True or false?
If for every $\mathbf{c} \in F^n$ there is a solution to the system $AB\mathbf{x}=\mathbf{c}$, then for every c $\in F^n$ there is a solution to the system $B\mathbf{x}=\mathbf{c}$. I intuitively understand that this is correct, but I am wondering if there is a formal way to prove it.
Thank you!
Edit: this solution is only applicable for square matrices. See Sean Lee's counter-example with non-square matrices above.
For a matrix $D$, the existence of a solution to $Dx=c$ for all $c\in F^n$ is equivalent to invertibility of $D$ (see this - this is actually only one direction but it is the one we want). In particular, here we get that $AB$ is invertible and by this, $B$ is invertible (and so is $A$).
Its not correct.
Take
$$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}, B = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} $$
Then
$$ AB = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} $$
Thus there will be a solution for the system $AB\bf{x} = \bf{c}, \ \forall \bf{c} \in F^n$ (the solution is just $ \bf{x} = \bf{c}$), however there will not necessarily be a solution for $B \bf{x} = \bf{c}$. For example, there is no solution for the following system: $$ B\bf{x} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \bf{x} = \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} $$
