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I want to prove that for matrices $A,B \in M_n (\mathbb K)$ where $\mathbb K \in \{\mathbb R, \mathbb C, \mathbb H\}$ if $AB = I$ then $BA = I$.

My proof is really short so I'm not sure it's right:

If $AB = I$ then $(BA)B = B$ and therefore $BA=I$?

learner
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    So you are saying that if you have two matrices $C$ and $D$ where $CD = D$, then $C$ always is $I$? – Thomas Jun 30 '14 at 16:06
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    Here you can find proof : http://math.stackexchange.com/questions/3852/if-ab-i-then-ba-i – agha Jun 30 '14 at 16:08
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    It’s not correct – see Thomas’ comment. And @agha: I think she or he is more interested in finding an own proof. – k.stm Jun 30 '14 at 16:08
  • You must use arbitrary matrix not a specific one – Kamster Jun 30 '14 at 16:18
  • For clarity, you should add the step $(BA)B=B(AB)$, and you just proved that $(BA)B=B$. If $B$ is not invertible, you cannot conclude $BA=I$. –  Jun 30 '14 at 17:01

3 Answers3

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The implication $(BA)B=B \Rightarrow BA=I$ is a little quick and not always true...

But observe that $$1=\det(BA)= \det(B)\det(A)$$ thus $B$ is invertible and it follows that $$BA= BA(BB^{-1}) = B(AB)B^{-1}=BB^{-1}=I.$$

Surb
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Consider $$ C = \pmatrix{1 & 0 \\ 0 & 0}, D =\pmatrix{1 & 0 \\ 0 & 0} $$ Here $CD = D$, but $C$ is not the identity. That is, your proof is not clear.

It has already been given in the comments above, but you can find a proof here: If $AB = I$ then $BA = I$

Community
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Thomas
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Your proof isn't correct: If in a ring with unit we have $ab=b\iff(a-1)b=0$ then we have not necessarily $a=1$ or $b=0$ unless we work on an integral domain (and of course $\mathcal M_n(\Bbb K)$ isn't an integral domain).

To prove this result notice that (by a classic result) that $AB=I$ imply that $A$ is surjective and $B$ is injective (since $I$ is bijective) and we conclude that $A$ and $B$ are invertible using the rank-nullity theorem.