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Determining whether a function is an inner product

  1. Nov 21, 2016 #1
    1. The problem statement, all variables and given/known data
    T/F: If ##T: \mathbb{R}^n \rightarrow \mathbb{R}^m## is a linear transformation and ##n>m##, then the function ##\langle v , w \rangle = T(v) \cdot T(w)## is an inner product on ##\mathbb{R}^n##

    2. Relevant equations


    3. The attempt at a solution
    The first three axioms of the inner product are straightforward. However, I am not sure how to show that ##\langle v , v \rangle = 0## iff ##v0##.

    Actually, maybe I have something. If ##\langle v , v \rangle = 0## then ##T(v) \cdot T(v) = ||T(v)||^2 \implies ||T(v)|| = 0 \implies T(v) = 0##. Thus, v is zero iff the null space of T is only zero. This is only the case when T is invertible. However, n > m, so T can't be invertible. Thus, v doesn't have to be 0, and thus we don't have an inner product space. Is this on the right track?
     
    Last edited: Nov 21, 2016
  2. jcsd
  3. Nov 21, 2016 #2

    fresh_42

    Staff: Mentor

    Are you sure you got it right? I would start to think about counterexamples. E.g. what if ##v \in \ker(T) - \{0\}##?
     
  4. Nov 21, 2016 #3
    I just added to my original post, so I am not sure if your post is after or before that post. Is my reasoning correct?
     
  5. Nov 21, 2016 #4

    fresh_42

    Staff: Mentor

    You still have a ##u## where there probably should be a ##v##. The equation with the norm could be used to show ##<v,v> \geq 0##, but what about my question?
     
  6. Nov 21, 2016 #5
    I'm not sure about your question, since I don't know whether I am right or wrong yet
     
  7. Nov 21, 2016 #6

    fresh_42

    Staff: Mentor

    What happens to vectors in the kernel of ##T##? And there are non-zero ones, because ##n >m##. What have you to be sure about? You have ##\langle v,v \rangle = 0 \Rightarrow T(v)=0##. Does ##v=0## follow?
     
  8. Nov 21, 2016 #7

    PeroK

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    I think you've successfully proved that it cannot be an inner product.
     
  9. Nov 21, 2016 #8
    ##v = 0## follows because n > m, which means that T is not invertible which means that ##Nul(T) \neq \{ 0 \}##. So it can't be an inner product, right?
     
  10. Nov 22, 2016 #9

    PeroK

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    Yes, that's right.
     
  11. Nov 22, 2016 #10

    fresh_42

    Staff: Mentor

    The argument was correct: How to find a ##v \neq 0## such that ##\langle v,v \rangle = 0## which contradicts the "iff" in the condition. But then with such a ##v##:
    (added and highlighted by me)
    If you want to be very exact, then ##n > m## means that ##T## cannot be injective.
    Injectivity of linear functions ##f## is equivalent to ##Nul(f) = \{0\}##.
     
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