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For a linear system, $x(t+1)=Ax(t)$, we know the condition for Lyapunov condition is $$A^{\top}PA-P<0.$$ This comes directly from the Lyapunov function $V(x)=x^{\top}Px$. Somehow, this condition has an equivalent one called 'observer form', which is $$AQA^{\top}-Q<0.$$ I wonder how this form is derived. Either from the system itself or from any linear algebraic transformation.

Extension: My original problem considers a switching system. Suppose we have a set of $A_i$, $i\in\{1,\cdots,n\}$. Further suppose there exists a common $P$, such that for all $i$, $$A_i^{\top}PA_i-P<0.$$ (Due to the existence of a common PD $P$, no matter how the system switches, it is stable.) Then is it equivalent to say: there exists a common PD $Q$, such that for all $i$, $$A_iQA_i^{\top}-Q<0.$$

Thanks.

Xuan Wang
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  • $A$ is stable iff $A^T$ is stable, maybe? – copper.hat May 05 '21 at 06:21
  • Yes, you are right. But I'm dealing with a switched system, and require the condition to hold for a set of $A$'s, which is a little more complicated. – Xuan Wang May 05 '21 at 06:34
  • Sorry, there's not enough information for me to guess what you are looking for. – copper.hat May 05 '21 at 06:44
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    Suppose we have a set of $A_i$, $i\in{1,\cdots,n}$. Suppose there exist a common $P$, such that for all $i$, $A_i^{\top}PA_i-P<0$. Then is it equvalient to say: there exist a COMMON $Q$, such that for all $i$, $A_iQA_i^{\top}-Q<0$? – Xuan Wang May 05 '21 at 07:13
  • Interesting question, I do not know, but would suspect not. There is a standard example (which escapes me now) of a time varying linear system that is unstable but each 'frozen' system (that is fix $A(t_0)$) is stable. I suspect one could create an example with a finite number of $A_k$s that has similar behaviour. – copper.hat May 05 '21 at 07:21
  • Intuitively I would say yes, since the observer form is related to the covariance of state (also see Kalman filter). Since a common $P$ implies that the system is asymptotically stable under arbitrary switching. And when interpreting the distribution associated with the covariance as a particle cloud then according to the common $P$ all those particles should converse to the origin implying the covariance goes to zero. However, the only thing I am not sure about is that the covariance isn't static, thus there might not be a direct link between it and $Q$. – Kwin van der Veen May 07 '21 at 21:44

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Ok, I found the answer. It can be simply proved by Schur complement and LMI.

Since $P$ is PD, $$A_i^{\top}PA_i-P<0 ~\Leftrightarrow~ \begin{bmatrix}P^{-1}& A_i\\A_i^{\top} & P \end{bmatrix}>0$$ Now, let $Q=P^{-1}$, one has $$\begin{bmatrix}Q& A_i\\A_i^{\top} & Q^{-1} \end{bmatrix}>0 ~\Leftrightarrow~ A_iQA_i^{\top}-Q<0 $$

Xuan Wang
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