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Math Help - Inner Products

  1. #1
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    Inner Products

    Let V be an inner product space. Prove that if v and w are vectors in V such that <v,x> = <w,x> for every x in V, then v = w.
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    Quote Originally Posted by mathbug View Post
    Let V be an inner product space. Prove that if v and w are vectors in V such that <v,x> = <w,x> for every x in V, then v = w.
    Let \{v_1,...,v_n\} be an orthonormal basis of V, then:

    v = \sum_{i=1}^n <v,v_i>v_i

    w = \sum_{j=1}^n <w,v_j>v_j

    now you finish.
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    So then <v,x> = sum <v,vi> [<vi,x>] = sum <w,vj> [<vj,x>] = <w,x> Then

    <v,v1>[<v1,x>] - <w,v1>[<v1,x>] +..+ <v,vn>[<vn,x>] - <w,vn>[<vn,x>]=0

    So then where am I supposed to go from here?
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  4. #4
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    Quote Originally Posted by mathbug View Post
    Let V be an inner product space. Prove that if v and w are vectors in V such that <v,x> = <w,x> for every x in V, then v = w.
    So you know that \langle v-w,x\rangle = 0 for every x. In particular, you could put x = v-w ... .
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  5. #5
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    Quote Originally Posted by mathbug View Post
    So then <v,x> = sum <v,vi> [<vi,x>] = sum <w,vj> [<vj,x>] = <w,x> Then

    <v,v1>[<v1,x>] - <w,v1>[<v1,x>] +..+ <v,vn>[<vn,x>] - <w,vn>[<vn,x>]=0

    So then where am I supposed to go from here?
    Opalg's solution is obviously nicer, however to complete this one:

    Since <v,x> = <w,x> ~ \forall x \in V, we get that <v,v_i> = <w,v_i> ~ \forall 1 \leq i \leq n.

    But, note that:

    w = \sum_{i=1}^n <w,v_i>v_i \Rightarrow w = \sum_{i=1}^n <v,v_i>v_i = v
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  6. #6
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    Quote Originally Posted by Defunkt View Post
    Opalg's solution is obviously nicer, however to complete this one:

    Since <v,x> = <w,x> ~ \forall x \in V, we get that <v,v_i> = <w,v_i> ~ \forall 1 \leq i \leq n.

    But, note that:

    w = \sum_{i=1}^n <w,v_i>v_i \Rightarrow w = \sum_{i=1}^n <v,v_i>v_i = v
    But how can you say that <v,v_i> = <w,v_i> for all x in V? If we had

    <v,v1>[<v1,x>] - <w,v1>[<v1,x>] +..+ <v,vn>[<vn,x>] - <w,vn>[<vn,x>]=0

    then we could factor and get

    [<v1,x>][<v,v1>-<w,vi>] +...+ [<vn,x>][<v,vn>-<w,vn>]=0

    I get that if <vi,x> were linearly independent, we would have that <w,v_i>=<v,v_i> for all x in V, but how can we show that?
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  7. #7
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    Quote Originally Posted by mathbug View Post
    But how can you say that <v,v_i> = <w,v_i> for all x in V? If we had

    <v,v1>[<v1,x>] - <w,v1>[<v1,x>] +..+ <v,vn>[<vn,x>] - <w,vn>[<vn,x>]=0

    then we could factor and get

    [<v1,x>][<v,v1>-<w,vi>] +...+ [<vn,x>][<v,vn>-<w,vn>]=0

    I get that if <vi,x> were linearly independent, we would have that <w,v_i>=<v,v_i> for all x in V, but how can we show that?
    I think you're making this too complicated --

    We are given that for every x \in V, <v,x> = <w,x>.
    In particular, if we let x=v_1, we get that <v,v_1> = <w,v_1>. Now let x = v_2, then <v,v_2> = <w,v_2> and so on: for every 1 \leq i \leq n, we get that <v,v_i>=<w,v_i>.
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  8. #8
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    Oh! You're right, I do tend to make things too complicated. That answer is very simple! But makes so much sense... I completely understand now. Thank you so much for all your help!
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