Eigenvalues of a Permutation? I'm stuck on the following problem for my Linear Algebra class:
Let $\pi:\{1, \ldots, n \} \rightarrow \{1, \ldots, n\}$ be a bijective map (permutation). Let $f_{\pi}:\mathbb{R}^n \rightarrow \mathbb{R}^n$ be defined by $f_{\pi}(x_1, \ldots, x_n) := (x_{\pi(1)},\ldots,x_{\pi(n)})$. Determine the set of eigenvalues of $f_{\pi}$.
I know that in general for a square matrix $A$, the set of real eigenvalues of $A$ is given by $\{\lambda \in \mathbb{R}\mid \det(A- \lambda I) = 0\}.$
I've figured out that the matrix representation of $f_{\pi}$ with respect to the standard ordered basis will have rows that contain all $0$s except for a $1$ in one column; this column will be different for each row. However, I can't figure out how to get the eigenvalues from there. Please advise.
 A: To elaborate a bit on Igor Rivin's answer:
Cycle decompositions work as follows.  Suppose you have:
$$
\begin{array}{c} 1 \mapsto 4 \\ 2\mapsto6 \\ 3\mapsto1 \\ 4\mapsto3 \\ 5\mapsto5 \\ 6\mapsto2 \end{array}$$
Then you have three cycles:
$$
\begin{array}{c}
1\mapsto4\mapsto3\mapsto1 \\[10pt]
2\mapsto 6\mapsto2 \\[10pt]
5\mapsto5
\end{array}
$$
They're "cycles" because the return to their starting points.  The cycle decomposition is just the list of cycles.  The cycle decomposition would usually be written something like this: $(1,4,3),\  (2,6),\ (5)$.
Now consider the vector $(a,0,a,a,0,0)$.  The vectors in the 1st, 4th, and 3rd positions are the same, so this vector would be mapped to itself.  It is thus an eigenvector with eigenvalue $1$.  You get one of those for each cycle.
Now consider
$$
(a,0,b,c,0,0)\mapsto (b,0,c,a,0,0)\mapsto (c,0,a,b,0,0)\mapsto(a,0,b,c,0,0),
$$
returning to the starting point.  This is a $120^\circ$ rotation.  It therefore has an eigenvalue whose cube is $1$, thus $\cos120^\circ+i\sin120^\circ= e^{i\pi/3}$.
Think about $\left(1,0,e^{-i2\pi/3},e^{-i\pi/3},0,0\right)$, and think about why I wrote the components in that order.  There's also a reason why I put a minus sign there, which I missed the first time.
A: Hint: use the cycle decomposition of your permutation. 
EDIT See Michael Hardy's answer for a careful explanation of this method.
A: For an analysis of the case over $\mathbb{C}$, see Michael Hardy's answer.
Hint: compare $\|x\|$ to $\|f_{\pi}(x)\|$.  What does this tell you about the eigenvalues of $f_\pi$?

Example: take $f(x,y) = (y,x)$. We note that $f(1,1) = (1,1)$, corresponding to an eigenvalue of $1$, and $f(1,-1) = -(1,-1)$, corresponding to an eigenvalue of $-1$.
