# Linear algebra identities of inverse matricies/transpose

## 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.

## 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

Last edited:

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