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Thread: Prove that function is a vector norm.

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

    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.


    ------------------------------------------------------------------------------------
    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 $
    ------------------------------------------------------------------------------------

    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.
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  2. #2
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    Quote Originally Posted by Mollier View Post
    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.


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

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

    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.
    .
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  3. #3
    Member Mollier's Avatar
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    Since you write that $\displaystyle ||x||_W=||Wx||=0 $, does that not automatically say that $\displaystyle ||x||_W=0 $? I guess if it was anything else it would look wierd.
    $\displaystyle 1 = ||Wx|| = 0 $

    Thanks mate!
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    Quote Originally Posted by Mollier View Post
    Since you write that $\displaystyle ||x||_W=||Wx||=0 $, does that not automatically say that $\displaystyle ||x||_W=0 $?


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

    What you need though is to prove that $\displaystyle \|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.
    $\displaystyle 1 = ||Wx|| = 0 $

    Thanks mate!
    .
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  5. #5
    Member Mollier's Avatar
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    Alright,

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

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

    Is this any better sir?

    Thanks!
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    Quote Originally Posted by Mollier View Post
    Alright,

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

    Since $\displaystyle (x_iw_i)^2 \geq 0, \quad for\; i=1,...,n $, we have that $\displaystyle x=0$.
    The $\displaystyle w_i$'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 $\displaystyle W=(w_{ij})\,,\,\,x=\begin{pmatrix}x_1\\x_2\\...\\x _n\end{pmatrix}$ , then $\displaystyle Wx=\begin{pmatrix}\sum\limits_{k=1}^nw_{1k}x_k\\\s um\limits_{k=1}^nw_{2k}x_k\\...\\\sum\limits_{k=1} ^nw_{nk}x_k\end{pmatrix}$ , and then

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

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

    Of course, you do NOT need all the messy calculations above, but only to use that $\displaystyle W$ is non-singular (you haven't used this!)...
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  7. #7
    Member Mollier's Avatar
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    Quote Originally Posted by tonio View Post
    Note: I'm assuming, since you didn't say otherwise, that the norm used in $\displaystyle \|Wx\|$ 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!
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    Quote Originally Posted by Mollier View Post
    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" $\displaystyle \|u\|_W$ by means of an OLD one $\displaystyle \|Wx\|$ ...! 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 $\displaystyle \|Wx\|$ , and that's why you HAVE to use non-singularity of $\displaystyle W$ 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 $\displaystyle WX$ all along.

    Tonio
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  9. #9
    Member Mollier's Avatar
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    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
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