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Chapter 6: Problem 2

If \(\mathcal{A}\) is a Markov sequence of finite rank \(n\), then \(\mathcal{A} \equiv 0\) if and only if $$ \mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0 . $$

Short Answer

Expert verified
A Markov sequence \(\mathcal{A}\) of finite rank \(n\) is equivalent to the zero sequence if and only if \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0\). This can be shown by proving that (1) if \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0\), then every element of the sequence is \(0\), (2) if the sequence is equivalent to 0, the first \(n\) elements must be \(0\). This bidirectional proof establishes the equivalence.

Step by step solution

01

Proving "if" direction: Assume \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0\)

Since \(\mathcal{A}\) is a Markov sequence of finite rank \(n\), we know that all the elements of the sequence can be expressed as a linear combination of the first \(n\) elements of the sequence. That is, \[ \mathcal{A}_{k} = c_1 \mathcal{A}_{k-1} + c_2 \mathcal{A}_{k-2} + \cdots + c_n \mathcal{A}_{k-n} \] for some constants \(c_1, c_2, \ldots, c_n\). Now, we are given that \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0\). Using this in the above equation, we get \[ \mathcal{A}_{k} = c_1 \cdot 0 + c_2 \cdot 0 + \cdots + c_n \cdot 0 = 0 \] for all \(k \ge n\). Thus, we have shown that if \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0\), then every element in the sequence is \(0\), i.e., \(\mathcal{A} \equiv 0\).
02

Proving "only if" direction: Assume \(\mathcal{A} \equiv 0\)

Now, we need to show that if the given Markov sequence \(\mathcal{A}\) is equivalent to \(0\), i.e., \(\mathcal{A} \equiv 0\), then \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0\). Since \(\mathcal{A}\equiv 0\), every element of the sequence is \(0\). Thus, the first \(n\) elements of the sequence are also \(0\), meaning \[ \mathcal{A}_{1} = \mathcal{A}_{2} = \ldots = \mathcal{A}_{n} = 0. \] By proving both directions, we conclude that \(\mathcal{A}\) is equivalent to the zero sequence if and only if \(\mathcal{A}_{1}=\mathcal{A}_{2}=\ldots=\mathcal{A}_{n}=0.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Finite Rank
In the context of Markov sequences, 'finite rank' refers to the idea that a sequence has a limited number of initial elements that determine the rest of the sequence. Specifically, a Markov sequence of finite rank means that after a certain point, each element of the sequence can be expressed as a combination of a finite number (rank) of preceding elements. To illustrate with a tangible example, if a sequence has a rank of 3, you can determine the fourth and subsequent elements by only knowing the first three.

This property is significant when working with sequences in mathematics because it implies a form of predictability and structure. A finite rank allows us to describe complex sequences with more straightforward expressions, making it practical in fields such as economics, statistics, and computer science.
Linear Combination
A linear combination refers to a mathematical construct where an element is formed by adding together multiples of certain elements, often using constant multipliers or coefficients. This concept is fundamental in areas such as linear algebra and is applied extensively in the analysis of sequences and series.

In our exercise, a linear combination is used to express any element of the Markov sequence beyond the initial elements that define its rank. For example, if you're shopping with a particular budget and you spend portions of it on different items, the total amount you've spent is a linear combination of your expenditures on individual items, with each item's cost acting as a coefficient.
Zero Sequence
A zero sequence in mathematics is a sequence where every element is zero. It is essential in various proofs and theoretical discussions because it serves as a neutral element in many operations, much like the number 0 does in basic arithmetic.

In our exercise scenario, proving that a Markov sequence is equivalent to a zero sequence under a specific condition is similar to demonstrating that a certain process or system consistently yields no result or change. This can be powerful in proving the absence of effect or in validating symmetries and other properties in mathematical systems.
Proof Techniques
Proof techniques are various methods used to establish the truth of mathematical statements. Some common techniques include direct proof, proof by contradiction, and proof by induction. In our exercise, we use direct proof鈥攁 logical argument that demonstrates a statement is true by assuming a set of initial conditions and logically showing that these lead to the statement's conclusion.

Direct proof is akin to a 'cause-effect' argument. You start with the cause (the initial conditions) and show that this invariably leads to the effect (the conclusion of the proof). These proof techniques are vital skills for mathematicians, as they establish the foundation for developing mathematical theory securely.

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Most popular questions from this chapter

Show that the result does not extend to analytic systems by considering the system $$ x^{+}=\frac{1}{2} x, \quad y=\sin x, $$ which is observable but not in finite time.

Observability form realizations are in general not controllable, except if \(p=1\). In this case, if \(\left(\alpha_{1}, \ldots, \alpha_{n}\right)\) give a minimal recursion then by Corollary \(6.6 .6\) this realization must be minimal. In fact, $$ \mathbf{R}_{n}(A, B)=\mathcal{H}_{n, n} $$ which has rank \(n\). Of course, for \(p>1\) it is possible to reduce any such realization to a controllable one using the Kalman decomposition.

Let \(\mathcal{A}\) be any recursive Markov sequence. Consider the transposed sequence $$ \mathcal{A}^{\prime}:=\mathcal{A}_{1}^{\prime}, \mathcal{A}_{2}^{\prime}, \ldots $$ This satisfies a recursion with the same coefficients \(\alpha_{i}\) 's. For these coefficients we let \((A, B, C)\) be the observability form realization of \(\mathcal{A}^{\prime}\). Since \(C A^{i-1} B=\mathcal{A}_{i}^{\prime}\) for all \(i\), also \(B^{\prime}\left(A^{\prime}\right)^{i-1} C^{\prime}=\mathcal{A}_{i}\) for all \(i\), and the system \(\left(A^{\prime}, C^{\prime}, B^{\prime}\right)\) is a controllable realization of \(\mathcal{A}\). We have obtained the controllability form realization of \(\mathcal{A}:\) $$ A=\left(\begin{array}{ccccc} 0 & 0 & \cdots & 0 & \alpha_{1} I \\ I & 0 & \cdots & 0 & \alpha_{2} I \\ 0 & I & \cdots & 0 & \alpha_{3} I \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & I & \alpha_{n} I \end{array}\right) \quad B=\left(\begin{array}{c} I \\ 0 \\ 0 \\ \vdots \\ 0 \end{array}\right) \quad C=\left(\begin{array}{llll} \mathcal{A}_{1} & \mathcal{A}_{2} & \cdots & \mathcal{A}_{n} \end{array}\right) $$ of dimension \(m n\).

A continuous-time system is observable on \([\sigma, \tau]\) (or observable in time \(T\), or observable) if and only if it is final-state observable on \([\sigma, \tau]\) (respectively, in time \(T\), final-state observable). Proof. Necessity always holds, so pick a final-state observable system and any two distinct events \((x, \sigma) \neq(z, \sigma)\). Let \(\omega\) be so that it final-state distinguishes the events \((x, \sigma)\) and \((z, \sigma) ;\) without loss of generality, assume \(\omega \in \mathcal{U}^{(\sigma, \tau)}\). If \(\bar{\lambda}_{x}^{\sigma, \tau}(\omega) \neq \bar{\lambda}_{z}^{\sigma, \tau}(\omega)\), then some restriction \(\omega_{[\sigma, t)}\) distinguishes between them, and there is nothing left to prove. So assume that $$ \phi(\tau, \sigma, x, \omega)=\phi(\tau, \sigma, z, \omega) . $$ Let \(\xi(t)=\phi(t, \sigma, x, \omega)\) and \(\zeta(t)=\phi(t, \sigma, z, \omega)\). Then \(\xi\) and \(\zeta\) are two solutions of \(\dot{x}=f(x, \omega)\) defined on all of \([\sigma, \tau]\) and having the same value at time \(t=\tau\). By uniqueness of solutions, \(\xi \equiv \zeta\), contradicting the assumption that \(\xi(\sigma)=\) \(x \neq z=\zeta(\sigma)\). This gives the implication on any fixed interval \([\sigma, \tau]\), and the other implications follow from here.

We consider again the parity check example discussed in Example 2.3.3. In particular, we shall see how to prove, using the above results, the last two claims in Exercise 2.3.4. The behavior to be realized is \(\lambda(\tau, 0, \omega)=\) $$ \begin{cases}1 & \text { if } \omega(\tau-3)+\omega(\tau-2)+\omega(\tau-1) \text { is odd and } 3 \text { divides } \tau>0 \\ 0 & \text { otherwise }\end{cases} $$ and we take the system with $$ x:=\\{0,1,2\\} \times\\{0,1\\} $$ and transitions $$ \mathcal{P}((i, j), l):=(i+1 \bmod 3, j+l \bmod 2) $$ for \(i=1,2\) and $$ \mathcal{P}((0, j), l):=(1, l) \text {. } $$ The initial state is taken to be \((0,0)\), and the output map has \(h(i, j)=1\) if \(i=0\) and \(j=1\) and zero otherwise. (The interpretation is that \((k, 0)\) stands for the state " \(t\) is of the form \(3 s+k\) and the sum until now is even," while states of the type \((k, 1)\) correspond to odd sums.) This is clearly a realization, with 6 states. To prove that there is no possible (time-invariant, complete) realization with less states, it is sufficient to show that it is reachable and observable. Reachability follows from the fact that any state of the form \((0, j)\) can be obtained with an input sequence \(j 00\), while states of the type \((1, j)\) are reached from \(x^{0}\) using input \(j\) (of length one) and states \((2, j)\) using input \(j 0\). Observability can be shown through consideration of the following controls \(\omega_{i j}\), for each \((i, j):\) $$ \omega_{01}:=0, \omega_{00}:=100, \omega_{10}:=10, \omega_{11}:=00, \omega_{21}:=0, \omega_{20}:=0 . $$ Then, \(\omega_{01}\) separates \((0,1)\) from every other state, while for all other pairs \((i, j) \neq\) \((0,1)\), $$ \lambda_{(i, j)}\left(\omega_{\alpha \beta}\right)=1 $$ if and only if \((i, j)=(\alpha, \beta)\).
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