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I am reviewing some Basic Linear Algebra. They define: A Matrix A is invertible if and only if there exists some matrix B such that $AB=I$ and $BA=I$.

So, in a slightly separate line of thought, I was wracking my brain trying to see if it is always the case that if $AB=I$ then $BA \neq I$. Essentially I was looking for a counter example. I wanted to show that if $AB=I$ then it is not enough to say that B is the inverse of A. I couldn't find an example. So it made me wonder if you could prove if $AB=I$ then $BA=I$ for basic matrix algebra.

I don't know know if this is possible. I appreciate any insights and or corrections. Similarly if you can provide one counter example then that would be most helpful also!!

Kind thanks MathExchange :)

cpmcintyre
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    These are possible ($AB=I$ and $BA\neq I$) when $A$ is not square. For a square matrix $A$, $AB=I$ implies $BA=I$.. – hardmath Aug 20 '17 at 13:41
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    @hardmath correct. At least for the finite dimensional cases. Otherwise consider the left/right shift operator. (But finite dim seems to be implied by the question) – lalala Aug 20 '17 at 13:44
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    @lalala: Yes, I assume "Matrix $A$" means a finite number of rows and columns. – hardmath Aug 20 '17 at 13:48
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    yes, deinitely for finite dimensional matrix. – cpmcintyre Aug 20 '17 at 15:37

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For a square matrix $A$: if there is a matrix $B$ such that $AB = I$, then it will also be the case that $BA = I$. Some proofs of this fact are presented here.

As for non-square matrices, we can say the following: if $A$ is $m \times n$ with $m > n$, then there can be no matrix $B$ such that $AB = I$. When $m \geq n$, there exists a matrix $B$ such that $BA = I$ if and only if $A$ has "full column-rank", which is true if and only if the transformation associated with $A$ is injective. If $m < n$, there can be no matrix $B$ such that $BA = I$. When $m \leq n$, there exists a matrix $B$ such that $AB = I$ if and only if $A$ has "full column-rank", which is true if and only if the transformation associated with $A$ is surjective.

Ben Grossmann
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