4
$\begingroup$

I denote vector norms with doulbe bars and matrix norms with triple bars.

It is well known that the vector norm $L_2$ i.e. $\| x \|_2 = \sqrt{x^\top x}$ induces the matrix norm $||| \cdot |||_2$, which is the largest singular value of a matrix.

Consider the weighted norm, i.e. $\| x \|_W = \sqrt{x^\top W x} = \| W^\frac12 x\|_2$, where $W$ is some diagonal matrix of positive weights.

What is the matrix norm induced by the vector norm $\| \cdot \|_W$ ?

Does it have a formula like $||| \cdot |||_W = |||F \cdot |||_2$ for some matrix $F$?

$\endgroup$
8
$\begingroup$

In general, any symmetric positive definite $W$ induces the norm $\|x\|_W=\sqrt{x^TWx}=\|W^{1/2}x\|_2$ on $\mathbb{R}^n$.

The matrix norm induced by $\|\cdot\|_W$ can be related to the spectral norm as $$ \|A\|_W =\max_{x\neq 0}\frac{\|Ax\|_W}{\|x\|_W} =\max_{x\neq 0}\frac{\|W^{1/2}Ax\|_2}{\|W^{1/2}x\|_2} =\max_{y\neq 0}\frac{\|W^{1/2}AW^{-1/2}y\|_2}{\|y\|_2} =\|W^{1/2}AW^{-1/2}\|_2 $$ (using the substitution $y:=W^{1/2}x$).

$\endgroup$
  • 1
    $\begingroup$ This is only the case if A is square and you want to apply the same weight to both the numerator and denominator. In the case where you want different weights or the matrix A is not square, you need to use the general equation for a weighted matrix norm given in math.stackexchange.com/q/394237. $\endgroup$ – Brandon J. DeHart Jan 11 '18 at 18:33

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.