On page 3 of the EE363 notes on linear matrix inequalities, the following Riccati inequality is listed
$$ 0 \leq A^TPA+Q-P-A^TPB(R+B^PB)^{-1}B^TPA, \qquad P \geq 0 $$
I know that the Schur's complement can be used to convert inequalities into LMI's, but I don't understand how they achieved the solution:
$$\begin{bmatrix}R+B^TPB & B^TPA & 0\\A^TPB&A^TPA-P&0\\ 0&0&P\end{bmatrix}\geq0$$
I believe they are applying the Schur's complement multiple times, because the Schur's complement is written as a 2x2 matrix (from "A Course in Robust Control Theory- A Convex Approach"):
$$(Ax+b)^T(Ax+b)-c^Tx-d<0$$
which converts into:
$$\begin{bmatrix}c^Tx+d & (Ax+b)^T\\Ax+b&I\end{bmatrix}\geq0$$
I find it more confusing as I search throughout other texts as the form appears to change over and over depending on the problem that is solved. How do you form the LMI of the Riccati inequality? And where does the 3rd column and row come from? I would prefer the inclusion of all the steps necessary however trivial despite knowing linear algebra, but I think I am missing a few "basic" tricks beyond Schur's complement.
The third row comes from the constraint $P\ge0$. Note the zero off-diagonal term. The rest of the LMI simply comes from a Schur complement with respect to the term $-A^TPB(R+B^PB)^{-1}B^TPA$.
Edit. More details as requested. If you have two inequalities of the form $M(x)>0$ and $N(x)>0$ which need to be both satisfied for some $x$, then this is equivalent to say that
$$\begin{bmatrix}M(x) & 0\\0 & N(x)\end{bmatrix}>0.$$
In your case, the problem consists of the Riccati inequality and the condition that $P\ge0$. A Schur complement on the Riccati inequality (assuming $R+B^TPB>0$, note the typo here in the main post) yields the LMI
$$\begin{bmatrix}R+B^TPB & B^TPA\\A^TPB&A^TPA+Q-P\end{bmatrix}\geq0$$
to which we adjoin the condition $P\ge 0$ to get
$$\begin{bmatrix}R+B^TPB & B^TPA & 0\\A^TPB&A^TPA+Q-P&0\\ 0&0&P\end{bmatrix}\geq0.$$