If a system $Ax = b$ has unique solution $x\in\mathbb{R}^{n}$, why is $x$ also the unique solution in $\mathbb{C}^{n}$? Trying to prove this but running into roadblocks; why do we know that since it is unique in the reals, that it is also a unique solution in the complex numbers?
 A: Say $A z = A (u+iv) = A u + i A v = b$. We conclude $Au = b$ and $A v = 0$. Now, necessarily $v = 0$,  otherwise we have at least two solutions $x$ and $x+v$. Also we must have $u=x$, since $u$ is a solution from above. We get $z = u + i v= x + i 0 = x$.
$\bf{Added:}$ We use that the matrix $A$ has real entries. Here is a counterexample with $A \in M_2(\mathbb{C})$.
The system
$$\left (\begin{matrix}
    1 & i \\
    i & -1 
    \end{matrix}\right) \cdot \left (\begin{matrix}
    x_1 \\
    x_2 
    \end{matrix}\right)= \left (\begin{matrix}
    0 \\
    0 
    \end{matrix}\right) $$
has a unique real solution $x=0$, but infinitely many complex solution $t\cdot (1,i)$.
A: Let us first assume $b=0$. In this case, the unique solution is the trivial solution $x=0$. By the invertible matrix theorem, $A$ is invertible in $M_n(\mathbb{R})$ so there exists an inverse $A^{-1} \in M_n(\mathbb{R}) \subset M_n(\mathbb{C})$. Therefore we have an inverse in the real numbers. Hope this helps and you can perhaps generalize this sort of argument.
