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?
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5Why does $BA\neq E$ imply $BAB\neq B$? (Hint. It does not.) – Sangchul Lee Jun 17 '17 at 03:29
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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.
clark
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if B = 0 then AB ≠ E. Is there other counterexamples? I understood what's the fallacy but I need some counterexamples. – Ris Jun 17 '17 at 05:18
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@Ris in any counter example $AB$ cannot equal $E$ since for all such counter examples $B$ cannot be a bijection. – clark Jun 17 '17 at 05:29
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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.
user76568
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