Cokernel compatibility equations ($Ax=b\text{ solvable}⟺y^{\ast }b=0\forall y\in \ker A^{\ast }$)

For a matrix $A$ over the complex numbers with the standard Hermitian inner product, the linear system $Ax=b$ is solvable exactly whenThe star is the conjugate transpose; for matrices over the real numbers, use the transpose. Necessity follows from $y^{\ast }Ax=(A^{\ast }y{)}^{\ast }x=0$. For sufficiency, $(\mathrm{im}A{)}^{\perp }=\ker A^{\ast }$, and taking the orthogonal complement again recovers $\mathrm{im}A$ in finite dimension. This is the cokernel obstruction expressed as scalar equations. ^[ref]

For $A=\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right)$, $\ker A^T=\mathrm{span}\{(0,1)\}$, so the single compatibility equation is $b_2=0$. If it holds, the kernel supplies the remaining input freedom; if it fails, changing the input cannot remove the obstruction.