Chapter 6: Problem 5
A (time-invariant, complete, initialized) system is minimal if and only if it is canonical.
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Chapter 6: Problem 5
A (time-invariant, complete, initialized) system is minimal if and only if it is canonical.
These are the key concepts you need to understand to accurately answer the question.
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The rank of the Markov sequence \(\mathcal{A}\) is \(\sup _{s, t} \operatorname{rank} \mathcal{H}_{s, t}(\mathcal{A}) .\)
If two controllable triples are similar, then the similarity must be given by the formulas obtained in the proof of Theorem 27, so in particular similarities between canonical systems are unique. This is because if \(T\) is as in Definition \(6.5 .8\), then necessarily $$ T^{-1} A^{i} B=\widetilde{A}^{i} \widetilde{B} $$ for all \(i\), so it must hold that \(T^{-1} \mathbf{R}=\widetilde{\mathbf{R}}\), and therefore $$ T=\mathbf{R} \widetilde{\mathbf{R}}^{\\#} $$ In particular, this means that the only similarity between a canonical system and itself is the identity. In the terminology of group theory, this says that the action of \(G L(n)\) on triples is free.
Calculate a canonical realization, and separately calculate the rank of the Hankel matrix, for each of these examples with \(m=p=1\) : 1\. The sequence of natural numbers \(1,2,3, \ldots\) 2\. The Fibonacci sequence \(0,1,1,2,3,5,8, \ldots\)
When \(\mathbb{K}=\mathbb{R}\) or \(\mathbb{C}\), one may use complex variables techniques in order to study realizability. Take for simplicity the case \(m=p=1\) (otherwise one argues with each entry). If we can write \(W_{\mathcal{A}}(s)=P(s) / q(s)\) with \(\operatorname{deg} P<\) \(\operatorname{deg} q\), pick any positive real number \(\lambda\) that is greater than the magnitudes of all zeros of \(q\). Then, \(W_{\mathcal{A}}\) must be the Laurent expansion of the rational function \(P / q\) on the annulus \(|s|>\lambda\). (This can be proved as follows: The formal equality \(q W_{\mathcal{A}}=P\) implies that the Taylor series of \(P / q\) about \(s=\infty\) equals \(W_{\mathcal{A}}\), and the coefficients of this Taylor series are those of the Laurent series on \(|s|>\lambda\). Equivalently, one could substitute \(z:=1 / s\) and let $$ \widetilde{q}(z):=z^{d} q(1 / z), \widetilde{P}(z):=z^{d} P(1 / z), $$ with \(d:=\operatorname{deg} q(s)\); there results the equality \(\widetilde{q}(z) W(1 / z)=\widetilde{P}(z)\) of power series, with \(\widetilde{q}(0) \neq 0\), and this implies that \(W(1 / z)\) is the Taylor series of \(\widetilde{P} / \widetilde{q}\) about 0 . Observe that on any other annulus \(\lambda_{1}<|s|<\lambda_{2}\) where \(q\) has no roots the Laurent expansion will in general have terms in \(s^{k}\), with \(k>0\), and will therefore be different from \(W_{\mathcal{A} .)}\) Thus, if there is any function \(g\) which is analytic on \(|s|>\mu\) for some \(\mu\) and is so that \(W_{\mathcal{A}}\) is its Laurent expansion about \(s=\infty\), realizability of \(\mathcal{A}\) implies that \(g\) must be rational, since the Taylor expansion at infinity uniquely determines the function. Arguing in this manner it is easy to construct examples of nonrealizable Markov sequences. For instance, $$ \mathcal{A}=1, \frac{1}{2}, \frac{1}{3 !}, \frac{1}{4 !}, \ldots $$ is unrealizable, since \(\mathcal{W}_{\mathcal{A}}=e^{1 / s}-1\) on \(s \neq 0\). As another example, the sequence $$ \mathcal{A}=1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots $$ cannot be realized because \(\mathcal{W}_{\mathcal{A}}=-\ln \left(1-s^{-1}\right)\) on \(|s|>1\).
If there are any \(S, T \in \mathcal{T}_{+}, T>0\), such that $$ I^{S}=I^{S+T}, $$ then necessarily \(I^{S}=I\).
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