Prove that if a product$ AB$ of $n\times n$ matrices is invertible, so are the factors $A$ and $B$. I'm really struggling with "simple" proofs, and I could only come up with this. I googled for the solution and know how to do it now, but I first want to know if there's anything wrong with my solution?
$$
(AB)^{-1}=B^{-1}*A^{-1}.
$$ For this equation to hold, $B$ must be invertible and so must $A$. //
 A: First: your solution isn't actually a solution. By writing $A^{-1}$ you are assuming that $A$ is invertible, which is what you want to prove.
One way to do this would be to note that
$$
\det(AB) = \det(A)\det(B).
$$
If $AB$ is invertible, then $\det(AB)\neq 0$, and so $\det(A)\neq 0$ and $\det(B)\neq 0$.
Another way: If $AB$ is invertible, then there is a $C$ such that 
$$
(AB)C = I
$$
That is
$$
A(BC) = I 
$$
So $BC$ is a right inverse of $A$, and so $A$ is invertible. Likewise $B$ is invertible. (See this question/answers for a bit more: If $AB = I$ then $BA = I$).
A: Suppose $AB$ is invertible. Then there is a matrix $C$ such that $C(AB)=I$ and $(AB)C=I$. Rearranging the parenthesis, we have $(CA)B=I$ and $A(BC)=I$, which shows $A$ and $B$ are invertible. 
A: What makes this proof tick is the following: If $T:V\to W$ is a linear transformation between finite dimensional vector spaces having the same dimension, then $T$ is injective iff $T$ is surjective iff $T$ is bijective. (This result follows immediately from the rank-nullity equation). 
Thinking of matrices as linear transformations, the claim is equivalent to: Given finite dimensional vector spaces $U,V,W$, all having the same dimension, and linear transformations $T:U\to V$ and $S:V\to W$, prove that if $S\circ T:U\to W$ is invertible then both $T$ and $S$ are invertible. Here invertible simply means that the function is injective and surjective. 
As $S\circ T$ is injective it follows immediately that $T$ is injective. As $S\circ T$ is surjective it follows immediately that $T$ is surjective. The result now follows from the above mentioned result. 
A: I would like to offer another perspective for how to go about proving this which might provide an intuition that perhaps the other answers might lack for some.
Given that $A\in R^{n\times n}$ and $B\in R^{n\times n}$, for $AB$ to be invertible,
$rank(AB) = n$ must be true by definition of an invertible matrix.
We know that this inequality is always true:
$rank(AB) \leq \min (rank(A), rank(B))$
this implies that:
$n \leq \min(rank(A), rank(B))$
Therefore, it is self evident that the rank of an $n\times n$ matrix cannot exceed $n$, so we must have the equality that 
$n = min(rank(A), rank(B))$ which is the same as saying 
$rank(A) = rank(B) = n$
This implies that the matrices $A$ and $B$ are full rank. All full rank matrices are invertible. Hence, $A$ and $B$ must be invertible.
Now that we know that $A$ and $B$ are invertible, suppose that $C = AB$. Then:
$
AB = C\\
A = CB^{-1}\\
B^{-1} = C^{-1}A= (AB)^{-1}A\\\\
B^{-1}A^{-1} = (AB)^{-1}
$
I do hope this was useful. If there are any questions or concerns, please let me know. I will leave it as an excerse for you to prove that $rank(AB) \leq min(rank(A), rank(B))$.
