How to prove that $AB$ is invertible if and only if $A$ is invertible? Let $A$ be a matrix and $B$ an invertible matrix. Show that $AB$ is invertible if and only if $A$ is invertible.
I know how to do this using determinants, but how else could you prove this?

Let $A$ and $B$ be $n\times n$ matrices. $B$ is invertible, so its rank is $n$.
If $A$ is invertible, then its rank is $n$. Then the rank of $AB$ is also $n$. Therefore $AB$ is invertible.
If $AB$ is invertible, then its rank is $n$. The rank of $A=ABB^{-1}$ is therefore also $n$ (the identity matrix has full rank).
We use the fact that multiplying two $n\times n$ matrices that have full rank results in a full rank $n\times n$ matrix. That is so, because each matrix will map $n$ linearly independent vectors onto another set of $n$ linearly independent vectors.
