LMI result for quadratic stability of norm-bounded differential inclusions I am trying to understand how equation (5.14) is derived in the textbook Linear Matrix Inequalities in System and Control Theory by Boyd et al. Specifically, the result deals with the quadratic stability of norm-bounded linear differential inclusions, which are defined as:

$\bullet$ NORM-BOUND LDIs: NLDIs are described by
  \begin{equation}
\dot{x} = A x + B_p, \qquad q = C_q x + D_{qp} p, \qquad p = \Delta(t) q, \qquad ||\Delta(t)|| \leq 1,
\end{equation}
  which we will rewrite as
  \begin{equation}
\dot{x} = A x + B_p, \qquad p^T p \leq (C_q x + D_{qp} p)^T (C_q x + D_{qp} p). \tag{5.2}
\end{equation}
  We assume well-posedness, i.e., $|| D_{qp} || < 1$.

The following result is presented for quadratic stability:

Using the $\mathcal{S}$-procedure, we find that quadratic stability of (5.2) is equivalent to the existence of $P$ and $\lambda$ satisfying
  \begin{equation}
\begin{aligned}
&P > 0, \qquad \lambda \geq 0, \\
\\
&\begin{bmatrix} A^T P + P A + \lambda C_q^T C_q & P B_p + \lambda C_q^T D_{qp} \\ (P B_p + \lambda C_q^T D_{qp})^T & - \lambda (I - D_{qp}^T D_{qp}) \end{bmatrix} < 0 .
\end{aligned}
\tag{5.12}
\end{equation}

And another result for quadratic stability is presented:

With the new variables $Q = P^{−1}$, $\mu = 1/\lambda$, quadratic stability of NLDIs is also equivalent to the existence of $\mu$ and $Q$ satisfying the LMI
  \begin{equation}
\begin{aligned}
&\mu \geq 0, \qquad Q > 0, \\
\\
&\begin{bmatrix} A Q + Q A^T + \mu B_p B_p^T & \mu B_p D_{qp}^T + Q C_q^T \\ (\mu B_q D_{qp}^T + Q C_q^T)^T & - \mu (I - D_{qp} D_{qp}^T \end{bmatrix} < 0 . 
\end{aligned}
\tag{5.14}
\end{equation}

I understand how (5.12) is derived, but is not clear to me how (5.14) is derived. I'm guessing it is derived from (5.12) through some sort of congruence transformation. Or maybe it was derived by re-applying the S-procedure to (5.2) in a different way.
 A: I figured this one out myself. Leaving out the subscripts of $B_p$, $C_q$ and $D_{qp}$ for clarity, first apply the following congruence transformation to (5.12)
\begin{align*}
\begin{bmatrix} A^T P + P A + \lambda C^T C & P B + \lambda C^T D \\ (PB + \lambda C^T D)^T & -\lambda(I - D^T D) \end{bmatrix} &< 0 \\
\begin{bmatrix} P^{-1} & 0 \\ 0 & I \end{bmatrix}
\begin{bmatrix} A^T P + P A + \lambda C^T C & P B + \lambda C^T D \\ B^T P + \lambda D^T C & -\lambda(I - D^T D) \end{bmatrix}
\begin{bmatrix} P^{-1} & 0 \\ 0 & I \end{bmatrix} &< 0 \\
\begin{bmatrix} P^{-1} A^T P + A + \lambda P^{-1} C^T C & B + \lambda P^{-1} C^T D \\ B^T P + \lambda D^T C & -\lambda(I - D^T D) \end{bmatrix}
\begin{bmatrix} P^{-1} & 0 \\ 0 & I \end{bmatrix} &< 0 \\
\begin{bmatrix} P^{-1} A^T + A P^{-1} + \lambda P^{-1} C^T C P^{-1} & B + \lambda P^{-1} C^T D \\ B^T + \lambda D^T C P^{-1} & -\lambda(I - D^T D) \end{bmatrix} &< 0 \\
\begin{bmatrix} Q A^T + A Q + \lambda Q C^T C Q & B + \lambda Q C^T D \\ B^T + \lambda D^T C Q & -\lambda(I - D^T D) \end{bmatrix} &< 0 .
\end{align*}
Next, expand the matrix using the Schur complement, which is a valid step since $-\lambda I < 0$
\begin{align*}
\begin{bmatrix} Q A^T + A Q & B \\ B^T & -\lambda I \end{bmatrix} - \begin{bmatrix} Q C^T \\ D^T \end{bmatrix} (- \lambda I) \begin{bmatrix} C Q & D \end{bmatrix} &< 0 \\
\begin{bmatrix} Q A^T + A Q & B & Q C^T \\ B^T & -\lambda I & D^T \\ C Q & D & - \frac{1}{\lambda} I \end{bmatrix} &< 0 .
\end{align*}
Next, apply another congruence transformation, which effectively swaps rows and columns
\begin{align*}
\begin{bmatrix} I & 0 & 0 \\ 0 & 0 & I \\ 0 & I & 0 \end{bmatrix}
\begin{bmatrix} Q A^T + A Q & B & Q C^T \\ B^T & -\lambda I & D^T \\ C Q & D & - \frac{1}{\lambda} I \end{bmatrix}
\begin{bmatrix} I & 0 & 0 \\ 0 & 0 & I \\ 0 & I & 0 \end{bmatrix} &< 0 \\
\begin{bmatrix} Q A^T + A Q & B & Q C^T \\ C Q & D & - \frac{1}{\lambda} I \\ B^T & -\lambda I & D^T  \end{bmatrix}
\begin{bmatrix} I & 0 & 0 \\ 0 & 0 & I \\ 0 & I & 0 \end{bmatrix} &< 0 \\
\begin{bmatrix} Q A^T + A Q & Q C^T & B \\ C Q & - \frac{1}{\lambda} I & D \\ B^T & D^T & -\lambda I \end{bmatrix} &< 0 .
\end{align*}
Finally, apply the Schur complement again
\begin{align*}
\begin{bmatrix} Q A^T + A Q & Q C^T \\ C Q & - \frac{1}{\lambda} I \end{bmatrix} -
\begin{bmatrix} B \\ D \end{bmatrix} (-\lambda I )^{-1}
\begin{bmatrix} B^T & D^T \end{bmatrix} &< 0 \\
\begin{bmatrix} Q A^T + A Q + \frac{1}{\lambda} B B^T & Q C^T + \frac{1}{\lambda} B D^T \\ C Q + \frac{1}{\lambda} D B^T & - \frac{1}{\lambda} I + \frac{1}{\lambda} D D^T \end{bmatrix} &< 0 \\
\begin{bmatrix} A Q + Q A^T + \mu B B^T & \mu B D^T + Q C^T \\ (\mu B D^T + Q C^T)^T & - \mu (I - D D^T) \end{bmatrix} &< 0
\end{align*}
which gives the result.
