# Thread: Prove that the function ||*||_w is a norm.

1. ## Prove that the function ||*||_w is a norm.

Hi.

Problem:
Prove that if $\displaystyle W$ is an arbitrary nonsingular matrix, the function $\displaystyle ||\cdot||_W$ defined by (3.3) is a vector norm.

(3.3) $\displaystyle ||x||_W = ||Wx||$.

Attempt:
I know that in if a function is a norm it has to have three properties.
(1) $\displaystyle ||x||=0 \;\textrm{if and only if}\; x=0$,
(2) $\displaystyle ||x||+||y||=||x+y||$ and
(3) $\displaystyle ||\alpha x|| = \alpha ||x||$

(1)

Let $\displaystyle x=0$.
Since $\displaystyle ||x||_W=||Wx||=||\sum^n_{i=1}w_ix_i||$ and every $\displaystyle x_i=0$, we have that $\displaystyle ||x||_W=0$.

Let $\displaystyle ||x||_W=0$.
Since $\displaystyle ||x||_W=||\sum^n_{i=1}w_ix_i||$ and $\displaystyle W$ is nonsingular such that $\displaystyle w_i\neq 0$ for $\displaystyle 1\leq i\leq n$, we have that $\displaystyle x_i=0$ and so $\displaystyle x=0$.

(3)

\displaystyle \begin{aligned} |\alpha| ||x||_W=&\;|\alpha| ||Wx||\\ =&\;|\alpha| ||\sum^n_{i=1}w_ix_i||\\ =&\; |\alpha| \left(\sum^n_{i=1}(w_ix_i)^p\right)^{1/p}\\ =&\; \left( \sum^n_{i=1}(\alpha w_i x_i)^p\right)^{1/p}\\ =&\; ||\alpha x||_W \end{aligned}

(2)

I do not know how to prove that the Triangle Inequality holds. Hints are greatly appreciated.

Thanks.

2. ## Triangular inequality

Once you have the proof for triangular inequality for any two vectors, the result should automatically follow.

Note that Wx and Wy are themselves vectors.

So ||W(x+y)|| = ||Wx+Wy|| <= || Wx|| + ||Wy||

I guess thats it ..

3. So basically ;

$\displaystyle ||x||_W+||y||_W=||Wx||+||Wy|| \geq ||Wx+Wy|| = ||x+y||_W$ ?

4. ## Triangular inequality

Ya i guess so...