Consider a $n\times m$ matrix $A$ that satisfies $$ Ax=b $$ for $m\times 1$ vectors $x$ and $b$. Then $$ Ax-b=\mathbf{0} $$ What are the conditions on $A$ so that I can write the previous system as $$ A(x-c)=\mathbf{0} $$ for some vector $c$ and where $Ac=b$?
If $A$ was a square matrix, it would be enough to guarantee its invertibility, since we could take $c=A^{-1}b$. I wonder what can be said in the $n\times m$ case.
In order to have a vector $c$ with $Ac=b$, you need $b$ to be in the column space of $A$ (since an expression of the form $Ac$ is a linear combination of the columns of $A$.)