using Schur's complement and Young's inequality to reduce matrix algebraic equation to LMI This questions concerns the practical implementation of schur's complement and Young's inequality.
Consider the following 
\begin{align}
\begin{pmatrix} \begin{pmatrix}
\mathbb{A}_{\mathbb{Z}} & \mathcal{O}\\
\mathcal{O} & \mathbb{A}_{\mathbb{P}} 
\end{pmatrix} + \begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix} \begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix} & \begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}\\
\begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}^T & -\mu \mathbb{I}_{q}
\end{pmatrix} + \underbrace{\begin{pmatrix}
\overline{B} \overline{K}\\
\mathcal{O}\\
\mathcal{O}
\end{pmatrix}}_{V^T} \underbrace{\begin{pmatrix}
\mathcal{O} & \tilde{B} & \mathcal{O}
\end{pmatrix}}_U + \begin{pmatrix}
\mathcal{O}\\
\tilde{B}^T\\ 
\mathcal{O}
\end{pmatrix} \begin{pmatrix}
\left(\overline{B} \overline{K} \right)^T & \mathcal{O} & \mathcal{O}
\end{pmatrix} \leq 0
\end{align}
Now to avoid any bilinear coupling between terms, the young's inequality is used, as given by
\begin{equation}
X^T \hspace{1mm} Y + Y^T \hspace{1mm} X \leq \dfrac{1}{2} \hspace{1mm} \left( X + S \hspace{1mm} Y \right)^T \hspace{1mm} S^{-1} \hspace{1mm} \left( X + S \hspace{1mm} Y \right)
\end{equation}
where $S$ is a symmetric positive definite matrix. To apply the inequality, we use $S = \dfrac{1}{2} \mathbb{I}_n$ for the term $\begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix} \begin{pmatrix}
\mathbb{Z} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}
\end{pmatrix}$ and $S = \epsilon \mathbb{Z}$ for the term $V^TU+U^TV$. Now, the intent is to use the young's inequality and convert the above algebraic equation in LMI using Schur's complement. I am unsure of the proper way and cannot reduce the equation to the requisite LMI. Given: $\mathbb{Z}$ is a $n \times n$ symmetric positive definite matrix.
Can someone walk me through the subsequent steps?
 A: The only term that is not linear in the unknown variables is the quadratic term in $\mathbb{Z}$, so it might be easier to replace that with another variable and add another inequality
$$
\mathbb{Z}^2 \leq \mathbb{X}, \tag{1}
$$
such that the initial in equality can be written as
\begin{align}
\begin{pmatrix} \begin{pmatrix}
\mathbb{A}_{\mathbb{Z}} & \mathcal{O}\\
\mathcal{O} & \mathbb{A}_{\mathbb{P}} 
\end{pmatrix} + \begin{pmatrix}
\mathbb{X} & \mathcal{O}\\
\mathcal{O} & \mathcal{I}_{n}^2
\end{pmatrix} & \begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}\\
\begin{pmatrix}
D_{11} & D_{12}\\
D_{21} & D_{22}
\end{pmatrix}^\top & -\mu\,\mathbb{I}_{q}
\end{pmatrix} + \begin{pmatrix}
\overline{B} \overline{K}\\
\mathcal{O}\\
\mathcal{O}
\end{pmatrix} \begin{pmatrix}
\mathcal{O} & \tilde{B} & \mathcal{O}
\end{pmatrix} + \begin{pmatrix}
\mathcal{O}\\
\tilde{B}^\top\\ 
\mathcal{O}
\end{pmatrix} \begin{pmatrix}
\left(\overline{B} \overline{K} \right)^\top & \mathcal{O} & \mathcal{O}
\end{pmatrix} \leq 0, \tag{2}
\end{align}
which is linear in all variables. One can then use Schur's complement on $(1)$ which allows us to rewrite it also as an LMI
$$
\begin{pmatrix}
I & \mathbb{Z} \\
\mathbb{Z} & \mathbb{X}
\end{pmatrix} \geq 0, \tag{3}
$$
with $I$ an $n$ by $n$ identity matrix, $\mathbb{Z} = \mathbb{Z}^\top > 0$ and $\mathbb{X} = \mathbb{X}^\top > 0$.
So the combination of $(2)$ and $(3)$ should be equivalent to your initial inequality and are all LMI's.
