Originally Posted by

**Mollier** **Problem:**

Prove that if W is an arbitrary nonsingular matrix, the function $\displaystyle ||\cdot ||_W$ defined by:

$\displaystyle ||x||_W=||Wx|| $ is a vector norm.

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**Attempt:**

The only thing I can think of is to go through the conditions a vector norm must satisfy, and check that it holds for a 2-norm..

__1. condition:__

$\displaystyle

||\alpha Wx||=|\alpha | ||Wx|| = ||\alpha \sum^n_{i=1}w_ix_i||\\

$

If I look at the 2-Norm, I get:

$\displaystyle ||\alpha Wx||_2 = \left( \alpha^2 \sum^n_{i=1}|w_ix_i|^2\right)^{1/2}=|\alpha | \left( \sum^n_{i=1}|w_ix_i|^2\right)^{1/2} = |\alpha | \;||Wx|| $

__2. condition:__

$\displaystyle ||Wx+Wy|| \leq ||Wx|| + ||Wy|| $

$\displaystyle ||Wx|| + ||Wy|| = \sqrt{(||Wx|| + ||Wy||)^2} = \sqrt{||Wx||^2+2||Wx||\;||Wy||+||Wy||^2} \geq $ $\displaystyle \sqrt{||Wx||^2+||Wy||^2} $

$\displaystyle = \sqrt{(w_1x_1)^2+...+(w_nx_n)^2+(w_1y_1)^2+...(w_n y_n)^2} = ||Wx + Wy||$

__3. condition:__

$\displaystyle ||Wx|| \geq 0 $

Since $\displaystyle (w_ix_i)^2 \geq 0 $ and $\displaystyle i=1,...,n$ we have $\displaystyle ||Wx|| = \left( \sum^n_{i=1}|w_ix_i|^2 \right)^{1/2} \geq 0 $

__4. condition:__

$\displaystyle ||Wx||=0 \; iff \; Wx=0 $

If $\displaystyle ||Wx||=0,\; and \; (w_ix_i)^2 \geq 0 \; then \; Wx=0 $

Here you need to prove that $\displaystyle \|x\|_W=\|Wx\|=0\Longleftrightarrow \|x\|_W=0$ ...here is where you need very strongly the condition that $\displaystyle W$ is a non-singular matrix.

Tonio

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I am pretty sure that this should be done without using any specific norm, but I do not know how.

Thank you for your time.