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