let-A-1-1-1-1-1-calculate-A-n-2-determine-e-A-and-e-A-

Question Number 60501 by prof Abdo imad last updated on 21/May/19

l e t A = (\begin{matrix} 1 1 \\ 1 1 \end{matrix})

1) c a l c u l a t e A^{n}

2) d e t e r m i n e e^{A} a n d e^{- A} .

Commented by maxmathsup by imad last updated on 21/May/19

w e P_{c} (A) = | \begin{matrix} 1 - x 1 \\ 1 1 - x \end{matrix} | = {(1 - x)}^{2} - 1 = x^{2} - 2 x + 1 - 1 = x^{2} - 2 x

c a y l e y h a m i l t o n t h e o r e m g i v e P_{c} (A) = 0 \Rightarrow A^{2} - 2 A = 0 \Rightarrow A^{2} = 2 A \Rightarrow

A^{3} = 2 A^{2} = 2^{2} A \Rightarrow A^{n} = 2^{n - 1} A \Rightarrow A^{n} = (\begin{matrix} 2^{n - 1} 2^{n - 1} \\ 2^{n - 1} 2^{n - 1} \end{matrix}) n ⩾ 1

2) w e h a v e e^{A} = \sum_{n = 0}^{\infty} \frac{A^{n}}{n!} = I + \sum_{n = 1}^{\infty} \frac{1}{n!} (\begin{matrix} 2^{n - 1} 2^{n - 1} \\ 2^{n - 1} 2^{n - 1} \end{matrix})

= I + (\begin{matrix} \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \\ \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} \end{matrix})

\sum_{n = 1}^{\infty} \frac{2^{n - 1}}{n!} = \frac{1}{2} \sum_{n = 1}^{\infty} \frac{2^{n}}{n!} = \frac{1}{2} {\sum_{n = 0}^{\infty} \frac{2^{n}}{n!} - 1} = \frac{1}{2} {e^{2} - 1} \Rightarrow

e^{A} = (\begin{matrix} 1 0 \\ 0 1 \end{matrix}) + (\begin{matrix} \frac{e^{2} - 1}{2} \frac{e^{2} - 1}{2} \\ \frac{e^{2} - 1}{2} \frac{e^{2} - 1}{2} \end{matrix})

e^{A} = (\begin{matrix} \frac{1 + e^{2}}{2} \frac{e^{2} - 1}{2} \\ \frac{e^{2} - 1}{2} \frac{1 + e^{2}}{2} \end{matrix})

a l s o w e c a l c u l a t e e^{- A} b y f o l o w i n g t h e s a m e s t e p s a n d u s e

e^{- A} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} A^{n}}{n!}

Answered by Prithwish sen last updated on 21/May/19

Is, A^{n} = (\begin{matrix} 2^{n - 1} 2^{n - 1} \\ 2^{n - 1} 2^{n - 1} \end{matrix})