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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.

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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.

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  • $\begingroup$ Little side note: it has to be V(x)>0, not < $\endgroup$ Jun 12 at 19:26
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    $\begingroup$ @OpticalResonator thanks! I've corrected it $\endgroup$
    – Chryron
    Jun 12 at 19:48
  • $\begingroup$ @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. $\endgroup$
    – Chryron
    Jun 13 at 12:34
  • $\begingroup$ @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? $\endgroup$
    – Chryron
    Jun 13 at 12:40
  • $\begingroup$ Let us continue this discussion in chat. $\endgroup$
    – AJN
    Jun 13 at 12:47

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