Q:

Linear Matrix Inequality - "HOW TO"

I have a tough time understanding how to use Linear matrix Inequality to solve simple inequality problems. I would appreciate a simple "How to" on the following examples. 
$$ \bar PA + A \bar P - W^TB^T - BW + \sigma I + \bar P^2 \lt 0 $$ 

$P$ is unknown matrix $m \times m$ (semi definite)
$A$ is known m*m matrix (representing dynamics of a LTI system)
$B$ is known $m \times 1$ vector 
$\sigma$ is unknown scalar
$I$ is the $m \times m$ identity matrix
$W$ should be unknown as well (?)

My question is, how to make it into a matrix, regarding Linear Matrix Inequality. The result is shown below, I would like to know, how to get there from the above mentioned inequality.
$$\begin{bmatrix} \sigma  I + \bar P A + A \bar P - W^T B^T - BW &&& \bar P\\ \bar P &&& -I\end{bmatrix} \lt 0 $$
Another example:

inequalit:

$$ W^TW \le c_1I $$

the LMI matrix is then: 

$$\begin{bmatrix} -c_1I & W \\ W^T & -I\end{bmatrix} \lt 0 $$
One more example:

inequalit:

$$ c_2^{-1}I \le \bar P^2 $$

the LMI matrix is then: 

$$\begin{bmatrix} - \bar P & I \\ I & -c_2 \bar P \end{bmatrix} \lt 0 $$
All I need is a simple explanation, how to go from one to another. I see some pattern, but I would like to make sure, or I would like to understand how to do it (not just guess).

A:

The repeating pattern in all these approaches is an attempt to get rid of the quadratic terms in order to obtain the LMI's. This is possible through  the Schur complement approach. Specifically, the following equivalence holds
$$M>0 \quad and \quad M-X^TR^{-1}X>0 \Leftrightarrow \left[\matrix{M & X^T\\X & R}\right]>0$$

