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?
$\endgroup$ 33 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$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).
$\endgroup$ 3 $\begingroup$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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