# Math Help - Prove that function is a vector norm.

1. ## Prove that function is a vector norm.

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

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

If I look at the 2-Norm, I get:
$||\alpha Wx||_2 = \left( \alpha^2 \sum^n_{i=1}|w_ix_i|^2\right)^{1/2}=|\alpha | \left( \sum^n_{i=1}|w_ix_i|^2\right)^{1/2} = |\alpha | \;||Wx||$

2. condition:
$||Wx+Wy|| \leq ||Wx|| + ||Wy||$
$||Wx|| + ||Wy|| = \sqrt{(||Wx|| + ||Wy||)^2} = \sqrt{||Wx||^2+2||Wx||\;||Wy||+||Wy||^2} \geq$ $\sqrt{||Wx||^2+||Wy||^2}$

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

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

4. condition:
$||Wx||=0 \; iff \; Wx=0$
If $||Wx||=0,\; and \; (w_ix_i)^2 \geq 0 \; then \; Wx=0$
------------------------------------------------------------------------------------

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

2. Originally Posted by Mollier
Problem:
Prove that if W is an arbitrary nonsingular matrix, the function $||\cdot ||_W$ defined by:
$||x||_W=||Wx||$ is a vector norm.

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

If I look at the 2-Norm, I get:
$||\alpha Wx||_2 = \left( \alpha^2 \sum^n_{i=1}|w_ix_i|^2\right)^{1/2}=|\alpha | \left( \sum^n_{i=1}|w_ix_i|^2\right)^{1/2} = |\alpha | \;||Wx||$

2. condition:
$||Wx+Wy|| \leq ||Wx|| + ||Wy||$
$||Wx|| + ||Wy|| = \sqrt{(||Wx|| + ||Wy||)^2} = \sqrt{||Wx||^2+2||Wx||\;||Wy||+||Wy||^2} \geq$ $\sqrt{||Wx||^2+||Wy||^2}$

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

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

4. condition:
$||Wx||=0 \; iff \; Wx=0$
If $||Wx||=0,\; and \; (w_ix_i)^2 \geq 0 \; then \; Wx=0$

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

Tonio

------------------------------------------------------------------------------------

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

3. Since you write that $||x||_W=||Wx||=0$, does that not automatically say that $||x||_W=0$? I guess if it was anything else it would look wierd.
$1 = ||Wx|| = 0$

Thanks mate!

4. Originally Posted by Mollier
Since you write that $||x||_W=||Wx||=0$, does that not automatically say that $||x||_W=0$?

Of course it does since: it's the same!! You wrote "doesn't that (i.e., $\|x\|_W=\|Wx\|=0$) say that $\|x\|_W=0$ ?" ....

What you need though is to prove that $\|x\|_W=0\Longrightarrow x=0$ , and you still haven't done this

Tonio

I guess if it was anything else it would look wierd.
$1 = ||Wx|| = 0$

Thanks mate!
.

5. Alright,

Say $||x||_W=\sum^n_{i=1}|x_iw_i|^2=0$.

Since $(x_iw_i)^2 \geq 0, \quad for\; i=1,...,n$, we have that $x=0$.
The $w_i$'s are the columns of a nonsingular matrix, so they can not be zero-vectors..

Is this any better sir?

Thanks!

6. Originally Posted by Mollier
Alright,

Say $||x||_W=\sum^n_{i=1}|x_iw_i|^2=0$.

Since $(x_iw_i)^2 \geq 0, \quad for\; i=1,...,n$, we have that $x=0$.
The $w_i$'s are the columns of a nonsingular matrix, so they can not be zero-vectors..

Is this any better sir?

Thanks!

Not really...and you need to check the other parts where you used this:

if $W=(w_{ij})\,,\,\,x=\begin{pmatrix}x_1\\x_2\\...\\x _n\end{pmatrix}$ , then $Wx=\begin{pmatrix}\sum\limits_{k=1}^nw_{1k}x_k\\\s um\limits_{k=1}^nw_{2k}x_k\\...\\\sum\limits_{k=1} ^nw_{nk}x_k\end{pmatrix}$ , and then

$\|x\|_W=\|Wx\|=\left(\right)^{1\slash 2}=\left(\sum\limits_{j=1}^n\left(\sum\limits_{k=1 }^nw_{jk}x_k\right)^2\right)^{1\slash 2}$ , which is not, of course, what you wrote...

Note: I'm assuming, since you didn't say otherwise, that the norm used in $\|Wx\|$ is the standard, euclidean one: $\|x\|=^{1\slash 2} = \left(x_1^2+\ldots +x_n^2\right)^2$ , or if we're in a complex space, then $\|x\|=\left(|x_1|^2+\ldots +|x_n|^2\right)^{1\slash 2}$

Of course, you do NOT need all the messy calculations above, but only to use that $W$ is non-singular (you haven't used this!)...

7. Originally Posted by tonio
Note: I'm assuming, since you didn't say otherwise, that the norm used in $\|Wx\|$ is the standard, euclidean one
That's the thing, the problem asks to prove that the function is a vector norm. I do not think I should be using the 2-Norm to prove this..
But anyways, thanks a lot!

8. Originally Posted by Mollier
That's the thing, the problem asks to prove that the function is a vector norm. I do not think I should be using the 2-Norm to prove this..
But anyways, thanks a lot!

Well, but you're defining a new "norm" $\|u\|_W$ by means of an OLD one $\|Wx\|$ ...! What is THIS last norm? Unless said otherwise, I think most mathematician will assume it is the standard, euclidean norm = what you call the 2-norm...see?

That's why I wrote what I wrote for $\|Wx\|$ , and that's why you HAVE to use non-singularity of $W$ to prove positiveness! Otherwise ALL you've done works for ANY matrix, so why would they tell you this one is non-singular??
So check this, and then FIX the whole demonstration since you used an incorrect formula for $WX$ all along.

Tonio

9. Yes sir.
Actually, I have taken a few steps back to the CBS inequality and the triangle inequality. I will slowly work my way towards norms, and will post an updated solution to this problem as soon as I understand it better.
You've been great