# Jordan Normal Form of $A_{\pi}: \mathbb{C}^n \to \mathbb{C}^n$ given by $A_{\pi}(v) = A_{\pi}(v_1,...,v_n) = (v_{\pi(1)},...,v_{\pi(n)})$.

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.

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 $$t=0,1,\ldots,l_j-1$$.
• 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?
• 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. Sep 24 '19 at 16:09