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 February 7th, 2019, 09:46 AM #1 Newbie   Joined: Feb 2019 From: Earth Posts: 3 Thanks: 0 Transformation matrix of permutation I need help with the following: Let $\displaystyle \sigma \in S_n$ be a permutation of numbers $\displaystyle 1, \dots, n$ and $\displaystyle f:K^n \rightarrow K^n, \begin{pmatrix} x^1 \\ \vdots \\ x^n \end{pmatrix} \mapsto \begin{pmatrix} x^{\sigma(1)} \\ \vdots \\ x^{\sigma(n)} \end{pmatrix}$ a linear mapping. (i) Calculate the linear transformation matrix $\displaystyle A_\sigma$ of $\displaystyle f$ regarding the standard basis $\displaystyle \{e_1,\dots,e_n\}$ of $\displaystyle K^n$. (ii) Show that $\displaystyle \mathrm{det}(A_\sigma) = \mathrm{sgn}(\sigma)$. (iii) Show that $\displaystyle A^{n!} = 1\!\!1$, $\displaystyle 1\!\!1 := \begin{pmatrix} 1 & \cdots & 1 \\ \vdots & \ddots & \vdots \\ 1 & \cdots & 1 \end{pmatrix}$. Ideas: (i) $\displaystyle A_\sigma = \begin{pmatrix} e_{\sigma(1)}^T \\ \vdots \\ e_{\sigma(n)}^T \end{pmatrix} = \begin{pmatrix} e_{\sigma(1)} & \cdots & e_{\sigma(n)}\end{pmatrix}$. I'm not sure if I need to show the first equality and how to do it for the second? (ii) Let $\displaystyle a^i_j$ where $\displaystyle 1 \leq i,j \leq n$ be the $\displaystyle i$th row, $\displaystyle j$th column of matrix $\displaystyle A_\sigma$. We know that $\displaystyle \mathrm{det}(A_\sigma) = \sum_{\pi \in S_n} \mathrm{sgn}(\pi) \cdot a_1^{\pi(1)} \cdot a_2^{\pi(2)} \cdot \ldots \cdot a_n^{\pi(n)}$. [Leibniz formula] In (i) we see that the $\displaystyle j$th column of $\displaystyle A_\sigma$ is the unit vector $\displaystyle e_{\pi(j)}$, so that $\displaystyle a_1^{\pi(1)} \cdot a_2^{\pi(2)} \cdot \ldots \cdot a_n^{\pi(n)} = 1$ and thus $\displaystyle \mathrm{det}(A_\sigma) = \sum_{\pi \in S_n} \mathrm{sgn}(\pi)$. I don't know how to go from here... (iii) Matrix multiplication yields $\displaystyle A^2=\begin{pmatrix} 1 & 0 & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} =: I$. Therefore $\displaystyle A^3$ has $\displaystyle 1$s only in the diagonal and only where $\displaystyle A^1$ already has $\displaystyle 1$s, $\displaystyle 0$s everywhere else. That's all I could figure out... February 7th, 2019, 05:05 PM #2 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics Your description of (iii) is not correct. The symbol 1 in this setting is supposed to denote the identity matrix, not the matrix of all 1s. How to proceed on these problems also depends a lot on your background. If you have had a first course in group theory then these are all pretty easy. I assume you have already done (i) since it is trivial. I would start with 3 so that once its done you immediately obtain the result that $A$ is invertible and $A^{(n-1)!} = A^{-1}$. From here, you can easily show that $\left| \left| A \right| \right|_2 \leq 1$ so every eigenvalue of $A$ has absolute value $\leq 1$. But using the result from 3, you have that $A^{-1}$ is also a permutation matrix so its eigenvalues are also bounded by 1. Taken together, you have that every eigenvalue is either 1 or -1 so their product is also. I assume from here you can easily figure out how to determine which value it is. Tags matrix, permutation, transformation Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Blanco Linear Algebra 1 December 31st, 2015 11:41 AM solrob Linear Algebra 9 September 26th, 2013 10:00 AM tomtong Algebra 0 June 10th, 2013 05:53 PM Niko Bellic Linear Algebra 5 January 3rd, 2013 11:32 AM aliya Linear Algebra 1 August 10th, 2010 12:39 PM

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