Similar setup to this previous question of mine
Let $R \in \mathbb{R}^{n \times n}$ be positive definite and $\phi \in \mathbb{R}^{p \times n}$. Assume that $p < n$ and that $\phi$ is rank $p$. Define the matrix
$$ R^\bot = R-R \phi^{\rm T} (\phi R \phi^{\rm T})^{-1} \phi R,$$
which is rank $n-p$ and positive semidefinite (proof in Lemma 3 and 4 of here) and also has the property that $R^\bot \phi = 0$. Since ${\rm rank}(R^\bot) = n-p$, $R^{\bot}$ has $n-p$ nonzero eigenvalues. I want to show that the nonzero eigenvalues of $R^\bot$ are bounded below by $\lambda_{\rm min}(R)$ and bounded above by $\lambda_{\rm max}(R)$ (min and max eigenvalues of $R$).
A proof for upperbound I think is pretty simple, as $R^\perp + R \phi^{\rm T} (\phi R \phi^{\rm T})^{-1} \phi R = R$, and both $R^\bot$ and $R \phi^{\rm T} (\phi R \phi^{\rm T})^{-1} \phi R$ are symmetric, positive semidefinite. So, Courant-Fischer min-max Theorem can be used to show that
$$ \lambda_{\rm max}(R^\bot) \le \lambda_{\rm max}(R).$$
I'm not sure how to approach the the other bound, as clearly $\lambda_{\rm min}(R^\bot) = 0$ but I am interested in the nonzero eigenvalues. I'm fairly certain that the nonzero eigenvalues of $R^\bot$ are, in fact, bounded below by $\lambda_{\rm min}(R)$ based on numerical tests but I could also be mistaken.
$R^{1/2}\phi^T(\phi R\phi^T)^{-1}\phi R^{1/2}$ is a rank-$p$ orthogonal projection. Hence $\Pi=I-R^{1/2}\phi^T(\phi R\phi^T)^{-1}\phi R^{1/2}$ is an orthogonal projection of rank $n-p$, $R^\perp=R^{1/2}\Pi R^{1/2}=(R^{1/2}\Pi)(\Pi R^{1/2})$ and it shares the same spectrum as $\Pi R\Pi$. It follows (from Courant-Fischer minimax principle or Cauchy's interlacing inequality) that the minimum nonzero eigenvalue of $R^\perp$ is bounded below by the smallest eigenvalue of $R$, which is positive.