Can a linear, inhomogeneous system of equations with real coefficients have a solution over the field of complex numbers but not over the field of real numbers?
My initial thought is "no" because if you can't find a solution to Ax=b, then that means A has no left inverse, real or complex. Also, I know that if the system is homogeneous and has a complex solution, then it must have a corresponding real solution. The problem is, I don't have any ideas for a rigorous way to prove this. Any help would be welcome!
If $A$ and $b$ are real and $A z = b$ where $z$ is complex, then $$A \overline{z} = \overline{A z} = \overline{b} = b$$ and $(z + \overline{z})/2$ is real with $A (z + \overline{z})/2 = b$.