A linear algebra question regarding the eigenvalues of the product of a diagonal matrix and a projection matrix I need to prove a statement in my research. The statement seems to be fundamental linear algebra, and numerical studies in MATLAB supported this statement, but I wasn't able to prove it after a few days of effort... Any insight would be greatly appreciated!
The problem is as follows:

Let $A$ be a $p$-by-$p$ diagonal matrix with distinct positive diagonal elements $a_1, \dots, a_p$. Let $Z := A^{1/2} 1_p$, where $1_p$ is a $p$-vector of ones. Let $$P := I - Z \left( Z^T Z \right)^{-1}Z^T$$ where $I$ is a $p$-by-$p$ identity matrix. Let $\xi_1,\dots,\xi_{p-1}$ be the $p-1$ nonzero eigenvalues of $AP$. Show that $1+t\xi_i$, $i=1,\dots,p-1$ are $p-1$ nonzero eigenvalues of $P+tAP$, where $t \in \mathbb R$ and $t \neq 0$.

 A: I was able to prove the statement and generalize it to the following lemma:
Lemma: Let A be a full-rank $p\times p$ symmetric matrix and P be a $p\times p$ projection matrix with rank $r<p$. Let $\xi_1, \dots, \xi_{r}$ be the $r$ nonzero eigenvalues of $AP$. Then for any $t\in\mathbb R$, $1 + t\xi_i$, $i=1, \dots, r$ are the $r$ nonzero eigenvalues of $(I_p + tA)P$.
Proof: 
If $t=0$, the result follows immediately from the fact that the eigenvalues of a projection matrix can be either 1 or 0.
If $t\neq 0$, we note that both $A$ and $P$ are symmetric matrices and thus $(AP)^T = PA$. Since transposing a matrix doesn't affect its eigen values, we study the eigenvalues of $PA$ and $P+tPA$. 
Let $X_i$ be the eigen-vector of $PA$ corresponding to $\xi_i$, $i=1,\dots, r$. Then $PAX_i=\xi_iX_i$. We see that 
$$PX_i = \frac{1}{\xi_i}PPAX_i = \frac{1}{\xi_i}PAX_i=X_i.$$
Therefore 
$$(P+tPA)X_i=PX_i+tPAX_i = X_i + t\xi_iX_i=(1 + t\xi_i)X_i$$
which shows that $(1 + t\xi_i)$ for $i=1, \dots,r$ are $r$ eigenvalues of $P+tPA$ with eigenvectors $X_1, \dots, X_{r}$.
