# PGRE | Eigenvalue Problem

## Homework Statement

The problem amounts to finding the eigenvalues of the matrix

|0 1 0|
|0 0 1|
|1 0 0|
(I have no idea how to set up a matrix in the latex format, if anyone can tell me that'd be great)

## Homework Equations

The characteristic equation for this matrix is

$\lambda^{3}=1$

## The Attempt at a Solution

The solution to this problem can be found on grephysics.net.
The characteristic equation can be solved by noting that
1=e$^{2\pi i}$

Using this fact, the eigenvalues as noted in the solution are
$\lambda_{n}=e^{\frac{2\pi i n}{3}}$, (n=1,2,3)

What I don't understand, is how one goes from

$\lambda^{3}=e^{2\pi i}$

to

$\lambda_{n}=e^{\frac{2\pi i n}{3}}$

If $\lambda^{3}=e^{2\pi i}$ then we can take both sides to the power of $\frac{1}{3}$ to get $\lambda=e^{\frac{2\pi i}{3}}$. But how can you just throw the n in the exponent and call these (n=1,2,3) the 3 eigenvalues?

vela
Staff Emeritus
Homework Helper

## Homework Statement

The problem amounts to finding the eigenvalues of the matrix

|0 1 0|
|0 0 1|
|1 0 0|
(I have no idea how to set up a matrix in the latex format, if anyone can tell me that'd be great)
Check out the brand new LaTeX FAQ!

What I don't understand, is how one goes from

$\lambda^{3}=e^{2\pi i}$

to

$\lambda_{n}=e^{\frac{2\pi i n}{3}}$

If $\lambda^{3}=e^{2\pi i}$ then we can take both sides to the power of $\frac{1}{3}$ to get $\lambda=e^{\frac{2\pi i}{3}}$. But how can you just throw the n in the exponent and call these (n=1,2,3) the 3 eigenvalues?
It's because $e^{2\pi n i}=1$ for all integer n. When you take the cube root of that, you find you get three distinct solutions, and the rest are repeats.

Ah yes, after using euler's formula it's much clearer now. Thanks Vela