18
$\begingroup$

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.

|cite|improve this question
$\endgroup$
  • 9
    $\begingroup$ You're assuming your conclusion when you write that $(AB)^{-1}=B^{-1}A^{1}$. $\endgroup$ – JSchlather Aug 3 '13 at 20:30
  • 5
    $\begingroup$ 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
  • 4
    $\begingroup$ Are $A,B$ square matrices or do we only assume that $AB$ is square? $\endgroup$ – Hagen von Eitzen Aug 3 '13 at 20:34
  • 1
    $\begingroup$ Assume $AB$ is square. $\endgroup$ – user60887 Aug 3 '13 at 20:35
  • 1
    $\begingroup$ It is not necessarily true that $(AB)^{-1}=B^{-1}A^{-1}$. $\endgroup$ – AnonSubmitter85 Aug 3 '13 at 20:35
39
$\begingroup$

$\;AB\;$ invertible $\;\implies \exists\;C\;$ s.t.$\;C(AB)=I\;$ , but using associativity of matrix multiplication:

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

|cite|improve this answer
$\endgroup$
  • 5
    $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • $\begingroup$ 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
11
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
7
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Yes but in a later section. $\endgroup$ – user60887 Aug 3 '13 at 20:37
  • 3
    $\begingroup$ It should be $\det X \ne 0$. $\endgroup$ – Boris Novikov Aug 3 '13 at 20:47
3
$\begingroup$

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).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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
3
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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.

|cite|improve this answer
$\endgroup$
1
$\begingroup$

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

|cite|improve this answer
$\endgroup$

protected by user99914 Nov 10 '15 at 12:19

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.