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 :)
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.