let A = (((1 −1)),((1 1)) ) 1) calculate A^(−1) and A^(−2) 2) calculate A^n 3) find e^A and e^(−A)


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Question Number 93415 by abdomathmax last updated on 13/May/20

$l e t A = (\begin{matrix} 1 - 1 \\ 1 1 \end{matrix})$ $1) c a l c u l a t e A^{- 1} a n d A^{- 2}$ $2) c a l c u l a t e A^{n}$ $3) f i n d e^{A} a n d e^{- A}$
Commented by prakash jain last updated on 13/May/20

$\det (A - λ I) = \| \begin{matrix} 1 - λ & - 1 \\ 1 & 1 - λ \end{matrix} \|$ $= {(1 - λ)}^{2} + 1 = 0$ $λ = 1 - i, 1 + i (i = \sqrt{- 1})$ $Eigenvector for λ = 1 - i$ $(A - λ I) [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = 0$ $([\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}] - [\begin{matrix} 1 - i & 0 \\ 0 & 1 - i \end{matrix}]) [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = 0$ $[\begin{matrix} i & - 1 \\ 1 & i \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = 0$ ${i x}_{1} - x_{2} = 0$ $x_{1} + {i x}_{2} = 0$ $x_{1} = 1, x_{2} = i$ $([\begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix}] - [\begin{matrix} 1 + i & 0 \\ 0 & 1 + i \end{matrix}]) [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = 0$ $[\begin{matrix} - i & - 1 \\ 1 & - i \end{matrix}] [\begin{matrix} y_{1} \\ y_{2} \end{matrix}] = 0$ $- {i y}_{1} - y_{2} = 0$ $y_{1} - {i y}_{2} = 0$ $y_{1} = 1, y_{2} = - i$ $S = [\begin{matrix} 1 & 1 \\ - i & i \end{matrix}]$ $S^{- 1} = [\begin{matrix} 1 / 2 & i / 2 \\ 1 / 2 & - i / 2 \end{matrix}]$ $S^{- 1} A S = [\begin{matrix} 1 + i & 0 \\ 0 & 1 - i \end{matrix}]$ $A = S [\begin{matrix} 1 + i & 0 \\ 0 & 1 - i \end{matrix}] S^{- 1}$ $A^{n} = S [\begin{matrix} {(1 + i)}^{n} & 0 \\ 0 & {(1 - i)}^{n} \end{matrix}] S^{- 1}$ $A^{n} = [\begin{matrix} 1 & 1 \\ - i & i \end{matrix}] [\begin{matrix} {(1 - i)}^{n} & 0 \\ 0 & {(1 + i)}^{n} \end{matrix}] [\begin{matrix} 1 / 2 & i / 2 \\ 1 / 2 & - i / 2 \end{matrix}]$ $= \frac{1}{2} [\begin{matrix} {(1 - i)}^{n} + {(1 + i)}^{n} & - i {({(1 - i)}^{n} - (1 + i))}^{n} \\ i ({(1 - i)}^{n} - {(1 + i)}^{n}) & {(1 - i)}^{n} + {(1 + i)}^{n} \end{matrix}]$
Commented by mathmax by abdo last updated on 13/May/20

$t h a n k y o u s i r .$
Commented by mathmax by abdo last updated on 13/May/20

$1) p_{c} (x) = d e t (A - x I) = \| \begin{matrix} 1 - x - 1 \\ 1 1 - x \end{matrix} \| = {(1 - x)}^{2} + 1$ $= x^{2} - 2 x + 2 c s y l e y h a m i l t o n t h e o r e m \Rightarrow A^{2} - 2 A + 2 I = 0 \Rightarrow$ $A (A - 2 I) = - 2 I \Rightarrow A \times (- \frac{1}{2}) (A - 2 I) = I \Rightarrow A i s i n v e r s i b l e$ $a n d A^{- 1} = - \frac{1}{2} A + I = (\begin{matrix} - \frac{1}{2} \frac{1}{2} \\ - \frac{1}{2} - \frac{1}{2} \end{matrix}) + (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$ $A^{- 1} = (\begin{matrix} \frac{1}{2} \frac{1}{2} \\ - \frac{1}{2} \frac{1}{2} \end{matrix}) a n d A^{- 2} = {(A^{- 1})}^{2}$
Commented by mathmax by abdo last updated on 13/May/20
$2) we have P_c (x) =det(A−xI) =x^2 −2x +2 P_c (x)=0 ⇔(x−1)^2 +1 =0 ⇔(x−1)^2 =−1 ⇒x =1+i or x=1−i let divide x^n by P_c (x) ⇒x^n =Q(x)P_c (x) +u_n x +v_n ⇒(1+i)^n =u_n (1+i)+v_n ⇒(1+i−1+i)u_n =(1+i)^n −(1−i)^n (1−i)^n =u_n (1−i) +v_n ⇒u_n =(1/(2i)){(1+i)^n −(1−i)^n } =(1/(2i)){ ((√2))^n e^((inπ)/4) −((√2))^n e^(−i((nπ)/4)) } =((√2))^n sin(((nπ)/4)) (1+i)^n +(1−i)^n =2u_n +2v_n ⇒u_n +v_n =(1/2){ (1+i)^n +(1−i)^n } =((√2))^(n ) cos(((nπ)/4)) ⇒v_n =((√2))^n cos(((nπ)/4))−((√2))^n sin(((nπ)/4)) we have Pc(A)=0 ⇒A^n =u_(n ) A +v_n I =((√2))^n sin(((nπ)/4)) (((1 −1)),((1 1)) ) +((√2))^n (cos(((nπ)/4))−sin(((nπ)/4)) (((1 0)),((0 1)) ) A^n = (((((√2))^n cos(((nπ)/4)) −((√2))^(n ) sin(((nπ)/4)))),((((√2))^n sin(((nπ)/4)) ((√2))^n cos(((nπ)/4) )) )$
$2) w e h a v e P_{c} (x) = d e t (A - x I) = x^{2} - 2 x + 2$ $P_{c} (x) = 0 \Leftrightarrow {(x - 1)}^{2} + 1 = 0 \Leftrightarrow {(x - 1)}^{2} = - 1 \Rightarrow x = 1 + i o r x = 1 - i$ $l e t d i v i d e x^{n} b y P_{c} (x) \Rightarrow x^{n} = Q (x) P_{c} (x) + u_{n} x + v_{n}$ $\Rightarrow {(1 + i)}^{n} = u_{n} (1 + i) + v_{n} \Rightarrow (1 + i - 1 + i) u_{n} = {(1 + i)}^{n} - {(1 - i)}^{n}$ ${(1 - i)}^{n} = u_{n} (1 - i) + v_{n}$ $\Rightarrow u_{n} = \frac{1}{2 i} {{(1 + i)}^{n} - {(1 - i)}^{n}} = \frac{1}{2 i} {{(\sqrt{2})}^{n} e^{\frac{i n π}{4}} - {(\sqrt{2})}^{n} e^{- i \frac{n π}{4}}}$ $= {(\sqrt{2})}^{n} s i n (\frac{n π}{4})$ ${(1 + i)}^{n} + {(1 - i)}^{n} = 2 u_{n} + 2 v_{n} \Rightarrow u_{n} + v_{n} = \frac{1}{2} {{(1 + i)}^{n} + {(1 - i)}^{n}}$ $= {(\sqrt{2})}^{n} c o s (\frac{n π}{4}) \Rightarrow v_{n} = {(\sqrt{2})}^{n} c o s (\frac{n π}{4}) - {(\sqrt{2})}^{n} s i n (\frac{n π}{4})$ $w e h a v e P c (A) = 0 \Rightarrow A^{n} = u_{n} A + v_{n} I$ $= {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) (\begin{matrix} 1 - 1 \\ 1 1 \end{matrix}) + {(\sqrt{2})}^{n} (c o s (\frac{n π}{4}) - s i n (\frac{n π}{4}) (\begin{matrix} 1 0 \\ 0 1 \end{matrix})$ $A^{n} = (\begin{matrix} {(\sqrt{2})}^{n} c o s (\frac{n π}{4}) - {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) \\ {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) {(\sqrt{2})}^{n} c o s (\frac{n π}{4} \end{matrix})$
Commented by mathmax by abdo last updated on 13/May/20

$\Rightarrow A^{n} = (\begin{matrix} {(\sqrt{2})}^{n} c o s (\frac{n π}{4}) - {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) \\ {(\sqrt{2})}^{n} s i n (\frac{n π}{4}) {(\sqrt{2})}^{n} c o s (\frac{n π}{4}) \end{matrix})$
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