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Given a directed graph $G$ with outdegree $k$ for all vertices and no directed 2-cycles, show that $G$ must have some nonreal eigenvalue.

I can show it for $k=1$, where it is easy to explicitly calculate the eigenvalue. But already for $k=2$, I guess some other idea is required.

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    how do you do it for $k=1$? – 2017-01-30
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    Maybe we can prove something about the characteristic polynomial? – 2017-01-30
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    for $k=1$ the characteristic polynomial has the form $x^a(x^b\pm 1)$ with $b>2$ so it must have non-real roots. – 2017-01-30
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    Yes, basically for $k=1$, each connected component of the graph is a cycle plus some branches hanging off it. And then the eigenvalues will be the $\ell^{th}$ roots of unity where $\ell$ is the length of the cycle. For $k>1$, I was thinking of decomposing the graph into $k$ graphs each with $k=1$, but I don't see how of different graphs the eigenvalues will "compose" when we do this. – 2017-01-30

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