Given a linear system $Ax=b$, for $A\in\mathbb{R}^{n\times n}$, does the exact solution $x$ exist if $b~\bot~ Ker(A^T)$, ie. vector $b$ is orthogonal to each vector of the null-space of $A^T$?
If one states that $b\in im(A)$, does that mean that the linear combination of the columns of $A$ may yield $b$. In such case, a solution to $Ax=b$ exists iff $b\in im(A)$.
So, if $b\in im(A)$, then $b~\bot~Ker(A^T)$. Could one state the opposite direction? I consider $A$ general square matrix; what would happen if $A$ is symmetric?