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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.

Rodrigo de Azevedo
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    If $A^{-1}$ denotes the inverse of $A$, then by definition of inverse $A^{-1}A = AA^{-1} = I$. – Gibbs Mar 30 '20 at 12:32
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    $A^{-1}$ is by definition an inverse on both sides. There are left inverses and right inverses, but they are not written as $A^{-1}$. – Paul Mar 30 '20 at 12:32
  • Something to think about: $$\begin{align}\mathbf A\mathbf A^{-1}&=\mathbf I\\mathbf A\mathbf A^{-1}\mathbf A&=\mathbf I\mathbf A=\mathbf A\\mathbf A(\mathbf A^{-1}\mathbf A)&=\mathbf A\mathbf I\end{align}$$ recalling associativity. Is the identity element unique? ;) – J. M. ain't a mathematician Mar 30 '20 at 12:43
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    That depends on what $A^{-1}$ means. See If $AB = I$ then $BA = I$. – user1551 Mar 30 '20 at 13:08
  • @j.m. is a poor mathematician. In the last argument you used that A*A(inverse)=I , which has not been shown yet –  Mar 30 '20 at 13:51
  • "which has not been shown yet" - isn't that what you were assuming in the first place? – J. M. ain't a mathematician Mar 30 '20 at 14:22
  • If $A$ and $B$ are finite matrices with $AB = I$, then we must have $B A = I$ also. This is because the finite-dimensionality together with $AB = I$ forces $A$ and $B$ to be bijections, and hence each is invertible as a linear map between vector spaces. This fails in the infinite-dimensional setting: we might have $AB = I$, but neither $A$ nor $B$ is invertible. – Joppy Mar 30 '20 at 14:34
  • @Joppy you need that $A$ and $B$ are square as well. As alluded to earlier, you can have non-square matrices $A$ and $B$ of sizes $m\times n$ and $n\times m$ such that $AB=I_m$. Take $(1~0)$ and $\binom{1}{0}$ for instance which multiply to give $(1)$, the $1\times 1$ identity matrix, but multiplying in the other order doesn't. $f\circ g$ a bijection doesn't directly imply $f$ and $g$ are both bijections. – JMoravitz Mar 30 '20 at 14:37
  • @JMoravitz yes that’s true, I forgot to mention that since when talking about invertible matrices I only ever have square matrices in mind! But it’s a good point. – Joppy Mar 30 '20 at 14:39

2 Answers2

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

R.Ishwariya
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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

Usman Abbas
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