Condition number of a diagonal matrix

Question

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 $$||A||\cdot||A^{-1}||=\frac{\max_i|\lambda_i|}{\min_i|\lambda_i|}?$$

Answer 1

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.

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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 $$|\lambda_i| \|X\| = \|\lambda_i X\| = \|A X\| \le \|A\| \|X\|.$$

Answer 2

This is not true in general!

Consider the norm $\|x\|_M := \|Mx\|_\infty$ where M is some invertible matrix.

Then $\|A\|_M = \| M A M^{-1}\|_\infty$ so $\kappa(A)_M = \| M A M^{-1}\|_\infty\| M A^{-1} M^{-1}\|_\infty$.

Take

$ D = \begin{pmatrix}1&0\\0&2\end{pmatrix}. $

Then $\kappa(D)_\infty = 2$ but if

$ M = \begin{pmatrix}1&2\\3&4\end{pmatrix} $

$\kappa(D)_M = \|\begin{pmatrix}4&-1\\6&-1\end{pmatrix}\|_\infty\|\begin{pmatrix}-0.5&0.5\\-3&2\end{pmatrix}\|_\infty = 35$.

It is true for any p-norm however.