User Name Remember Me? Password

 Linear Algebra Linear Algebra Math Forum

June 21st, 2019, 12:59 AM   #1
Newbie

Joined: Jun 2019
From: London

Posts: 13
Thanks: 0

eigenvalues

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 λ1, λ2, ..., λ5 be the proper numbers of the permutation of the order5. Find λ ∗ = min | λi |. With Gaussian elimination, i found that λ = 1. However there must be 5 eigenvalues and there must be complex values. Hw are they being calculated?
Attached Images Снимок экрана 2019-06-21 в 11.51.37.jpg (12.2 KB, 0 views) June 21st, 2019, 03:38 AM   #2
Math Team

Joined: May 2013
From: The Astral plane

Posts: 2,265
Thanks: 932

Math Focus: Wibbly wobbly timey-wimey stuff.
Quote:
 Originally Posted by mathodman25 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 λ1, λ2, ..., λ5 be the proper numbers of the permutation of the order5. Find λ ∗ = min | λi |. With Gaussian elimination, i found that λ = 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 Tags eigenvalues Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post matamat19 Linear Algebra 5 April 15th, 2019 03:51 PM Luiz Linear Algebra 2 September 24th, 2015 05:32 PM bonildo Linear Algebra 4 June 8th, 2012 07:02 PM MacLaurin Real Analysis 1 July 18th, 2009 07:44 PM alpacino Calculus 1 February 24th, 2009 02:10 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      