#Title:  Valores e vectores proprios em dimencao 2

#Let:

n  e_11 = [ 1,  2,  3,  -2,  -1,  -3,  4 ] ;

n  e_12 = [ 1,  2,  3,  -2,  -1,  -3,  4 ] ;

n  e_21 = [ 1,  2,  3,  -2,  -1,  -3,  4 ] ;

n  e_22 = [ 1,  2,  3,  -2,  -1,  -3,  4 ] ;

n  m1 = [ 5,  2,  3,  -2,  -2,  -3,  4 ] ;

n  m2 = [ 5,  2,  3,  -2,  -2,  -3,  4 ] ;

cn  a = #m1 * #e_22 * # e_11 - #m2 * #e_12 * # e_21 ;

cn  b = #m2 * #e_21 * # e_11 - #m1 * #e_11 * # e_21 ;

cn  c = #m1 * #e_22 * # e_12 - #m2 * #e_12 * # e_22 ;

cn  d = #m2 * #e_21 * # e_12 - #m1 * #e_11 * # e_22 ;

cn  del = #e_11 * # e_22 -  #e_21 * # e_12 ;

cn  lambda_1 = #del * #m1 ;

cn  lambda_2 = #del * #m2 ;

cn  f = - (#a + #d) ;

cn  g = - (#b * #c) ;


#Question:

\noindent Encontre os valores e vectores pr\'oprios da matriz: 

$$
A=\left(
\begin{array}{cc}
a & b\\
c & d
\end{array}
\right).
$$

#Suggestion:

\noindent Os valores $\lambda_1$ e $\lambda_2$ pr\'oprios s\~ao solu\c c\~oes da equa\c c\~ao caracter\'\i stica
$$
{\rm det}(A-\lambda I)=
\{\rm det}
\left(
\begin{array}{cc}
a-\lambda & b\\
c & d-\lambda
\end{array}
\right) =0.
$$
Os vectores pr\'oprios correspondentes $e_1=(e_{11},e_{12})$ e $e_2=(e_{21},e_{22})$ s\~ao vectores n\~ao triviais que verificam as equa\c c\~oes 
$Ae_k=\lambda_k e_k$, $k=1,2$, respectivamente.


#Solution:

\noindent A equa\c c\~ao caracter\'\i stica ${\rm det}(A-\lambda I)=0$ \'e
$$
\lambda^2 + #f \lambda + #g =0.
$$
As suas raizes s\~ao $\lambda_1 = #lambda_1$ e $\lambda_2 = #lambda_2$. Das equa\c c\~oes
\begin{eqnarray*}
&& #a * e_{11} + #b * e_{12} = #\lambda_1 e_{11},\\
&& #a * e_{21} + #b * e_{22} = #\lambda_1 e_{21},
\end{eqnarray*}
encontramos $e_1=(#e_11,#e_12)$ e $e_2=(#e_21,#e_22)$. 


#Result:

$$
\lambda_1 = #lambda_1,
$$
$$
e_1=(#e_11,#e_12),
$$
$$
\lambda_2 = #lambda_2,
$$
$$
e_1=(#e_21,#e_22).
$$


#Verification:

eq ( #e_11 * #e_12resposta - #e_12 * #e_11resposta = 0 ) ;

eq ( #e_21 * #e_22resposta - #e_22 * #e_21resposta = 0 ) ;