eigenvalues of fitted value?

Question

Let a $m\times n$ full-column matrix be $A$. Also, denote $m\times m$ diagonal matrix as $D$.

Define $P(A)D=A(A'A)^{-1}A'D$ as the orthogonal projection of $D$ onto the column space of $A$, or "fitted value" in regression term. Note that $(A'A)$ is full rank, so $P(A)$ always exists.

I'm wondering how the eigenvalues of $P(A)D$ are determined. Can we use the fact that $D$ is already a diagonal matrix?

Answer 1

It's not true that the positive eigenvalues of $P(A)D$ are diagonal elements of $D$. Consider e.g. $$P(A) = \pmatrix{8/9 & -2/9 & -2/9\cr -2/9 & 5/9 & -4/9\cr -2/9 & -4/9 & 5/9\cr}, \ D = \pmatrix{a & 0 & 0\cr 0 & b & 0\cr 0 & 0 & c}$$ Then $P(A)D$ has characteristic polynomial $${t}^{3}- \left(\frac89 a +\frac59 b + \frac59 c\right)t^2+ \frac{4ac+bc+4ab}{9} t $$ and none of $a,b,c$ are roots of this.