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Proving that T(u).T(V)=u.v if an only if A^T =A^-1

  1. Nov 20, 2014 #1
    Let u and v be vectors in R^n , and let T be a linear operator on R^n. Prove that T(u).T(v)= u.v if and only if A^T=A^-1 where A is the standard matrix for T.


    What I have so far is --> If T is a linear operator on R^n, then ; T:R^n --> R^n
    and if u and v be the vectors then ; T(u).T(V) = Au.Av
    Where A is the standard matrix of T
    { and so we want Au.Av=u.v, which needs to be if and only if A^T=A^-1}

    Therefore first we find what it means if A^T=A^-1
    ----> if A^T=A^-1 then multiplying both by A --> AA^T= I
    Therefore the matrix A is and orthogonal matrix/
    And so from the last statement , if Au.Av= u.v if and only if A^T=A^-1
    Then --> Au.Av=u.v if and only if A is an orthogonal matrix.
    Therefore we must prove that Au.Av= u.v means that this is only true if the matrix A is orthogonal.

    This is where I am stuck I'm not sure how to prove this.
    I'm not sure if this next part is necessary but I found that

    u.v = u1v1+u2v2...UnVn

    and that T(u)=A(u) such that a(11)u(1), a(12)u(2),..,a(1n)u(1)
    a(21)u(1), a(22)u(2)...,a(2n)u(2)
    .................................................
    ...............................................
    a(n1)u(1), a(n2)u(2)....a(nn)u(n)

    and same for T(v)=A(v) such that
    a(11)v(1), a(12)v(2),..,a(1n)v(1)
    a(21)v(1), a(22)v(2)...,a(2n)v(2)
    .................................................
    ...............................................
    a(n1)v(1), a(n2)v(2)....a(nn)v(n)

    Im not sure if I need to develop T(u).T(v) further then A(u).A(v)..

    I would really appreciate some help on this one, thanks
     
  2. jcsd
  3. Nov 20, 2014 #2

    Fredrik

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    Are you sure that you're allowed to use ##Tu\cdot Tv=Au\cdot Av## without proof? If you use this, you don't even have to use the definition of "standard matrix for T". The simplest way to proceed is to use that ##x\cdot y=x^Ty##.

    This is a valid approach even if you're not allowed to use ##Tu\cdot Tv=Au\cdot Av## without proof. You just have to do the proof as well. This is pretty straightforward if you expand u and v in the standard basis, and use the definitions of dot product, matrix multiplication, and "standard matrix for T".

    Moderators: I moved the thread here from the linear algebra forum.
     
  4. Nov 20, 2014 #3
    I only meant for example; Tu in matrix form would be [A][x] like -->a(11)u(1), a(12)u(2),..,a(1n)u(1)
    a(21)u(1), a(22)u(2)...,a(2n)u(2)
    .................................................
    ...............................................
    a(n1)u(1), a(n2)u(2)....a(nn)u(n) ....Because A is the standard matrix of T. is that not the right way?

    Also, I don't really understand what I am suppose to do with x.y=x^Ty
     
  5. Nov 21, 2014 #4

    Fredrik

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    OK, you have a point. The observation you just made makes it obvious that ##Tu\cdot Tv=Au\cdot Av##. So I guess it's not necessary to do the proof I had in mind:
    \begin{align}
    &Tu\cdot Tv =\sum_{i,j} u_i v_j Te_i\cdot T e_j =\sum_{i,j,k,l} u_i v_j (Te_i)_k (Te_j)_l e_k\cdot e_l\\
    &=\sum_{i,j,k,l} u_i v_j A_{ki} A_{lj}\delta_{kl} =\sum_{i,j,k} u_i v_j A_{ki} A_{kj}=\sum_k(Au)_k (Av)_k =Au\cdot Av
    \end{align}

    Use it to rewrite both sides of ##u\cdot v=Au\cdot Av##. You know how to deal with ##(Au)^T##, right?
     
    Last edited: Nov 21, 2014
  6. Nov 24, 2014 #5
    no sorry, I'm unsure where to go from "Therefore we must prove that Au.Av= u.v means that this is only true if the matrix A is orthogonal."
     
  7. Nov 24, 2014 #6

    Fredrik

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    If you rewrite ##u\cdot v=Au\cdot Av## using ##x\cdot y=x^Ty##, you should almost immediately see that the equality holds if and only if ##A## is orthogonal.

    The general formula for the transpose of a product is ##(AB)^T=B^TA^T##. This follows almost immediately from the definitions:
    $$((AB)^T)_{ij}=(AB)_{ji}=\sum_k A_{jk}B_{ki} =\sum_k (A^T)_{kj} (B^T)_{ik} =\sum_k (B^T)_{ik} (A^T)_{kj}=(B^TA^T)_{ij}.$$
     
  8. Nov 24, 2014 #7
    ok, so are you meaning for vector u to be x and vector v to be y?
     
  9. Nov 24, 2014 #8

    Fredrik

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    When you rewrite ##u\cdot v##, yes. Of course, that's not the only thing you need to rewrite.
     
    Last edited: Nov 24, 2014
  10. Nov 26, 2014 #9
    hmm.. ok
     
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