let the matrix A = (((1 2)),((0 −3)) ) 1) calculate A^n for n integr 2) find e^A and e^(−A) .


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Question Number 74019 by mathmax by abdo last updated on 17/Nov/19

$l e t t h e m a t r i x A = (\begin{matrix} 1 2 \\ 0 - 3 \end{matrix})$ $1) c a l c u l a t e A^{n} f o r n i n t e g r$ $2) f i n d e^{A} a n d e^{- A} .$
Commented by mathmax by abdo last updated on 18/Nov/19
$the caracteristic polynom of A is p(x)=det(A−xI) = determinant (((1−x 2)),((0 −3−x)))=−(3+x)(1−x) =(x+3)(x−1) the propers value is λ_1 =−3 and λ_2 =1 V(λ_1 )=ker(A+3I)={u/(A+3I)u=0} u ((x),(y) ) (A+3I)u=0 ⇔ (((4 2)),((0 0)) ) ((x),(y) ) =0 ⇒4x+2y =0 ⇒2x+y=0 ⇒ y=−2x ⇒(x,y)=(x,−2x)=x(1,−2) ⇒V(−3)=vet(e_1 ) with e_1 (1,−2) V(λ_1 )=Ker(A−I)={u /(A−I)u=0} (A−I)u=0 ⇒ (((0 2)),((0 −4)) ) ((x),(y) )=0 ⇒y=0 ⇒(x,y)=x(1,0) ⇒P = (((1 1)),((−2 0)) ) and D = (((−3 0)),((0 1)) ) A =P DP^(−1) ⇒ A^n =PD^n P^(−1) we have p_c (P)= determinant (((1−x 1)),((−2 −x)))=−x(1−x)+2 =x^2 −x +2 cayley hamilton ⇒p^2 −p +2I =0 ⇒p(p−I) =−2I ⇒ p{−(1/2)(p−I)} =I ⇒p^(−1) =(1/2){ I−p} =(1/2){ (((1 0)),((0 1)) ) − (((1 1)),((−2 0)) )} =(1/2) (((0 −1)),((2 1)) ) A^n =(1/2) (((1 1)),((−2 0)) ) ((((−3)^n 0)),((0 1)) ) (((0 −1)),((2 1)) ) =(1/2) ((((−3)^(n ) 1)),((−2(−3)^n 0)) ) (((0 −1)),((2 1)) ) =(1/2) ((( 2 1−(−3)^n )),((0 2(−3)^n )) ) = (((1 ((1−(−3)^n )/2))),((0 (−3)^n )) )$
$t h e c a r a c t e r i s t i c p o l y n o m o f A i s$ $p (x) = d e t (A - x I) = \| \begin{matrix} 1 - x 2 \\ 0 - 3 - x \end{matrix} \| = - (3 + x) (1 - x)$ $= (x + 3) (x - 1) t h e p r o p e r s v a l u e i s λ_{1} = - 3 a n d λ_{2} = 1$ $V (λ_{1}) = k e r (A + 3 I) = {u / (A + 3 I) u = 0}$ $u (\begin{matrix} x \\ y \end{matrix}) (A + 3 I) u = 0 \Leftrightarrow (\begin{matrix} 4 2 \\ 0 0 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow 4 x + 2 y = 0 \Rightarrow 2 x + y = 0 \Rightarrow$ $y = - 2 x \Rightarrow (x, y) = (x, - 2 x) = x (1, - 2) \Rightarrow V (- 3) = v e t (e_{1}) w i t h$ $e_{1} (1, - 2)$ $V (λ_{1}) = K e r (A - I) = {u / (A - I) u = 0}$ $(A - I) u = 0 \Rightarrow (\begin{matrix} 0 2 \\ 0 - 4 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow y = 0 \Rightarrow (x, y) = x (1, 0)$ $\Rightarrow P = (\begin{matrix} 1 1 \\ - 2 0 \end{matrix}) a n d D = (\begin{matrix} - 3 0 \\ 0 1 \end{matrix})$ $A = P {D P}^{- 1} \Rightarrow A^{n} = {P D}^{n} P^{- 1}$ $w e h a v e p_{c} (P) = \| \begin{matrix} 1 - x 1 \\ - 2 - x \end{matrix} \| = - x (1 - x) + 2 = x^{2} - x + 2$ $c a y l e y h a m i l t o n \Rightarrow p^{2} - p + 2 I = 0 \Rightarrow p (p - I) = - 2 I \Rightarrow$ $p {- \frac{1}{2} (p - I)} = I \Rightarrow p^{- 1} = \frac{1}{2} {I - p}$ $= \frac{1}{2} {(\begin{matrix} 1 0 \\ 0 1 \end{matrix}) - (\begin{matrix} 1 1 \\ - 2 0 \end{matrix})} = \frac{1}{2} (\begin{matrix} 0 - 1 \\ 2 1 \end{matrix})$ $A^{n} = \frac{1}{2} (\begin{matrix} 1 1 \\ - 2 0 \end{matrix}) (\begin{matrix} {(- 3)}^{n} 0 \\ 0 1 \end{matrix}) (\begin{matrix} 0 - 1 \\ 2 1 \end{matrix})$ $= \frac{1}{2} (\begin{matrix} {(- 3)}^{n} 1 \\ - 2 {(- 3)}^{n} 0 \end{matrix}) (\begin{matrix} 0 - 1 \\ 2 1 \end{matrix})$ $= \frac{1}{2} (\begin{matrix} 2 1 - {(- 3)}^{n} \\ 0 2 {(- 3)}^{n} \end{matrix}) = (\begin{matrix} 1 \frac{1 - {(- 3)}^{n}}{2} \\ 0 {(- 3)}^{n} \end{matrix})$
Commented by mathmax by abdo last updated on 18/Nov/19

$2) e^{A} = \sum_{n = 0}^{\infty} \frac{A^{n}}{n!} = (\begin{matrix} \sum_{n = 0}^{\infty} \frac{1}{n!} \sum_{n = 0} \frac{1 - {(- 3)}^{n}}{2 n!} \\ 0 \sum_{n = 0}^{\infty} \frac{{(- 3)}^{n}}{n!} \end{matrix})$ $= (\begin{matrix} e \frac{1}{2} (e - e^{- 3}) \\ 0 e^{- 3} \end{matrix})$
Commented by mathmax by abdo last updated on 18/Nov/19

$e^{- A} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} A^{n} = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} (\begin{matrix} 1 \frac{1 - {(- 3)}^{n}}{2} \\ 0 {(- 3)}^{n} \end{matrix})$ $= (\begin{matrix} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n!} \frac{1 - {(- 3)}^{n}}{2} \\ 0 \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(- 3)}^{n}}{n!} \end{matrix})$ $= (\begin{matrix} e^{- 1} \frac{1}{2} (e^{- 1} - e^{3}) \\ 0 e^{3} \end{matrix})$
	Answered by mind is power last updated on 17/Nov/19
	$det(A−xI_2 )=0 let f(x,y)=(x+2y,−3y) ⇒(1−x)(−3−x)=0⇒x∈{1,−3} x=1 f(x,y)=(x,y)⇒y=0 e=(1,0) f(x,y)=−3(x,y)⇒4x+2y=0⇒y=−2x e_2 =(1,−2) A=PDP^(−1) P= (((1 1)),((0 −2)) ),P^− =−(1/2) (((−2 −1)),((0 1)) ) A=−(1/2) (((1 1)),((0 −2)) ). (((1 0)),((0 −3)) ). (((−2 −1)),(( 0 1)) ) A^n =ePD^n P^− e^A =Σ(A^k /(k!))=P(Σ_(k=0) ^(+∞) (D^k /(k!)).)P^− =P. (((e 0)),((0 e^(−3) )) ).P^− e^(−A) =P(Σ_(k=0) ^(+∞) (((−D)^k )/(k!)))P^− =P. (((e^(−1) 0)),((0 e^3 )) ) P^−$
	$\det (A - {xI}_{2}) = 0$ $let f (x, y) = (x + 2 y, - 3 y)$ $\Rightarrow (1 - x) (- 3 - x) = 0 \Rightarrow x \in {1, - 3}$ $x = 1$ $f (x, y) = (x, y) \Rightarrow y = 0$ $e = (1, 0)$ $f (x, y) = - 3 (x, y) \Rightarrow 4 x + 2 y = 0 \Rightarrow y = - 2 x$ $e_{2} = (1, - 2)$ $A = {PDP}^{- 1}$ $P = (\begin{matrix} 1 1 \\ 0 - 2 \end{matrix}), P^{-} = - \frac{1}{2} (\begin{matrix} - 2 - 1 \\ 0 1 \end{matrix})$ $A = - \frac{1}{2} (\begin{matrix} 1 1 \\ 0 - 2 \end{matrix}) . (\begin{matrix} 1 0 \\ 0 - 3 \end{matrix}) . (\begin{matrix} - 2 - 1 \\ 0 1 \end{matrix})$ $A^{n} = {ePD}^{n} P^{-}$ $e^{A} = Σ \frac{A^{k}}{k!} = P (\underset{k = 0}{\sum^{+ \infty}} \frac{D^{k}}{k!} .) P^{-} = P . (\begin{matrix} e 0 \\ 0 e^{- 3} \end{matrix}) . P^{-}$ $e^{- A} = P (\underset{k = 0}{\sum^{+ \infty}} \frac{{(- D)}^{k}}{k!}) P^{-} = P . (\begin{matrix} e^{- 1} 0 \\ 0 e^{3} \end{matrix}) P^{-}$
	Commented by abdomathmax last updated on 18/Nov/19

	$t h a n k y o u s i r .$
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