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In the "Lyapunov Stability" chapter of the text by Khalil, there is an example on how to solve a Lyapunov Equation. Here equation $(3.12)$ is $PA+A^TP=-Q$.

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How is the $3\times 3$ matrix on the left-side of the first matrix equation formed?

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If you write the equation $PA+A^TP=-Q$ in coordinates you will get $$ \begin{bmatrix}p_1 & p_2\\p_2 & p_3\end{bmatrix}\begin{bmatrix}0 & -1\\1 & -1\end{bmatrix}+\begin{bmatrix}0 & 1\\-1 & -1\end{bmatrix}\begin{bmatrix}p_1 & p_2\\p_2 & p_3\end{bmatrix}=\begin{bmatrix}\color{red}{2p_2} & \color{green}{-p_1-p_2+p_3}\\-p_1-p_2+p_3 & \color{blue}{-2p_2-2p_3}\end{bmatrix}=\begin{bmatrix}\color{red}{-1} & \color{green}{0}\\0 & \color{blue}{-1}\end{bmatrix} $$ which is equivalent to the system $$ \begin{cases} \color{red}{2p_2}&=\color{red}{-1},\\ \color{green}{-p_1-p_2+p_3}&=\ \ \ \color{green}{0},\\ \color{blue}{-2p_2-2p_3}&=\color{blue}{-1}. \end{cases} $$ The $3\times 3$ matrix is exactly the system matrix.

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