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PGRE | Eigenvalue Problem

  • #1

Main Question or Discussion Point

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

[itex]\lambda^{3}=1[/itex]

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[itex]^{2\pi i}[/itex]

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

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

[itex]\lambda^{3}=e^{2\pi i}[/itex]

to

[itex]\lambda_{n}=e^{\frac{2\pi i n}{3}}[/itex]


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

Answers and Replies

  • #2
vela
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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

[itex]\lambda^{3}=e^{2\pi i}[/itex]

to

[itex]\lambda_{n}=e^{\frac{2\pi i n}{3}}[/itex]


If [itex]\lambda^{3}=e^{2\pi i}[/itex] then we can take both sides to the power of [itex]\frac{1}{3}[/itex] to get [itex]\lambda=e^{\frac{2\pi i}{3}}[/itex]. But how can you just throw the n in the exponent and call these (n=1,2,3) the 3 eigenvalues?
It's because [itex]e^{2\pi n i}=1[/itex] for all integer n. When you take the cube root of that, you find you get three distinct solutions, and the rest are repeats.
 
  • #3
Ah yes, after using euler's formula it's much clearer now. Thanks Vela
 

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