Given
$$\lambda \left( Daa^TD \right) = \begin{cases} 0 & \text{with multiplicity}~ d-1 \\ 1 & \text{with multiplicity}~1 \end{cases}$$
is it possible to compute $\lambda_{\min}$ and $\lambda_{\max}$ of $D-Daa^TD$ where $D$ is a diagonal matrix with $D_{ii}\geq 0$ and $a\in \mathbb{R}^d$ column vector.
Thanks.
The only information we have about $M = Daa^TD$ is that $M$ is a rank-$1$ positive definite symmetric matrix with trace $1$ whose image lies inside the image of $D$. This is not enough to say much at all about the eigenvalues of $D-Daa^TD$.
Let $d_{\max},d_{\min}$ denote the maximum and minimum diagonal entries of $D$. Using Weyl's inequalities (or a similar, weaker result), we can show that $$ d_{\max} - 1 \leq \lambda_{\max} \leq d_{\max}, \qquad d_{\min} - 1 \leq \lambda_{\min} \leq d_{\min} . $$ These inequalities are strict, and it is possible to find a vector $a$ that produces any desired $\lambda_{\max}$ (or $\lambda_{\min}$) within the interval defined above.