We are given a vector $z$ and a matrix $A$. We know that $C = AA^T$ and C is a symmetric, positive definite matrix.
Furthermore we are given the equation: $R = \frac{1}{z^Tz} Az(Az)^T$
What can we say about the new matrix $R$ (Eigenvectors, Eigenvalues, trace etc.)? How does the first part of the term $\frac{1}{z^Tz}$ relate to second one $Az(Az)^T$ and to $R$? Is it possible to express $R$ in terms of $C$?
In my understanding $\frac{1}{z^Tz}$ is some kind of normalization factor, bringing the vectors $z$ to unit length.
Define the vectors $$\eqalign{ u &= \frac{z}{\|z\|},\quad x=Au\in{\mathbb R}^n\\ }$$ Then $R = xx^T$ is a rank-one matrix and therefore has only one non-zero eigenvalue, which is simple to calculate. $$Rx = xx^Tx = \lambda x \quad\implies\lambda = x^Tx = \frac{z^TA^TAz}{z^Tz}$$ The other $(n-1)$ eigenvalues equal zero, and the corresponding eigenvectors are
any set of $(n-1)$ linearly independent vectors which are orthogonal to $x$.