If $A A^{-1} = I$, does that automatically imply $A^{-1} A = I$?

Question

Or, in other words, if $A A^{-1} = I$ is satisfied, is there a possibility that $A^{-1} A \neq I$?

My suspicion is the answer is no.

Answer 1

Short Answer

Yes AA^-1 = A^-1A = I when the Det(A)$\neq$0 and A is a square matrix.

Long Answer

A matrix is basically a linear transformation applied to some space. For the sake of simplicity I will assume that we are in a 2D plane having 2 basis vectors i^{^} and j^{^} each having the magnitude of 1 with coordinates (1,0) and (0,1) respectively. The columns of the matrix tells where the corresponding basis vectors lands after the transformation.

A^-1A means that first we apply A transformation then we apply A^-1 transformation. When we apply A transformation we reach some plane having some different basis vectors but after apply A^-1 we again reach to the plane have basis i^{^} (0,1) and j^{^} (1,0). It means that after applying A^-1 we reach to the transformation which does nothing. The identity transformation $ \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ \end{bmatrix}$. It is basically the plane we started before applying A transformation.

That's why A^-1A = I.

This is same as AA^-1. It means that we first apply the A^-1 transformation which will take as to some plane having different basis vectors. If we think what is the inverse of A^-1 ? We are basically asking that what transformation is required to get back to the Identity transformation whose basis vectors are i^{^}(1,0) and j^{^}(0,1). Transformation A will do the job, since in this case after applying the A transformation we will reach to the identity transformation.

That's why AA^-1 = I

Answer 2

The result is not true for non-square matrices, since their left and right inverses are not equal.