This question is from a system theory test without answers or solutions:
Let the following two cases be given
$A) \quad A=\begin{bmatrix}-2&1\\-1&0\end{bmatrix} \quad $and$ \quad C=\begin{bmatrix}1&-1 \end{bmatrix} \qquad B) \quad A=\begin{bmatrix}1&2\\2&1 \end{bmatrix} \quad $and$ \quad C=\begin{bmatrix}1&0\end{bmatrix}$
and consider the following two statements
$\quad \textbf{1.}$ There exists a solution $W \succ 0$ to
$\quad \qquad WA+A^TW-C^TC \prec 0$
$\quad \textbf{2.}$ There exists a solution $P \succ 0$ to
$\quad \qquad PA+A^TP+C^TC=0$
$\qquad $that is given by the integral expression
$\quad \qquad P:=\int_{0}^{\infty}(e^{At})^TC^TCe^{At}dt$
Which of the statements
My approach:
First for case $A)$ we check statement $1$
$\begin{bmatrix}W_1&W_2\\W_2&W_3\end{bmatrix}\begin{bmatrix}-2&1\\-1&0 \end{bmatrix}+\begin{bmatrix}-2&-1\\1&0\end{bmatrix}\begin{bmatrix}W_1&W_2\\W_2&W_3\end{bmatrix}-\begin{bmatrix}1&-1\\-1&1\end{bmatrix}=0$
We multiply the matrices and bring $\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$ to the other side of the $=$ sign. Which gives:
$\begin{bmatrix}-4W_1-2W_2&W_1-2W_2-W_3\\W_1-2W_2-W_3&2W_2\end{bmatrix}=\begin{bmatrix}1&-1\\-1&1\end{bmatrix}$
In order to solve this we construct the matrix:
$\left[ \begin{array}{ccc|c}-4&-2&0&1\\1&-2&-1&-1\\0&2&0&1\end{array}\right] \rightarrow \left[ \begin{array}{ccc|c}1&0&0&-\frac{1}{2}\\0&1&0&\frac{1}{2}\\0&0&1&-\frac{1}{2}\end{array} \right]\quad$and thus $\quad W=\begin{bmatrix}-\frac{1}{2}&\frac{1}{2}\\\frac{1}{2}&-\frac{1}{2}\end{bmatrix}$
Which is not positive definite (the eigenvalues are $0$ and $-1$), so for $A$ statement 1 is false.
The existence of a positive definite $P$ can be checked in a similar way. So the rest is just a repetition of the steps taken above.
Is this the correct way to solve these questions?
Theorem 12.5 from the book Linear Systems Theory by Joao Hespanha (page 115) may be used. Here, you need for example to check the controllability of $(A,C)$, see points 5 and 6 in the theorem in the picture below.