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Inner product space

  1. Aug 1, 2008 #1

    Defennder

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    1. The problem statement, all variables and given/known data
    Let V be an inner product space. Then for x,y,z [tex]\in[/tex] V and c[tex]\in[/tex]F, where F is a field denoting either R or C, prove that

    <x,x> = 0 if and only if x=0.

    Notes on notation:
    Here <x,y> denotes the inner product of vectors x and y on some vector space V.


    2. Relevant equations
    <x,y+z> = <x,y> + <x,z>
    <cx,y> = c<x,y>
    And a few others.



    3. The attempt at a solution
    It seems pretty straightforward to prove the converse, namely that x=0 implies <0,0> = 0, like this:

    <-x+x,0> = <-x,0> + <x,0>
    <0,0> = <0,0> + <0,0>
    <0,0> = 0.

    But how do I prove the "forward" conjecture? I know that x=0 iff for some y[tex]\in[/tex] V x+y = y, but I can't start with x, only <x,x> = 0, and I don't see how to "extract" x such that I can show x+y = y.

    Any thoughts?
     
  2. jcsd
  3. Aug 1, 2008 #2

    radou

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    <x | x> = 0 iff x = 0 is one of the inner product "axioms", i.e. the properties which the inner product must have in order to be one. Unless I'm missing something, there's nothing to prove.
     
  4. Aug 1, 2008 #3

    Defennder

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    The question was from one of the theorems for an inner product space. And the author left it as an excercise to the reader.
     
  5. Aug 1, 2008 #4

    HallsofIvy

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  6. Aug 1, 2008 #5

    Dick

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    Right. E.g. the inner product for 4 vectors in special relativity satisfies all the axioms you've stated so far, but is has null (lightlike) vectors such that <x,x>=0.
     
    Last edited by a moderator: Aug 1, 2008
  7. Aug 1, 2008 #6

    Defennder

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    Here's some more the book provided:
    [tex]<x+z,y> = <x,y> + <z,y>[/tex]
    [tex]<x,cy> = \bar{c}<x,y>[/tex] Here the bar denotes complex conjugation.

    [tex]\overline{<x,y>} = <y,x>[/tex]
    [tex]<x,x> > 0 \ \mbox{if} \ x \neq 0[/tex]
    Note that the second ">" represents "greater than". But I find this to be odd, since F is treated as either R or C, and the complex numbers are not ordered in any way. The book I'm using is Linear Algebra 3rd Edn by Stephen Friedberg, Arnold J. Insel, and Lawrence Spence. The axioms can be found in the beginning of the chapter on inner product spaces, chap 6.

    Well in this case, it is given that x=0, and we know that -x is the additive inverse of x in F, so -x+x = 0. Using the first axiom I just quoted above, this breaks down to <-x,0> + <x,0>. And since x=0 and -0 = 0, both reduce to <0,0>. By the law of additive cancellation, <0,0> cancels on both sides, leaving one <0,0> on one side with 0 on the other side.

    <0,0> + 0 = <-x+x,0>
    <0,0> + 0 = <-x,0> + <x,0>
    <0,0> + 0 = <0,0> + <0,0>
    0 = <0,0>

    Is this a hint for proving <x,x> = 0 implies x=0 or the one just above?
     
  8. Aug 1, 2008 #7

    Dick

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    The first line shows <x,x> must be real even over a complex vector space, so ordering is not a problem. I think you know <0,0>=0. Then <x,x>=0 -> x=0, right? Look at the contrapositive, x not equal to zero implies <x,x> not equal to zero.
     
  9. Aug 1, 2008 #8

    Defennder

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    Hey that's right. I can't I missed that.

    But how does that prove <x,x> = 0 is impossible unless x=0?
     
  10. Aug 1, 2008 #9

    Dick

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    P->Q and (not Q)->(not P) are logically equivalent. P is "<x,x>=0", Q is "x=0". P->Q is what you are trying to prove. (not Q)->(not P) is "x not 0 implies <x,x> not 0". One of your axioms is "x not 0 implies <x,x>>0".
     
  11. Aug 1, 2008 #10

    Defennder

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    Oh I see. Sheesh. I thought that was unrelated to this problem until I understood the 3rd axiom. Thanks!
     
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