let give the matrice A= (((1 2 )),((2 1)) ) calculate A^n then find e^A .


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Question Number 28533 by abdo imad last updated on 26/Jan/18

$l e t g i v e t h e m a t r i c e A = (\begin{matrix} 1 2 \\ 2 1 \end{matrix})$ $c a l c u l a t e A^{n} t h e n f i n d e^{A} .$
Commented by abdo imad last updated on 28/Jan/18

$w e h a v e A = I + 2 J w i t h I = (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$ $a n d J = (\begin{matrix} 0 1 \\ 1 0 \end{matrix})$ $w e h a v e J^{2} = I \Rightarrow J^{2 n} = I a n d J^{2 n + 1} = J$ $A = I + 2 J w i t h t h e c o n d i t i o n I J = J I$ $A^{n} = {(2 J + I)}^{n} = \sum_{k = 0}^{n} C_{n}^{k} {(2 J)}^{k}$ $= \sum_{k = 0}^{n} 2^{k} C_{n}^{k} J^{k} = Σ (_{k = 2 p} . . .) + \sum_{k = 2 p + 1} (. . .)$ $= \sum_{p = 0}^{[\frac{n}{2}]} 2^{2 p} C_{n}^{2 p} I + \sum_{p = 0}^{[\frac{n - 1}{2}]} 2^{2 p + 1} C_{n}^{2 p + 1} J$ $= (\begin{matrix} \sum_{p = 0}^{[\frac{n}{2}]} 2^{2 p} C_{n}^{2 p} 0 \\ 0 \sum_{p = 0}^{[\frac{n}{2}]} 2^{2 p} C_{n}^{2 p} \end{matrix})$ $+ (\begin{matrix} 0 \sum_{p = 0}^{[\frac{n - 1}{2}]} 2^{2 p + 1} C_{n}^{2 p + 1} \\ \sum_{p = 0}^{[\frac{n - 1}{2}]} 2^{2 p + 1} C_{n}^{2 p + 1} 0 \end{matrix})$ $= (\begin{matrix} x_{n} y_{n} \\ y_{n x_{n}} \end{matrix})$ $w i t h x_{n} = \sum_{p = 0}^{[\frac{n}{2}]} 2^{2 p} C_{n}^{2 p} a n d y_{n} = \sum_{p = 0}^{[\frac{n - 1}{2}]} 2^{2 p + 1} C_{n}^{2 p + 1} .$ $e^{A} = \sum_{n = 0}^{\infty} \frac{A^{n}}{n!} = (\begin{matrix} Σ \frac{x_{n}}{n!} Σ \frac{y_{n}}{n!} \\ Σ \frac{y_{n}}{n!} Σ \frac{x_{n}}{n!} \end{matrix})$
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