SIAM J. CONTROL
Vol. 12, No. 1, February 1974
REGULATION AND INTERNAL STABILIZATION
IN LINEAR MULTIVARIABLE SYSTEMS*
W. M.
WONHAM?
AND
J. B.
PEARSON:
Abstract. For the multivariable control system described by
Ax
+ Bu,
y= Cx, z
Dx,
Fx
constructive necessary and sufficient conditions are given for the existence of state feedback u
such that (i) Ker F Ker C (observability constraint), (ii) D exp [t(A + BF)] 0 as
(output
regulation), and (iii) any unstable modes of A + BF are either uncontrollable or unobservable at y
(internal stability). It is assumed that Ker C is A-invariant, or equivalently that an observer or dynamic
compensator is utilized. A common application is treated, and sensitivity is considered for a simple
example.
1. Introduction. In this paper we continue the discussion in [1] on output
regulation for the system
:(t)-- Ax(t) + Bu(t),
y(t)- Cx(t),
z(t)- Dx(t).
(1)
(2)
We refer to [1] for notation and a general description of the problem. As in
we regard y(. as the observed variable and z(. as the variable to be regulated.
We assume that either a dynamic observer as in [1], or a dynamic compensator
Ker C, we have Aff
as in [2], is utilized; equivalently, writing dV"
For regulation of z(. ), it is required to find a feedback map F :f- such
’
that
f/(A + BF) Ker D.
(3)
To respect the observability constraint, F must satisfy the condition
Ker F
(4)
= ’.
Necessary and sufficient conditions for the existence of F subject to (3) and (4)
were given in
In this paper we impose the additional requirement that F stabilize all the
unstable modes of A which are both controllable and observable. Precisely,
regard Y’/dV as the state space of the system (1), (2) made observable by reduction
modf’. The controllable, observable subspace is then ((AN)+ V)/V. We
require that the map induced on ((AN) + V)/f by the closed loop system
map A + BF be stable. Equivalently, any observable, unstable modes of A / BF
":
Received by the editors August l, 1972, and in revised form March 13, 1973.
Department of Electrical Engineering, University of Toronto, Toronto, Canada, M5S 1A4. This
research was supported in part by the National Research Council of Canada under Grant A-7399.
Department of Electrical Engineering, Rice University, Houston, Texas 77001.
6
w.M. WONHAM AND J. B. PEARSON
must be uncontrollable;that is,
f+(A + BE)+ V <A> +
1"3
4/’__ 0.
It is natural to call a system in which F has been chosen to satisfy (5) internally
stable. In this way we are led to formulate the following.
REGULATOR PROBLEM WITH INTERNAL STABILIZATION (RPIS). Given the maps
D :Y
and a subspace Y c 35 with AU ,A/, .find
B :i
A :f
F: f l such that
Ker F =
(6)
,
(7)
,
,
f+(A + BF)
(<A> + l/)
and
f+(A + BF) Ker D.
Here it is easily checked that (7) is equivalent to (5).
The "restricted regulator problem" (RRP) was defined in [1] and is identical
to RPIS except that the internal stability requirement (7) is dropped. We remark
that RRP may be solvable when RPIS is not:that is, internal stability need not
be compatible with output regulation, as shown by examples in [1] and [2].
In 2 we provide a first set of necessary and sufficient conditions that RPIS
be solvable. While not constructive, theyare exploited to show that in this problem,
dynamic compensation in the sense of [1] is redundant. In 3 we give constructive
necessary and sufficient conditions in the case sls 0, and in 4 extend them to
the general case. An application is discussed in 5. Finally, in 6 we indicate
how to deal with the sensitivity problem which may arise when the map A is
subject to small perturbations.
2. Solution of RPIS: general considerations.
THEOREM 1. RPIS is solvable if and only if there exists a subspace /" f
such that
(9)
(0)
(11)
U
Ker D
n A-1(’/
I-
"3
),
f+(A) N v + A(# N V)
V N (<A> + 4/) V,
,
and
(12)
f+(A)
<A’> +
We observe that conditions (9), (10) and (12) are equivalent to solvability
of RRP, as shown by Theorem of [1]; only condition (11) is new. The proof
follows exactly the same lines as in the theorem cited, and so need only be sketched.
f+(A + BF). Since AV" c
Proof. Suppose RP!Sis solvable and put
and Ker F V’,-we have by Lemma 2 of [1],
f+(A + BF) N
/
f+(A) N
Then (9)--(11) follow immediately from (6)-(8), and (12) follows from the general
7
LINEAR MULTIVARIABLE SYSTEMS
identity [1, Lemma 4],
(A) + f+(A)= (A) + f+(A + BF).
such that
Conversely, if (9) and (10) are true, there exists Fo:f
Ker Fo dV" and (A + BFo)U
Write Ao A + BFo. By (12) and (13),
(13)
.
’
f+(Ao)
(14)
Now
f+(Ao) VI dV +(A) 91 U c //.
(15)
Just as in the proof of Theorem of [1], (14) and (15) imply the existence of
F1:2’
has all the properties required.
if one exists,
While Theorem does not indicate how to find a suitable
then
no
solution can be
it is well suited to showing that if RPIS is not solvable,
obtained by broadening the assumptions to include the possibility of state space
extension, that is, dynamic compensation. We may interpret this result as a
deterministic "separation theorem" which asserts that, after insertion of a dynamic
observer, no further dynamic signal processing is required to achieve the stated
design objectives, if these objectives can be achieved at all.
To be precise, introduce extended spaces and maps exactly as in 1, 3].
the
notation used there, we now define the "extended regulator-problem with
In
internal stabilization" (ERPIS) as that of finding suitable fa (that is, d(fa)) and
then Fe
Oe, such that
:re
Ker Fe
e+
f A --k B
Fe N
< A e > -k- ,IV’)
and
(Ae + BeFe)
Ker D (
THEOREM 2. ERPIS is solvable only if RPIS is solvable.
Proof. If ERPIS is solvable, Theorem implies the existence of
such that
e c (Ker D
-
f.) f/ A-1(
,e+( Ae) M / + Ae( N )
(16)
fe
, + ’a),
(7)
Ve,
(18)
(19)
Here we have used the facts (cf. [1, 3]) that e V, Ker D
Ker D
and (AeNe) (A)
Let P :re fe be the projection on f along f, and define
Pe. It is
enough to show that / has the properties (9)-(12), and this requires only the
-
It should, however, be borne in mind that in the present problem formulation, no explicit account
is taken of the sensitivity of the synthesis to parameter perturbations; and if this is done, additional
dynamic elements may sometimes be used to advantage ( 6).
8
W. M. WONHAM AND J. B. PEARSON
application of P to both sides of the relations (16)-(19). By definition of P and
Ae, PAe AeP and Ae]Sf A. Using these facts and rewriting (16) as
e c KerD @ a, Ae’e c e +’there follows
c KerD, Ac +,
which is equivalent to (9). Next, the obvious relation
(Ae)
(20)
+(A)
a,
,
together with (19), establishes (12). To verify (11) from (18) we use the following
general result for a map P and subspaces
[3, Proposition A.5]:
(21)
P(M
)
(PM)
(P)
if and only if
(M+) KerP=M
,,
(22)
KerP+
,
,
KerP.
With Ker P
M U and
<A> + f, + (22) follows at once,
and then (21) applied to (18) yields (ll). It remains to check (10) from (17). By
(20),
*(A)
+(A)
and so
f+(A) f’l Y
(23)
c
P/
Also, by (21), (22) we shall have
(24)
P(e f )
Vf
,
,
provided
(25)
As for (25), let
with Ve
(+)N
gN
,
+vN
,.
e and n e A/. Then
ne7,, +
and by (18), Ve A/’. Therefore x e A/ f’l 5f, 0, that is,
( + ) 0,
proving (25). Then (24) is true, and (17) yields
(26)
PAe( f"l .IV’)= A("U f-I A/’).
Finally, (10) results from (23) and (26).
v
.
x,,
3. Constructive solution of RPIS when U 0. We first recall from [3]
c
and [4] certain properties of (A, B)-invariant subspaces, that is, subspaces
such that Ac
+ Let_F(V)be the set of maps F:- such that
’
.
LINEAR MULTIVARIABLE SYSTEMS
Then _F(C)4:
is (A,B)-invariant. We
if and only if
(A + BF)C
denote by C* the largest (in the sense of inclusion) (A,B)-invariant subspace
c Ker D,
C* sup {//"
Ker D [’1 A- 1(// + )}
and by N* the largest controllability subspace [4] contained in Ker D. Then
and _FffU*) E(N*). If F e_F(U *) and AF A + BF, let AF be the
N* c
map induced in f/N* by Av. Then the restriction $vI(C*/N*) is independent
of the choice of F e _F(C*).
Let Arelative
and AN N
The subspace N decomposes
5e and N 5
to A if there exists a subspace 5e such that ASe
A conis given in the
structive necessary and-sufficient condition that N decompose
-
*
-.
--
-.
Appendix; it amounts to the well-known fact that decomposability is equivalent
to the existence of a solution to a simple linear matrix equation.
THEOREM 3. Let
O. Then RPIS is solvable if and only if
+(A) (A[) + *,
(27)
and in W/N*, with F _F(U*), the subspace
(/* N f+(AF) N (A}
decomposes the subspace
+
(* N +(A) + *)/*
relative to the map induced by Ay in
It is clear that this solvability criterion is constructive, as //*, N* are computabl by simple algorithms [4 an (arbitrary) F E _F(/*) is readily constructed
[4] and decomposability is verifiable by a transformation of basis and rank checks
(Appendix).
Proof. (If). Let F _F(V*). By (27) and Lemma 4 of
f+(AF) (A + /*.
As the subspace on the right is Ay-invariant, we have
(28)
(A[5 I’-I f+(AF) +
,T+(AF)
* I"1 f+(Av),
Also, by the assumption of decomposability, there exists a subspace
such that
(29)
(30)
AF#
N*
f"
/* (-1
+(AF)
and
(31)
+ N*
* N +(Av)
N*
We remark that with
(32)
U*
N (A) N +(Av) + N*
N*
-N--’"
fixed, (29)-(31) hold for all F F_(U*). By (31),
* N 5+(Av) * N (A) N f+(AF) +
,
10
W. M. WONHAM AND J. B. PEARSON
and by (28) and (32),
f+(Av)
(33)
Clearly,
(A[) + U* f’)f+(Av) (A[) + U.
(34)
Also, by (31),
so that
# ) (A’)
(35)
Finally, let
such that
.*.
AF denote the map induced by A in f/*, and choose F _F(f*)
r(Avl*)
Then N* decomposes t relative to
on F) such that
Av; that is, there exists
(depending
A
(36)
and
(37)
From (33)-(37), we conclude that
/
f’l <A[>
0,
+(A)
and it follows by Theorem that RPIS is solvable.
(Only if). If RPIS is solvable, Theorem supplies a subspace U such that
and
f+(A)
0 we have U f’l
(38)
0, so that
Since U fq (A[)
=
it is clear that F _F(U) f) _F(*) can be chosen such that
(38) and Lemma 4 of [1] we have
*
.
From this
* *
AvI* is stable. By
Av-invariant there follows
5+(Av) = <A> f] f+(Av)( / f’l 5+(Av)
= <A> f-I f+(Av) + (* f’l f+(Av) = 5+(Av),
Since all the subspaces here are
and therefore
/* f-)
f+(Av)
c
(A> f’l f+(AF)
/" ["l
5J’+(AF).
LINEAR MULTIVARIABLE SYSTEMS
* and using U* = U, we obtain
* f) f+(Av)= (A) f] +(AF) f"l q/* U I’) +(AF)
-, say.
5
Intersecting both sides with
(39)
Let P’W
W/* be the canonical projection. By the stability of AvI*,
we have
(5 @ -) f)
(Sg @ -) f’l Ker P
*
=0
5e f) Ker P
_
and therefore
(40)
(pc)
p(3 CI 3-)
(p-)
-
f3 Ker P,
O.
By (39) and (40) we have, finally,
P[(* Iq 5f+(Av)]
P[(A5 f"l f+(Av) f"l //*] if3 P[// I"l +(Av)],
a decomposition of the type required.
that
Remark. The foregoing proof of sufficiency made no essential use of the fact
is actually the supremal element of the family of subspaces
*
{/-Y"
c
Ker D f-I
A-1( + )}.
*
The sole reason for stating Theorem 3 in terms of
is that this element of
is readily computable algorithmically, and so the obtained conditions are constructive. It is clear from the proof that the conclusion of Theorem 3 is valid
provided the stated conditions hold for some element e_, with
replaced
by the largest controllability subspace (c.s.) c
.
*
4. Constructive solution of RPIS for arbitrary V’. It is not difficult to extend
Theorem 3 to the general case. Suppose first that RPIS is solvable with the map F.
Since Ker F = U, we have by Lemma 2 of [11] that
f+(A)
(41)
is (A
f] U
f+(A + BF) f’l U
+ BF)-invariant. Let
..,
be the canonical projection, and let bars designate the maps induced in 5f. As
q/exists uniquely such that FP
F, and it is easily seen
Ker F = Ker P, F"
--.
exists uniquely such
and
(41),
that P Ker F Ker Similarly, by (8)
"q/
Ker/.
B
define
by B PB.
Ker
D
and
P
Finally,
that DP D,
Now A
+ BF
"
+/ff so (by [5, Lemma 2]),
Pf + (A
+ BF)
+(
+ BF).
12
W. M. WONHAM AND J. B. PEARSON
Also
[f+(A + BF) + (A[5 + A/] f’l Ker P
g*(A) V
f+(A + BF)
n Ker P + ((A5 + g’) n Ker P.
With these observations, we may project both sides of (6)-(8) to obtain
(42)
(43)
(44)
Ker F
A/,
.
+( + BF) ((A5 + AT)c
+( + BF) Ker
,
Automatically
+($) N 17
or equivalently,
(45)
A/" c
f-(A).
We have shown that if RPIS is solvable, so is the reduced problem (42)-(44)
in f, and (45) is true as well. Conversely, suppose
f + (A) gl A/"
(46)
Ker D,
and that F" f --, exists such that (42)-(44) are true. Define F FP. By reversing
the steps which led to (42)-(44), one can routinely verify that (6)-(8) are true,
i.e., that RPIS is solvable. We therefore have the following.
LEMMA 1. RPIS is solvable if and only if the reduced problem (42)--(44) is
solvable under assumption (46).
Next we show that in (43) we may set V" 0.
LEMMA 2. If (42)--(46) are true, then
.
r+(. + BE) (A)
(47)
Conversely, if (46) and (47) hold, so does (43).
Proof. By (43),
,+( + BF) fl (<A> + AT) f I"1 f+(fi,
A7 N
+($)
+ BF)
by (42)
--0.
Conversely, the left side of (43) can be written
+( + BF) I") [<A> I"1 +($ + BF) + <A> I"1 -($ + BF) + N7 ["1 r-(fi.)]
+ (, + BF)fl <.#>
=0.
By Lemmas and 2, the solvability of RPIS is equivalent to solvability of the
reduced problem (42), (44), (4?) under assumption (46). Our next result implies
13
LINEAR MULTIVARIABLE SYSTEMS
that condition (42) is redundant. For simplicity of notation, we temporarily drop
bars.
be such that
LEMMA 3. Let Fo :
f+(A + BFo) CI
There exists Fx :
O.
ll such that
Ker F
(48)
(49)
<AN>
= f-(A),
f+(A + BF)= f+(A + BFo).
Proof. In this proof, primes are used as indices. The lemma will be proved
in three steps. First, let P+ :<AN>
<AN> be the projection on <AN> 13 f+(A)
along <A]N> f3 f-(A), and write A + AI[<A]> VI Y’+(A)],B + P+B. Since
<A> f-) +(A) P+<A>
(A+,B +) is controllable,
we have that
A+
+
B +F +
so there exists F +
:Y’+(A)
is stable. Choose 5f- such that
<A>
f’l f-(A) 0) 5-
such that
f-(A),
5e
5e- and (A>
and then
such that 5e
according to
’
5f. Now define
F’o’f
F)I[<A[> fq 3+(A)]-- F +,
F)I[<A> (3 f-(A) 53 O.
Write
A’o
(50)
A
+ BF’o and F
F. It is then clear that
Ker F f-(A),
f+(A)) f-) <A’> 0,
Fo
+(A’o + BF)
<A[>
O.
As the second step, we claim there exists an F’x’f
such that
Ker F’
(A[>,
f+(A’o + BF’,)= f+(A’o + BF’) (= f+(A
(51)
(52)
+ BFo)).
For this let
F’II<A>
O,
-
F’If+(A) + BF’)-- Fglf+(A) + BUg),
and let
F’II- be defined arbitrarily on
some complement
+ (A) (= f+(A) + (A))in 5f. Write
A A’o + B F, A’a
A’o + B F’x.
Since f + (A) is A-invariant, and
A’If+(A’)
A[f+(A),
of
f+(A’o + BF))
14
W. M. WONHAM AND J. B. PEARSON
there follows by Lemma 2 of[l],
f+(A’x) f-I 35+(A’)
f+(Ao),"
so that
35 +
35 + A )
(53)
A’
Similarly, we have
35+(A’o) 0 (A)
35+(A’) f’l (A)
O.
By Lemma 4 of 1], there results
<A[> ( 35+(A’g)
(54)
<A[> ( 35+(A’),
and (52) follows at once from (53) and (54).
As the last step, we prove the existence of F’ "35
’
35-(A)+ (A),
Ker F’;
(55)
(56)
such that
35+(A’o + BF’)= 35+(A’o + BF’).
A bar will denote a subspace or induced map in 35
35/(AJ). Let P’35 --, 35
+(fi.) ( 2-(ft.), and by (51), F’ F]P
be the canonical projection. We have
for some F’ "35 --, ’. Define F1 F1P, where
Fail-(A)
0.
Then
Uer F
P Ker f
("
(A),
so that
35-(A)
c
Ker F1
and (55) is true. Also if x e 35+(A’o
+ (AN)
Ker
+ BF’) then Px e 2+(fi.), so
Fx FIPX F’IPX
F’lX;
therefore
(A’o + BF’)If+(A’o + BF’)= (A’o + BF’)]35+(A’o + BF’I),
and there follows
(57)
35 + A’o
+ B F’
35+(A’o + B F’
Similarly,
(A’o + BF’)I<A>
(A’o +
and so
35+(A’o + BF) f-I (A)
35+ (A o + BF’) 0 (A)
O.
LINEAR MULTIVARIABLE SYSTEMS
15
This means
,Y+(A’o + BF’)’
+ (A o + BF)
C +(A’o + BF;)
, f + (A’o + BF’a),
(58)
and (56) follows by (57) and (58).
It remains only to define
F{) +
F
F’.
Then (48) follows by (50) and (55); statement (49) follows by (52) and (56).
It is now easy to prove our main result. For this we revert to the notation
introduced at the beginning of this section.
THEOREM 4. In the general case dV" # O, RPIS is solvable if and only if (i)
f+(A) f"l
(59)
A/" c
Ker D,
and (ii),
f/[f+(A) ffl 4/’], the reduced problem
in the factor space
that is, there exists Fo
(60)
"
is solvable,
such that
.g+( + BFo)c Ker B
and
16)
+
( + BFo)
gl
<A)
O.
Of course, the reduced problem (ii) is formally identical to that of Theorem 3.
Proof. (If). Suppose the reduced problem (RP) defined by (60) and (61) is
solvable. Lemma 3 applied to RP yields a map F’f --, such that
’
Ker F
f-(A),
and F satisfies (44) and (47). Since f-(A) W we are assured that (42) is true
as well. As already noted, Lemmas and 2 now imply that RPIS is solvable.
(Only if). The necessity of (59)is immediate from (9)and (10); the necessity
of (60) and (61) follows by Lemmas and 2.
5. Application" regulation in the presence of step disturbances. As a simple
application of Theorem 3, consider the system
fm
2
Z
A1X1
2_
A3x2 + Bau,
-’-0,
OlX + O2x 2.
We assume that y x and (A, B) is controllable. The equations represent a
controllable plant subjected to step disturbances which enter both dynamically
and directly at the regulated output, a situation common in industrial process
control.
In basis-flee terms, our assumptions amount to the following"
(62)
(63)
4/’= 0,
Im A
c
(A>.
16
W. M. WONHAM AND J. B. PEARSON
We now have our next theorem.
THEOREM 5. Subject to the assumptitons (62) and (63) RPIS
only if
(64)
(A[) + Ker D N A-I f.
is solvable
if and
Proof. (If). Exploiting the remark after the proof of Theorem 3, let
7 Ker D n A-M.
(65)
From (64) it is clear, first, that
+(A) = (A) +
.
Also, as AUc M, there exists an F6_F(/) such that AF[/
A F A + BF. Then
Ker A F ’+(AF)
0, where
SO
C N f+(Av) N <A[>
C N <A[>.
According to Theorem 4.3 of [4], the supremal c.s.
in
is given by
The second condition of Theorem 3 (with U, o in place of U*, 2*) will thus be
satisfied if (7 (A[))/( 7) decomposes /( 7) relative to the map
induced by A v in 7/( /7). Since AFIt7 0 this is trivial, and the result follows.
(Only if). Let f f/(A[), and now use bars for subspaces and induced
maps in f. By (63), A 0. Suppose RPIS is solvable by the map F. Since AF A
for all F, and since Y’+(AF)N (AN)= 0, we know that f+(AF)= Ker AF.
Now Ker AF A-I for any F, so
n
n
n
n
f+(AF)
Ker D N A-1.
By Lemma 3 of [1], there results
f+(A) = (A) + Ker D N A-,
and therefore
f
(A) + f+(A)
(A[5 + Ker D N A-
6. Dynamic compensation and sensitivity. Even if RPIS is solved, it need not
be true that output regulation is maintained if internal parameters deviate slightly
from their nominal values. However, it is often possible to achieve insensitivity
by application of the theory to a suitably augmented version of the original
problem. We illustrate the method by a trivial example, deferring a general
treatment to a future article. Consider
2c
ax + u,
y-’-(X1,X2)’
"2
0,
Z-- X
X2,
17
LINEAR MULTIVARIABLE SYSTEMS
where x1, x 2 are scalars. RPIS is solvable with u
fix1
.qt_
f, +a<0,
f2= -(a+fl).
If a changes to a + e, with (f, f2) fixed, the condition f +
f2x2
a
where
+ e < 0 for internal
stability is preserved for small e, but
z(oe)
-(a + e + fx)- ex2(0+),
so output regulation fails.
Heuristically, replace ex by a new variable x3 with 23 0. That is, x3 should
satisfy the unstable differential equation induced in f/(A);since only
is assumed to change, this equation does not depend on Now consider RPIS
for the augmented system
.
2
ax + X 3
22
23
/ U,
O,
Z
(Xl, X2)"
Xl
Here the pair (C, A) is observable; an observer for w
y
9(;2"
Xl
x3 is given by
k= -w+(a+ 1)yl +u.
(66)
Also, RPIS is solvable; one solution is u f’x, with
f3= -1;
f2=l,
fl=-l-a,
this can be implemented as
(67)
-(2 + a)yl + Y2 + w.
Returning to the original perturbed system,
u
21 --(a + e)x +
u,
utilize (66) and (67) to find the transfer function
(s)/2(s)- --s(s /
x2(0+)/s, clearly z(t) --, 0 as
)/[(s
/
1) 2
--, oe for all sufficiently small
With 22(s)
To summarize, parametric insensitivity is achieved by additional integrators,
associated with an observer for suitably chosen fictitious external disturbances.
--
Appendix" on decomposability relatiye to a given subspaee. 2 We assume
and AN N ,Y-. By
are given, with A-c
that subspaces N’,-c
we can and do assume that
restricting A to
Let J :’ --. be the natural injection, e the identity on N, and A1 AIN.
It is a standard fact that decomposes 2( relative to A if and only if there exists
a map Q:Y" ---, such that
,
(A.1)
(A.2)
QJ
QA
1,
AQ.
While no originality is claimed for the following discussion, it seems not to be explicit in most
textbooks on linear algebra.
18
W. M. WONHAM AND J. B. PEARSON
Indeed, if (A.1) and (A.2) hold, set 5 Ker Q. Then if x e f, x JQx + (1
f. Also,
-JQ)x; since Q(1 JQ)x 0, we have x e ? + 5, so that + 5
xefqS implies x= lax=QJx=0, hence N=0. Finally, Qx=O
f with A
Conversely if @
implies QAx AQx 0,so A
let Q be the natural projection on N along
is an arbitrary complement of N in f.
Now let N
f, where
In a compatible basis, A and J have matrices
..
(A.3)
A
,
j
A2
By (A.3), the relations (A.1) and (A.2) are equivalent to
and
(A.4)
AQ2
QzA2
A3
O.
Thus to check whether decomposes f, it is enough to verify whether the linear
matrix equation (A.4) has a solution Q z, a computational problem which in
principle is straightforward.
Of greater theoretical interest is the following result, which can be inferred
from [6, p. 199, Theorem 13] together with [7, p. 97, Ex. 6].
PROPOSITION. decomposes f if and only if the elementary divisors of AI,
together with those of the induced map A in f/, give all the elementary divisors
of A.
In (A.3), A is the matrix of AI and A 2 that of A. The proposition thus solves
the existence problem for (A.4). As a well-known special case, (A.4) has a solution
(which is even unique) if the spectra of A and A 2 are disjoint.
Aeowledgment. It is a pleasure to acknowledge stimulating discussions
with E. J. Davison and S-H Wang, and with our graduate students, O. A. Sebakhy
and T. L. Bratton.
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[1]
I2]
[3]
[4]
[5]
[6]
[7]
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S. P. BHATTACHARYYA AND J. I. PEARSON, Error systems and the servomechanism problem, Princeton
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A. S. MORSE AND W. M. WONHAM, Decoupling and pole assignment by dynamic compensation,
this Journal, 8 (1970), pp. 317-337.
--, Decoupling and pole assignment in linear multivariable systems: a geometric approach,
this Journal, 8 (1970), pp. 1-18.
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