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This question already has an answer here:

Let $A,B$ be $m\times m$ matrices such that $AB$ is invertible. Show $A,B$ invertible.

Here is what I have attempted:

We have $AB$ invertible so there exists an $m\times m$ matrix $C$ such that $C(AB)=(AB)C=I_{m}$. Since matrix multiplication is associative we immediately have a right and left inverse respectively: $(CA)B=A(BC)=I_{m}$. I know the fact that I have square matrices needs to play a roll and gaurentees me that the product will be, in general, defined. I have tried to multiply $CA$ and $BC$ on other sides but that didn't get me anywhere.

I've asked myself what would happen if they weren't square and its obvious, by definition, that they would not be invertible but I am not seeing a way to use the fact that they are to gaurentee ourselves a general inverse.

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marked as duplicate by Arnaud D., Lord Shark the Unknown, Brian Borchers, BCLC, Parcly Taxel Mar 10 '18 at 23:44

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  • $\begingroup$ I just now wanted to try this: Since $A,B$ are matrices they correspond to linear transformations $T,U$ as does their composition $TU$. Since $TU$ is invertible we know $T,U$ invertible and so the corresponding matrices are invertible. Does this work? I think it is at least missing a detail or two... $\endgroup$ – Aaron Zolotor Oct 29 '15 at 22:00
  • $\begingroup$ Do you know that a (necessary and) sufficient condition for invertibility of a square matrix is having a one sided inverse? $\endgroup$ – egreg Oct 29 '15 at 22:18
  • $\begingroup$ I'm not sure that I do. I've been pondering a way to show that if it is true $\endgroup$ – Aaron Zolotor Oct 29 '15 at 22:19
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    $\begingroup$ It depends on the tools you have available. If you know about linear maps and associated matrices, it's easy. There are also purely matrix theory proofs. $\endgroup$ – egreg Oct 29 '15 at 22:21
  • $\begingroup$ Excuse me? Do I know you? $\endgroup$ – Aaron Zolotor Oct 23 '18 at 17:08
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Okay, so I figured this out.

We have a left and right inverse for our matrices $B$ and $A$ respectively. Since $A,B$ both correspond to linear transformations from a space of dimension n to another of dimension n we know that they are functions which possess a right and left inverse respectively, and thus are onto and one-to-one respectively. Therefore they are both one-to-one and onto since having one property gives the other when we are dealing with vector spaces of finite and equal dimension. Since $A,B$ both correspond to invertible transformations we know they are invertible. Q.E.D.

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$AB$ is invertible so $\det(AB)\neq 0$. We have $\det(AB)=\det(A)\det(B)$; since it is nonzero it follows that neither $\det(A)$ or $\det(B)$ are zero, so $A$ and $B$ are both invertible.

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Recall that a linear map $M$ can only lower the dimension of a subspace. By this I mean that if $V \subset \mathbb R ^m$ has dimension $n$ then $M(V)$ has dimension at most $n$.

Recall a matrix is invertible if and only if it has full rank, if and only if its null-space is empty, if and only if it is surjective, if an only if the image of the whole space has maximum dimension.

Since $AB \colon \mathbb R ^m \to \mathbb R ^m$ is invertible we have $AB(\mathbb R ^m) = \mathbb R ^m$. Then if $B$ is non-invertible it is not surjective so $B(\mathbb R ^m)$ has dimension less than $m$, and so $AB(\mathbb R ^m) = A(B(\mathbb R ^m))$ has dimension less than $m$, a contradiction. If $A$ is non-invertible but $B$ is then $AB(\mathbb R ^m) = A(\mathbb R ^m)$ has dimension less than $m$, again a contradiction.

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If $AB$ is invertible, then there exists $D$ s.t. $ABD=I$

Then $A$ is invertible with inverse $BD$.

It remains to show that $B$ is invertible.

Now we have

$$ABD=I$$ $$\to BD=A^{-1}$$ $$\to B=A^{-1}D^{-1}$$

Then $B$ is invertible as it is the product of invertible matrices...wait, is that given?

QED if yes. Idk o/w.

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