Can I compare real and complex eigenvalues?

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?

Solution 1:

In general, when talking about "largest" eigenvalue, we are usually talking about largest in absolute value (or magnitude,) where $|a+bi|=\sqrt{a^2+b^2}$.

This means sometimes that there isn't one eigenvalue that is "largest", because two different eigenvalues can have the same absolute value.

As mentioned by others, complex numbers are not themselves ordered.

As mentioned in the comments below, if you know a matrix has only real eigenvalues, then the question of "largest" and "smallest" eigenvalues will depend on the context.

The "largest" eigenvalue for a matrix $A$ is often interesting, particularly when it is unique, because then for large $n$, $A^n$ is dominated by the action on the eigenvectors for those values. This is useful for putting bounds on $A^n\mathbf v$.

Solution 2:

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 $\mathbb{R}$.