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We can define a binary relation on the complex numbers as follows.

  • $z \leq z'$ iff there exists $x \in [0,\infty)$ with $z+x=z'.$

Equivalently,

  • $z \leq z'$ iff $\mathrm{Re}(z) \leq \mathrm{Re}(z')$ and $\mathrm{Im}(z) = \mathrm{Im}(z')$.

It is straightforward to show that $\leq$ is a partial order on $\mathbb{C}$, extending the usual (total) order on $\mathbb{R},$ and that the following holds for all $z,z' \in \mathbb{C}.$

If $z \leq z',$ then:

  1. Given $w \in \mathbb{C}$, we have $w+z \leq w+z'$.
  2. Given $r \in [0,\infty)$, we have $rz \leq rz'.$
  3. $-z' \leq -z$.
  4. $\bar{z} \leq \bar{z'}$

Despite these formulae, I find it hard to believe that the relation $\leq$ could actually be useful. There just aren't enough points comparable to each other.

Conjecture. The partial order $\leq$ on $\mathbb{C}$ lacks a (non-trivial) use or application.

Question. Does anyone know of a counterexample to this "conjecture"?

goblin GONE
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  • this seems like a solid conjecture. – Ittay Weiss Mar 19 '14 at 10:13
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    I think applications of this partial order exist but we just simply don't refer to it by defining a partial order because it's much easier to say "$\mbox{Im}(z)=\mbox{Im}(z')$ and $\mbox{Re}(z)\leq\mbox{Re}(z')$ therefore..." – Dan Rust Mar 19 '14 at 10:43
  • @DanielRust, I'd be interested to hear where that condition shows up. – goblin GONE Mar 19 '14 at 12:35
  • I think you'll find a lot of applications in hyperbolic geometry where the geodesics in the upper half plane which have one point at infinity are precisely the intersection of the upper half plane with the connected components of the above defined partial order (in the induced topology). – Dan Rust Mar 19 '14 at 12:44
  • @DanielRust okay, thanks. – goblin GONE Mar 19 '14 at 12:45

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For a non-negative matrix, i.e. a matrix with all entries greater or equal to zero, the set of eigenvalues has a very special form. This is explained in the Perron-Frobenius theorem:

http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem

One consequence of this theorem is e.g.: Let $A$ be an irreducible non-negative $n \times n$ matrix with period $h$ and spectral radius $\rho(A) = r$, then $A$ has exactly $h$ (where $h$ is the period) complex eigenvalues with absolute value $r$. Each of them is a simple root of the characteristic polynomial and is the product of $r$ with an $h$th root of unity.

For $\rho(A)=1$, this means that you can decompose the vector space in a part where the action of the matrix tends to zero and a part where the action of the matrix is periodic.

This classical result of Perron and Frobenius was subsequently generalized to infinite dimensional systems (ordered topological vector spaces, Banach lattices). A standard reference is:

H.H. Schaefer, Banach lattices and positive operators , Springer (1974).

Whenever you need eigenvalues (spectral theory) you usually work over $\mathbb{C}$ (as opposed to $\mathbb{R})$ equipped with the order you describe. Non-negative matrices (operators) then have a "nice" set of eigenvalues.

Uwe Stroinski
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