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Proving a fact about inner product spaces

  1. Apr 6, 2017 #1
    1. The problem statement, all variables and given/known data
    Let ##V## be a vector space equipped with an inner product ##\langle \cdot, \cdot \rangle##. If ##\langle x,y \rangle = \langle x, z\rangle## for all ##x \in V##, then ##y=z##.

    2. Relevant equations


    3. The attempt at a solution
    Here is my attempt. ##\langle x,y \rangle = \langle x, z\rangle## means that ##\langle x,y \rangle - \langle x, z\rangle = 0##, and so ##\langle x, y - z\rangle = 0##. This is as far as I've gotten. I don't see how to deduce that ##y - z = 0##.
     
  2. jcsd
  3. Apr 6, 2017 #2

    fresh_42

    Staff: Mentor

    It's hidden in the term "inner product". In the sense used here, it is non-degenerated, which means exactly this: ##\langle x,y \rangle=0 \; \forall \; x \in V \Longrightarrow y=0##. This is true for the usual dot product like in real vector spaces. It is not automatically true for an arbitrary bilinear form. So in concrete cases, one has to check the properties of the "inner product" used.
     
  4. Apr 6, 2017 #3
    So are you saying that ##\langle x,y \rangle=0 \; \forall \; x \in V \Longrightarrow y=0## is a property of inner product spaces in general?
     
  5. Apr 6, 2017 #4

    fresh_42

    Staff: Mentor

    No, I'm not saying this, because I'm used to deal with bilinear forms, that do not obey this rule and I don't see why they shouldn't be called inner product, too. So I am saying: look at your definition. This is even more important in the complex case, in which the product isn't symmetric anymore. (See https://en.wikipedia.org/wiki/Inner_product_space#Alternative_definitions.2C_notations_and_remarks.)

    The convention however is, that inner products have to be positive definite, that is ##\langle x,y \rangle \geq 0## and ##\langle x,x \rangle=0 \Longleftrightarrow x=0##. With this convention we get in your case ##\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0 \Longrightarrow y-z=0##.
     
  6. Apr 6, 2017 #5
    Could you explain the implication ##\forall_{x}\; \langle x,y-z \rangle=0 \Longrightarrow \langle y-z,y-z \rangle=0##?
     
  7. Apr 6, 2017 #6

    fresh_42

    Staff: Mentor

    Yes.
    ... still for all ##x\in V##, so especially for ##x:=y-z##.
     
  8. Apr 7, 2017 #7
    As fresh_42 said, it holds for all ## x## including ## x:=y-z##. There is an axiom of scalar product, that gives you the answer of what ##y-z## can be if its length is zero.
     
  9. Apr 7, 2017 #8

    Math_QED

    User Avatar
    Homework Helper

    Don't you mean ##<x,x> \geq 0##?
     
  10. Apr 7, 2017 #9

    fresh_42

    Staff: Mentor

    Yes, of course. Copy and paste error.
     
  11. Apr 9, 2017 #10

    WWGD

    User Avatar
    Science Advisor
    Gold Member

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