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Homework Help: Linear algebra identities of inverse matricies

  1. Jul 20, 2010 #1
    1. The problem statement, all variables and given/known data
    Left Inversion in Rectangular Cases. Let A[tex]^{-1}_{left}[/tex] = (A[tex]^{T}[/tex]A)[tex]^{-1}[/tex]A[tex]^{T}[/tex] show A[tex]^{-1}_{left}[/tex]A = I.

    This matrix is called the left-inverse of A and it can be shown that if A [tex]\in[/tex] R[tex]^{m x n}[/tex] such that A has a pivot in every column then the left inverse exists.

    Right Inversion in Rectangular Cases. Let A[tex]^{-1}_{right}[/tex] = A[tex]^{T}[/tex](AA[tex]^{T}[/tex])[tex]^{-1}[/tex]. Show AA[tex]^{-1}_{right}[/tex] = I.

    This matrix is called the right-inverse of A and it can be shown that if A [tex]\in[/tex] R[tex]^{m x n}[/tex] such that A has a pivot in every row then the right inverse exists.

    2. Relevant equations



    3. The attempt at a solution
    I tried the left part and this is what I did:
    A[tex]^{-1}_{left}[/tex] / (A[tex]^{T}[/tex]A)[tex]^{-1}[/tex]= A[tex]^{T}[/tex]
    A[tex]^{-1}_{left}[/tex](A[tex]^{T}[/tex]A) = A[tex]^{T}[/tex]
    A[tex]^{-1}_{left}[/tex]A = A[tex]^{T}[/tex]( A[tex]^{T}[/tex])[tex]^{-1}[/tex] = I

    I'm not sure if this is correct or not, so I want to see if I have the right idea. I know that A*A[tex]^{-1}[/tex] = I so I thought this would work. Also isn't the right one the exact same thing? Or do I have to do that one a different way? Oh and can someone also explain the concept of left and right inverse. I don't really understand it. Thanks
     
  2. jcsd
  3. Jul 20, 2010 #2
    Sorry it's not showing right at all. I'll try to make it clearer. Al is A.

    Attempt:
    Al^(-1) / (A^T * A)^(-1) = A^T
    Al^(-1) * (A^T * A) = A^T
    Al^(-1) * A = A^T * (A^T)^(-1) = I
    therefore, Al^(-1) * A = I

    Hopefully that made what I was trying to do more clear

    Can someone please help?
     
  4. Jul 20, 2010 #3

    hunt_mat

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    Homework Helper

    Don't write divide when dealing with matrices, use the inverse notation.
    Write the following to compute the inverse of [tex]A^{T}A[/tex]
    [tex]
    (A^{T}A)^{-1}(A^{T}A)=I
    [/tex]
    Multiply on the right by the appropriate stuff to find the expression for the inverse and then use this in the definition of the left inverse. I should come out in the wash.
     
  5. Jul 20, 2010 #4
    So, I did the (A[tex]^{T}[/tex]A)[tex]^{-1}[/tex](A[tex]^{T}[/tex]A) = I
    and because of the property AA[tex]^{-1}[/tex] = I the left side is I. So, is this proff enough for Aleft? Is Aright basically the same thing then?
     
  6. Jul 21, 2010 #5

    hunt_mat

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    I was trying to get you to show that
    [tex]
    (A^{T}A)^{-1}=A^{-1}(A^{T})^{-1}
    [/tex]
    Then use this in the definition of the left inverse to compute that
    [tex]
    A_{left}^{-1}A=I
    [/tex]
     
  7. Jul 21, 2010 #6

    HallsofIvy

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    Are you allowed to assume that [itex]A^{-1}[/itex] and [itex](A^T)^{-1}[/itex] exist?
     
  8. Jul 21, 2010 #7

    hunt_mat

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    Hmm, most likely not! my bad...
     
  9. Jul 21, 2010 #8

    hunt_mat

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    And Spiffy, I think you're done in your proof.

    Mat
     
  10. Jul 21, 2010 #9
    oh ok, thank you. Would I do the same thing for Aright? I'm paranoid because it seems the same but i'm asked about it as well so I expect it to be different.
     
  11. Jul 21, 2010 #10

    hunt_mat

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    Homework Helper

    I think you'll be fine. Nothing to worry about.

    Mat
     
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