Linear algebra identities of inverse matricies/transpose

[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_{left}.

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?