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Math Help - Help with a proof about inner product space

  1. #1
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    Help with a proof about inner product space

    Hi guys.
    I need help with the following:

    Let V be a vector space over \mathbb{c}.
    Let B=\left \{ v_1, v_2,...,v_n \right \} be a base for V.

    a. let v\in V such that for every i: <v_i,v>=0. prove that v=0.
    b. let u,w\in V such that for every i: <v_i,u>=<v_i,w>. prove that u=w.


    this is what I tried so far:
    a.
    let v_j\in B. since B is base for V, then:
    <v_j,v>=<v_j,\sum_{i=1}^{n}\alpha_iv_i>=\overline{  <\sum_{i=1}^{n}\alpha_iv_i,v_j>}=\sum_{i=1}^{n} \alpha_ i \overline{<v_i,v_j>}=...

    not sure what's next, though. what am I missing here?
    Am I even in the right direction?

    Thanks in advanced!
    Last edited by Stormey; May 22nd 2013 at 01:36 PM.
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  2. #2
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    Re: Help with a proof about inner product space

    Quote Originally Posted by Stormey View Post
    Let V be a vector space over \mathbb{c}.
    Let B=\left \{ v_1, v_2,...,v_n \right \} be a base for V.

    a. let v\in V such that for every i: <v_i,v>=0. prove that v=0.
    b. let u,w\in V such that for every i: <v_i,u>=<v_i,w>. prove that u=w.

    For part a):
    Recall that <v,v>=0\text{ if and only if }v=0.

    Now because v\in V then v = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}}

    You finish the proof of part a.
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  3. #3
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    Re: Help with a proof about inner product space

    Thanks for the help, Plato.

    Quote Originally Posted by Plato View Post
    For part a):
    Recall that <v,v>=0\text{ if and only if }v=0.

    Now because v\in V then v = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}}

    You finish the proof of part a.
    how come?
    let's say j=2, how do I use this fact here:

    \alpha_1<v_2,v_1>+\alpha_2<v_2, v_2>+\alpha_3<v_2,v_3>+...+ \alpha_n <v_2,v_n>=0
    Last edited by Stormey; May 22nd 2013 at 11:07 PM.
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    Re: Help with a proof about inner product space

    Quote Originally Posted by Stormey View Post
    let's say j=2, how do I use this fact here:
    \alpha_1<v_2,v_1>+\alpha_2<v_2, v_2>+\alpha_3<v_2,v_3>+...+ \alpha_n <v_2,v_n>=0

    \left( {\forall j} \right)\left[ {\left\langle {v,{v_j}} \right\rangle  = 0} \right]
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  5. #5
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    Re: Help with a proof about inner product space

    Quote Originally Posted by Plato View Post
    \left( {\forall j} \right)\left[ {\left\langle {v,{v_j}} \right\rangle  = 0} \right]
    I'm sorry, I guess I'm missing something here.

    I'll start over, and use some other letter to clarify my question.
    u is some vector in vector space V, such that <v_i,u>=0 for every i ( v_i\in B, and B is base).
    how do I go from: "a dot product of vector with himself is 0 iff it is 0 itself" to: if the dot product of u (which is some vector in V) with each of the basis vectors is 0, then u is 0"?

    thanks.
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    Re: Help with a proof about inner product space

    Quote Originally Posted by Stormey View Post
    I'm sorry, I guess I'm missing something here.
    I'll start over, and use some other letter to clarify my question.
    u is some vector in vector space V, such that <v_i,u>=0 for every i ( v_i\in B, and B is base).

    That changes nothing of what I have posted above.
    \{v_1,v_2,\cdots,v_n\} is a basis and the vector u has the property that \forall j<u,v_j>=0~.

    Using the basis u = \sum\limits_{k = 1}^n {{\alpha _k}{v_k}} ,
    so \left\langle {u,u} \right\rangle  = \left\langle {u,\sum\limits_{k = 1}^n {{\alpha _k}{v_k}} } \right\rangle  = \sum\limits_{k = 1}^n {{\alpha _k}\left\langle {u,{v_k}} \right\rangle }  = 0
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  7. #7
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    Re: Help with a proof about inner product space

    Thank you!
    now it is clear.
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