# 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:

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?

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!
• Thanks for this good answer. Regarding 2) you say "Taking the complex conjugates...", it is not clear to me why we can take them. Can you please explain this. – Matthias Dec 10 '18 at 10:56
• First split into $n$ complex equations $a_{i1} z_1 + \cdots + a_{in} z_n = 0$ and now those equalities in $\Bbb C$ can just be conjugated on both sides as usual to get $\overline{a_{i1}} \,\overline{z_1} + \cdots + \overline{a_{in}}\, \overline{z_n} = 0$. So this is just the fact that $w=0$ is equivalent to $\bar w=0$ in $\Bbb C$. – Christoph Dec 10 '18 at 11:09
• I think the technique is clear, but it is not clear to my why "... in $\mathbb{C}$ can just be conjugated...". So I'm not insistent for pedantic reasons, but I want to know your trains of thoughts. Is it like an automatism... here are complex numbers what can I do with them (e.g. dissociate imaginary from real parts, conjugate...) And now you know that in this particular situation the conjugation ist the technique to go with? – Matthias Dec 10 '18 at 11:20
• If $x=y$ and $f$ is any function, then $f(x)=f(y)$. Apply this to the situation where $x$ and $y$ are complex numbers and $f(x)=\bar x$. You get that $x=y$ implies $\bar y=\bar x$. Now since $\overline{\bar x}= x$, you can also go the other way. It is the obvious thing to do in this situation because you know $Az=0$ and you want to somehow get an equation involving $\bar z$. – Christoph Dec 10 '18 at 11:36

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 $$Au=Av=0$$.

• But you can't pick $a+bi$ such that for a given complex vector $u$ the scalar multiple $(a+bi)u$ is real, can you? (Try $u=(1,i)$) – Christoph Dec 10 '18 at 10:46
• @Christoph That was not really what I was trying to say. I will read the question and answer and check. – AnyAD Dec 10 '18 at 10:52
• I see, you wanted to produce a non-trivial complex solution so you assume $u$ is real in that part? – Christoph Dec 10 '18 at 10:53
• @Christoph Thank you for pointing that out. I've added 'real' in the answer. – AnyAD Dec 10 '18 at 10:55
• @AnyAD. Thanks for your answer. Why is it that $Au=-iAv \implies Au=Av=0$? – Matthias Dec 10 '18 at 11:23