If-A-4-3-1-0-use-the-fact-that-A-2-4A-3I-2-and-mathematical-induction-to-prove-A-n-3-n-1-2-A-3-3-n-2-I-if-n-1-

Question Number 40352 by scientist last updated on 20/Jul/18

I f A = [\begin{matrix} 4 - 3 \\ 1 0 \end{matrix}] u s e t h e f a c t t h a t

A^{2} = 4 A - 3 I_{2} a n d m a t h e m a t i c a l i n d u c t i o n t o p r o v e

A^{n} = \frac{(3^{n} - 1)}{2} A + \frac{3 - 3^{n}}{2} I i f n ⩾ 1

Commented by prof Abdo imad last updated on 20/Jul/18

t h e c a r a c t e r i s t i c p o l y n o m o f A i s

p_{c} (A) = d e t (A - x I) = | \begin{matrix} 4 - x - 3 \\ 1 - x \end{matrix} |

= - x (4 - x) + 3 = x^{2} - 4 x + 3

k a y l e y h a m i l t o n t h e o r e m g i v e

A^{2} - 4 A + 3 I = 0 \Rightarrow A^{2} = 4 A - 3 I \Rightarrow

l e t p r o v e b y r e c u r r e n c e t h a t

A^{2 n} = \frac{3^{2 n} - 1}{2} A + \frac{3 - 3^{2 n}}{2} I

t h e e q u a l i t y i s t r u e f o r n = 0

A^{2 n + 2} = A^{2} {\frac{3^{2 n} - 1}{2} A + \frac{3 - 3^{2 n}}{2} I}

= (4 A - 3 I) {\frac{3^{2 n} - 1}{2} A + \frac{3 - 3^{2 n}}{2} I}

= 2 (3^{2 n} - 1) A^{2} + 2 (3 - 3^{2 n}) A - 3 \frac{3^{2 n} - 1}{2} A

- 3 \frac{3 - 3^{2 n}}{2} I

= 2 (3^{2 n} - 1) (4 A - 3 I) + \frac{4 (3 - 3^{2 n}) - 33^{2 n} + 3}{2} A

- 3 \frac{3 - 3^{2 n}}{2} I

= 8 (3^{2 n} - 1) A - 6 (3^{2 n} - 1) I + \frac{15 - 7 3^{2 n}}{2} A

- 3 \frac{3 - 3^{2 n}}{2} I

= \frac{16 (3^{2 n} - 1) + 15 - 73^{2 n}}{2} A - \frac{12 (3^{2 n} - 1) + 3 (3 - 3^{2 n})}{2} A

= \frac{9 . 3^{2 n} - 1}{2} A - \frac{9 3^{2 n} - 3}{2} I

= \frac{3^{2 n + 2} - 1}{2} A + \frac{3 - 3^{2 n + 2}}{2} I t h e r e l a t i o n i s t r u e

a t t e r m (n + 1) a l s o w e m u s t p r o v e b y r e c u r r e n c e

t h a t

A^{2 n + 1} = \frac{3^{2 n + 1} - 1}{2} A + \frac{3 - 3^{2 n + 1}}{2} I .