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Math Help - 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 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)

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

    (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 ;

    ||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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