# 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?

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 $A^n\mathbf v$.

• One needs to be careful with one issue: when talking about the smallest eigenvalue of a Hermitian matrix, this may either mean the “farthest down” value (e.g. with the ground-state energy of a QM Hamiltonian – you can always gauge this to be positive, but it's quite common to have all energies negative) or indeed the smallest-absolute value. – leftaroundabout Apr 19 '16 at 14:39

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 $\mathbb{R}$.

• You completely missed the point of the question. – Najib Idrissi Apr 22 '16 at 11:26
• Yes and no, Thomas Andrews' answer already answers the point of the question, I merely wanted to give this simple argument why no natural order on $\mathbb{C}$ exists that extends the order of $\mathbb{R}$. – Mathematician 42 Apr 22 '16 at 12:54

You can't, in the sense that there is no natural total order in $\mathbb{C}$.

• But I'm asked to do it. This is an assignment. – mavavilj Apr 19 '16 at 13:55
• Well is $(0,1)$ or $(1,0)$ larger? You need to boil things down to real numbers in some way to compare complex values. Either you have miscalculated the eigenvalues, or you have misunderstood the question. – Matias Heikkilä Apr 19 '16 at 13:56
• There are total orders on $\mathbb{C}$ (e.g. the lexicographic order, for which $1 > i > -i > -1$). I don't know what you mean by "natural", though it is true that $\mathbb{C}$ cannot be made into an ordered field. Anyway, in the context of eigenvalues it's the absolute value that counts, as Thomas Andrews says. – Najib Idrissi Apr 19 '16 at 14:01
• Yes there are total orders on $\mathbb{C}$, but which one you would use if you were asked to compare $a \in \mathbb{C}$ and $b \in \mathbb{C}$? Could you be sure that your friend uses the same order if given the same task? If $a, b \in \mathbb{R}$ instead, we tend to use the usual order relation, and there would be no problem. That's what I mean by natural. But this is not very interesting, since the original question should have been read as "compare the absolute values". Good to know that convention in the future, though. – Matias Heikkilä Apr 20 '16 at 9:02