Eigenvalues of the product of these two matrices 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 :)
 A: 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 $\lambda \leq 1$.
