Linear algebra identities of inverse matricies/transpose

  • Thread starter SpiffyEh
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Homework Statement


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.

Homework Equations





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
 
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Answers and Replies

  • #2
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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:

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