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Question Number 41141 by Kishan Daroga last updated on 02/Aug/18

If X= [(3,(−4)),(1,(−1)) ], the value of X^n is

$$\mathrm{If}\:{X}=\begin{bmatrix}{\mathrm{3}}&{−\mathrm{4}}\\{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix},\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{X}^{{n}} \mathrm{is} \\ $$

Commented by math khazana by abdo last updated on 04/Aug/18

let A = (((3 −4)),((1 −1)) ) the caracteristic polynome of A is p(x)=det(A−xI) = determinant (((3−x −4)),((1 −1−x))) =(x−3)(x+1) +4 =x^2 +x−3x−3 +4 =x^2 −2x +1 =(x−1)^2 so 1 is double proper value for A v(1)=ker(A−I)={u /(A−I)u=0} let u ((x),(y) ) ⇒ (((2 −4)),((1 −2)) ) ((x),(y) ) =0 ⇒ {_(x−2y=0) ^(2x−4y=0) ⇒x=2y ⇒(x,y)=(2y,y)=y(2,1) so v(1)=D_e with vector e(2,1) its clear that A is not diagonalisable by cayley hamilton theorem give A^2 −2A +I =0⇒ A^2 =2A −I ⇒ A^3 =(2A−I)A=2A^2 −A =2(2A−I)−A =3A −2I for that we must determine the sequences u_n and v_n with verify A^n =u_n A +v_n I ⇒ A^(n+1) =u_(n+1) A +v_(n+1) I =(u_n A +v_n I)A =u_n A^2 +v_n A =u_n (2A−I) +v_n A =(2u_n +v_n )A −u_n I ⇒ u_(n+1) = 2u_n +v_n and v_(n+1) =−u_n ...be continued...

$${let}\:{A}\:=\:\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:−\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}\:\:{the}\:{caracteristic}\:{polynome}\:{of} \\ $$$${A}\:{is}\:{p}\left({x}\right)={det}\left({A}−{xI}\right)\:=\begin{vmatrix}{\mathrm{3}−{x}\:\:\:\:\:\:\:−\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}−{x}}\end{vmatrix} \\ $$$$=\left({x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)\:+\mathrm{4}\:={x}^{\mathrm{2}} \:+{x}−\mathrm{3}{x}−\mathrm{3}\:+\mathrm{4} \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{1}\:=\left({x}−\mathrm{1}\right)^{\mathrm{2}} \:{so}\:\mathrm{1}\:{is}\:{double}\:{proper} \\ $$$${value}\:{for}\:{A} \\ $$$${v}\left(\mathrm{1}\right)={ker}\left({A}−{I}\right)=\left\{{u}\:/\left({A}−{I}\right){u}=\mathrm{0}\right\} \\ $$$${let}\:{u}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\Rightarrow\begin{pmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:−\mathrm{4}}\\{\mathrm{1}\:\:\:\:\:\:\:\:\:\:−\mathrm{2}}\end{pmatrix}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\mathrm{0}\:\Rightarrow \\ $$$$\left\{_{{x}−\mathrm{2}{y}=\mathrm{0}} ^{\mathrm{2}{x}−\mathrm{4}{y}=\mathrm{0}} \:\Rightarrow{x}=\mathrm{2}{y}\:\Rightarrow\left({x},{y}\right)=\left(\mathrm{2}{y},{y}\right)={y}\left(\mathrm{2},\mathrm{1}\right)\right. \\ $$$${so}\:{v}\left(\mathrm{1}\right)={D}_{{e}} \:\:\:{with}\:{vector}\:{e}\left(\mathrm{2},\mathrm{1}\right)\:{its}\:{clear}\:{that} \\ $$$${A}\:{is}\:{not}\:{diagonalisable}\:\:{by}\:{cayley}\:{hamilton} \\ $$$${theorem}\:{give}\:{A}^{\mathrm{2}} \:−\mathrm{2}{A}\:+{I}\:=\mathrm{0}\Rightarrow \\ $$$${A}^{\mathrm{2}} \:=\mathrm{2}{A}\:−{I}\:\Rightarrow\:{A}^{\mathrm{3}} \:=\left(\mathrm{2}{A}−{I}\right){A}=\mathrm{2}{A}^{\mathrm{2}} −{A} \\ $$$$=\mathrm{2}\left(\mathrm{2}{A}−{I}\right)−{A}\:=\mathrm{3}{A}\:−\mathrm{2}{I}\:\:{for}\:{that}\:{we}\:{must} \\ $$$${determine}\:{the}\:{sequences}\:{u}_{{n}} \:{and}\:{v}_{{n}} \:{with} \\ $$$${verify}\:\:{A}^{{n}} \:\:={u}_{{n}} \:{A}\:+{v}_{{n}} {I}\:\Rightarrow \\ $$$${A}^{{n}+\mathrm{1}} \:={u}_{{n}+\mathrm{1}} {A}\:+{v}_{{n}+\mathrm{1}} {I}\:\:=\left({u}_{{n}} {A}\:+{v}_{{n}} {I}\right){A} \\ $$$$={u}_{{n}} {A}^{\mathrm{2}} \:+{v}_{{n}} {A}\:={u}_{{n}} \left(\mathrm{2}{A}−{I}\right)\:+{v}_{{n}} {A} \\ $$$$=\left(\mathrm{2}{u}_{{n}} +{v}_{{n}} \right){A}\:−{u}_{{n}} {I}\:\Rightarrow\: \\ $$$${u}_{{n}+\mathrm{1}} =\:\mathrm{2}{u}_{{n}} \:+{v}_{{n}} \:\:\:{and}\:{v}_{{n}+\mathrm{1}} =−{u}_{{n}} \:\:\:\:...{be}\:{continued}... \\ $$

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