Consider symmetric positive definite matrices $X_{f} = (C^TQ_fC)^{-1}$, $~~ X_{g} = (C^TQ_gC)^{-1}$, $~~X_{fp} = (C^T(Q_f+Q_p)C)^{-1}$, $~~ X_{gp} = (C^T(Q_g+Q_p)C)^{-1}$, where $C \in \mathbb{R}^{n \times m}$ is a full column tall matrix ($n > m$). (additional info: $X$ are covariance matrices for me; Size of $C$ is such that $X \succ \mathbf 0$)
$Q$ ($\in \mathbb{R}^{n \times n}$) is a diagonal matrix. In particular, $Q_f = diag(q_1,...,q_n)$; $Q_g = diag(q_1,...,q_{n-3}, 0,0,0)$; $Q_p = diag(p_1,...,p_n)$; $q_i$'s and $p_i$'s are positive real values (additionally, in my case $q_i >> p_i$ and $p_i < 1$).
Can we prove that $\mathbf{eig(X_{f} X_{g}^{-1}) \leq eig(X_{fp}X_{gp}^{-1})}$ ?
My attempt to prove it is as follows (one can ignore this):
Here, I am trying to relate $eig(X_{fp}X_{gp}^{-1})$ with $eig(X_{fp}X_{g}^{-1})$ and then with $eig(X_{f}X_{g}^{-1})$. I have proved $eig(X_{fp}X_{g}^{-1}) \leq eig(X_{fp}X_{gp}^{-1})$, but my approach lead to inconclusive relationship between $eig(X_{fp}X_{g}^{-1})$ and $eig(X_{f}X_{g}^{-1})$.
In fact, numerically I know that, $~~eig(X_{fp}X_{g}^{-1}) \leq eig(X_{f} X_{g}^{-1}) \leq eig(X_{fp}X_{gp}^{-1})$, but I am not able to prove this.
$eig(X_{fp}X_{gp}^{-1}) $
$= eig(X_{fp}^\frac{1}{2}X_{gp}^{-1}X_{fp}^\frac{1}{2})~~~~$ [similar matrices]
$= eig(X_{fp}^\frac{1}{2}(C^T(Q_g+Q_p)C)X_{fp}^\frac{1}{2})$
$= eig(X_{fp}^\frac{1}{2}(C^TQ_gC) X_{fp}^\frac{1}{2} + \underbrace{X_{fp}^\frac{1}{2} (C^TQ_pC)X_{fp}^\frac{1}{2})}_{\textrm{symmetric positive definite}}$
$\geq eig(X_{fp}^\frac{1}{2}(C^TQ_gC) X_{fp}^\frac{1}{2}~~~~$ [Weyl Theorem: $\lambda_k(A)+\lambda_1(B) \leq \lambda_k(A+B)$, when $A$ and $B$ are Hermitian ]
$= eig(X_{fp}X_{g}^{-1})$
Now, $eig(X_{fp}X_{g}^{-1})$
$= eig(X_{g}^\frac{-1}{2}X_{fp}X_{g}^\frac{-1}{2})~~~~$
$= eig(X_{g}^\frac{-1}{2}(X_{f} - X_{f}A^T(Q_p^{-1}+AX_{f}A^T)^{-1}AX_{f})X_{g}^\frac{-1}{2})~~~~$ [$X_{fp} = (C^TQ_fC+C^TQ_pC)^{-1}$ and using Sherman–Morrison formula]
$= eig(X_{g}^\frac{-1}{2}X_{f} X_{g}^\frac{-1}{2} - \underbrace{ X_{g}^\frac{-1}{2} ( X_{f}A^T(Q_p^{-1}+AX_{f}A^T)^{-1}AX_{f})X_{g}^\frac{-1}{2})}_{\textrm{I can show that it is positive definite}}~~~~$
$\geq eig(X_{f} X_{g}^{-1}) + \textrm{some negative value}~~~~$ [Weyl Theorem]
But this is inconclusive concerning my given problem statement.
Numerically, I know that all $eig(X_{f} X_{g}^{-1})$ are sandwiched between $eig(X_{fp}X_{g}^{-1})$ and $eig(X_{fp}X_{gp}^{-1})$. $~~eig(X_{fp}X_{g}^{-1}) \leq eig(X_{f} X_{g}^{-1}) \leq eig(X_{fp}X_{gp}^{-1})$
Please help. Thanks in advance.