\(\Sigma\) is \(\delta\)-sampled controllable if and only if the pair \(\left(e^{\delta A}, B\right)\) is controllable and \(A\) has no eigenvalues of the form \(2 k \pi i / \delta\) for any nonzero integer \(k\).

Let \(\Sigma\) be a continuous-time linear system, analytically varying on \(\mathcal{I}\). Prove that if \(\Sigma\) is controllable on any nontrivial subinterval \([\sigma, \tau]\) then $$ \operatorname{rank}\left[B_{0}(t), B_{1}(t), \ldots, B_{n-1}(t)\right]=n $$ for almost all \(t \in \mathcal{I}\). (Hint: First prove that if rank \(M^{(k)}(t)=\operatorname{rank} M^{(k+1)}(t)\) for \(t\) in an open interval $J \subseteq \mathcal{I}\(, then there must exist another subinterval \)J^{\prime} \subseteq J$ and analytic matrix functions $$ V_{0}(t), \ldots, V_{k}(t) $$ on \(J^{\prime}\) such that $$ M_{k+1}(t)=\sum_{i=0}^{k} M_{i}(t) V_{i}(t) $$ on \(J^{\prime} .\) Conclude that then rank $M^{(k)}(t)=\operatorname{rank} M^{(l)}(t)\( for all \)l>k\( on \)J^{\prime}$. Argue now in terms of the sequence $\left.n_{k}:=\max \left\\{\operatorname{rank} M^{(k)}(t), t \in \mathcal{I}\right\\} .\right)$

The following statements are equivalent for \(L, W\) as above: (a) \(L\) is onto. (b) \(L^{*}\) is one-to-one. (c) \(W\) is onto. (d) \(\operatorname{det} W \neq 0\). (e) \(W\) is positive definite. Consider again the situation in Example 3.5.1. Here \(L\) is onto iff the matrix $$ W=\int_{\sigma}^{\tau} k(t)^{*} k(t) d t>0 . $$ Equivalently, \(L\) is onto iff \(L^{*}\) is one-to-one, i.e., there is no \(p \neq 0\) in \(X\) with \(k(t) p=0\) for almost all $t \in[\sigma, \tau)$, (3.18) or, with a slight rewrite and \(k_{i}:=i\) th column of \(k^{*}\) : \(\left\langle p, k_{i}\right\rangle=0\) for all \(i\) and almost all $t \Rightarrow p=0 .$ (3.19)

Consider, as an example, the system \(\Sigma\) corresponding to the linearized pendulum \((2.31)\), which was proved earlier to be controllable. In Appendix C.4 we compute $$ e^{t A}=\left(\begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array}\right) $$ Thus, for any \(\delta>0\), $$ \mathbf{R}\left(e^{\delta A}, B\right)=\left(\begin{array}{ll} 0 & \sin \delta \\ 1 & \cos \delta \end{array}\right) $$ which has determinant \((-\sin \delta)\). By Lemma \(3.4 .1, \Sigma\) is \(\delta\)-sampled controllable iff $$ \sin \delta \neq 0 \text { and } 2 k \pi i \neq \pm i \delta \text {, } $$ i.e., if and only if \(\delta\) is not a multiple of \(\pi\). Take, for instance, the sampling time \(\delta=2 \pi\). From the explicit form of \(e^{t A}\), we know that \(e^{\delta A}=I\). Thus, $$ A^{(\delta)}=A^{-1}\left(e^{\delta A}-I\right)=0 $$ so \(G=0\). This means that the discrete-time system \(\Sigma_{[\delta]}\) has the evolution equation $$ x(t+1)=x(t) $$ No matter what (constant) control is applied during the sampling interval $[0, \delta]$, the state (position and velocity) is the same at the end of the interval as it was at the start of the period. (Intuitively, say for the linearized pendulum, we are acting against the natural motion for half the interval duration, and with the natural motion during the other half.) Consider now the case when \(\delta=\pi\), which according to the above Lemma should also result in noncontrollability of \(\Sigma_{[\delta]} .\) Here $$ F=e^{\delta A}=-I $$ and $$ A^{(\delta)}=A^{-1}\left(e^{\delta A}-I\right)=-2 A^{-1}=2 A $$ so $$ G=2 A B=\left(\begin{array}{l} 2 \\ 0 \end{array}\right) . $$ Thus, the discrete-time system \(\Sigma_{[6]}\) has the evolution equations: $$ \begin{aligned} &x_{1}(t+1)=-x_{1}(t)+2 u(t) \\ &x_{2}(t+1)=-x_{2}(t) \end{aligned} $$ This means that we now can partially control the system, since the first coordinate (position) can be modified arbitrarily by applying suitable controls \(u\). On the other hand, the value of the second coordinate (velocity) cannot be modified in any way, and in fact at times \(\delta, 2 \delta, \ldots\) it will oscillate between the values \(\pm x_{2}(0)\), independently of the (constant) control applied during the interval.

Let \(\Sigma\) be a controllable continuous-time linear system, and let \(Q\) be a real symmetric positive definite \(m \times m\) matrix. Pick any \(x, z \in X\), and \(\sigma, \tau \in \mathbb{R}\). Find a formula for a control $\omega \in \mathcal{L}_{m}^{\infty}(\sigma, \tau)\( which gives \)\phi(\tau, \sigma, x, \omega)=z$ while minimizing $$ \int_{\sigma}^{\tau} \omega(s)^{*} Q \omega(s) d s . $$ Prove that for each pair \(x, z\) there is a unique such control. Do this in two alternative ways: (1) Applying again the material about pseudoinverses, but using a different inner product in the set \(\mathcal{L}_{m}^{2}\). (2) Factoring \(Q=Q_{1}^{*} Q_{1}\) and observing that the same result is obtained after a change of variables.

If \(x\) is an equilibrium state, then $$ \mathcal{R}^{S}(x) \subseteq \mathcal{R}^{S+T}(x) $$ for each \(S, T \in \mathcal{T}_{+}\).