I am currently working on the following problem: $$ \textbf{Y} \in \mathbb{R}^{n \times q}, \textbf{X} \in \mathbb{R}^{n \times p} $$ $$ \textbf{Q} \in \mathbb{R}^{n \times n} (\hbox{symmetric, positive definite and invertible)} $$ Furthermore, $n > > p,q$.
I've managed to get to a stage where I'm really wanting to find some bounds of the eigenvalues of the following quantity. I'm really hoping to either prove or disprove that its eigenvalues are between $0$ and $1$.
I would like to find the eigenvalues of: \begin{equation} (\textbf{Y}^\top \textbf{Q}^{-1} \textbf{H}\textbf{Y})(\textbf{Y}^{\top} \textbf{Q}^{-1} \textbf{Y})^{-1} \end{equation}
Where $\textbf{Y}^{\top} \textbf{Q}^{-1} \textbf{Y} $ is invertible and all other matrices are of full rank.
$\textbf{H}$ is an $n \times n$ idempotent matrix such that $$\textbf{H} = \textbf{X}(\textbf{X}^\top \textbf{Q}^{-1} \textbf{X})^{-1} \textbf{X}^\top \textbf{Q}^{-1}$$
Any help would be appreciated! Thanks in advance :)
Let $U \Sigma U^\top$ be the singular value decomposition of $Q$. Introducing $$ S = Q^{-1/2} \equiv U \Sigma^{-1/2} U^\top\\ Z = S Y\\ W = S X $$ the problem reduces to studying eigenvalues of $$ A = (Z^\top S H S^{-1} Z)(Z^\top Z)^{-1}\\ H = S^{-1}W(W^\top W)^{-1}W^\top S. $$ Collecting together it gives $$ A = (Z^\top W(W^\top W)^{-1}W^\top Z)(Z^\top Z)^{-1}. $$ Eigenvalues of $A$ are solutions to following problem $$ \operatorname{det} \left(Z^\top W(W^\top W)^{-1} W^\top Z - \lambda Z^\top Z\right) = 0 $$ Let $(x,\lambda)$ be an eigenpair of the problem $$ Z^\top W(W^\top W)^{-1} W^\top Z x = \lambda Z^\top Z x\\ x^\top Z^\top W(W^\top W)^{-1} W^\top Z x = \lambda x^\top Z^\top Z x $$ Denote $y = Zx$ $$ y^\top W(W^\top W)^{-1}W^\top y = \lambda y^\top y. $$ Since $W^\top W$ is positive definite, the expression on the left is nonnegative. Thus $\lambda \geq 0$.
This eigenvalue $\lambda$ is bounded by $$ \lambda \leq \max_{y = Zx, x\in \mathbb{R}^q} \frac{y^\top W(W^\top W)^{-1}W^\top y}{y^\top y} \leq \max_{y \in \mathbb{R}^n} \frac{y^\top W(W^\top W)^{-1}W^\top y}{y^\top y} $$ Let $U\Sigma V\top$ be the SVD decomposition for $W$ (the different one I've used for $Q$). $$ U \in \mathbb{R}^{n\times n}\\ \Sigma \in \mathbb{R}^{n\times p}\\ V \in \mathbb{R}^{p\times p} $$ Now $$ \lambda \leq \max_{y \in \mathbb{R}^n} \frac{y^\top W(W^\top W)^{-1}W^\top y}{y^\top y} = \max_{y \in \mathbb{R}^n} \frac{y^\top U\Sigma V^\top(V\Sigma^\top \Sigma V^\top)^{-1}V\Sigma^\top U^\top y}{y^\top y} = \\ = \max_{z \in \mathbb{R}^n} \frac{z^\top \Sigma V^\top(V\Sigma^\top \Sigma V^\top)^{-1}V\Sigma^\top z}{z^\top z} $$ Note that $$ (V\Sigma^\top \Sigma V^\top)^{-1} = V(\Sigma^\top \Sigma)^{-1} V^\top $$ Substituting that gives $$ \lambda \leq \max_{z \in \mathbb{R}^n} \frac{z^\top \Sigma (\Sigma^\top \Sigma )^{-1}\Sigma^\top z}{z^\top z}. $$ Let $\Sigma$ be $$ \Sigma = \begin{pmatrix} \sigma_1 \\ & \sigma_2 \\ && \ddots \\ &&&\sigma_p\\ &&{\large 0} \end{pmatrix} $$ Direct computation shows that $\Sigma (\Sigma^\top \Sigma )^{-1}\Sigma^\top$ is $$ \Sigma (\Sigma^\top \Sigma )^{-1}\Sigma^\top = \begin{pmatrix} I_p & 0\\ 0 & 0 \end{pmatrix} $$ which finishes the proof that $\lambda \leq 1$.