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Question Number 91321 by  M±th+et+s last updated on 29/Apr/20
1)find eigen values and corresponding eigen vector of the matrix A= [((cos(θ)),(−sin(θ))),((sin(θ)),( cos(θ))) ] 2)solve 6y^2 dx−x(y+2x^2 )dy=0
$$\left.\mathrm{1}\right){find}\:{eigen}\:{values}\:{and}\:{corresponding} \\ $$$${eigen}\:{vector}\:{of}\:{the}\:{matrix} \\ $$$$ \\ $$$${A}=\begin{bmatrix}{{cos}\left(\theta\right)}&{−{sin}\left(\theta\right)}\\{{sin}\left(\theta\right)}&{\:\:\:\:\:{cos}\left(\theta\right)}\end{bmatrix} \\ $$$$ \\ $$$$\left.\mathrm{2}\right){solve} \\ $$$$ \\ $$$$\mathrm{6}{y}^{\mathrm{2}} {dx}−{x}\left({y}+\mathrm{2}{x}^{\mathrm{2}} \right){dy}=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 30/Apr/20
1) P(x) =det(A−xI) = determinant (((cosθ−x −sinθ)),((sinθ cosθ−x))) =(cosθ−x)^2 +sin^2 θ P(x)=0 ⇔(x−cosθ)^2 +sin^2 θ =0⇔(x−cosθ)^2 =(isinθ)^2 ⇒ x−cosθ =+^− isinθ ⇒x =cosθ +^− isinθ so the values are λ_1 =e^(iθ) and λ_2 =e^(−iθ) V(λ_1 )=Ker(A−λ_1 I) ={u /(A−λ_1 I)u =0} u ((x),(y) ) (A−λ_1 )u=0 ⇒ (((cosθ−e^(iθ) −sinθ)),((sinθ cosθ−e^(iθ) )) ) ((x),(y) )=0 ⇒ (((−isinθ −sinθ)),((sinθ −isinθ)) ) ((x),(y) )=0 ⇒ { ((−isinθ x−sinθy =0)),((sinθx −isinθy =0)) :} ⇒ { ((isinθ x+sinθ y =0)),((sinθ x−isinθ y=0 ⇒sinθ x =isinθ y)) :} if sinθ ≠0 ⇒x =iy ⇒(x,y)=(iy,y) =y e_1 with e_1 ((i),(1) ) V(λ_2 ) =Ker(A−λ_2 I) ={u /(A−λ_2 I)u =0} (A−λ_2 I)u =0 ⇒ (((cosθ −e^(−iθ) −sinθ)),((sinθ cosθ −e^(−iθ) )) ) ((x),(y) )=0 ⇒ (((isinθ −sinθ)),((sinθ isinθ)) ) ((x),(y) )=0 ⇒ { ((isinθ x−sinθy =0)),((sinθ x +isinθ y =0 )) :} ⇒sinθ x =−isinθ y so if sinθ ≠0 we get x=−iy ⇒ (x,y) =(−iy,y)=ye_2 with e_2 (((−i)),(1) ) We see V(λ_1 ) =Δ_e_1 dim(V(λ_1 ))=1 V(λ_2 )=Δ_(e_2 ) and dim(V(λ_2 ))=1 A is diagonalisable....
$$\left.\mathrm{1}\right)\:{P}\left({x}\right)\:={det}\left({A}−{xI}\right)\:=\begin{vmatrix}{{cos}\theta−{x}\:\:\:\:\:\:\:\:\:−{sin}\theta}\\{{sin}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\theta−{x}}\end{vmatrix} \\ $$$$=\left({cos}\theta−{x}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta\: \\ $$$${P}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\left({x}−{cos}\theta\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta\:=\mathrm{0}\Leftrightarrow\left({x}−{cos}\theta\right)^{\mathrm{2}} \:=\left({isin}\theta\right)^{\mathrm{2}} \:\Rightarrow \\ $$$${x}−{cos}\theta\:=\overset{−} {+}{isin}\theta\:\Rightarrow{x}\:={cos}\theta\:\overset{−} {+}{isin}\theta\:\:{so}\:{the}\:{values}\:{are} \\ $$$$\lambda_{\mathrm{1}} ={e}^{{i}\theta} \:\:{and}\:\lambda_{\mathrm{2}} ={e}^{−{i}\theta} \\ $$$${V}\left(\lambda_{\mathrm{1}} \right)={Ker}\left({A}−\lambda_{\mathrm{1}} {I}\right)\:=\left\{{u}\:/\left({A}−\lambda_{\mathrm{1}} {I}\right){u}\:=\mathrm{0}\right\} \\ $$$${u}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:\:\:\:\:\:\:\left({A}−\lambda_{\mathrm{1}} \right){u}=\mathrm{0}\:\Rightarrow\begin{pmatrix}{{cos}\theta−{e}^{{i}\theta} \:\:\:\:\:\:\:\:\:−{sin}\theta}\\{{sin}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\theta−{e}^{{i}\theta} }\end{pmatrix}\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\mathrm{0}\:\Rightarrow \\ $$$$\begin{pmatrix}{−{isin}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:−{sin}\theta}\\{{sin}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{isin}\theta}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\mathrm{0}\:\Rightarrow\begin{cases}{−{isin}\theta\:{x}−{sin}\theta{y}\:=\mathrm{0}}\\{{sin}\theta{x}\:−{isin}\theta{y}\:=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\begin{cases}{{isin}\theta\:{x}+{sin}\theta\:{y}\:=\mathrm{0}}\\{{sin}\theta\:{x}−{isin}\theta\:{y}=\mathrm{0}\:\Rightarrow{sin}\theta\:{x}\:={isin}\theta\:{y}}\end{cases} \\ $$$${if}\:{sin}\theta\:\neq\mathrm{0}\:\Rightarrow{x}\:={iy}\:\Rightarrow\left({x},{y}\right)=\left({iy},{y}\right)\:={y}\:{e}_{\mathrm{1}} \:\:\:\:{with}\:{e}_{\mathrm{1}} \begin{pmatrix}{{i}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${V}\left(\lambda_{\mathrm{2}} \right)\:={Ker}\left({A}−\lambda_{\mathrm{2}} {I}\right)\:=\left\{{u}\:/\left({A}−\lambda_{\mathrm{2}} {I}\right){u}\:=\mathrm{0}\right\} \\ $$$$\left({A}−\lambda_{\mathrm{2}} {I}\right){u}\:=\mathrm{0}\:\Rightarrow\begin{pmatrix}{{cos}\theta\:−{e}^{−{i}\theta} \:\:\:\:\:\:\:\:−{sin}\theta}\\{{sin}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{cos}\theta\:−{e}^{−{i}\theta} }\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\mathrm{0}\:\Rightarrow \\ $$$$\begin{pmatrix}{{isin}\theta\:\:\:\:\:\:\:\:\:−{sin}\theta}\\{{sin}\theta\:\:\:\:\:\:\:\:\:\:\:\:\:\:{isin}\theta}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\mathrm{0}\:\Rightarrow\begin{cases}{{isin}\theta\:{x}−{sin}\theta{y}\:=\mathrm{0}}\\{{sin}\theta\:{x}\:+{isin}\theta\:{y}\:=\mathrm{0}\:}\end{cases} \\ $$$$\Rightarrow{sin}\theta\:{x}\:=−{isin}\theta\:{y}\:{so}\:{if}\:{sin}\theta\:\neq\mathrm{0}\:{we}\:{get}\:{x}=−{iy}\:\Rightarrow \\ $$$$\left({x},{y}\right)\:=\left(−{iy},{y}\right)={ye}_{\mathrm{2}} \:\:{with}\:{e}_{\mathrm{2}} \begin{pmatrix}{−{i}}\\{\mathrm{1}}\end{pmatrix} \\ $$$${We}\:{see}\:\:{V}\left(\lambda_{\mathrm{1}} \right)\:=\Delta_{{e}_{\mathrm{1}} } \:\:\:\:\:\:\:{dim}\left({V}\left(\lambda_{\mathrm{1}} \right)\right)=\mathrm{1} \\ $$$${V}\left(\lambda_{\mathrm{2}} \right)=\Delta_{{e}_{\mathrm{2}} \:} \:\:\:\:\:{and}\:{dim}\left({V}\left(\lambda_{\mathrm{2}} \right)\right)=\mathrm{1} \\ $$$${A}\:{is}\:{diagonalisable}…. \\ $$$$ \\ $$
Commented by  M±th+et+s last updated on 30/Apr/20
nice and correct work sir god bless you
$${nice}\:{and}\:{correct}\:{work}\:{sir}\:{god}\:{bless}\:{you} \\ $$
Commented by turbo msup by abdo last updated on 30/Apr/20
you are welcome .
$${you}\:{are}\:{welcome}\:. \\ $$

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