Consider this problem:

Let $\pi$ be a permutation of integers $\{1,2,...,n\}$. If $x = (x_1, x_2, ...,x_n)$ is a vector in $\mathbb{C}^n$, write $Ax = (x_{\pi(1)},x_{\pi(2)},...,x_{\pi(n)})$. Find the spectrum of A.

My thoughts:

1) Since $A$ is a permutation of $x$, one of these permutations must be the identity permutation, for which the eigenvalue should be $1$.

2) The only other eigenvalue should be $0$?? Since there shouldn't exist another $\lambda$ for which $Ax = \lambda x$ since the direction of $x$ isn't preserve in all other permutations?

Are my thoughts correct? What is the spectrum of $A$? And what would $A$ look like?


$0$ is never an eigenvalue because the matrix is invertible (because by switching columns and rows you eventually get the identity matrix, so that the dterminant of the matrix is $\pm 1$).

The permutation $\pi$ can be written as a product of disjoint cycles, $\pi=\gamma_1\gamma_2\ldots \gamma_r$, where each $\gamma_k$ is a cycle. Denote by $\Gamma_k$ the support of $\gamma_k$ and $g_k=|\Gamma_k|$. Let $\Delta=\lbrace 1,2,\ldots, n \rbrace \setminus \bigcup_k \Gamma_k$, and consider the subspaces $E_k={\mathsf{span}}(e_i|i\in \Gamma_k)$, $F={\mathsf{span}}(e_i|i\in \Delta)$. Then all those subspaces are invariant by $A$, and the total space is the direct sum of them.

On $F$, $A$ acts as the identity (with eigenvalue $1$). On $E_k$, $A$ acts a linear operator with characteristic polynomial $X^{g_k}-1$, its eigenvalues are the $g_k$-th roots of unity.

  • $\begingroup$ I guess you meant to write $\Gamma_k$ instead of $S_k$ and $F$ instead of $\Delta$? $\endgroup$ – levap Nov 5 '17 at 9:19
  • $\begingroup$ @levap Corrected, thanks. $\endgroup$ – Ewan Delanoy Nov 5 '17 at 9:36
  • $\begingroup$ I'm sorry my understanding of permutation is limited, but what is $\Gamma_k$ the "support"? $\endgroup$ – B.Li Nov 5 '17 at 22:08
  • 1
    $\begingroup$ @B.Li The support of $\gamma$ is the set of all values not fixed by $\gamma$ : $\lbrace i | \gamma(i)\neq i\rbrace$ $\endgroup$ – Ewan Delanoy Nov 5 '17 at 22:21

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