I have read that this inequality
$$ P - \sum_{j=0}^{m} \left( A^{(j)} + B^{(j)} K \right)^{T} P \left(A^{(j)} + B^{(j)} K\right) \succ 0 $$
can be reformulated into a large linear matrix inequality (LMI)
$$\left[\begin{array}{ccccc}S & \left(A^{(0)} S+B^{(0)} Y\right)^{T} & \left(A^{(1)} S+B^{(1)} Y\right)^{T} & \cdots & \left(A^{(m)} S+B^{(m)} Y\right)^{T} \\ \star & S & 0 & \cdots & 0 \\ \star & \star & S & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \star & \star & \star & \cdots & S\end{array}\right] \succ 0$$
by defining $S=P^{-1}$, and $Y=K P^{-1}$.
I know how to reformulate the inequality into a single LMI by using Schur complements, if the inequality does not take the sum of these terms. However, I do not know by which lemma this large LMI can be formulated.
Any hint would be appreciated. Many thanks in advance.
I'm gonna answer my question myself, thanks for Panos' hint.
According to the Schur complement, $$ X \succ 0 $$ iff $$ C \succ 0, A-B C^{-1} B^{\top}$$ where $$ X=\left[\begin{array}{cc} A & B \\ B^{\top} & C \end{array}\right]. $$
Let $A = S$, $B^{T} = \left[ \left(A^{(0)} S+B^{(0)} Y\right)^{T} \quad\left(A^{(1)} S+B^{(1)} Y\right)^{T} \quad \cdots \quad\left(A^{(m)} S+B^{(m)} Y\right)^{T} \right]$ and $C=\left[ \begin{array}{cccc} S & 0 & \cdots & 0 \\ \star & S & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \star & \star & \cdots & S \end{array} \right]$.
Then, $A - B^{T}C^{-1}B = S - \Sigma_i (A^{(i)}S + B^{(i)}Y)^{T}S^{-1}(A^{(i)}S + B^{(i)}Y)$. After inserting $S=P^{-1}, \text { and } Y=K P^{-1}$ into the equation above, we prove the first inequality is equivalent to the large LMI.