Linear algebra identities of inverse matricies

[SpiffyEh]

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

 

[SpiffyEh]

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?

 

[hunt_mat]

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.

 

[SpiffyEh]

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?

 

[hunt_mat]

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]

 

[HallsofIvy]

Are you allowed to assume that [itex]A^{-1}[/itex] and [itex](A^T)^{-1}[/itex] exist?

 

[hunt_mat]

Hmm, most likely not! my bad...

 

[hunt_mat]

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

Mat

 

[SpiffyEh]

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.

 

[hunt_mat]

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

Mat