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

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


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

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

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

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

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

    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 $\displaystyle <v,v>=0\text{ if and only if }v=0$.

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

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

    $\displaystyle \alpha_1<v_2,v_1>+\alpha_2<v_2, v_2>+\alpha_3<v_2,v_3>+...+ \alpha_n <v_2,v_n>=0$
    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:
    $\displaystyle \alpha_1<v_2,v_1>+\alpha_2<v_2, v_2>+\alpha_3<v_2,v_3>+...+ \alpha_n <v_2,v_n>=0$

    $\displaystyle \left( {\forall j} \right)\left[ {\left\langle {v,{v_j}} \right\rangle = 0} \right]$
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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
    $\displaystyle \left( {\forall j} \right)\left[ {\left\langle {v,{v_j}} \right\rangle = 0} \right]$
    I'm sorry, I guess I'm missing something here.

    I'll start over, and use some other letter to clarify my question.
    $\displaystyle u$ is some vector in vector space $\displaystyle V$, such that $\displaystyle <v_i,u>=0$ for every i ($\displaystyle v_i\in B$, and $\displaystyle B$ 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 $\displaystyle u$ (which is some vector in $\displaystyle V$) 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.
    $\displaystyle u$ is some vector in vector space $\displaystyle V$, such that $\displaystyle <v_i,u>=0$ for every i ($\displaystyle v_i\in B$, and $\displaystyle B$ is base).

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

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

    Thank you!
    now it is clear.
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