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I know this proof is short but a bit tricky. So I suppose that $AB$ is invertible then $(AB)^{-1}$ exists. We also know $(AB)^{-1}=B^{-1}A^{-1}$. If we let $C=(B^{-1}A^{-1}A)$ then by the invertible matrix theorem we see that since $CA=I$(left inverse) then $B$ is invertible. Would this be correct?

Edit Suppose $AB$ is invertible. There exists a matrix call it $X$ such that $XAB=I$. Let $C=XA$ Then $CB=I$ and it follows that $B$ is invertible by the invertible matrix theorem.

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    $\begingroup$ You're assuming your conclusion when you write that $(AB)^{-1}=B^{-1}A^{1}$. $\endgroup$ – JSchlather Aug 3 '13 at 20:30
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    $\begingroup$ The equation $(AB)^{-1} = B^{-1}A^{-1}$ presumes the existence of both $A^{-1}$ and $B^{-1}$. $\endgroup$ – Sangchul Lee Aug 3 '13 at 20:31
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    $\begingroup$ Are $A,B$ square matrices or do we only assume that $AB$ is square? $\endgroup$ – Hagen von Eitzen Aug 3 '13 at 20:34
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    $\begingroup$ Assume $AB$ is square. $\endgroup$ – user60887 Aug 3 '13 at 20:35
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    $\begingroup$ It is not necessarily true that $(AB)^{-1}=B^{-1}A^{-1}$. $\endgroup$ – AnonSubmitter85 Aug 3 '13 at 20:35
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$\;AB\;$ invertible $\;\implies \exists\;C\;$ s.t.$\;C(AB)=I\;$ , but using associativity of matrix multiplication:

$$I=C(AB)=(CA)B\implies B\;\;\text{is invertible and}\;\;CA=B^{-1}$$

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    $\begingroup$ this only shows that B has a left inverse. you cannot infer that B also has a right inverse. A has a right inverse but no left inverse. Only if A or B are square you can infer that. But this is neither stated by you nor by the OP. $\endgroup$ – Andreas H. Aug 4 '13 at 2:10
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    $\begingroup$ Unless otherwise stated I always assume $\;A,B\;$ are square. If the OP meant otherwise he can say. $\endgroup$ – DonAntonio Aug 4 '13 at 3:52
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    $\begingroup$ Especially since the theorem is false as stated if $A,B$ are not square. A non-surjective mapping composed with a non-injective mapping can be invertible- think of embedding $\mathbb R ^2$ into $\mathbb R ^3$ and then projecting back to $\mathbb R ^2$. The corresponding matrices will be $2 \times 3$ and $3 \times 2$ and thus not invertible, but the product will be $2 \times 2$ and invertible. $\endgroup$ – Devlin Mallory Aug 4 '13 at 9:14
  • $\begingroup$ Exactly @Devlin. I was about to remark that if not square a matrix can be left or right invertible yet, perhaps, not both. +1 $\endgroup$ – DonAntonio Aug 4 '13 at 9:16
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    $\begingroup$ Proving B also has a right inverse is easy if we assume that $A$ is also invertible. $$(AB)C=I \implies A(BC)=I \implies A^{-1}A(BC)=A^{-1} \implies BC=A^{-1} \implies BCA=A^{-1}A \implies B(CA)=I$$ $\endgroup$ – ViX28 Jun 5 '18 at 20:53
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A more basic argument, based on invertible $\iff$ nonsingular, is as follows. If $B$ were singular, there would be $x\ne 0$ with $Bx=0$, hence with $(AB)x=0$, whence $AB$ is likewise singular.

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Do you have the theorem that $X$ is invertible if and only if $\det X \neq 0$?

If so, then if $AB$ is invertible, $\det AB\ne 0$. But $\det AB = \det A\cdot \det B \ne 0$, so both $\det A$ and $\det B$ are nonzero, and therefore both $A$ and $B$ are invertible.

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  • $\begingroup$ Yes but in a later section. $\endgroup$ – user60887 Aug 3 '13 at 20:37
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    $\begingroup$ It should be $\det X \ne 0$. $\endgroup$ – Boris Novikov Aug 3 '13 at 20:47
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Thanking Ted Shifrin I completely revise my answer.

You want to prove that $B$ is invertible using only semigroup properties of matrices (i.e. multiplication, associativity and existence of the unity $I$).

You had proved that $B$ has a left inverse, $CB=I$. Now it should prove that there is a right inverse, $BD=I$. However, it is not true in semigroups, generally speaking. An example: so called bicyclic monoid $S=\langle a,b| ab=1\rangle$ (it is generated by $a,b$ with the defining relation $ab=1$). In $S$ the product $ab$ is invertible (since it equals $1$), $b$ has the left inverse $a$, but has not a right inverse.

Сonclusion: To prove what you want, it is not enough to use semigroup properties. One have to use either determinants or vectors (or something else).

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  • $\begingroup$ Analogously? What do you have in mind? $\endgroup$ – Ted Shifrin Aug 3 '13 at 21:02
  • $\begingroup$ @Ted Shifrin: Generally speaking, an element of a semigroup may have a left inverse and a right inverse, and they must not be equal. $\endgroup$ – Boris Novikov Aug 3 '13 at 21:10
  • $\begingroup$ You miss the point: How do you deduce that $B$ has a right inverse? $\endgroup$ – Ted Shifrin Aug 3 '13 at 21:11
  • $\begingroup$ @Ted Shifrin: You are right, thank you. I will edit the answer. $\endgroup$ – Boris Novikov Aug 3 '13 at 21:20
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This might be easier to do by thinking of $A$ and $B$ as linear operators rather than by matrix arithmetic. Assume $AB$ is invertible and assume that $B$ is not. Thus, $B$ either fails to be injective or fails to be surjective. If $B$ is not injective, then there is $x,y$ with $x \neq y$ such that $Bx=By$, and hence $ABx=ABy$, and $AB$ is not injective, which is a contradiction. Now, assume that $B$ is not surjective. Thus, its image must have dimension strictly less than the dimension of its domain. (Here is where we assume $A,B$ are square.) Thus, the composition $AB$ must have an image of dimension strictly less than the dimension of the domain of $B$, and thus cannot be surjective, another contradiction. Thus $B$ is injective and surjective and thus invertible.

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Take some arbitrary vector from the null space from $B$ and call this $\vec{w}$. Than $B\vec{w} = \vec{0}$ and so is $AB\vec{w} = \vec{0}$. We conclude that $\text{null}(B) \subseteq \text{null}(AB)$. Now since $AB$ is invertible we know that $\text{null}(AB) = \{ \vec{0} \}$, and therefore so is $\text{null}(B)$. Using the fundamental theorem of invertible matrices we conclude that $B$ is invertible as well.

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$n=\mathrm{rank}(AB)\le\mathrm{rank}(B)$, therefore $\mathrm{rank}(B)=n$.

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Let $A,B\in M_{n\times n}(F)$ and let $AB$ be invertible.

Suppose that $B$ is not invertible. The linear transformation $L_B:F^n\rightarrow F^n$, $L_B(x)=Bx$ is then not invertible as the matrix representation of $L_B$ is not invertible. However, since it is a linear transformation and the domain and codomain are each of the same finite dimension, it follows that $L_B$ is not injective and not surjective. As $L_B$ is not injective, there exists some $x\in F^n$ where $L_B(x)=Bx=0$ and $x\neq 0$.

However, $Bx=0$ means $$\left[(AB)^{-1}A\right]Bx=\left[(AB)^{-1}A\right]\cdot 0 \,\Rightarrow\, x=0$$ a contradiction. Therefore $B$ is invertible.

As $AB$ is invertible,

$$A\left[B(AB)^{-1}\right]=I_n.$$ Then $$(AB)^{-1}AB=I_n\,\Rightarrow\, (AB)^{-1}ABB^{-1}=B^{-1} \,\Rightarrow\, (AB)^{-1}A=B^{-1} \,\Rightarrow\, B(AB)^{-1}A=BB^{-1}$$ which means $$\left[B(AB)^{-1}\right]A=I_n$$ so $A$ is invertible.

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