Let $A$ be an $n$ by $n$ real matrix, and $b$ an $n$ by $1$ column vector with real entries. Prove that there exists an $n$ by $1$ column vector solution $x$ to the equation $Ax=b$, if and only if $b$ is in the orthocomplement of the kernel of the transpose of $A$.
So my idea was to first assume that there is such an $x$, so I want to show that $\langle b,y\rangle=0$ for all $y\in \mbox{ker}A^t$, but I dont know how to tackle it.
This is not homework nor anything, so I dont mind if you spoil it.