Let A be a diagonal matrix of size $N$. I am trying to find the eigenvalues of $A+P^{-1} \alpha A P$ where $\alpha$ is a real number. Here is my attempt:
\begin{equation} (A+P^{-1} \alpha A P)v=\lambda v, \end{equation} \begin{equation} (P^{-1} \alpha A P)v=(\lambda I - A) v, \end{equation} \begin{equation} (\alpha A P)v=P(\lambda I - A) v, \end{equation} \begin{equation} Lv=R v, \end{equation}
where $L=\alpha A P$ and $R=P(\lambda I - A)$. Assuming that $v\neq 0$, we can conclude that $L=R$ (?). Therefore, by equating corresponding elements of L and R:
\begin{equation} l_{ij}=\sum_{k=1}^{N}\alpha a_{ik}p_{kj}=\sum_{k=1}^{N}p_{ik}(\lambda 1_{kj} - a_{kj})=r_{ij} \end{equation}
Since A is a digonal matrix, $a_{ij} = 0$ for $i \neq j$. Hence,
\begin{equation} \lambda = a_{ii}+\alpha a_{jj} \end{equation}
Question 1: Is my calculation correct?
Question 2: How do I proceed from here to find the $N$ eigenvalues? Since $i$ and $j$ can be anything from $1$ to $N$, from this calculation, it looks like that there are $N^{2}$ eigenvalues! This is a sign that I did something wrong, but I cannot find where it went wrong.