My approach,
$A$ AND $B$ are $n \times n$ matrices such that,
$BA=I$
Now,
We can write, $A = AI$
$A= A (BA)$
$A= (AB)A$
$A(I-AB)=0$
$I-AB =0$
Hence, $AB=I$
I don't feel like what I'm doing is the correct way to go. Please help ! Thanks in advance.
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3$A=(AB)A \stackrel?\implies A(I-AB)=0$ and $AC=0 \stackrel?\implies C=0$ – azif00 Mar 03 '21 at 03:47
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1you need to justify going from A(I-AB)=0 to I-AB=0 – Simon Segert Mar 03 '21 at 03:47
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2The statement isn't true for linear transformations on infinite-dimensional vector spaces, so you must some use the fact that the matrices are $n\times n$. – saulspatz Mar 03 '21 at 03:57
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1See https://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – user1551 Mar 03 '21 at 03:59
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1Does this answer your question? If $AB = I$ then $BA = I$ If you need feedback on your attempt, then that is provided by the first two comments. – Sarvesh Ravichandran Iyer Mar 03 '21 at 04:10
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Your third step is wrong. You go from $A-ABA=0$ to $A(I-AB)=0$. But if you multiply that out, you get $A-AAB$, not $A-ABA$. The correct expression you can get is $(I-AB)A=0$ (by factoring $A$ on the right), but then you cannot left-multiply by $B$ to conclude that $I-AB=0$, so you get stuck. – Arturo Magidin Mar 03 '21 at 05:02