# Lyaponuv stability condition of linear systems for homogenous P in V(x) = x^T P x

I am currently learning about using Lyaponuv functions to find Linear Matrix Inequalities (LMIs) as conditions for stability of a linear time invariant system.

i.e. $$\dot{x}(t) = Ax(t)$$ is stable if there exists a function $$V(x)$$ such that $$V(x)>0$$ for all $$x \neq 0$$ and $$V(0) = 0$$ and $$\dot{V}(x) < 0$$ for all $$x \neq 0$$

Generally we have learned to do this with a quadratic Lyapunov function $$V(x) = x^TPx$$ where $$P\succ 0$$ and $$A^TP + PA \prec 0$$, but I just came across this footnote which I am struggling to understand.

The $$P$$ matrix also has to be symmetric which I am not sure why is true.

When performing convex optimisation on systems with Lyaponuv functions our lectures sometimes use the first LMIs i.e. $$P \succ 0$$ and other times we use $$P \succeq I$$ and I am not sure when and why the difference arises.

• Little side note: it has to be V(x)>0, not < Jun 12 at 19:26
• @OpticalResonator thanks! I've corrected it Jun 12 at 19:48
• @AJN thank you, that does answer my question as to why $P$ needs to be symmetric. And yes I do believe $(P-I)\succeq 0$ is the stricter condition but I'm not sure why or when we would use it. Jun 13 at 12:34
• @AJN I don't really have a proof for it which is why I was confused. It only says due to homogeneity in $P$, the $P \succ 0$ and $A^TP + PA \prec 0$ conditions turn into $P \succeq I$ and $A^TP + PA \preceq I$. Any idea what homogeneity in $P$ could be referring to? Jun 13 at 12:40
• – AJN
Jun 13 at 12:47