Let $\phi \in \mathbb{R}^{p \times n}$ and $R \in \mathbb{S}^{n \times n}_{++}$ (symmetric positive definite), where $n > p$. Assume that there exists $0 < \alpha < \beta$ such that $\alpha I \prec \phi \phi^{\rm T} \prec \beta I$ and $\gamma > 0$ such that $R \succ \gamma I$.
Is there some lower bound I can place on ${\rm Trace}(\phi R \phi^{\rm T})$ as a function of ${\rm Trace}(R)$? For example, some $\varepsilon > 0$ such that $\varepsilon \, {\rm Trace}(R) < {\rm Trace}(\phi R \phi^{\rm T})$?
I have that since $\alpha I \prec \phi \phi^{\rm T}$ and $R$ is positive definte, it follows that $\phi R \phi^{\rm T}$ is full rank, and therefore positive definite. Thus ${\rm Trace}(\phi R \phi^{\rm T}) > 0$. But I'm not sure if $ {\rm Trace}(\phi R \phi^{\rm T}) $ can be arbitrarily close to $0$.
No. The infimum of $\operatorname{tr}(\phi R\phi^T)$ is $p\alpha\gamma$. This can be proved easily by performing a singular value decomposition on $\phi$ or by applying von Neumann's trace inequality to $\operatorname{tr}(\phi^T\phi R)$. This infimum value doesn't depend on $\operatorname{tr}(R)$. Consider e.g. $\phi=\pmatrix{I_p&0}$ and $R=\pmatrix{I_p\\ &cI_{n-p}}$ where $c\ge1$ is arbitrarily large. We have $\operatorname{tr}(\phi R\phi^T)=p$ for all $c$ but $\operatorname{tr}(R)\to\infty$ as $c\to\infty$.