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

  1. #1
    Member Mollier's Avatar
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    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.
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  2. #2
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    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 ..
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  3. #3
    Member Mollier's Avatar
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    So basically ;

    $\displaystyle ||x||_W+||y||_W=||Wx||+||Wy|| \geq ||Wx+Wy|| = ||x+y||_W$ ?
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  4. #4
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    Triangular inequality

    Ya i guess so...
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