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Linear algebra identities of inverse matricies/transpose

  1. Jul 17, 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
     
    Last edited: Jul 17, 2010
  2. jcsd
  3. Jul 17, 2010 #2
    Sorry it's not showing right at all. I'll try to make it clearer. Al is A[tex]_{left}[/tex].

    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?
     
    Last edited: Jul 17, 2010
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