Permutations #2

17.02.2010 | by Peter

In uno omnia.*

– Athanasius Kircher

***

$a\in X_n$

$\sigma = \left(\sigma\left(1\right),\sigma\left(2\right),\cdots,\sigma\left(n\right)\right)$

$(a \; \sigma(a) \; \sigma^2(a) \; \cdots \; \sigma^{|a|-1}(a))$

$\sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ \sigma\left(1\right) & \sigma\left(2\right) & \cdots & \sigma\left(n\right) \end{pmatrix}$

$P \cdot \overline{x}= \begin{pmatrix} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0\\ \end{pmatrix} \cdot \begin{pmatrix}1 \\ 2 \\ 3 \\ 4 \\\end{pmatrix} = \begin{pmatrix}4 \\ 3 \\ 2 \\ 1 \\\end{pmatrix}$

$1^{b_1} 2^{b_2} 3^{b_3} \cdots n^{b_n}$

$\sigma = \begin{pmatrix} x_1 & x_2 & \cdots & x_n \\ \sigma\left(x_1\right) & \sigma\left(x_2\right) & \cdots & \sigma\left(x_n\right) \end{pmatrix}$

$\sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{pmatrix}$

$\overline{x} =\begin{pmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \\\end{pmatrix}$

$P_\sigma \cdot \begin{pmatrix}x_1 \\ x_2 \\ \vdots \\ x_n \\\end{pmatrix} = \begin{pmatrix} \sigma(x_1) \\ \sigma(x_2) \\ \vdots \\ \sigma(x_n) \\\end{pmatrix}$

$\sigma = \begin{pmatrix} 1 & 2 & \cdots & n \\ \sigma\left(1\right) & \sigma\left(2\right) & \cdots & \sigma\left(n\right) \end{pmatrix}$

$\sigma = \left(\sigma\left(1\right),\sigma\left(2\right),\cdots,\sigma\left(n\right)\right)$

$1^{b_1} 2^{b_2} 3^{b_3} \cdots n^{b_n}$

$\sigma_2 = \begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 2 & 1 \end{pmatrix}$

$a\in X_n$

Quod…

… erat…

… demonstrandum.

* All in one.

***

3 Comments to “Permutations #2”

Jochen Hein wrote on 21. February 2010, 16:15:06
fett
Peter wrote on 21. February 2010, 19:55:22
Spassibo!
Jochen Hein wrote on 22. February 2010, 6:23:35
Echt, and then glix denn next kiss.
Peter, geküsst!

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Permutations #2

3 Comments to “Permutations #2”

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