If $AB$ invertible $\implies$ $A,B$ invertible, Given that $A,B\in M_{n\times n}(F)$

Question

I know how to prove if $A,B$ invertible then $AB$ is invertible and $(AB)^{-1}=B^{-1}A^{-1}.$ But I'm not sure about how to prove the converse.

My attempt:

If $AB$ is invertible, then $B$ must be one-one and $A$ must be onto. Since $\mathcal L(V,W)\sim M_{n\times n}(F)$, $L_A:V\to W$, $L_B:V\to W$ and $\dim(V)=\dim(W)$ and they're finite dimension and linear, so one-one means onto and onto means one-one, so $A,B$ are invertible.

Is this correct and/or how to improve it?

Answer 1

Your proof is fine!

If one can prove $B$ is one-one and $A$ is onto, then the result follows from this fact:(as you mentioned)

Suppose $V$ is finite dimensional vector space. If $T \in L(V)$, then the following are equivalent:

(a) $T$ is invertible

(b) $T$ is one-one

(c) $T$ is onto

Suppose $AB$ is invertible, then $\exists C \in L(\Bbb{F}^n)$ so that $$(AB)C=C(AB)=I_n$$

Let $v\in F^n$ so that $Bv=0$.

$v=Iv=(CAB)v=CA(Bv)=CA(0)=0$,concluding $\text{null}\; B=\{0\}$ and so $B$ is invertible

Likewise, let $v \in F^n$ be arbitrary.

$v=Iv=(ABC)v=(AB)Cv=A(BCv)$, concluding $A$ is onto