# 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 $$t \neq 0$$.

• What does $1+t\xi_i$ mean? – Darth Vader Apr 13 at 19:29
• 1 + t$\xi$ is "the sum of one and the product of t and $\xi_i$" – Ken Apr 13 at 19:42
• $\xi$ is a vector and $1$ is a number- that's why I am confused by your notation. – Darth Vader Apr 13 at 19:49
• @DarthVader Each $\xi_i$ is a number (one of the $p - 1$ eigenvalues of $AP$). – user44191 Apr 13 at 19:51
• I see- I mistakenly read $\xi$ as eigenvector :). Sorry, – Darth Vader Apr 13 at 19:52

Lemma: Let A be a full-rank $$p\times p$$ symmetric matrix and P be a $$p\times p$$ projection matrix with rank $$r. Let $$\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}$$.