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Question Number 28258 by abdo imad last updated on 22/Jan/18
let give A = ( 1 1 ) ( 2 −1) find e^(A ) and e^(−tA) .
$${let}\:{give}\:{A}\:\:=\:\left(\:\mathrm{1}\:\:\:\:\:\:\:\mathrm{1}\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\mathrm{2}\:\:\:\:−\mathrm{1}\right) \\ $$$$\:\:{find}\:\:{e}^{{A}\:} \:\:\:\:{and}\:\:\:{e}^{−{tA}} .\:\: \\ $$
Commented by abdo imad last updated on 23/Jan/18
e^A = Σ_(n=0) ^∝ (A^n /(n!)) let find A^n the carscteristic polynomial of A is P_c (A)=det(A −XI)= determinant (((1−X 1)),((2 −1−X))) =−(1−X^2 ) −2= X^2 −3 =(X−(√3))(X +(√(3))) so the proper values of A?are λ_1 =(√3) and λ_2 =−(√3) V(λ_1 ) =ker(A −λ_1 I) u(x,y)∈V(λ_1 )⇔ (A −λ_1 I)u=0 (((1−(√3) 1)),((2 −2−(√3))) ) ((x),(y) ) = (((o )),(o) ) after cslculus we find V( λ_1 )= D_e_(1 ) with e_1 ((1),((−(1−(√3)))) ) V(λ_2 )=ker(A −λ_(2 ) I) u(x,y)∈V(λ_2 ) ⇔ (A −λ_2 I)u=0 ⇔ (((1+(√3) 1)),((2 −1+(√3))) ) ((x),(y) ) = ((0),(0) ) after calculus we find V(λ_2 )= D_e_2 with e_2 ((1),((−(1+(√(3))))) ) so the diagonal mstrix is D= ((((√3) 0 )),((0 −(√3_ ))) ) and the inversible matrix id P= (((1 1)),((−(1−(√3) −(1+(√3) ) )) ) we have P^(−1) =((tran(com P))/(detP)) . com(P)= (−1)^(i+j) a_(ij) we find P^(−1) = (1/(2(√3))) (((1+(√3) 1)),((−(1−(√(3))) −1)) ) we have A =PDP^(−1) ⇒ A^n = P D^n P^(−1) after all calculus we find A^n =(1/(2(√3))) ( ((√3) )^n (1+(√3))−(−(√3))^n (1−(√3)) ((√3))^n −(−(√3))^n ) ( 2 ((√3))^n −2(−(√3))^n −(1−(√3))((√3))^n +(1+(√3)) (−(√3))^n ) . we have e^A = Σ_(n=0) ^∝ (A^n /(n!)) and e^(tA) = Σ_(n=0) ^∝ A^n (t^n /(n!)) .
$${e}^{{A}} =\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{{A}^{{n}} }{{n}!}\:\:\:{let}\:\:{find}\:\:{A}^{{n}} \:{the}\:{carscteristic}\:{polynomial} \\ $$$${of}\:\:{A}\:\:{is}\:{P}_{{c}} \left({A}\right)={det}\left({A}\:−{XI}\right)=\begin{vmatrix}{\mathrm{1}−{X}\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}−{X}}\end{vmatrix} \\ $$$$=−\left(\mathrm{1}−{X}^{\mathrm{2}} \right)\:−\mathrm{2}=\:{X}^{\mathrm{2}} −\mathrm{3}\:=\left({X}−\sqrt{\mathrm{3}}\right)\left({X}\:+\sqrt{\left.\mathrm{3}\right)}\right. \\ $$$${so}\:{the}\:{proper}\:{values}\:{of}\:{A}?{are}\:\lambda_{\mathrm{1}} =\sqrt{\mathrm{3}}\:\:{and}\:\lambda_{\mathrm{2}} =−\sqrt{\mathrm{3}} \\ $$$${V}\left(\lambda_{\mathrm{1}} \right)\:={ker}\left({A}\:−\lambda_{\mathrm{1}} {I}\right) \\ $$$${u}\left({x},{y}\right)\in{V}\left(\lambda_{\mathrm{1}} \right)\Leftrightarrow\:\left({A}\:−\lambda_{\mathrm{1}} {I}\right){u}=\mathrm{0} \\ $$$$\begin{pmatrix}{\mathrm{1}−\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:−\mathrm{2}−\sqrt{\mathrm{3}}}\end{pmatrix}\:\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{{o}\:}\\{{o}}\end{pmatrix}\:\:\:{after}\:{cslculus}\:{we}\:{find} \\ $$$${V}\left(\:\lambda_{\mathrm{1}} \right)=\:{D}_{{e}_{\mathrm{1}\:} } \:\:{with}\:{e}_{\mathrm{1}} \begin{pmatrix}{\mathrm{1}}\\{−\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)}\end{pmatrix}\:\: \\ $$$${V}\left(\lambda_{\mathrm{2}} \right)={ker}\left({A}\:−\lambda_{\mathrm{2}\:} {I}\right) \\ $$$${u}\left({x},{y}\right)\in{V}\left(\lambda_{\mathrm{2}} \right)\:\:\Leftrightarrow\:\left({A}\:−\lambda_{\mathrm{2}} {I}\right){u}=\mathrm{0} \\ $$$$\Leftrightarrow\begin{pmatrix}{\mathrm{1}+\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:−\mathrm{1}+\sqrt{\mathrm{3}}}\end{pmatrix}\:\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix}\:{after}\:{calculus}\:{we}\:{find} \\ $$$${V}\left(\lambda_{\mathrm{2}} \right)=\:{D}_{{e}_{\mathrm{2}} } \:\:\:\:\:{with}\:\:{e}_{\mathrm{2}} \begin{pmatrix}{\mathrm{1}}\\{−\left(\mathrm{1}+\sqrt{\left.\mathrm{3}\right)}\right.}\end{pmatrix}\:\:\:{so}\:{the}\:{diagonal} \\ $$$${mstrix}\:{is}\:{D}=\:\begin{pmatrix}{\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\mathrm{0}\:}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:−\sqrt{\mathrm{3}_{} }}\end{pmatrix}\:\:{and}\:\:{the}\:{inversible} \\ $$$${matrix}\:{id}\:\:{P}=\:\:\:\begin{pmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\left(\mathrm{1}−\sqrt{\mathrm{3}}\:\:\:\:\:\:\:−\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\:\right)\:\:\:\:\right.}\end{pmatrix}\:\:{we}\:{have} \\ $$$${P}^{−\mathrm{1}} \:\:\:=\frac{{tran}\left({com}\:{P}\right)}{{detP}}\:\:.\:\:\:{com}\left({P}\right)=\:\left(−\mathrm{1}\right)^{{i}+{j}} \:{a}_{{ij}} \:\:{we}\:{find} \\ $$$${P}^{−\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\begin{pmatrix}{\mathrm{1}+\sqrt{\mathrm{3}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{−\left(\mathrm{1}−\sqrt{\left.\mathrm{3}\right)}\:\:\:\:\:\:\:−\mathrm{1}\right.}\end{pmatrix}\:\:\:{we}\:{have}\:{A}\:={PDP}^{−\mathrm{1}} \: \\ $$$$\Rightarrow\:\:{A}^{{n}} =\:{P}\:{D}^{{n}} {P}^{−\mathrm{1}} \:\:{after}\:{all}\:{calculus}\:{we}\:{find} \\ $$$$\:\:\:\:{A}^{{n}} \:\:\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:\:\left(\:\:\:\:\:\left(\sqrt{\mathrm{3}}\:\:\right)^{{n}} \left(\mathrm{1}+\sqrt{\mathrm{3}}\right)−\left(−\sqrt{\mathrm{3}}\right)^{{n}} \left(\mathrm{1}−\sqrt{\mathrm{3}}\right)\:\:\:\:\:\:\:\:\:\:\:\left(\sqrt{\mathrm{3}}\right)^{{n}} \:−\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:\:\:\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\:\:\:\mathrm{2}\:\left(\sqrt{\mathrm{3}}\right)^{{n}} \:\:\:−\mathrm{2}\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{1}−\sqrt{\mathrm{3}}\right)\left(\sqrt{\mathrm{3}}\right)^{{n}} \:\:+\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\:\left(−\sqrt{\mathrm{3}}\right)^{{n}} \:\:\:\:\:\right)\:\:\:.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${we}\:{have}\:\:{e}^{{A}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:\frac{{A}^{{n}} }{{n}!}\:\:{and}\:\:{e}^{{tA}} \:\:=\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\:{A}^{{n}} \:\:\:\frac{{t}^{{n}} }{{n}!}\:. \\ $$

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