Linear algebra identities of inverse matricies

[SpiffyEh]

Homework Statement

Left Inversion in Rectangular Cases. Let A^{-1}_{left} = (A^{T}A)^{-1}A^{T} show A^{-1}_{left}A = I.

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

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

This matrix is called the right-inverse of A and it can be shown that if A \in R^{m x n} 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^{-1}_{left} / (A^{T}A)^{-1}= A^{T}
A^{-1}_{left}(A^{T}A) = A^{T}
A^{-1}_{left}A = A^{T}( A^{T})^{-1} = 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^{-1} = 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 A^{T}A
<br /> (A^{T}A)^{-1}(A^{T}A)=I<br />
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^{T}A)^{-1}(A^{T}A) = I
and because of the property AA^{-1} = 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
<br /> (A^{T}A)^{-1}=A^{-1}(A^{T})^{-1}<br />
Then use this in the definition of the left inverse to compute that
<br /> A_{left}^{-1}A=I<br />

 

[HallsofIvy]

Are you allowed to assume that A^{-1} and (A^T)^{-1} 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