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

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

Hi.

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

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

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

(1)

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

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

(3)


\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 ;

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

4. ## Triangular inequality

Ya i guess so...