Eigenvector of matrix with all positive entries If 
$A=\begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix}$
has all positive real entries, satisfies $AX=Y$, for 
$X=\begin{bmatrix}
x \\
y \\
\end{bmatrix}$
, $Y$ is a scalar multiple of $X$, and $X$ is an eigenvector of $A$. Then prove there exists an eigenvector $X$ in the first and second quadrant. (That is, $x,y\geq0$ and $x\leq0,y\geq0$).
Through some computation, I have determined ratios for the entries of the eigenvector $X$ as:
$$\frac {y}{x}=\frac {\lambda-a}{b} \ \ \ \ \ \frac {y}{x}= \frac {c}{\lambda-d}$$
and
$$\frac {y}{x}= \frac {d-a\pm\sqrt{(a+d)^2-4(ad-bc)}}{2b}$$
I am not sure if the above formulas may help, but I am rather stuck on the above question. Any help would be much appreciated. Thanks in advance.
 A: You have almost got the answer! For the second eigen vector take $x=-1$ and note that $d-a - \sqrt {(a+d)^{2}-4(ad-bc)}$ is negative: this follows from the fact that $4(ad-bc)<4ad$. Similarly, for the first eigen vector take $x=1$ and choose the plus sign.
A: if eigenvaule are $ \lambda _{1}$ and $\lambda _{2}$,
we have
$\lambda _{1}+ \lambda _{2}=a+d   $
$\lambda _{1}\times  \lambda _{2}=ad-bc  $
$
    \begin{bmatrix}
    x_1  \\
    y_1  \\
    \end{bmatrix}                     
 \begin{bmatrix}
    x_2  \\
    y_2 \\
    \end{bmatrix}  $are two eigenvectors with $ \lambda _{1}$ and $\lambda _{2}$,
it implies that
$$
    \begin{bmatrix}
    x_1  \\
    y_1  \\
    \end{bmatrix}                     
 \begin{bmatrix}
    a & b  \\
    c & d  \\
    \end{bmatrix}=  \lambda _{1}   \begin{bmatrix}
    x_1  \\
    y_1  \\
    \end{bmatrix}   $$
means
$$
    \begin{bmatrix}
    x_1  \\
    y_1  \\
    \end{bmatrix}                     
 \begin{bmatrix}
    a- \lambda _{1}  & b  \\
    c & d- \lambda _{1}   \\
    \end{bmatrix}=  \lambda _{1}   \begin{bmatrix}
    0  \\
    0  \\
    \end{bmatrix}   $$
so
$\left ( a-\lambda _{1} \right )\cdot x_{1}+by_{1}=0$
similarly $\left ( a-\lambda _{2} \right )\cdot x_{2}+by_{2}=0$
$\frac{x_{1}}{y_{1}}=\frac{\lambda _{1}-a}{b}$,
$\frac{x_{2}}{y_{2}}=\frac{\lambda _{2}-a}{b}
$
then
$\frac{x_{1}\times x_{2} }{y_{1}\times y_{2}}=\frac{\lambda _{1}\cdot \lambda _{2}-a\left (\lambda _{1}+\lambda _{2}  \right )+a^{2}}{b^{2}}
$
then $\frac{x_{1}\times x_{2} }{y_{1}\times y_{2}}=-\frac{c}{b}< 0
$
We may assume $\frac{x_{1}}{y_{1}}> 0$,
then 
$$ \begin{bmatrix}
    x_1  \\
    y_1  \\
    \end{bmatrix} or 
 \begin{bmatrix}
   - x_1  \\
   - y_1  \\
    \end{bmatrix} 
$$ in the first quadrant,
$$ \begin{bmatrix}
    x_2  \\
    y_2  \\
    \end{bmatrix} or 
 \begin{bmatrix}
   - x_2  \\
   - y_2  \\
    \end{bmatrix} 
$$ in the second quadrant.
Other cases are same.
