# PGRE | Eigenvalue Problem

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1. Nov 3, 2011

### CanIExplore

1. The problem statement, all variables and given/known data
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)

2. Relevant equations
The characteristic equation for this matrix is

$\lambda^{3}=1$

3. 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?

2. Nov 3, 2011

### vela

Staff Emeritus
Check out the brand new LaTeX FAQ!

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.