TITLE: Any complex number can be the eigenvalue of some non-negative matrix
QUESTION [8 upvotes]: Let $z\in\Bbb C$. Show that there exists a non-negative matrix $A$ (with entries $\geq 0$) such that $z$ is an eigenvalue of $A$.
If $z$ is real, it is easy.
Since, $a\geq 0$ is an eigenvalue of 
$$\begin{pmatrix}
a&0\\
0&a
\end{pmatrix};$$
while $a<0$ is an eigenvalue of 
$$\begin{pmatrix}
0&-a\\
-a&0
\end{pmatrix}.$$
For complex $z$, we should need the rows of $A$ is greater than $3$...Con we construct it? Or could we prove the statement above by using some facts of non-negative matrices...

REPLY [3 votes]: As Byron Schmuland has pointed out in another thread, every point $a+ib$ inside the closed equilateral triangle with corners $1,\omega,\omega^2$ (where $\omega$ is a cube root of unity) can be realised as the eigenvalue of a doubly stochastic matrix of the form
$$
P=\begin{bmatrix}1-s-t&s&t\\ t&1-s-t&s\\ s&t&1-s-t \end{bmatrix},
$$
where $s=\frac{1-a}3+\frac{b}{\sqrt{3}}$ and $t=\frac{1-a}3-\frac{b}{\sqrt{3}}$.
The constraint that $a+ib$ lies inside the triangle formed by $1,\omega,\omega^2$ would make $s,t\ge0$ and $s+t\le1$. (Note that $P$ is also a circulant matrix, so that its whole spectrum can be expressed explicitly in terms of the coefficients on the first row.)
Consequently, every complex number $z$ is the eigenvalue of some nonnegative multiple of a doubly stochastic matrix.