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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 $F$?

ziutek
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1 Answers1

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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 $y:=W^{1/2}x$).

Algebraic Pavel
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    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 https://math.stackexchange.com/q/394237/. – Brandon J. DeHart Jan 11 '18 at 18:33