Structural Properties of the Unobservable Subspace

Abstract

The structural properties of the unobservable subspace are explored. In particular the canonical decomposition of the unobservable subspace as a direct sum of cyclic subspaces as well as the conditions for this subspace to be spectral for the system matrix is studied. These properties are applied to simple input-simple output (SISO) feedback systems by connecting the spectral decomposition of the unobservable subspace to the total cancellation of unobservable modes in the compensator with multiple transmission zeros in the plant.

1. Introduction

In control theory, a dynamical system is a processing element that transforms an input into an output both depending on time. When restricted to linear-invariant systems, a realization of is usually defined in the state-space by a quadruplet of matrices and a vector of internal states . This quadruplet describes a system of differential equations of the type , . In many occasions it is interesting to analyze the capability of inferring internal states by the knowledge of the outputs; this is called the observability of a system and was firstly introduced by American Hungarian scientist Rudolf E. Kalman, [1]. The set of unobservable states that can not be determined from the outputs has structure of -invariant subspace, .

The intersection of the unobservable subspace with another subspace is common in control theory. For example, in the Kalman decomposition it is possible to determine the -invariant controllable-unobservable subspace by choosing as the controllable subspace. Another example is found in the stabilizing solutions of the Riccati equation where is a stable invariant subspace [2]. In all those cases is -invariant which implies that is also -invariant; furthermore, it is usual to require a trivially -invariant intersection, . When is not an -invariant subspace or even not a subspace, something can be stated about the invariance of . For instance, the uncontrollable set is not a subspace but we can build the largest subspace contained in , which is not necessarily -invariant. If admits a spectral decomposition, turns out to be a controlled-invariant subspace since , where is the subspace generated by the columns of (for more details of conditioned invariant subspaces the reader is referred to [3, 4]). Furthermore, if has a spectral decomposition every subspace is a conditioned invariant subspace under with respect to ; that is, . In the pole placement by output injection it is usual to take an arbitrary subspace which is conditioned invariant; that is, for any matrix , . However this condition can be removed when is spectral. Thus, for an arbitrary matrix , and is -invariant (also this means that is conditioned invariant).

Recently it was shown that the unobservable subspace plays a central role in the well-posedness of a type of systems described by impulsive differential equations called reset control systems [5]. In fact the -invariance of the intersection of the unobservable with an arbitrary subspace seems to be key in the study of these systems.

Motivated for the above applications we analyze the canonical structure of the unobservable subspace as a direct sum of cyclic subspaces. Since the observability concept only involves matrices and , we restrict the problem to systems of the form , (sometimes called strictly proper systems). The main result is developed in Section 2 where the canonical decomposition is derived by resorting to Kalman’s decomposition [6]. Also we deal with the necessary and sufficient conditions for the unobservable subspace to be spectral for . Section 3 is devoted to show some interesting properties of the unobservable cyclic subspaces; additionally the necessary and sufficient conditions of -invariance for the intersection of the unobservable subspace with an arbitrary subspace are analyzed. Section 4 deals with the connection between -spectrality of the unobservable subspace and the existence of multiple transmission zeros in feedback systems.

Throughout the paper we will use the following notation. describes the ring of polynomials in the variable over . Given vectors , the vector space generated by these vectors is written as . The spectrum of a matrix , that is, the set of eigenvalues of , is denoted by . The eigenspace of with eigenvalue , that is, the set of eigenvectors of with the same eigenvalue , is indicated with and stands for the root subspace of ; it contains the vectors from any Jordan chain of corresponding to . is the geometric multiplicity of ; that is, .

The realization of a system is given by a triplet ; that is, , , and represents the observability matrix of ; that is, , where “” is the Kronecker product, , and is the th vector of the standard basis with at the th entry and the remaining entries set to . are the nullspace of the matrix and is an -cyclic subspace generated by . If , for and . The set of transmission zeros of is denoted by .

The set of unobservable modes is and conversely the set of observable modes is . Associated with these sets we have the subspaces and .

2. Canonical Decomposition of the Unobservable Subspace

Given a strictly proper feedback system we study the decomposition of the unobservable subspace as a direct sum of cyclic subspaces. In particular we are interested in determining necessary and sufficient conditions for this subspace to be -spectral, that is, when is a direct sum of the root spaces where .

In the literature it is shown that is -invariant and ; from this it is straightforward to write . If denotes the eigenspace for an unobservable mode , it is clear that . Since we draw . However, what is not so clear is the inclusion for unobservable modes . In general this result is not true for matrices without control structure as shown in the example below.

Example 1. Let be a linear system with matrices In this case and where is a vector of the standard basis. From the Popov-Belevitch-Hautus test (PBH test, [3]) it is easy to check that so . The generalized eigenvectors are and , which are not in the unobservable subspace. Henceforth .

Root spaces are decomposed in direct sum of cyclic subspaces so it is worth analyzing the conditions for an -cyclic subspace to be unobservable. As expected this depends on the unobservability of its generator.

Lemma 2. A necessary and sufficient condition for an -cyclic subspace to be unobservable is that .

Proof. Consider the following.
Sufficiency. Assume that . Let be a basis of . Since , for . For , depends on the vectors in which results into for .
Necessity. If , in particular .

The following theorem is central in our development and reveals that the unobservable subspace is actually a direct sum of -cyclic subspaces.

Theorem 3. can be decomposed into a direct sum of -cyclic subspaces.

Proof. is -invariant. From Kalman’s decomposition there exists an adapted basis such that is similar to a matrix via a matrix ; that is, . We assume that and adopt the notation for vectors embedded into . Let be an -cyclic subspace generated by (Krylov subspace); that is, where . Because is also a lower triangular matrix we can embed into through an isomorphism . Generators behave well with respect to similarity [7]; that is, is an -cyclic subspace with generator . As a result of this and . Finally, it is obvious that . In virtue of Lemma 2, . Thus, is the direct sum of cyclic subspaces with isomorphic to an -cyclic subspace .

To illustrate the application of Theorem 3 we provide the following example.

Example 4. Let be a linear system with matrices Now we derive Kalman’s decomposition of via a similarity transformation : The unobservable matrix has a generator and a cyclic subspace : Note that to check that is cyclic it is sufficient to observe that and .
We embed into by defining . Again we can check that is a generator of the cyclic subspace by the conditions and . As expected is isomorphic to since the embedding is injective and . Finally, is the generator of a cyclic subspace : In the last identity note that is an isomorphism and which results in . It is also easy to see that since .

Corollary 5. If admits a canonical decomposition in direct sum, then the unobservable subspace can be decomposed in direct sum: where with being a generalized eigenvector of .

It is worth emphasizing that the above direct decomposition of in cyclic subspaces may not include all the cyclic subspaces associated with Jordan blocks (from now on and for the sake of brevity -cyclic subspaces from the Jordan decomposition of will be referred to as Jordan -cyclic subspaces) where . This occurs when the Jordan cyclic subspace has an eigenvector such that or more generally when . Furthermore the cyclic subspaces of the decomposition in Theorem 3 may not coincide with those of the Jordan canonical decomposition. What is clear is that the cyclic subspaces in the decomposition of are included in Jordan -cyclic subspaces as stated in the following corollary.

In the following proposition we state necessary and sufficient conditions for to be -spectral.

Proposition 6 (spectral subspace). Given a system its unobservable subspace is a spectral subspace for if and only if every for all .

Proof. Assume that is spectral for ; that is, is a direct sum of root spaces for where . This decomposition means that if then . Since for all , we conclude that for all .
Now we prove the converse. Assume that for all and define , . We prove the inclusion . Because root spaces are maximal invariant subspaces it is clear that admits a decomposition in direct sum . We define the -invariant subspace . By reductio ad absurdum assume that . The linear transformation restricted to , , has an eigenvalue and an eigenvector . In addition we have the chain of inclusions which means that ; that is, is an unobservable mode of (restriction of to ). However, the modes of are those in . This contradiction proceeds from the fact that was nonempty. Henceforth, , and is spectral for .

3. Structural Properties of the Unobservable Subspace

We begin analyzing the -cyclic subspaces that appear in the canonical decomposition of , which will be called unobservable subspaces. We point out that every unobservable cyclic subspace lies in one Jordan -cyclic subspace, as the following corollary suggests.

Corollary 7. Let be an unobservable -cyclic subspace as in Theorem 3. Then there exists a Jordan -cyclic subspace such that .

Proof. The vector is in a Jordan -cyclic subspace; that is, there exists a generator such that for some eigenvalue . Since is -invariant then vectors are in . Therefore and .

Example 8. The matrix in Example 4 has only an -cyclic subspace with generator . has generalized eigenvectors , , and and eigenvector . As shown above with generator . It is easy to see that this generator is a linear combination of generalized eigenvectors of , . Thus .

Lemma 2 can also be proved as a consequence of Corollary 7. We only show the sufficiency since the necessity is trivial: implies for some unobservable generator . By definition of cyclic subspace, is the minimal subspace that contains the generator . Therefore , and in virtue of Corollary 7, for some generator . From the decomposition of in -cyclic subspaces it is known that if . However and then . As a result, , and .

That lemma reveals that a sufficient condition for a Jordan -cyclic subspace to match with in the decomposition of is justly . This is reasonable since in that case the Jordan -cyclic subspace should be in .

Not only is every unobservable -cyclic subspace in a Jordan -cyclic subspace as revealed in 2, but also the former inherits successive generalized eigenvectors from the latter. This means that we can easily build a basis for the unobservable -cyclic subspace by choosing a subchain of generalized eigenvectors, from the Jordan -cyclic subspace, ending up in an eigenvector. This idea is stated in the following lemma.

Lemma 9. If with and are generalized eigenvectors, that is, then .

Proof. Firstly let us write the eigenvector as a linear combination of the vectors in the basis of , that is, , and multiply it on the left by : Since are linearly independent it follows that for . Consequently depends linearly on .
As for we write and Again are linearly independent so for . Thus depends linearly on and .
Proceeding similarly we find that depends on for and we obtain the following triangular arrangement: From this it is evident that, for , depends linearly on and then . Owing to the fact that vectors are linearly independent they constitute a basis for . Henceforth, .

Example 10. In Example 8, , where Applying back substitution , , then and .

Example 11. Consider the system with We have the canonical Jordan decomposition of in cyclic subspaces: From Corollary 5 the unobservable subspace can be decomposed as Therefore, Note that and are eigenvectors of the Jordan cyclic subspaces and , respectively, while is an eigenvector of . The last fact is a consequence of Lemma 9 since and .

Above in Lemma 2 we have shown that every unobservable -cyclic subspace is determined by the unobservability of its generator. Now we state the converse result for an -cyclic subspace to be in the observable subspace. (It is very obvious that the set of observable states has no structure of subspace. However, we can consider the observable subspace as the orthogonal complement of the unobservable subspace, .) In this case the observability of the eigenvector of the -cyclic subspace determines the observability of the whole subspace. This is proved in the following lemma.

Lemma 12. If and then .

Proof. By reductio ad absurdum, assume that . Then and . We take an arbitrary eigenvector . Since , , but from Lemma 9   depends linearly on which means and thus . This is a contradiction since we assumed by hypothesis that .

Note that the converse is not always true; that is, does not necessarily imply . In general is not spectral for an arbitrary matrix as shown in the example below.

Example 13. Let be a linear system with matrices In this case and where is a vector of the standard basis. From the PBH test it is easy to check that so . The generalized eigenvectors are and , which are not in the unobservable subspace. Henceforth .

Finally we address the problem of the -invariance of the intersection of with an arbitrary subspace . This problem is interesting in the well-posedness of reset control systems.

Lemma 14. A subspace is -invariant if and only if is an -cyclic subspace.

Proof. The reader is referred to [8].

Corollary 15. A subspace is -invariant if and only if is a direct sum of -cyclic subspaces.

Proof. Assume that admits a decomposition in direct sum of -cyclic subspaces . It is evident that is -invariant if and only if is -invariant, and from Lemma 14 this occurs whenever is -cyclic.

Due to the unobservable subspace is -invariant and in virtue of Corollary 15 it follows that is a direct sum of -cyclic subspaces.

Corollary 16. Let be a subspace in . is -invariant if and only if is a direct sum of cyclic subspaces.

Proof. The A-invariance of is a consequence of Corollary 15.

4. Geometric Multiplicity of the Zeros in Feedback Systems

This section is devoted to prove that unobservable modes of unit feedback systems (whose components have minimal realizations) always have geometric multiplicity . The geometric multiplicity of a zero was firstly defined by Owens for minimal realizations, [9], and we extend the definition for any system.

Definition 17 (geometric multiplicity of a zero). Given a SISO system with a realization not necessarily minimal, the geometric multiplicity of is defined as the dimension of the nullspace of the pencil matrix .

If it is evident that , and the loss of full rank occurs when there exist a vector and a scalar such that Obviously if has a minimal realization (i.e., there does not exist zero/pole cancellations), is invertible which means that is uniquely determined by (since implies ) and the geometric multiplicity of is always . If has a nonminimal realization there exist two cases:(i): It is evident that and that . In virtue of the PBH test, the condition indicates that is an unobservable mode and thus the geometric multiplicity is the geometric dimension of the unobservable mode, .(ii): In this case and with . If the system has a proper realization, that is, , it follows that , and thus , and . It is known that is either or so that the geometric multiplicity is either or . If the system has no proper realization, that is, , is uniquely determined by according to the relation ; as a consequence the geometric multiplicity of is .

Now we focus on unit feedback systems where and are minimal realizations.

Lemma 18. Let be an unobservable mode of and assume that and have minimal realizations; then the geometric multiplicity of the zero is always 1.

Proof. Resorting to the PBH test, It follows that : By reductio ad absurdum if , . If , from the minimality of the realization for and the PBH criterion, , which is impossible since . In virtue of the minimality of , since if and only if , then leads to a contradiction since it was assumed that .
For to be a zero of it is necessary that to be invertible and this occurs so long as . Assume that ; the PBH criterion implies actually it is verified for all eigenvector in , which is absurd from the minimality of ; thus . Note that is effectively a zero of : since and from arguments of minimality along with the PBH criterion . Henceforth . Owing tothe fact that is a zero of , is uniquely determined by via the transformation .
implies and for SISO systems . Then there exist a vector and a scalar such that . As a result and linearly depends on : Henceforth the geometric multiplicity of as an eigenvalue of is always , and there is only a Jordan block associated with any .

5. Transmission Zeros and -Spectrality of the Unobservable Subspace

In this section we analyze when the unobservable subspace of a feedback SISO system is spectral for the system matrix. To this purpose consider a SISO unit feedback control system with single input and single output . Assume that the plant and the regulator have minimal realizations and , respectively: where , are internal states, stands for the error signal, and represents the control law. The closed-loop system can be written in the state-space by considering the state of dimension : We begin exploring the connection of an unobservable -cyclic subspace to the transmission zeros of the plant. Let be such an -cyclic subspace with generator and dimension . A basis of eigenvectors of is given by where . It is obvious that and . In virtue of Lemma 2, a necessary and sufficient condition for to be in is that . As result we proceed by checking the tower of generalized eigenvectors, , from top to bottom. To this end we need the following technical lemmas on zeros of SISO systems.

Lemma 19. A necessary and sufficient condition for to be unobservable is that .

Proof. Consider the following.
Sufficiency. Since has a minimal realization, if then for all is satisfied. Also the minimal realization of the plant means that implies , which results in the invertibility of . We can define so , and is an unobservable mode of .
Necessity. If is unobservable, in virtue of the PBH’s criterion there exists a vector such that . This condition signifies that .
If and , then , and implies , which is a contradiction. As a result, . However this is absurd because . Therefore, .

Lemma 20. If is unobservable then .

Proof. From Lemma 19  . The minimal realization of implies . Consider . The invertibility of guarantees that . On the other hand since . Henceforth .

From Lemma 20 and the PBH test it follows that with . Additionally implies . If has a minimal realization, (i.e., has only one Jordan block associated with ). Given that and it follows that is invertible, and then is uniquely determined by ; that is, . As a result and This means that and that we only need to account for one of the eigenvectors of to apply the PBH test. Given that we have only one Jordan block associated with , there exists only one -cyclic subspace .

Now we need the following technical lemma around the multiple zeros of a SISO system.

Lemma 21. A necessary and sufficient condition for to be a zero of multiplicity of the SISO system (with transfer function is

Proof. For more details the reader is referred to Appendices.
In the following proposition we state sufficient conditions for to be in in terms of the existence of a multiple zero in the plant.

Proposition 22. A sufficient condition for to be in is that should be a zero of with multiplicity .

Proof. Assume that with multiplicity . We proceed recurrently through the tower of generalized eigenvectors from top to bottom by connecting to through the following problem: As a result we have that is a function of : In virtue of Lemma 21, for , so . Henceforth . On the other hand is -invariant and ; as a consequence and for , which means . From Lemma 2  .

Because systems and are minimal it follows that they are controllable and observable. From the PBH test it is known that if is an unobservable mode with eigenvector then and . Now, we explore whether or not : To this purpose it is worth resorting to the external description of the feedback system by means of the closed-loop transfer function. We write the transfer functions of the controller and the plant as and , respectively (with the numerator and the denominator being coprimes), and the closed-loop transfer function from to as If then , and as a result (due to . However, we have shown above that which is in contradiction with the assumption of minimality of the transfer function . Therefore, .

If it is clear that is a root of the characteristic polynomial , and then . The minimality of the transfer function together with (or ) implies that . As a result, the lack of observability is caused by cancellations of zeros of with poles of .

Remark 23. Assume that we reduce the closed-loop transfer function as , where but . If is a root of the denominator of there exists a cancellation zero-pole. Additionally, and then which is impossible since and is a divisor of .

In some occasions can be greater than the algebraic multiplicity of as a pole of . The following example reveals this case.

Example 24. Consider a plant with an external description and an observable internal description : And let be a regulator system with transfer function and an observable state-space description : The interconnected system via feedback has the following state-space equations: The poles of are located at , ; that is, . In virtue of the Jordan canonical decomposition we can write where and . According to Lemma 9 where and . From the PBH test the mode is observable () and is nonobservable (). Furthermore, since , in virtue of Lemma 12 it follows that . However , and . This means that and we can not derive a spectral decomposition of as a direct sum of root spaces. Note that in this case we have the following closed-loop transfer function: In this case there exists a cancellation of two poles of the regulator with two zeros of the plant at , but the feedback generates two additional closed-loop poles at . Thus, we have a unique Jordan block of dimension and an unobservable invariant subspace . Henceforth, we conclude that .

Proposition 25. A necessary condition for to be in is to be a pole of with multiplicity .

Proof. If with , then for . This implies that ; that is, is a generalized eigenvector associated with in . Therefore is a pole of with multiplicity .

Example 26. We have one unobservable mode with three subcyclic subspaces , , and and the root subspace .

From Propositions 25 and 22 we conclude that a necessary and sufficient condition for to be in is that there should exist exactly cancellations between zeros of the plant and poles of the regulator. Owing to the fact that systems and are minimal it follows that they are controllable and observable. From the Popov-Belevitch-Hautus test (PBH test) it is known that if is an unobservable mode with eigenvector then and . Furthermore, the single output system is observable if and only if in the Jordan form matrix there is one Jordan block associated with each distinct eigenvalue and every entry of corresponding to the last column of each Jordan block is nonzero. For an unobservable mode , if we have that is -cyclic and as a consequence we can determine whether or not in terms of poles and zeros.

6. Discussion

The canonical structure of the unobservable subspace as a direct sum of cyclic subspaces (spectral decomposition) has been analyzed via Kalman’s decomposition. Necessary and sufficient conditions have been provided for the unobservable subspace to be spectral for the system matrix ; specifically, when a cyclic subspace has an eigenvector in the orthogonal complement of the unobservable subspace then it is strictly included in this subspace. This property does not always hold for the unobservable subspace, but rather we can state at most that the unobservable cyclic subspaces are included in a cyclic subspace from the Jordan decomposition. This analysis leads to determining the conditions of -invariance for the intersection of the unobservable subspace with an arbitrary subspace. Finally, spectral decomposition of the unobservable subspace is connected to the existence of multiple transmission zeros in feedback systems.

Appendices

A. Multiple Transmission Zeros

This section is devoted to present the proof of technical Lemma 21.

Proof. Consider the following.
Sufficiency. From the definition of SISO system in the state-space form the following identities are in order: Finding from (A.2) yields and by definition of zero of the system , Thus, Now folding (A.2) into the last expression leads to and as a consequence, On the other hand we have assumed that is a multiple zero so the following identity is immediate: In a similar way, and finding on the right hand of this equation we obtain Using (A.8) and (A.4), the output is simplified into that is, In virtue of the multiplicity of the zero we conclude that and repeating successively this procedure results in Necessity. The proof is straightforward. If for then for , and additionally , so that is a zero of the system .

Remark A.1. An alternative proof of this is based on the derivative of the inverse function of a linear transformation with matrix (this is a simple application of the chain rule): From this it follows that Let be the transfer function of the system and a zero of . Applying the last identity with we derive the following result: Now if we apply the same procedure to we obtain and so it is easy to check the identity For to be a zero of multiplicity of , it is necessary that for . According to (A.19) this is equivalent to which is the same expression that we delivered in the technical lemma.

B. Nonminimal Transfer Functions

In this section we illustrate how a nonminimal transfer function (reducible rational function) can be represented in the space-state form.

Example B.1. Let be an unobservable system with nonminimal transfer function . Its expansion in partial fractions exhibits two zero residues: We define the states for which leads to the relations . As a result, we can express in the Jordan form: As the vector of observations tends to ; that is, there exist states that do not contribute to the output, as expected from the unobservability of .

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.