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Let $\pi \in S_n$ be a permutation. Prove that $A_{\pi}: \mathbb{C}^n \to \mathbb{C}^n$ given by $A_{\pi}(v) = A_{\pi}(v_1,...,v_n) = (v_{\pi(1)},...,v_{\pi(n)})$. Show that $A_{\pi}$ is linear and find the Jordan Normal Form of $A$.

Is easy to show that $A_{\pi}$ is linear. I'm having problems with the last part. I know that to find this matrix, its enough to calculate $A_{\pi}(e_i)$ where $\{e_1,...,e_n\}$ is a basis of $\mathbb{C}^n$. I know that $\{e_1,ie_1,...,e_n,ie_n\}$ is a basis of $\mathbb{C}^n$ where $\{e_1,...,e_n\}$ is a canonical basis of $\mathbb{R}^n$. But since the image of $A_{\pi}$ is write as $n$-coordinate vector and $\pi$ can be any permutation, I'm having problems to calculate the associated matrix. I would like some help.

Assume that $\pi$ change the first and second coordinates. So

$$A_{\pi}(e_1) = (0,1,0,...,0) \longleftrightarrow e_2$$ $$A_{\pi}(e_2) = (1,0,0,...,0) \longleftrightarrow e_1$$ $$A_{\pi}(ie_1) \longleftrightarrow ie_2$$ $$A_{\pi}(ie_2) \longleftrightarrow ie_1$$

So, I suspect that in general $A_{\pi}(e_j) = e_{\pi(j)}$ and $A_{\pi}(ie_j) = ie_{\pi(j)}$. Then, probably the matrix associate to $A_{\pi}$ change the columns, but I'm not sure how to write this as $n \times n$ complex matrix.

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Let $e_1,e_2,\ldots,e_n\in\Bbb C^n$ be the standard basis vectors. Write $\pi$ as a product of disjoint cycles $\sigma_1,\sigma_2,\ldots,\sigma_k$ of lengths $l_1,l_2,\ldots,l_k$ (so $l_1+l_2+\ldots+l_k=n$). For $\sigma_j$ suppose $\sigma_j=(m_j^1\ \ m_j^2\ \ \ldots \ \ m_j^{l_j})$. Show that $W_j=\operatorname{span} \left\{e_{m^r_j}:r=1,\ldots,l_j\right\}$ is an invariant subspace of $A_\pi$, and $A_\pi\Big|_{W_j}:W_j\to W_j$ is diagonalizable with eigenvalues $e^{2\pi i t/l_j}$ for $t=0,1,\ldots,l_j-1$. Since $\Bbb C^n=\bigoplus_{j=1}^nW_j$ we conclude that $A_\pi$ is diagonalizable with eigenvalues $e^{2\pi i t/l_j}$ where $j=1,2,\ldots,k$ and $t=0,1,\ldots,l_j-1$.

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  • $\begingroup$ I have some questions about your answer. I got the idea, but to show that $A_{\pi}|_{W_j}$ is diagonalizable I need to find the matrix of $A_{\pi}|_{W_j}$ for get the characteristical polynomial, or is there other way? $\endgroup$
    – Greg
    Sep 24 '19 at 16:03
  • $\begingroup$ The matrix of $A_\pi\Big|_{W_j}$ with respect to the ordered basis $e_{m_j^1},e_{m_j^2},\ldots,e_{m_j^{l_j}}$ is simple: it is the $l_j\times l_j$-matrix $$\begin{pmatrix} 0&0&0&\dots&0&1\\ 1&0&0&\dots&0&0\\ 0&1&0 &\dots&0\\ \vdots&\vdots&\vdots&&\vdots&\vdots\\ 0&0&0&\dots&1&0\end{pmatrix}.$$ This is a circulant matrix. The wiki page on circulant matrices gives you the information about its eigenvalues and eigenvectors. $\endgroup$ Sep 24 '19 at 16:09

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