In this section, we consider the case of the system (1) with a singular pencil. The class of in this case is characterized by a uniquely defined element, , known as the complex Kronecker canonical form, see [39, 41, 44, 45], specified by the complete set of invariants of the singular pencil . This is the set of the elementary divisors (e.d.) and the minimal indices (m.i.). Unlike the case of the regular pencils, where the pencil is characterized only from the e.d., the characterization of a singular matrix pencil apart from the set of the determinantal divisors requires the definition of additional sets of invariants, the minimal indices. The distinguishing feature of a singular pencil is that either or and . Let , be the right and the left null space of a matrix respectively. Then the equations
(18)
(19)
where is the transpose tensor, have solutions in , , which are vectors in the rational vector spaces and respectively. The binary vectors and express dependence relationships among the columns or rows of respectively. Note that and are polynomial vectors. Let and . It is known, see [39, 41, 44, 45], that and as rational vector spaces are spanned by minimal polynomial bases of minimal degrees
(20)
and
(21)
respectively. The set of minimal indices and are known as column minimal indices (c.m.i.) and row minimal indices (r.m.i.) of respectively. To sum up, in the case of a singular pencil, we have invariants of the following type:
-
finite elementary divisors of the type ;
-
infinite elementary divisors of the type ;
-
column minimal indices of the type ;
-
row minimal indices of the type .
The Kronecker canonical form, see [39, 41, 44, 45], is defined by
(22)
where , are defined as in Section 2. The matrices , , and are defined by
(23)
where for
(24)
where for . The matrices , are defined as
(25)
where for
(26)
where for .
For algorithms about the computations of these matrices, see [39, 41, 44, 45].
Following the above given analysis, there exist nonsingular matrices P, Q with , such that
(27)
Let
(28)
where , , , and .
Lemma 3.1 The system (1) is divided into five subsystems:
(29)
the subsystem
(30)
the subsystem
(31)
the subsystem
(32)
and the subsystem
(33)
Proof Consider the transformation
Substituting the previous expression into (1), we obtain
Whereby multiplying by P and using (27), we arrive at
(35)
Moreover, let
where , , , and . Taking into account the above expressions, we arrive easily at the subsystems (29), (30), (31), (32), and (33). The proof is completed. □
Solving the system (1) is equivalent to solving subsystems (29), (30), (31), (32) and (33). The solutions of the systems (29), (30) are given by (10) and (12) respectively; see Propositions 2.1 and 2.2.
Proposition 3.1 The subsystem (31) has infinite solutions and can be taken arbitrarily
Proof If we set
by using (23), (24), the system (31) can be written as
(37)
Then, for the non-zero blocks, a typical equation from (37) can be written as
(38)
or, equivalently,
or, equivalently,
or, equivalently,
(39)
The system (39) is a regular-type system of difference equations with equations and unknowns. It is clear from the above analysis that in every one of the subsystems one of the coordinates of the solution has to be arbitrary by assigned total. The solution of the system can be assigned arbitrarily
The proof is completed. □
Proposition 3.2 The solution of the system (32) is unique and is the zero solution
(40)
Proof From (25), (26) the subsystem (32) can be written as
(41)
Then for the non-zero blocks, a typical equation from (41) can be written as
(42)
or, equivalently,
or, equivalently,
or, equivalently,
(43)
We have a system of +1 difference equations and unknowns. Starting from the last equation, we get the solutions
which means that the solution of the system (32) is unique and is the zero solution. The proof is completed. □
Proposition 3.3 The subsystem (33) has an infinite number of solutions that can be taken arbitrarily
Proof It is easy to observe that the subsystem
does not provide any non-zero equations. Hence all its solutions can be taken arbitrarily. The proof is completed. □
We can now state the following theorem.
Theorem 3.1 Consider the system (1) with a singular pencil and known boundary conditions of type (2). Then the boundary value problem (1)-(2) is consistent if and only if:
-
1.
-
2.
the column minimal indices are zero, i.e.,
-
3.
(46)
Furthermore, when the boundary value problem (1)-(2) is consistent, it has a unique solution if and only if
-
1.
-
2.
(48)
In this case the unique solution is given by the formula
(49)
where
C
is the unique solution of the algebraic system
(50)
In any other case the system has infinite solutions.
Proof First we consider that the system has non-zero column minimal indices and non-zero row minimal indices. By using the transformation (34), the solutions of the subsystems (29), (30), (31), (32) and (33) are given by (10), (12), (36), (40) and (44) respectively. Note that from Proposition 2.1 the solution (10) exists if and only if
Furthermore, if
Since is unknown, it can be replaced with the unknown vector C. Then
or, equivalently,
Since and can be taken arbitrarily, it is clear that the general singular discrete time system for every suitable defined boundary condition has an infinite number of solutions. It is clear that the existence of the column minimal indices is the reason that the systems (31) and consequently (33) exist. These systems as shown in Propositions 3.1 and 3.3 have always infinite solutions. Thus a necessary condition for the system to have a unique solution is not to have any column minimal indices which are equal to
In this case the Kronecker canonical form of the pencil has the following form:
(51)
Then the system (1) is divided into three subsystems (29), (30), (32) with solutions (10), (12), (40) respectively. Thus
or, equivalently,
The solution exists if and only if
or, equivalently,
or, equivalently,
For the above algebraic system, there exists at least one solution if and only if
The algebraic system (50) contains equations and p unknowns. Hence the solution is unique if and only if
and
where C is then the unique solution of (50). The uniqueness of C can be proved as follows. If we assume that the algebraic system has two solutions and , then
and
or, equivalently,
But the matrix is left invertible since it is assumed to have p linear independent columns and and hence
The unique solution is then given from (49). The proof is completed. □
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