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

Question

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:

1. Can somebody show me an example for a homogenous system of linear equations with real coefficients with a non-trivial complex solution?
2. if my idea is correct, why can I assume that $\bar{z_1}$ is a solution too?
3. if my idea is not correct, where is my mistake?

Many thanks in advance

Answer 1

1. 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.
2. 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)
3. It is correct!