(\[ref:ideal\]) as (t)&=&A\_e(t)-B(t)(x(t)), e(0)=e\_0,     \[err:sys\] where $\tilde{W}(t)\triangleq \hat{W}(t)-W \in \IR^{(s+n)\times m}$ is the weight error and $e_0 \triangleq x_0-x_\rom{r_0}$. The update law given by (\[update:law\]) can be derived by using Lyapunov analysis by considering the Lyapunov function candidate (e,)&=& eP e + \^[-1]{}  ( \^)( \^).   \[lyap:cand\] Note that $\mathcal{V}\bigl(0,0\bigl)=0$, $\mathcal{V}\bigl(e,\tilde{W}\bigl)>0$ for all $(e,\tilde{W})\neq(0,0)$, and $\mathcal{V}\bigl(e,\tilde{W}\bigl)$ is radially unbounded. Now, differentiating (\[lyap:cand\]) and then using (\[update:law\]) and (\[err:sys\]) yields (e(t),(t))=-e(t) R e(t)0, which guarantees that the system error $e(t)$ and the weight error $\tilde{W}(t)$ are Lyapunov stable, and hence, are bounded for all $t \in \overline{\IR}_+$. Since $\sigma\bigl(x(t)\bigl)$ is bounded for all $t \in \overline{\IR}_+$, it follows from (\[err:sys\]) that $\dot{e}(t)$ is bounded, and hence, $\ddot{\mathcal{V}}\bigl(e(t),\tilde{W}(t)\bigl)$ is bounded for all $t \in \overline{\IR}_+$. Now, it follows from Barbalat’s lemma \[, Lemma 8.2\] that \_[t]{}(e(t),(t))=0, which consequently shows that $\lim_{t\rightarrow\infty}e(t)=0$.

*Remark 3.1*. Although $\lim_{t\rightarrow\infty}e(t)=0$, the state vector $x(t)$ can be far different from $x_\rom{r}(t)$ during transient time (learning phase), unless a high learning rate $\gamma$ is used in the update law (\[update:law\]). To see this, we first let $x_\rom{r_0}=x_0$ in (\[ref:ideal\]), and hence, $e(0)=0$ in (\[err:sys\]). Note that this condition is realizable since it is assumed that the state vector $x_\rom{p}(t)$ 