A permutation matrix of order n is a matrix of size n X n, composed of 0 and 1, that the sum (in the field of real numbers) of elements for each of its columns and each row is equal to 1. Let Î»1, Î»2, ..., Î»5 be the proper numbers of the permutation of the order5. Find Î» âˆ— = min | Î»i |. With Gaussian elimination, i found that Î» = 1. However there must be 5 eigenvalues and there must be complex values. Hw are they being calculated?

The eigenvalue equation is \(\displaystyle \lambda ^5 - 1 = 0\). This immediately gives a real root of 1. To get the others you can do it the hard way:

\(\displaystyle \lambda ^5 - 1 = ( \lambda - 1)(\lambda ^4 + \lambda ^3 + \lambda ^2 + \lambda + 1) = 0\)

We would have to solve the equation \(\displaystyle \lambda ^4 + \lambda ^3 + \lambda ^2 + \lambda + 1 = 0\), which is a quartic equation. It looks ugly and it is ugly. There's a way to do it but I don't recommend it, myself. I've never been able to do one yet.

So let's go to the geometry. In this case (the 5th roots of unity) the roots start at \(\displaystyle \lambda _0 = 1\) and progress in a circle in the complex plane, at evenly spaced intervals over the unit circle.

So we have 4 more roots: \(\displaystyle \lambda _1 = e^{i (1 \cdot 2 \pi /5 )}\), \(\displaystyle \lambda _2 = e^{i (2 \cdot 2\pi /5)}\), etc. Or more compactly:

\(\displaystyle \lambda _n = e^{i (n \cdot 2 \pi/ 5 )}\) where n = 1, 2, 3, 4.

This can be a handy trick to have in your toolbelt.

-Dan