Let $$A=\begin{bmatrix}\alpha &0\\0&\alpha^{-1}\end{bmatrix}, B=\begin{bmatrix}\beta&-1\\0&\beta \end{bmatrix},$$ where $\alpha, \beta>0.$
I want to understand why the line spanned by (1,0) is the common (attracting) fixed point for $A,B$ on the projective space $\mathbb{P}$. Thanks in advance.
My attempt: I can show that the line spanned by (1,0) is an eigenspace of $A$ and $B$, but I want to understand how it works on the projective space.
Saying that $(x:y)$ is a fixed point for a matrix $A$ in the projective space is exactly saying that if $\begin{pmatrix}x'\cr y'\end{pmatrix}:= A\begin{pmatrix}x\cr y\end{pmatrix}$, then $(x':y')=(x:y)$ which means that there exists an nonzero $\lambda$ such that $\begin{pmatrix}x'\cr y'\end{pmatrix}= \lambda\begin{pmatrix}x\cr y\end{pmatrix}$. This is equivalent to asking $A\begin{pmatrix}x\cr y\end{pmatrix}= \lambda\begin{pmatrix}x\cr y\end{pmatrix}$ for some nonzero $\lambda$.
In other words, $(x:y)$ is a fixed point for $A$ if and and only if $\begin{pmatrix}x\cr y\end{pmatrix}$ is an eigenvector corresponding to a nonzer eigenvalue of $A$.
Now the only eigenvalue of $B$ is $\beta$ and the corresponding eigenspace is the line generated by $\begin{pmatrix}1\cr 0\end{pmatrix}$. It is then easy to see that any element of this line is also an eigenvector for $A$, and we are done.