:

  1. For any [submultiplicative] matrix norm $\|\cdot\|$, we have $\|A\| \ge \max_i |\lambda_i|$. (See below.)
  2. Since subordinate norms are submultiplicative matrix norms, this inequality holds in the setting of your question.
  3. Moreover, by considering $x$ being the standard basis vectors, we see that we actually have the equality $\|A\| = \max_i |\lambda_i|$. Can you conclude from here?

Proof of Claim 1: Let $\|\cdot \|$ be a [submultiplicative] matrix norm. Let $x$ be a $\lambda_i$-eigenvector, and let $X$ be the $n \times n$ matrix whose columns are all $x$. Then $$|\lambda_i| \|X\| = \|\lambda_i X\| = \|A X\| \le \|A\| \|X\|.$$

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  • $\begingroup$ What if the norm is not the Euclidean norm? $\endgroup$ – Right Oct 24 '17 at 2:24

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