What is the eigenvalue problem for the given matrix and how can it be solved?

[CanIExplore]

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]

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!

https://www.physicsforums.com/showthread.php?t=546968

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.

 

[CanIExplore]

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