Eigenvector of matrix with all positive entries

Question

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.

Answer 1

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.

Answer 2

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.