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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.

Michael Hardy
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justin
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3 Answers3

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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.

Michael Hardy
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    When you wrote $(c,0,a,b)$, did you mean to write $(c,0,a,b,0,0)$? Also, how did you conclude that the rotation is $120^{\circ}$? – justin Dec 08 '13 at 21:59
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    Typo fixed. If it's a rotation at all, it has to be $120^\circ$ because it returns to the starting point after exactly three iterations. – Michael Hardy Dec 08 '13 at 23:06
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Hint: use the cycle decomposition of your permutation.

EDIT See Michael Hardy's answer for a careful explanation of this method.

Igor Rivin
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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 $-1$.

Ben Grossmann
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  • I think the norm of each is the same....and we're just working over the reals...so since $|x| = | f_{\pi}(x)|$, we can conclude that if $x$ is an eigenvector, then the corresponding eigenvalue is $1$ or $-1$? – justin Dec 08 '13 at 21:56
  • Yeah, you've got it. From there, you'd need example(s) to show that it can be both $1$ and $-1$. See my latest update. – Ben Grossmann Dec 08 '13 at 22:00
  • Alright, that sounds good! – justin Dec 08 '13 at 22:01
  • The eigenvalue is a complex number whose norm is $1$. If it's real then it's $1$ or $-1$, but it might not be real. In that case, it's $e^{i\theta}$ for some real $\theta$, since those are the only complex numbers whose norm is $1$. – Michael Hardy Dec 08 '13 at 23:53
  • That depends on whether your definition of eigenvalues includes complex numbers. From the phrasing of the question, it seemed that OP was only looking for real eigenvalues. Your explanation of how to find the set of complex eigenvalues is enlightening, but seems beyond the demands of an introductory linear algebra course. – Ben Grossmann Dec 09 '13 at 00:29
  • @MichaelHardy I am also aware that in the complex case, the restriction $|\lambda| = 1$ that my observation produces is a necessary but not sufficient condition on $\lambda$ to be a potential eigenvalue since, as your answer demonstrates, we must have $\lambda = e^{\theta i}$ where $\theta$ is some rational multiple of $\pi$. – Ben Grossmann Dec 09 '13 at 00:32
  • @justin regarding your suggested edit: if we consider all the real eigenvalues of permutations for a fixed $n \geq 2$: all permutations will have an eigenvalue of $1$, but some will also have an eigenvalue of $-1$. This is true whether or not $n$ is odd or even. – Ben Grossmann Dec 11 '13 at 02:36
  • For a more precise answer as to which permutations have $-1$ as an eigenvalue, we'll have to follow Michael Hardy's reasoning. A permutation will have eigenvalue $-1$ if and only if, when that permutation is decomposed into disjoint cycles, one of those cycles has even length. – Ben Grossmann Dec 11 '13 at 02:38
  • @justin also, feel free to unaccept my answer and accept Michael's, if you feel that his answer is more deserving. He was certainly more thorough. – Ben Grossmann Dec 11 '13 at 03:09
  • @Omnomnomnom okay, thanks for letting me know. I wasn't sure what the right etiquette would be. – justin Dec 11 '13 at 03:45