Having a homogenous system of linear equations with real coefficients with a non-trivial complex solution than there is a real solution too. I'm struggeling a bit with this proof.

Suppose we have a homogenous system of linear equations with real coefficients with a non-trivial complex solution than there is a real solution too. 

This looks really simple and my first thought was...
In our course we proved that, if $\alpha,\beta \in L(A,0) \implies \alpha + \beta \in L(A,0)$. 
So let $z_1 = a+ib, z_2 = \bar{z_1} = a-ib$ both $\in L(A,0)$.
Than the sum of them $z_1+z_2 = a+ib + a-ib = 2a \in L(A,0)$. Since $a\in\mathbb{R}$ there is a real solution too. 
Additionally i figured out, that the complex conjugation is a field homorphism. 
My questions are:


*

*Can somebody show me an example for a homogenous system of linear equations with real coefficients with a non-trivial complex solution?

*if my idea is correct, why can I assume that $\bar{z_1}$ is a solution too?

*if my idea is not correct, where is my mistake?


Many thanks in advance
 A: *

*Let $z=a+ib\in\Bbb C^n$ with with $a,b\in\Bbb R^n$. When $Az=0$ for a real matrix $A$, you can split this into real and imaginary part to see that $Az=0$ is equivalent to $Aa=0$ and $Ab=0$. Hence, you can combine any two real solutions as $a+ib$ to obtain a (non-trivial) complex solution.

*Taking the complex conjugate of $Az=0$ gives $\bar A\bar z=0$ and for real $A$ this just gives $A\bar z=0$, hence $z$ is a solution if and only if $\bar z$ is a solution. (This also follows from 1. since $-b$ is a solution iff $b$ is one)

*It is correct!

A: The set of solutions of $Ax=0$ is a subspace (of the correspondinf vector space), so all the linear combinations of 'solution' vectors are again solutions of $Ax=0$. Fot $A $ real, we can consider the solution set/space either as a subspace (of the vector space we're working in) with scalars from $\mathbb{R}$ or $\mathbb{C}$. Considering it over the complex number field, again any scalar multiple of a real solution vector $u$ is a solution. (Clearly if $Au=0$ then  $A((a+ib)u)=0$.
Conversely if $A(u+iv)=0$ ($u,v $ real) then $Au=-iAv$ so $Au=Av=0$.
