I would like to know how large the error ball of $\|x\|$ is when using the Lyapunov function $x^\top Px$:
Assumption: I have an almost linear closed-loop system $\dot{x}=(A-BK)x+\epsilon(x)$ with small perturbation $\epsilon$. Suppose I've defined *the Lyapunov function $V=x^\top Px$ and managed to show $W=\sqrt{V}< b_{error}$ using the Comparison lemma. Thus now we know $\sqrt{x^\top Px}< b_{error}$.
*Note that $P$ is positive definite satisfying the Riccati equation of the linear closed-loop system $A-BK$.
Question: What I want to know is the error ball of $\|x\|$, not $\sqrt{x^\top Px}$. How can I go to the error ball of $\|x\|$ from $b_{error}$?
I guess, the decomposition $P=\Lambda^\top \Lambda$ and $w=\Lambda x$ would work: $\sqrt{x^\top Px}$= $\sqrt{w^\top w}=\|w\|<b_{error}$, meaning $\|\Lambda^{-1}x\|<b_{error}$. If I can somehow lower bound this inequality by |x|, like $c\|x\|<\|\Lambda^{-1}x\|<b_{error}$, I can prove |x| converges to $b_{error}/c$. But I'm not sure what can be $c$.