Prove that if $AB$ is invertible then $B$ is invertible.

Question

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.

Answer 1

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

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

Answer 2

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.

Answer 3

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).

Answer 4

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.

Answer 5

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.