$AB = E$ then $BA = E$ proof?

Question

Let $A$ and $B$ be a square matrix with same size. If $AB = E$ and $BA \neq E$ $BAB \neq EB$ so $B \neq B \Rightarrow$ a contradiction. Where is this solution wrong?

Answer 1

You are arguing that $BA\neq E \Rightarrow BAB \neq EB$. This is not generally true for any matrix $B$ for instance what if $B=0$ then $BAB=0=EB$.

Viewing matrices as maps, you would like to say $ g \neq h$ then $g\circ f\neq h \circ f $. For that, to necessarily hold for any $g$ and $h$, $f$ must be injective and surjective on the domain of $g$ and $h$.

So, to make your argument work you need first to establish that $AB=E$ implies that $B$ is a bijection. From there, assuming $BA\neq E$ would indeed give a contradiction by arguing as you did.

Answer 2

You are treating $\neq$ as being dual to $=$. It is not always so... It is simply logical negation of $=$ in equations. Not an equation. That's it.