on the solvability of the positive real lemma equations

Systems & Control Letters 47 (2002) 211–219www.elsevier.com/locate/sysconle

On the solvability of the positive real lemma equations�

Augusto Ferrantea ; ∗, Luciano Pandol+b

aDipartimento di Elettronica e Informatica, Universit�a di Padova, via Gradenigo 6=A, 35131 Padova, ItalybDipartimento di Matematica, Politecnico di Torino, Corso Duca degli Abruzzi 24, I-10129 Torino, Italy

Received 20 October 2001; received in revised form 20 April 2002

Abstract

In this paper, we consider the classical equations of the positive real lemma under the sole assumption that the state matrixA has unmixed spectrum: �(A) ∩ �(−A) = ∅. Without any other system-theoretic assumption (observability, reachability,stability, etc.), we derive a necessary and su4cient condition for the solvability of the positive real lemma equations.c© 2002 Elsevier Science B.V. All rights reserved.

Keywords: Linear systems; Frequency characteristic; Positive real lemma; Lur’e equations; Spectral factorization

1. Introduction

Let A∈Rn×n; B∈Rn×m; C ∈Rn×m and D∈Rm×m

be given matrices. The following set of equations isof fundamental importance in a number of classicalproblems in systems and control theory and in estima-tion theory.

A′P + PA=−QQ′; (1a)

PB = C − QW; (1b)

W ′W = D + D′ (1c)

(an apex always denotes transposition). A central andextensively investigated topic is, therefore, that of

� This research was supported in part by the Italian Ministerodell’Universit>a e della Ricerca Scienti+ca e Tecnologica. It +ts theprogram of GNAMPA and the project Identi.cation and Controlof Industrial Systems.

∗ Corresponding author.E-mail addresses: [email protected] (A. Ferrante),

[email protected] (L. Pandol+).

+nding necessary and/or su4cient conditions forthe existence of a triple (P = P′; Q;W ) solvingEqs. (1).Eqs. (1), known asLur’e equations orPositive Real

Lemma equations, have been introduced in the forties,see [9], in the contest of the problem of absolute sta-bility. It is well known that Eqs. (1) are strictly con-nected with a positivity issue. Indeed, the followingnecessary condition is well known and easily proved:if system (1) is solvable, then

(i!):=C′(i!I − A)−1B − B′(i!I + A′)−1C+D + D′¿ 0 ∀!∈R: (2)

The function was introduced by Popov in 1960s(see [14]) but, it can be traced back at least to 1930sin the context of dissipative networks, see [2]. Thematrix (i!) admits a rational extension (s) to thecomplex plane. Clearly, (s) = ′(−s) so that if (2)holds, (s) is a spectral density function.The relations between Eqs. (1) and the positivity

condition (2) is one of the cornerstone of the mod-ern systems theory as illustrated in the crucial papers[8,16], see also [1]. Moreover, it is an important topic

0167-6911/02/$ - see front matter c© 2002 Elsevier Science B.V. All rights reserved.PII: S 0167 -6911(02)00189 -5

212 A. Ferrante, L. Pandol. / Systems & Control Letters 47 (2002) 211–219

in Nudel’man approach to interpolation problems, see[11].In classical papers, it is proved that condition (2)

implies solvability of Eqs. (1), under the additionalassumption that the pair (A; B) is controllable, see [8]or [15] for a recent proof based on new ideas. Sowe have the following classical result: if we associateto the positivity condition (2) the additional assump-tion of controllability, then problem (1) is solvable.In fact, simple examples show that the controllabilityassumption is far from being necessary, and recentlyinterest arose again on this classical problem, see thepapers [3,6,10,13], in an attempt to weaken as much aspossible the additional condition of controllability. Inparticular, in the paper [13] a necessary and su4cientcondition for the solvability of (1) was derived underthe assumption that A is stable, i.e. supRe �(A)¡ 0.

In this paper, we extend that condition to the classof those matrices A which satisfy the condition

�(A) ∩ �(−A′) = ∅: (3)

We recall that condition (3) is necessary and su4-cient in order that, for every matrix Q, the Lyapunovequation

A′P + PA=−QQ′ (4)

admits a solution P=P′. In such case, this solution isunique, see e.g. [5, Vol. 1, Chapter 8].The condition given in [13] is in terms of a spectral

factor of (s). If condition (2) holds, an extensionof the classical Fejer–Riesz Theorem due to Youla[17] implies the existence of a rational matrix N (s)which is holomorphic together with its right inverse inRe s¿ 0, and which satis+es (i!)=N ′(−i!)N (i!).This is the spectral factorization of (i!) and it wasuseful in the paper [13] since also the resolvent matrix(sI − A)−1 was assumed to be regular in Re s¿ 0.Now we are not assuming this condition any longer,so that we need to replace the spectral factorizationwith a diLerent factorization, which is adapted to thespectrum of A. This is presented in the next section,where also the main result is stated. The proof is inSection 3. The +nal Section 4 contains an extensionwhich may have an interest. In fact, the case whenker A �= {0} and the geometric and algebraic multiplic-ity of the zero eigenvalue coincide, is considered andsolved.

2. Main result

In this section, we present a necessary and su4cientcondition characterizing solvability of Eqs. (1) undercondition (3). To this aim, we +rst set notations andderive some auxiliary results.Let zi be the eigenvalues of the matrix A. Since

A is real, the set {zi} is symmetric with respect tothe real axis; moreover, due to the unmixing property(3), −zi is not an eigenvalue of A: thus, in particu-lar, none of the zi lies on the imaginary axis. We +xsmall circles Ci, described only once in the positivedirection, around each point −zi. We shall denote by� the union of the circles Ci and by � the union ofthe open regions �i circumscribed by the Ci, and wede+ne∫

�f(s) ds:=

∑i

∫Ci

f(s) ds:

The circles Ci are assumed to be so small that:

1. (cl�i)∩ (cl�j)=∅; ∀i �= j and (cl�i)∩ (cl�∗j )=

∅; ∀i; j, with �∗j denoting the mirrored reMection

of �j with respect to the imaginary axis.2. No point zi is encircled.3. If z0 �∈ {−zi} is a point for which the rank of the

spectral density drops below its normal rank (i.e.such that the rank of (z0) is strictly smaller thanthe rank of (s) considered as a rational matrixfunction [7, p. 373]) then z0 is not encircled.We have the following result.

Proposition 1. Let (s) be de.ned by (2). If (i!)¿ 0 ∀!∈R; then there exists a real rationalmatrix function N (s) such that

1. we have:

(i!) = N ′(−i!)N (i!):

2. N (s) is analytic in �.3. If (s) is not identically zero thenN (s) has a rightinverse that is analytic in � (if (s) is identicallyzero; then N (s) is zero too).Given any function N (s) satisfying 1–3 (hereafter

addressed as �-spectral factor) and a rational matrixfunction L(s) analytic in � and such that

(i!) = L′(−i!)L(i!);

A. Ferrante, L. Pandol. / Systems & Control Letters 47 (2002) 211–219 213

then there exists a rational matrix function V (s),analytic in � and such that

L(s) = V (s)N (s): (5)

The proof of the above proposition follows ver-batim the lines of the spectral factorization theoremproved by Youla [17, Theorem 2] where the region ofanalyticity was chosen to be the right-half plane in-stead of �. Indeed, a �-spectral factor N (s) may beconstructed by the same algorithm described in [17]based on the Smith–McMillan canonical form for ra-tional matrix functions [7, Section 6.5.2]. We observethat in the case considered by Youla, the region of an-alyticity together with the region obtained by specularreMection with respect to the imaginary axis, coversthe whole complex plane with the only exception ofthe points of the imaginary axis. For this reason thespectral factor obtained by Youla is essentially unique,i.e. unique up to multiplication on the left side by aconstant orthogonal matrix.In the case of Proposition 1, since the closure of

� ∪ (− N�) does not cover the whole complex plane,the �-spectral factor is not essentially unique. Indeed,for each pair (z;−z) of real zeros in the Smith–McMillan canonical form of (s), if {z;−z}∩{zi}=∅,the �-spectral factor N (s) may be selected to beright-invertible in either z or −z. Similar consider-ations can be made for pairs of complex conjugatezeros.Now, letN (s) be any �-spectral factor of the spectral

density (s) de+ned by (2). Moreover, let z0 be aneigenvalue of the matrix A and denote by Jz0 ;� the �thJordan chain of z0 (the Jordan chains are supposed tobe enumerated in any order). Let r be the length ofJz0 ;�, so that

Jz0 ;� = {v0; v1; : : : ; vr−1}with

Av0 = z0v0; Avi = z0vi + vi−1 0¡i6 r − 1:

We introduce the block matrix

Nz0 ;�:=

N0 0 0 : : : 0N1 N0 0 : : : 0...

. . .Nr−1 Nr−2 Nr−3 : : : N0

; (6)

where r is the length of the Jordan chain that we areconsidering and

Nh =1h!

dh

dshN ′(−z0) =

[1h!

dh

dshN ′(−s)

]|z0

;

i.e. h!Nh is the hth derivative of the function #(s) =N ′(−s), computed at s=z0. Note that (s), and hencealso N (s) and Nz0 ;�, depends only on the original dataand are computable starting from the given matricesA; B, C and D.We are now ready to state our main result whose

proof is in the next section.

Theorem 2. Let A; B; C and D be real matrices asdescribed above and assume that �(A)∩ �(−A′)= ∅.If Eqs. (1) are solvable then (i!)¿ 0 for each !and the following condition holds for every Jordanchain Jz0 ;� of the matrix A:

col[C′v0; C′v1; : : : ; C′vr−1]∈ imNz0 ;�: (7)

Conversely, let (i!) be nonnegative for each !and let condition (7) hold for every Jordan chain ofA. Then, Eqs. (1) are solvable. Condition (7) doesnot depend on the speci.c �-spectral factor N (s) of (s).

Remark 3. Condition (7) is automatically satis+edby the unobservable Jordan chains of A; and theunobservable part of the system [A; B; C; D] does notcontribute to (i!). Hence; the presence of the un-observable part in the realization [A; B; C; D] has noinMuence neither on condition (2) nor on the solv-ability of system (1).

Remark 4. When (s) is identically zero; N (s) issuch and; hence; Nz0 ;� is the zero matrix for any z0 and�. Therefore; we have the following corollary

Corollary 5. Let (i!) be identically zero. Then;Eqs. (1) is solvable if and only if C = 0.

Remark 6. Theorem 2 states the equivalence betweensolvability of Eqs. (1) and conditions (2) and (7). Aprocedure for practical veri+ability of such conditionsis outlined below.

The positivity condition (2) is classical and a hugeamount of literature has been produced on this topic. It


may be eLectively checked by resorting to the proce-dure illustrated in [17]: the +rst step consists in com-puting the Smith–McMillan canonical form of (s):

(s) = C(s)D(s)F(s): (8)

Necessary conditions for the positivity of are: (i) allthe diagonal elements di(s), i=1; 2; : : : ; r, of the diag-onal matrix D(s) satisfy the condition di(s)=di(−s),and (ii) all the pure imaginary poles and zeros of di(s),i=1; 2; : : : ; r, have even multiplicity. If these two con-ditions hold, from decomposition (8) one can com-pute an elementary polynomial matrix M (s)=M

′(−s)

[17]. Moreover, a condition guaranteeing the positiv-ity condition (2) is

M (i!)¿ 0 ∀!∈R: (9)

A procedure to check condition (9) was derived in[12] and may be found in [17].An alternative procedure based on a state space re-

alization of (s) may be found in [4]. To this regardnote that positivity is a property of (s) and not of itsrealization, so that, starting from the given realization(cf. left-hand side of (2)), it is convenient to computea minimal realization which can be done by standardlinear algebra techniques.Checking condition (7) requires the computation

of Nz0 ;� and hence the computation of the �-spectralfactor N (s). This can be done by following the sameprocedure employed for the construction of the canon-ical spectral factor (i.e. a spectral factor analytic withits inverse in the left half plane) in [17]. More pre-cisely, in [17] the key step to compute the canonicalspectral factor is Lemma 4 where it is proved that thediagonal matrix D(s) in the Smith–McMillan canon-ical form (8) of (s) may be factored as D(s) ='((−s))(−s))(s)((s), with ' being a constant diag-onal matrix whose diagonal elements are either ±1,((s) being a diagonal polynomial matrix analytic to-gether with its inverse in the left-half complex plane,and )(s) being a diagonal polynomial matrix ana-lytic together with its inverse in the whole complexplane except the imaginary axis. From this factoriza-tion, it is straightforward to obtain a factorization ofthe form D(s) = '((−s))(−s))(s)((s) with ((s) be-ing a diagonal polynomial matrix analytic togetherwith its inverse in � (note that, by unmixing condition(3), )(s) is analytic together with its inverse in �).This is the only modi+cation needed in the procedure

illustrated in [17] to obtain a �-spectral factor. The laststep needed to check condition (7) is the computationof the Jordan structure of A, i.e. the computation of itseigenvalues and generalized eigenvectors.

3. The proof

3.1. Preliminary observations

First observe that, in view of assumption (3), givenany matrix Q with n rows, the Lyapunov equation (4)admits a unique solution

P =12*i

∫�(sI + A′)−1QQ′(sI − A)−1 ds: (10)

Therefore, the system of Eqs. (1) may be equivalentlyrewritten as[12*i

∫�(sI + A′)−1QQ′(sI − A)−1 ds

]B = C − QW;

(11a)

W ′W = D + D′: (11b)

Indeed, (10) provides an obvious bijection betweensolutions (P;Q;W ) of (1) and solutions (Q;W ) of(11).

3.2. Proof of necessity

Let (P = P′; Q;W ) be a solution of Eqs. (1) andde+ne the functions

E(s):=Q′(sI − A)−1B; L(s):=E(s) +W: (12)

We note that L(s) is analytic in �. We prove:

Lemma 7. Let Eqs. (1) be solvable and let L(s) bethe function de.ned in (12). The function L(s) is an-alytic in � and satis.es L′(−s)L(s) = (s).

Proof. We write the Lyapunov equation (4) as

P(A− sI)−1 + (sI + A′)−1P=(sI + A′)−1QQ′(A− sI)−1

and we use PB = C − QW . We obtain

[C′ −W ′Q′](sI − A)−1B − B′(sI + A′)−1[C − QW ]=E′(−s)E(s):

We sumW ′W to both sides and we get the results.


We note explicitly that we are not asserting that thefactor L(s) in (12) is �-spectral. Moreover, if (i!)is identically zero then L(s) = E(s) is also zero.In order to prove the inclusion (7) we now +x an

eigenvector v0 of A. Let z0 be the corresponding eigen-value. We see from (11a) that

v′0C = v′0QW +12*i

∫�

1s + z0

v′0QE(s) ds = v′0QW

+v′0QE(−z0) = v′0QL(−z0):

The special matrix L(s) constructed above may not be�-spectral. However, L(s) is a factor of (s) and isanalytic in �. Thus, in view of (5), we have

C′v0 ∈ im L′(−z0) =imN ′(−z0)V ′(−z0) ⊆ imN ′(−z0);

which is the +rst component of the desired inclusionin (7).Repeating the above procedure for each vector of

all the Jordan chains in the same spirit of [13] we seethat condition (7) holds.

3.3. Proof of su>ciency

In the case when (s) is identically zero, condition(7) clearly reduces toC=0, so that (P;Q;W )=(0; 0; 0)is immediately seen to be a solution of Eqs. (1). Then,from now on, we may assume that (s) is not identi-cally zero.The condition (i!)¿ 0 is now an assumption,

and we know the existence of a �-spectral factor N (s).To show the existence of a solution (P;Q;W ) ofEqs. (1), let

W = N (∞): (13)

Clearly, such W satis+es Eq. (1c) (that is the same of(11b)) so that it is now su4cient to show that thereexists a matrix Q satisfying (11a). For the proof ofexistence of such Q that is, of course, the di4cult part,we de+ne the strictly proper rational matrix function

E(s) = N (s)−W = N (s)− N (∞):

We +rst prove that, under condition (7), we can +ndQ which solves

C =12*i

∫�(sI + A′)−1QE(s) ds + QW: (14)

Then, we shall show that

E(s) = Q′(sI − A)−1B (15)

with Q being the solution of (14).In order to better clarify the spirit of the proof of the

existence of a matrix Q solving (14), we +rst assumethat the matrix A is diagonalizable. For each eigen-vector v0 of A, let z0 be the corresponding eigenvalue.Moreover, let u0 be a vector (whose existence is guar-anteed by condition (7)) such that: C′v0 =N ′(−z0)u0.Finally, let Q be the matrix such that for any eigen-vector v0 the relation

v′0Q = u′0holds: note that, such relations uniquely specify Q.Once +xed such Q we have to check that it satis+es(14). To this end, it is clearly su4cient to show thatfor each v0 the relation

v′0C = v′0

[12*i

∫�(sI + A′)−1QE(s) ds + QW

]

holds. Indeed, we have

v′0

[12*i

∫�(sI + A′)−1QE(s) ds + QW

]

=12*i

∫�

1s + z0

v′0QE(s) ds + v′0QW

= v′0QN (−z0) = u′0N (−z0) = v′0C:

The proof in the general case is analogous exceptthat for any eigenvalue z0, instead of the eigenvectorsv0, we have to consider each one of the matrices Vobtained by stacking together the vectors of a Jordanchain:

V = [ v0 v1 · · · vr−1 ]

with

Av0 = z0v0; Avi = z0vi + vi−1; 0¡i6 r − 1:

Let ui, i=0; 1; : : : ; r − 1, be vectors (whose existenceis guaranteed by condition (7)) such that

col[C′v0; C′v1; : : : ; C′vr−1]=Nz0 ;�col[u0; u1; : : : ; ur−1]:

Moreover, let

U = [ u0 u1 · · · ur−1 ]:

Finally, let Q be the matrix such that for any matrixV of Jordan chains the relation

V ′Q = U ′


holds (as in the previous case, such relations uniquelyspecify Q).To show that such Q solves Eq. (14), we use the

following recurrence:

(sI + A)−1vk =1

s + z0vk − 1

s + z0(sI + A)−1vk−1;

(sI + A)−1v0 =1

s + z0v0;

that yields

(sI + A)−1vk =k∑

j=0

(−1)k−j

(s + z0)k−j+1 vj:

We also de+ne the shift matrix T ′ such that

VT ′ = [ 0 v0 v1 · · · vr−2 ]:

so that

(sI + A)−1V =r−1∑j=0

(−1)j

(s + z0)j+1VT ′j:

We are now ready to check that Q satis+es (14). Tothis end, it is clearly su4cient to prove the followingequality for each V :

V ′C = V ′[

12*i

∫�(sI + A′)−1QE(s) ds + QW

]:

Indeed, we have

V ′[

12*i

∫�(sI + A′)−1QE(s) ds + QW

]

=r−1∑j=0

12*i

∫�

(−1)j

(s + z0)j+1 TjV ′QE(s) ds + V ′QW

=r−1∑j=0

(−1)j

j!dj

dsjVQN (s)|s=−z0

=

u′0N′0

u′1N′0 + u′0N

′1

u′2N′0 + u′1N

′1 + u′0N

′2

...

u′r−1N′0 + u′r−2N

′1 + · · ·+ u′0N

′r−1

=

v′0Cv′1Cv′2C...

v′r−1C

= V ′C:

It remains to show that (15) holds. To this aim weshall need the following result.

Lemma 8. Let N (s) be a �-spectral factor of (s)and let 0(s) be a strictly proper rational matrixfunction analytic in C\{zi}. The function 0(s)is identically zero if and only if for every zi andevery nonnegative integer k the residuum at zi of(s − zi)kN ′(−s)0(s) is zero.

Proof. We +x our attention to one of the points zi; thatwe denote by z0. Let � be the order of the pole of 0(s)at z0. This is also the order of the pole of N ′(−s)0(s)because N ′(−s) has a regular left inverse at z0.We recall the formula for the product of the matrices

N ′(−s) =+∞∑i=0

Ni(s − z0)i ; 0(s) =+∞∑j=−�

0j(s − z0)j:

The product is+∞∑h=−�

[h+�∑r=0

Nr0h−r

](s − z0)h:

We multiply by (s− z0)k and we equate the residuumto zero. Of course, it is su4cient to assume 06 k6 �.We obtain the triangular system�−k−1∑r=0

Nr0−k−1−r = 0:

The diagonal entry is always equal to N0 = N ′(−z0),which is left invertible. It follows that every 0j withnegative j is zero. Repeating this argument for everycandidate pole z0 we see that 0(s) is strictly properand without poles; hence it is zero.

Both the matrices in (15) are strictly proper andE(s) is analytic in C\{zi}. In order to prove equality,we apply the previous lemma to their diLerence, forevery eigenvalue of A. To this aim, we shall need thefollowing equality that can be easily checked:

C′(sI − A)−1B − B′(sI + A′)−1C

= E′(−s)E(s) +W ′E(s) + E

′(−s)W: (16)

Now, let us +x one eigenvalue z0 of A and a circleC0, centered at z0 and not intersecting any of the Ci.We assume that the radius of C0 is smaller then thatof C0 so that if s belongs to C0 then −s belongs to


the interior of the disk C0. This will be used in thecomputations below.We multiply both sides of (16) by (s − z0)k . Inte-

gration along C0 gives

12*i

∫C0

(s − z0)kC′(sI − A)−1B ds

=12*i

∫C0

(s − z0)k E′(−s)E(s) ds

+12*i

∫C0

W ′(s − z0)k E(s) ds

=Res[N ′(−s)(s − z0)k E(s); s = z0]: (17)

Now we use equality (14). We get

C′(sI − A)−1B

=12*i

∫�E′(z)Q′(zI + A)−1(sI − A)−1B dz

+W ′Q′(sI − A)−1B

=W ′Q′(sI − A)−1B

+12*i

∫�E′(z)Q′ 1

s + z{(zI + A)−1

+(sI − A)−1}B dz:

Integration along C0 gives

12*i

∫C0


=12*i

∫C0

12*i

(s − z0)k∫

�E′(z)Q′

× 1s + z

[(zI + A)−1 + (sI − A)−1]B dz ds

+Res[W ′(s − z0)kQ′(sI − A)−1B; s = z0]:

We compute the integral as

12*i

∫C0

12*i

(s − z0)k∫

�E′(z)Q′ 1

s + z[(zI + A)−1

+ (sI − A)−1]B dz ds

=12*i

∫�

12*i

E′(z)Q′

∫C0

(s − z0)k

× 1s + z

[(zI + A)−1 + (sI − A)−1]B ds dz

=12*i

∫�E′(z)Q′ 1

2*i

∫C0

(s − z0)k

s + z

×(sI − A)−1 ds B dz

=12*i

∫C0

12*i

∫�E′(z)

1s + z

dz

×(s − z0)kQ′(sI − A)−1B ds

=12*i

∫C0

E′(−s)(s − z0)kQ′(sI − A)−1B ds

=Res[E′(−s)(s − z0)kQ′(sI − A)−1B; s = z0]:

Therefore,

12*i

∫C0


=Res[N ′(−s)(s − z0)kQ′(sI − A)−1B; s = z0](18)

and, comparing with (17),

Res[N ′(−s)(s − z0)k E(s); s = z0]

=Res[N ′(−s)(s − z0)kQ′(sI − A)−1B; s = z0]:

This holds for every z0 and every k. Hence, therequired equality (15) follows from Lemma 8. Thiscompletes the proof.

4. The case when 0 is a nondefective eigenvalue

We extend the previous results to the case when0∈ �(A), with the geometric and algebraic multiplic-ities of the 0 eigenvalue being the same.It is well known that the solutions to system (1) are

not invariant under coordinate transformations; but,the mere existence of the solution is invariant. Hencewe can assume that the matrix A is block diagonal,

A=[0 00 A1

]: (19)

We note that if T is a coordinate transformation inthe state space Rn then P is transformed to T ′PT andQQ′ to T ′QQ′T . Hence, symmetry and positivity arepreserved even if T is not orthogonal. We assume thatthe unmixing property (3) is satis+ed by A1. The ma-trices B and C and the candidate solutions P and Q tothe Lyapunov equation (4) can be written accordingly


in block diagonal form and Eqs. (1) looks like[0 PA1

A′1P

′A′1P1 + P1A1

]=−

[S0 SS′S1

];

[P0 PP′P

] [B0

B1

]=[C0

C1

]− QW;

where

S =

[S0 SS′S1

]

denotes QQ′ in block form. It is seen from above thatS0 is zero so that also S must be zero becauseQQ′¿ 0.Hence the Lyapunov equation has the form[

0 PA1

A′1P

′A′1P1 + P1A1

]=−

[0 00 S1

]; where S1¿ 0:

The condition A′1P = 0 implies P = 0 because 0 is

not an eigenvalue of A1. Moreover, we can write Q as

Q =[

0Q1

]

with Q1Q′1 = S1. Hence, a solution exists if and only

if it is possible to solve the following problems:

1. The problem P0B0 = C0, with the constraintsP0 = P′

0;2. The problem A′

1P1 + P1A1 = Q1Q′1, P1B1 = C1 −

Q1W .

The solution of Problem 2 was studied in the pre-vious sections. We examine Problem 1.

Proposition 9. Problem 1 admits a solution if andonly if

B′0C0 = C′

0B0; (20)

ker B0 ⊆ kerC0: (21)

Proof. Necessity is obvious. Concerning su4ciency;we observe that the coordinate systems in Rm and inker A can be chosen at will; provided that we onlyperform orthogonal coordinate transformations (whichdo not change the property that P is symmetric). Weuse orthogonal coordinate transformations T andT inRm and in Rn; respectively. The transformations T andT are chosen in such a way that the +rst columns of Tare a basis of [ker B0]⊥ and the remaining ones are a

basis of ker B0; the +rst columns of T are a basisof im B0 and the remaining ones a basis of [im B0]⊥.In the new coordinate system; the matrix B0 is trans-formed to

TB0T =[B0 00 0

]

and B0 is invertible. The transformations beingorthogonal; we can assume that B0 was given in theabove form from the outset. The matrix C0 takes acompatible block form and condition (21) impliesthat its second block column is zero:

C0 =[C0 0C1 0

]:

Condition (20) shows that C0B−10 is symmetric. In

fact; in the chosen basis condition (20) assumes theform B

′0C0 = C

′0B0 that yields

C0B−10 = (B

′0)

−1C′0:

Once that the matrices have been written in thisform we see that a symmetric solution P0 to P0B0=C0

is

P0 =

[C0B

−10 (B

−10 )′C

′1

C1B−10 0

]:

5. Concluding remarks

A complete characterization of the solvability of thepositive real lemma Eqs. (1) has been derived underthe sole assumption that the matrix A has unmixedspectrum. This in particular implies that no eigenvalueis purely imaginary. The condition was then extendedto the case when ker A �= {0}, provided that (=0 is anondefective eigenvalue of A .

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on the solvability of the positive real lemma equations

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