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

  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!
  • $\begingroup$ 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. $\endgroup$ – Matthias Dec 10 '18 at 10:56
  • $\begingroup$ 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 $\Bbb C$. $\endgroup$ – Christoph Dec 10 '18 at 11:09
  • $\begingroup$ 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? $\endgroup$ – Matthias Dec 10 '18 at 11:20
  • $\begingroup$ 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 $\bar z$. $\endgroup$ – 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 $Au=Av=0$.

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

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