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


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Question Number 91321 by M±th+et+s last updated on 29/Apr/20

$1) f i n d e i g e n v a l u e s a n d c o r r e s p o n d i n g$ $e i g e n v e c t o r o f t h e m a t r i x$ $A = [\begin{matrix} c o s (θ) & - s i n (θ) \\ s i n (θ) & c o s (θ) \end{matrix}]$ $2) s o l v e$ $6 y^{2} d x - x (y + 2 x^{2}) d y = 0$
Commented by mathmax by abdo last updated on 30/Apr/20
$1) P(x) =det(A−xI) = determinant (((cosθ−x −sinθ)),((sinθ cosθ−x))) =(cosθ−x)^2 +sin^2 θ P(x)=0 ⇔(x−cosθ)^2 +sin^2 θ =0⇔(x−cosθ)^2 =(isinθ)^2 ⇒ x−cosθ =+^− isinθ ⇒x =cosθ +^− isinθ so the values are λ_1 =e^(iθ) and λ_2 =e^(−iθ) V(λ_1 )=Ker(A−λ_1 I) ={u /(A−λ_1 I)u =0} u ((x),(y) ) (A−λ_1 )u=0 ⇒ (((cosθ−e^(iθ) −sinθ)),((sinθ cosθ−e^(iθ) )) ) ((x),(y) )=0 ⇒ (((−isinθ −sinθ)),((sinθ −isinθ)) ) ((x),(y) )=0 ⇒ { ((−isinθ x−sinθy =0)),((sinθx −isinθy =0)) :} ⇒ { ((isinθ x+sinθ y =0)),((sinθ x−isinθ y=0 ⇒sinθ x =isinθ y)) :} if sinθ ≠0 ⇒x =iy ⇒(x,y)=(iy,y) =y e_1 with e_1 ((i),(1) ) V(λ_2 ) =Ker(A−λ_2 I) ={u /(A−λ_2 I)u =0} (A−λ_2 I)u =0 ⇒ (((cosθ −e^(−iθ) −sinθ)),((sinθ cosθ −e^(−iθ) )) ) ((x),(y) )=0 ⇒ (((isinθ −sinθ)),((sinθ isinθ)) ) ((x),(y) )=0 ⇒ { ((isinθ x−sinθy =0)),((sinθ x +isinθ y =0 )) :} ⇒sinθ x =−isinθ y so if sinθ ≠0 we get x=−iy ⇒ (x,y) =(−iy,y)=ye_2 with e_2 (((−i)),(1) ) We see V(λ_1 ) =Δ_e_1 dim(V(λ_1 ))=1 V(λ_2 )=Δ_(e_2 ) and dim(V(λ_2 ))=1 A is diagonalisable....$
$1) P (x) = d e t (A - x I) = \| \begin{matrix} c o s θ - x - s i n θ \\ s i n θ c o s θ - x \end{matrix} \|$ $= {(c o s θ - x)}^{2} + {s i n}^{2} θ$ $P (x) = 0 \Leftrightarrow {(x - c o s θ)}^{2} + {s i n}^{2} θ = 0 \Leftrightarrow {(x - c o s θ)}^{2} = {(i s i n θ)}^{2} \Rightarrow$ $x - c o s θ = \bar{+} i s i n θ \Rightarrow x = c o s θ \bar{+} i s i n θ s o t h e v a l u e s a r e$ $λ_{1} = e^{i θ} a n d λ_{2} = e^{- i θ}$ $V (λ_{1}) = K e r (A - λ_{1} I) = {u / (A - λ_{1} I) u = 0}$ $u (\begin{matrix} x \\ y \end{matrix}) (A - λ_{1}) u = 0 \Rightarrow (\begin{matrix} c o s θ - e^{i θ} - s i n θ \\ s i n θ c o s θ - e^{i θ} \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow$ $(\begin{matrix} - i s i n θ - s i n θ \\ s i n θ - i s i n θ \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow {\begin{cases} - i s i n θ x - s i n θ y = 0 \\ s i n θ x - i s i n θ y = 0 \end{cases}$ $\Rightarrow {\begin{cases} i s i n θ x + s i n θ y = 0 \\ s i n θ x - i s i n θ y = 0 \Rightarrow s i n θ x = i s i n θ y \end{cases}$ $i f s i n θ \neq 0 \Rightarrow x = i y \Rightarrow (x, y) = (i y, y) = y e_{1} w i t h e_{1} (\begin{matrix} i \\ 1 \end{matrix})$ $V (λ_{2}) = K e r (A - λ_{2} I) = {u / (A - λ_{2} I) u = 0}$ $(A - λ_{2} I) u = 0 \Rightarrow (\begin{matrix} c o s θ - e^{- i θ} - s i n θ \\ s i n θ c o s θ - e^{- i θ} \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow$ $(\begin{matrix} i s i n θ - s i n θ \\ s i n θ i s i n θ \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = 0 \Rightarrow {\begin{cases} i s i n θ x - s i n θ y = 0 \\ s i n θ x + i s i n θ y = 0 \end{cases}$ $\Rightarrow s i n θ x = - i s i n θ y s o i f s i n θ \neq 0 w e g e t x = - i y \Rightarrow$ $(x, y) = (- i y, y) = {y e}_{2} w i t h e_{2} (\begin{matrix} - i \\ 1 \end{matrix})$ $W e s e e V (λ_{1}) = Δ_{e_{1}} d i m (V (λ_{1})) = 1$ $V (λ_{2}) = Δ_{e_{2}} a n d d i m (V (λ_{2})) = 1$ $A i s d i a g o n a l i s a b l e . . . .$
Commented by M±th+et+s last updated on 30/Apr/20

$n i c e a n d c o r r e c t w o r k s i r g o d b l e s s y o u$
Commented by turbo msup by abdo last updated on 30/Apr/20

$y o u a r e w e l c o m e .$
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