First, $A$ and $B$ are square matrices

So to prove if $AB$ is invertible, then $A$ is invertible:

I let $C=(AB^{-1})B$

Then $CA=(AB^{-1})AB=I$

And $C=A^{-1}$ , so A is invertible.

But how do I prove it the other way around?

I'm pretty sure this would require that $B$ also be invertible for it to be true.

Is there a quick way to disprove that if $A$ is invertible then $AB$ is invertible?

  • 1
    $\begingroup$ To disprove something it would be enough to give a counterexample. $\endgroup$ – hardmath Dec 16 '16 at 11:49
  • 1
    $\begingroup$ For a trivial counterexample to "$A$ invertible $\Rightarrow$ $B$ invertible", take $B$ to be the zero matrix. $\endgroup$ – lokodiz Dec 16 '16 at 12:05
  • $\begingroup$ It looks like you mean $(AB)^{-1}$ above rather than $AB^{-1}$. $\endgroup$ – anomaly Aug 7 '17 at 18:40

The proof that if $A$ and $B$ are invertible, then $AB$ is invertible can be done more elegantly if you know these two results:

$(1)$. $\det{AB} = (\det (A))*(\det(B)).$
$(2)$. A matrix $B$ is invertible if and only if $\det(B) \neq 0$.

Proof: Suppose that both $A$ and $B$ are invertible. Then $\det(A) \neq 0$ and $\det(B) \neq 0$. Now by $(1)$, $\det(AB) \neq 0$, so by $(2)$, $AB$ is invertible.


I don't quite get your question since the claim is false. What if $B$ is the zero matrix?

Implication of $(AB)^{-1} \implies det(A) \neq 0$ is trivially true $$(AB)^{-1} = C = B^{-1}A^{-1} = C \implies A^{-1} = BC = B(AB)^{-1}$$

But the point is, the converse does not apply (if $A$ is invertible then $AB$ is invertible).

  • $\begingroup$ The question wanted a way to disprove that if A is invertible then AB is invertible, I'll edit the question now. $\endgroup$ – TweezerCube Dec 16 '16 at 15:15
  • $\begingroup$ And doesn't $B$ is the zero matrix disprove that? $\endgroup$ – q.Then Dec 16 '16 at 15:16
  • $\begingroup$ Yep, it does through counterexample $\endgroup$ – TweezerCube Dec 16 '16 at 15:18

Consider the square matrices as linear maps $\mathbb{R}^n \to \mathbb{R}^n$. These three conditions are equivalent:

  • $A$ is invertible,
  • $\operatorname{im} A = \mathbb{R}^n$,
  • $\operatorname{ker} A = 0$.

Since $\operatorname{im} AB \subseteq \operatorname{im} A$, we have $$ \operatorname{im} AB = \mathbb{R}^n \Rightarrow \operatorname{im} A = \mathbb{R}^n. $$ Due to the equivalent conditions mentioned above: $$ AB \text{ is invertible } \Rightarrow A \text{ is invertible}. $$

You could use the third condition to show that $B$ is also invertible.


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