Consider the matrix $A\in\mathbb{R}^{n\times n}$ with all eigenvalues inside the unit circle and the symmetric positive definite matrix $P\in\mathbb{R}^{n\times n}$ satisfying $ APA^\top-P+I=0 $. Also consider the matrices $ \Delta \in \mathbb{R}^{n\times (n+m)} $ and $ Q\in \mathbb{R}^{T\times n} $, the full row rank matrix $ W\in\mathbb{R}^{(n+m)\times T} $ with $ T>m+n $, and the matrix $ E_k\in \mathbb{R}^{T\times T} $ defined as follows \begin{equation} E_k:=\begin{bmatrix} 0 & 0 \\ 0 & I_k \end{bmatrix},\quad \forall \,0<k<T. \end{equation} I need to show that there exists some matrix $ R\in \mathbb{R}^{T\times n} $ such that for any $ k $, at least one the following problems is feasible(I have colored the difference between the problems to specify it): \begin{equation*} \left\{\begin{split} &\Sigma_1 P^{-1} \Sigma_1^\top +A \Sigma_1^\top +\Sigma_1 A^\top\leq 0\\ &\Sigma_1=\Delta W \color{red}{E_k} (Q+R)\\ &WR=0 \end{split}\right. \end{equation*} \begin{equation*} \left\{\begin{split} &\Sigma_2 P^{-1} \Sigma_2^\top +A \Sigma_2^\top +\Sigma_2 A^\top\leq 0\\ &\Sigma_2=\Delta W \color{red}{(E_k-I)} (Q+R)\\ &WR=0 \end{split}\right. \end{equation*}
My attempt: I tried to approach the problem by partitioning $ W=[W_1\ W_2] $ with the assumption that $ W_1\in \mathbb{R}^{(n+m)\times (T-k)} $ has rank $ n+m $ (I assumed that $ T-k\geq n+m $ to make this assumption). Then, I will have $ 0=WR=[W_1\ W_2] \begin{bmatrix} R_1 \\ R_2 \end{bmatrix} $ that implies $ -W_1 R_1= W_2 R_2 $. So for any $ R_2 $, I can find some $ R_1 $ to satisfy this relation (this is due to having $ W_1 $ as full rank matrix). Thus, $ R_2 $ is a free variable. Then, I can use the structure of $ E_k $ to write $ \Sigma_1=\Delta W {E_k} (Q+R)=\Delta W (Q_2+R_2) $ where $ Q_2 $ is suitable partition of $ Q $ as $ Q=\begin{bmatrix} Q_1 \\ Q_2 \end{bmatrix} $. This means that I can choose $ R_2=-Q_2 $ to get $ \Sigma_1=0 $ and satisfy the first problem.
The main issue with my approach is that still I have to assume that $ W_1 $ is full rank, but this is not guaranteed. So I was wondering if there is any other way to solve the problem without this assumption.