TITLE: Can I compare real and complex eigenvalues?
QUESTION [21 upvotes]: I'm calculating the eigenvalues of the matrix    $\begin{pmatrix} 2 &0 &0& 1\\
     0 &1& 0& 1\\
     0 &0& 3& 1\\
    -1 &0 &0 &1\end{pmatrix}$,
which are $1$,$3$, $\frac{3}{2}+\sqrt{3}i$ and $\frac{3}{2}-\sqrt{3}i$.
I wish to recognize the biggest and smallest of these. But how can I compare real and complex numbers?

REPLY [3 votes]: Suppose that there was an order on $\mathbb{C}$ compatible with the natural order on $\mathbb{R}$. Then either $i>0$ or $i<0$. Assume that $i>0$, then multiplying this inequality by $i$ we find that $-1=i^2>0$, but that's a contradiction. Thus $i<0$. But then multiplying by $i$ we get $-1=i^2>0$ (the inequality flipped since we multiplied by a negative number). In both cases we arrive at a contradiction. Hence there's no order on $\mathbb{C}$ compatible with $\mathbb{R}$.