Are all eigenvalues of $((C^TQ_fC)^{-1} (C^TQ_gC))$ lie in $(0,1]$? Consider a matrix $X_f = (C^TQ_fC)^{-1}$ and $X_g = (C^TQ_gC)^{-1}$, where $C \in \mathbb{R}^{n \times m}$ is a full column tall matrix ($m < n$).
$Q$ ($\in \mathbb{R}^{n \times n}$) is a diagonal matrix. In particular, $Q_f = diag(q_1,...,q_n)$, $Q_g = diag(q_1,...,q_{n-3}, 0,0,0)$. $q_i$'s are positive real values.
Considering $C_i^T$ as the $i$th row of matrix $C$, we can write,
$C^TQ_fC = C^TQ_gC + \underbrace{q_{n-2}C_{n-2}C_{n-2}^T}_{\textrm{rank-1 matrix}}$ $+ \underbrace{q_{n-1}C_{n-1}C_{n-1}^T}_{\textrm{rank-1 matrix}}$ $+ \underbrace{q_{n}C_{n}C_{n}^T}_{\textrm{rank-1 matrix}}$
Note that $C$ and $Q$ have been chosen such that $C^TQ_fC$ and $C^TQ_gC$ are always positive definite (and symmetric too).
Can we prove that all eigenvalues of $X_fX_g^{-1}$ lie in $(0,1]$??
My understanding is as follows:
Using Sherman–Morrison formula, We can express,
$X_fX_g^{-1} = ((C^TQ_gC)^{-1} - P)(C^TQ_gC) = I - P(C^TQ_gC)$
where $P$ is appropriately obtained matrix after applying Sherman–Morrison formula to express $X_f$. I can show that $P$ is a positive definite matrix for the above mentioned case. Now, to prove all eigenvalues of $X_fX_g^{-1}$ lie in $(0,1]$, I need to prove all eigenvalues of $P(C^TQ_gC)$ lie in $[0,1)$. Here, I am stuck.
Please help. Thanks in advance.
 A: There are some issues here, e.g.  if $m=n-1$ then $X_g = (C^TQ_gC)^{-1}$ doesn't exist.  But I proceed on the assumption, stated by OP that $X_g \succ \mathbf 0$.
Symmetry is to be preferred so instead focus on the similar matrix $X_f^\frac{1}{2}X_g^{-1}X_f^\frac{1}{2}\succ \mathbf 0$
(i.e. it is congruent to $X_g^{-1}$ and hence positive definite)
1.) to prove that $\lambda_{\max}\big(X_f^\frac{1}{2}X_g^{-1}X_f^\frac{1}{2}\big)\leq 1$ it suffices to consider the inverse problem of proving  $\lambda_{\min}\big(X_f^\frac{-1}{2}X_gX_f^\frac{-1}{2}\big)\geq 1$ and observe that $\big(X_f^\frac{-1}{2}X_gX_f^\frac{-1}{2}\big)$ has the same spectrum as  $\big(X_g^\frac{1}{2}X_f^{-1}X_g^\frac{1}{2}\big)$.  So we want to prove $\lambda_{\min}\big(X_g^\frac{1}{2}X_f^{-1}X_g^\frac{1}{2}\big)\geq 1$
2.)
A nice way to proceed involves using OP's decomposition
$X_f^{-1} $
$=C^TQ_fC $
$= C^TQ_gC + \underbrace{q_{n-2}C_{n-2}C_{n-2}^T}_{\textrm{rank-1 matrix}}$ $+ \underbrace{q_{n-1}C_{n-1}C_{n-1}^T}_{\textrm{rank-1 matrix}}$ $+ \underbrace{q_{n}C_{n}C_{n}^T}_{\textrm{rank-1 matrix}}$
$=C^TQ_gC+\mathbf x'_1\mathbf x_1^T+\mathbf x'_2\mathbf x_2^T++\mathbf x'_3\mathbf x_3^T$
now multiply on left and right by $X_g^\frac{1}{2}$
$X_g^\frac{1}{2}X_f^{-1}X_g^\frac{1}{2}$
$=\big(C^TQ_gC\big)^\frac{-1}{2}C^TQ_fC\big(C^TQ_gC\big)^\frac{-1}{2} $
$=\big(C^TQ_gC\big)^\frac{-1}{2}\Big(C^TQ_gC+\mathbf x'_1\mathbf x_1^T+\mathbf x'_2\mathbf x_2^T++\mathbf x'_3\mathbf x_3^T\Big)\big(C^TQ_gC\big)^\frac{-1}{2}$
$ =I+\mathbf y'_1\mathbf y_1^T+\mathbf y'_2\mathbf y_2^T+\mathbf y'_3\mathbf y_3^T$
the conclusion then follows immediately from either
(a) interlacing
(b) observing that
$\lambda_{\min}\big(X_g^\frac{1}{2}X_f^{-1}X_g^\frac{1}{2}\big)$
$=\min_{\Vert \mathbf z\Vert_2=1}\mathbf z^T\big(X_g^\frac{1}{2}X_f^{-1}X_g^\frac{1}{2}\big)\mathbf z$
$=\min_{\Vert \mathbf z\Vert_2=1}\mathbf z^TI\mathbf z+\mathbf z^T\big(\mathbf y'_1\mathbf y_1^T +\mathbf y'_2\mathbf y_2^T+\mathbf y'_3\mathbf y_3^T\big)\mathbf z$
$\geq \Big(\min_{\Vert \mathbf z\Vert_2=1}\mathbf z^TI\mathbf z\Big)+\Big(\min_{\Vert \mathbf z'\Vert_2=1}(\mathbf z')^T\big(\mathbf y'_1\mathbf y_1^T +\mathbf y'_2\mathbf y_2^T+\mathbf y'_3\mathbf y_3^T\big)\mathbf z'\Big)$
$= 1 + 0$
$=1$
