On the existence of common quadratic Lyapunov functions for second-order linear time-invariant discrete-time systems

INTERNATIONAL JOURNAL OF ADAPTIVE CONTROL AND SIGNAL PROCESSINGInt. J. Adapt. Control Signal Process. 2002; 16:729–751 (DOI: 10.1002/acs.720)

On the existence of common quadratic Lyapunov functions forsecond-order linear time-invariant discrete-time systems

Mehmet Akar and Kumpati S. Narendran,y

Department of Electrical Engineering, Yale University, P.O. Box 208267, New Haven, CT 06520-8267, U.S.A.

SUMMARY

Necessary and sufficient conditions for the existence of a common quadratic Lyapunov function (CQLF)are given for two second-order linear time-invariant discrete-time systems. These conditions are laterextended to an arbitrary number of systems. The conditions are readily verifiable both analytically andgraphically. The paper also provides a constructive procedure for computing a CQLF when it exists.Copyright # 2002 John Wiley & Sons, Ltd.

1. INTRODUCTION

The problem of finding a common quadratic Lyapunov function (CQLF) for linear discrete-time and continuous-time systems has been studied by many researchers. In this paper, ourinterest is only in discrete-time systems for which solutions have appeared for some special cases[1–6]. In Reference [1], Narendra and Balakrishnan provided the CQLF for pairwisecommutative matrices by solving a sequence of Stein equations, which was also noted later inReference [2]. Mori et al. [3], and Shorten and Narendra [4] derived conditions for the existenceof a CQLF for matrices that are simultaneously upper (or lower) trianguralizable. The relationbetween Lie-algebra and stability of the switched linear system was first noted by Gurwitz inReference [7], and later it was shown by Liberzon et al. that a CQLF exists if the Lie-algebragenerated by the matrices is solvable. Another somewhat related result was given by Ooba andFunahashi in Reference [6], where the authors proposed special classes of matrices for whichCQLF vðxÞ ¼ xTPx exists, where P is diagonal.

As discussed above, the available results on CQLF are quite restrictive. For the first time inReference [8], Shorten and Narendra derived necessary and sufficient conditions for theexistence a CQLF for a pair of second-order linear continuous-time systems, and described aconstructive procedure for finding a CQLF when it exists. These results were later extended tothe existence of CQLF for an arbitrary number of second-order systems [9]. In this paper,

Received 24 July 2001Revised 28 September 2001

Accepted 28 May 2002Copyright # 2002 John Wiley & Sons, Ltd.

yE-mail: [email protected]

Contract/grant sponsor: Office of Naval Research; contract/grant number: N00014-97-1-0948

nCorrespondence to: Kumpati Narendra, Department of Electrical Engineering, Yale University, P.O. Box 208267, NewHaven, CT 06520-8267, U.S.A.

discrete-time versions of the results in References [8, 9] are presented. In particular, necessaryand sufficient conditions are derived for the existence of a CQLF for finitely many second-orderlinear discrete-time systems. The solutions in References [8, 9], as well as those in the presentpaper, are important not only because they are complete, but also because they provide insightinto the solution of the CQLF problem in higher dimensions, which will have a major impact oncontrol theory.

The organization of this paper is as follows. In Section 2, some mathematical results arepresented. To improve readability, the proofs of these preliminary results are relegated to theappendix. In Section 3, the main result for two second-order systems is stated and a detailedproof is provided. Section 4 addresses some computational aspects of the problem which are ofpractical interest. Numerous examples are included in Section 5 to illustrate the main result ofSection 3. The extension of the results to three and more discrete-time systems is furtherdiscussed in Section 6 and finally, concluding remarks are given in Section 7.

2. MATHEMATICAL PRELIMINARIES

Consider N LTI discrete-time dynamical systems

SAi : xðk þ 1Þ ¼ AixðkÞ; i 2 N¼4 f1; 2; . . . ;Ng ð1Þ

where Ai; i 2 N are constant Schur matrices inRn�n (i.e., matrices with eigenvalues in the interiorof the unit circle in the complex plane). The objective of the paper is to determine necessary andsufficient conditions for the existence of a CQLF of the form, vðxÞ ¼ xTPx; P ¼ PT > 0 such thatthe first time differences of vðxÞ along the trajectories of each system, SAi ; i 2 N are strictlynegative. It is well known that any linear time-invariant dynamical system, xðk þ 1Þ ¼ AxðkÞ;where A is a constant Schur matrix, has a Lyapunov function xTPx where P satisfies the Steinequation

ATPA� P ¼ �Q; P ¼ PT > 0; Q ¼ QT > 0 ð2Þ

In view of (2), it follows that our interest is in determining necessary and sufficient conditionsfor the existence of symmetric positive definite matrices P which simultaneously satisfy the Steinequations

ATi PAi � P ¼ �Qi; i 2 N ð3Þ

for some symmetric positive definite matrices Qi; i 2 N: If such a solution exists, then we will saythat P is a Stein solution for the set of matrices fA1;A2; . . . ;ANg:

While our ultimate objective is to determine the existence of CQLFs for systems defined in Rn

for arbitrary values of the integer n; we confine our attention in this paper to the case n ¼ 2: InLemma 1, several results are collected, some of which apply to the general case (i.e., any positiveinteger n). These results are summarized as nine items (i)–(ix). Comments are made following theitems to indicate their relevance to the problem considered. Items (i)–(vi) apply to the generalcase where Ai 2 Rn�n; and the last three items apply to the special case n ¼ 2: These nine itemsprovide a framework within which the existence of CQLFs can be studied. The proofs of items(i), and (iv)–(vi) are straightforward and are not included. Proofs of the remaining items areprovided in the appendix.

Copyright # 2002 John Wiley & Sons, Ltd. Int. J. Adapt. Control Signal Process. 2002; 16:729–751

M. AKAR AND K. S. NARENDRA730

Lemma 1

(i) P is a Stein solution for A 2 Rn�n (it satisfies Equation (2)) if and only if it is a Steinsolution for all cA; jcj41:

(ii) If P is a Stein solution for A 2 Rn�n; then it is also a Stein solution for all Ak ; k 2 Zþ

(where Zþ is the set of positive integers).(iii) If P is a Stein solution for both A1;A2 2 Rn�n; then it is also a Stein solution for A1A2:(iv) From (ii) and (iii), it follows that if P is a Stein solution for A1 and A2; it is also a Stein

solution for An11 A

n22 A

n31 . . .Ank

2 ; where n1; n2; n3; . . . ; nk are non-negative integers.(v) Let T 2 Rn�n be a non-singular matrix. If P is a Stein solution for both A1;A2 2 Rn�n;

then T TPT is a Stein solution for both T�1A1T and T�1A2T :The consequence of (v) is that appropriate coordinate transformations of the problem

stated in (9) and (10) can be made to simplify the choice of the Lyapunov function.(vi) Let A 2 Rn�n and let I þ A be invertible. Then the following algebraic identity holds

ATPA� P ¼ ðI þ AÞT½P � P ðI þ AÞ�1 � ðI þ AÞ�TP �ðI þ AÞ ð4Þ

Similarly, if I � A is invertible, the following identity holds

ATPA� P ¼ ðI � AÞT½P � P ðI � AÞ�1 � ðI � AÞ�TP �ðI � AÞ ð5Þ

The equivalent representations given in (4) and (5) are found to be convenient for thederivation of the conditions given in the following sections.

(vii) Let %AA ¼ adjðAÞ; be the classical adjoint of A 2 R2�2: Then P is a Stein solution for A ifand only if it is a Stein solution for %AA:

(viii) Let A 2 R2�2 be a Schur matrix, and denote A ¼ ½aij�: Then, the locus of the points (p12;p22) defined by

detðATPA� P Þ ¼ 0; P ¼1 p12

p12 p22

" #

describes an ellipse if and only if a21=0; and a parabola if and only if a21 ¼ 0:(ix) Let A1;A2 2 R2�2 be Schur matrices, and denote A1 ¼ ½aij�; and A2 ¼ ½ %aaij�:

(a) If a21 ¼ %aa21 ¼ 0; then A1 and A2 have a diagonal Stein solution

P ¼ diag½1;p22�; where p22 > maxa212

ð1� a211Þð1� a222Þ;

%aa212ð1� %aa211Þð1� %aa222Þ

� �

(b) If ja11 � a22j51� a11a22 þ a12a21 and %aa12 ¼ %aa21 ¼ 0; then A1 and A2 have acommon diagonal Stein solution, P ¼ diag½1; ja12=a21j�:

2.1. The CQLF vðxÞ ¼ xTPx

The existence of a CQLF vðxÞ ¼ xTPx for each system SAi ; i 2 N; also implies the exponentialstability of the time-varying system

xðk þ 1Þ ¼ AðkÞxðkÞ; AðkÞ 2 A¼4 fA1;A2; . . . ;ANg ð6Þ

This observation follows directly from Lyapunov’s theory. However, it is worth noting that theconverse of the above statement is not true (In Example 1 in Section 5, it is shown that the


QUADRATIC LYAPUNOV FUNCTIONS 731

switching system is stable even when a CQLF does not exist). This does not undermine theimportance of the CQLF problem in any way, because using the necessary and sufficientconditions given in References [10, 11], one still needs to solve a CQLF problem for anexpanded system in order to show the stability of (6) (see continuation of Example 1 in Section6). Moreover, the CQLF problem comes up in stability analysis of other classes of dynamicalsystems which are composed of linear subsystems, e.g., Takagi–Sugeno fuzzy systems and linearparameter-varying systems.

2.2. Matrix Pencils H ða;A1;A2Þ and Gða;A1;A2Þ

In this paper, the following two matrix pencils play a central role in the derivation of the mainresult presented in Section 3:

H ða;A1;A2Þ ¼ ½0:5I � Gða;A1;A2Þ��1½0:5I þ Gða;A1;A2Þ�; a 2 ½0; 1� ð7Þ

Gða;A1;A2Þ ¼ a½0:5I � ðI þ A1Þ�1� þ ð1� aÞ½0:5I � ðI þ A2Þ

�1�; a 2 ½0; 1� ð8Þ

In both cases, the parameter a defining the pencils lies in the closed interval [0,1]. The matrixpencil, H ða;A1;A2Þ; is said to be Schur, if the eigenvalues of the matrix for every value ofa 2 ½0; 1� lie in the interior of the unit circle. Similarly, the matrix pencil, Gða;A1;A2Þ; is said to beHurwitz, if the eigenvalues of Gða;A1;A2Þ for every value a 2 ½0; 1� lie in the open left half of thecomplex plane. From relation (7), it follows that the matrix pencil H ða;A1;A2Þ is Schur if andonly if the matrix pencil Gða;A1;A2Þ is Hurwitz. The following properties of the pencils,H ða;A1;A2Þ and Gða;A1;A2Þ; are found to be important in the proof of the main result given inSection 3.

Lemma 2

(i) Given that A1;A2 2 R2�2 are Schur, the matrix pencil H ða;A1;A2Þ is Schur if and only ifH ða;�A1;�A2Þ is Schur. The same also holds for H ða;A1;�A2Þ and H ða;�A1;A2Þ:

(ii) Given two pencils, Gða;A1;A2Þ and Gða;A1;�A2Þ; where A1;A2 2 R2�2 are Schur, itfollows that either one or both of them have real eigenvalues for some a 2 ½0; 1�:

(iii) If A1;A2 2 R2�2 are Schur or have complex eigenvalues on the unit circle, and Gða;A1;A2Þhas a zero eigenvalue for some a ¼ a0 2 ð0; 1Þ with the corresponding eigenvector x0; thenx0 cannot be an eigenvector of either A1 or A2:

The form of the Schur matrices A1 and A2 determines the procedure used to determine theCQLF in the main result given in Section 3. For some specific cases, a CQLF can be obtaineddirectly. In other cases a constructive procedure is needed. Lemma 3 distinguishes between thesevarious cases.

Lemma 3

Given A1;A2 2 R2�2; let H ða;A1;A2Þ and H ða;A1;�A2Þ be Schur pencils. Consider the followingcases where either A1 or A2 has two zero eigenvalues and one of the following conditions holds:



(i) Either A1 or A2 is a null matrix. (ii) A non-singular transformation matrix T exists such that

T�1A1T ¼0 1

0 0

" #; T�1A2T ¼

0 a12

0 0

" #

In these cases, a CQLF can be computed directly for SA1and SA2

: For all other cases, there existconstants c1 and c2; jc1j; jc2j > 1; so that the matrix pencil H ða;B1;B2Þ with B1 ¼ c1A1 andB2 ¼ c2A2 have the following properties: (i) H ða;B1;B2Þ is Schur or has complex eigenvalues onthe unit circle. (ii) The pencil H ða;B1;B2Þ � I is singular for a single value of a ¼ a0 2 ½0; 1�:

3. MAIN RESULT FOR TWO SYSTEMS

In this section, we consider two second-order systems SA1and SA2

; and state necessary andsufficient conditions for the existence of a symmetric positive definite matrix P whichsimultaneously satisfies the Stein equations

AT1 PA1 � P ¼ �Q1 ð9Þ

AT2 PA2 � P ¼ �Q2 ð10Þ

for some symmetric positive definite matrices Q1 and Q2:

Theorem 1

SA1and SA2

with A1 and A2 2 R2�2 have a CQLF if and only if the matrix pencils H ða;A1;A2Þand H ða;A1;�A2Þ are Schur.

Proof

Necessity: Let SA1and SA2

have a CQLF, vðxÞ ¼ xTPx; P ¼ PT > 0; so that Equations (9) and(10) are simultaneously satisfied for some symmetric positive definite matrices Q1 and Q2: Bydirect substitution for H ða;A1;A2Þ in terms of Gða;A1;A2Þ from (7) and using Lemma 1(vi), it canbe shown that HTða;A1;A2ÞPH ða;A1;A2Þ � P is negative definite for all a 2 ½0; 1�: Hence, thematrix pencil H ða;A1;A2Þ is Schur. Similarly, the Schur stability of three other matrix pencilsH ða;A1;�A2Þ; H ða;�A1;A2Þ; and H ða;�A1;�A2Þ can be derived using the same procedure (withminor modifications). However, it should be noted that demonstrating the Schur stability ofonly two pencils, H ða;A1;A2Þ and H ða;A1;�A2Þ is adequate (Lemma 2(i)).

Sufficiency: In this case, we assume that the pencils H ða;A1;A2Þ; and H ða;A1;�A2Þ are Schur,and demonstrate that a CQLF exists for A1 and A2: We also show constructively how such aLyapunov function can be realized. In what follows, a general procedure is described fordetermining the CQLF. However, there are two specific situations in which the generalprocedure is not applicable. These correspond to cases where one of the matrices (say A1) orboth matrices A1 and A2 have two zero eigenvalues. In these situations, a CQLF can becomputed in a straightforward manner. We consider these special cases first before proceedingto the general case.

In the first special case, the matrix A1 is a null matrix. In the second special case, both A1; andA2 have two zero eigenvalues and a non-singular matrix T exists such that T�1A1T and T�1A2T



both have upper triangular form. In the first case, any Lyapunov function of SA2is a Lyapunov

function of SA1: In the second case, it is known that a CQLF exists by Lemma1(ix)a.

In what follows, we assume that the two matrices, A1 and A2; do not belong to the classesdescribed above. For such matrices, by Lemma 3, there exist constants, c1 and c2 (jc1j; jc2j > 1),such that the pencil H ða; c1A1; c2A2Þ has an eigenvalue unity for a single value of a ¼ a0 2 ½0; 1�;and is Schur or has complex eigenvalues on the unit circle for all other values. The proof ofsufficiency is different for the two cases where a0 2 ð0; 1Þ; and a0 2 f0; 1g; and hence, these aretreated separately.

Case 1: a0 2 ð0; 1Þ: Let c1 and c2 be two constants (jc1j; jc2j > 1), and B1 ¼4c1A1; B2 ¼

4c2A2 be

two matrices such that the pencil H ða;B1;B2Þ is Schur (or has complex eigenvalues on the unitcircle) for all a 2 ½0; 1� except a ¼ a0 2 ð0; 1Þ: Depending on the computed constants, c1 and c2;the matrices B1 and B2 are either Schur or have complex eigenvalues on the unit circle. If both B1

and B2 have complex eigenvalues on the unit circle, then it follows (from the proof of Lemma 3)that there exists a non-singular matrix, T ; such that the matrices, A1 and A2 can be transformedinto the form

T�1A1T ¼a11 a12

�a12 a11

" #; T�1A2T ¼

%aa11 %aa12

� %aa12 %aa11

" #

where a211 þ a21251; %aa211 þ %aa21251: Hence, vðxÞ ¼ xTx is a Lyapunov function for the transformedsystems, ST�1A1T and ST�1A2T which implies that P ¼ ðTT TÞ�1 satisfies Equations (9) and (10)simultaneously. Excluding the above specific case where both B1 and B2 have complexeigenvalues on the unit circle, in the sequel we will assume (with no loss of generality) that B1 isSchur.

Let H ða0;B1;B2Þx0 ¼ x0; i.e., x0 is the eigenvector corresponding to the unit eigenvalue ofH ða0;B1;B2Þ: This can be equivalently expressed as

Gða0;B1;B2Þx0 ¼ 0 ð11Þ

or x0 is the eigenvector of Gða0;B1;B2Þ corresponding to a zero eigenvalue. Let q1 be a vectororthogonal to ðI þ B1Þ

�1x0; or

qT1 ðI þ B1Þ�1x0 ¼ 0 ð12Þ

As shown below, the vectors q1 and BT1 q1 span R2 or the matrix ½q1;BT

1 q1� is of full rank. If½q1;BT

1 q1� is singular, q1 is an eigenvector of BT1 ; and a constant l 2 R exists such that

BT1 q1 ¼ lq1 ð13Þ

Since qT1 ðI þ B1Þ�1x0 ¼ 0 by (12), (13) implies that qT1B1ðI þ B1Þ

�1x0 ¼ 0; or B1ðI þ B1Þ�1x0 and

ðI þ B1Þ�1x0 are in the same direction (or B1ðI þ B1Þ

�1x0 ¼ l1ðI þ B1Þ�1x0 for some non-zero

constant l1). However, this results in a contradiction by Lemma 2(iii). Hence q1 and BT1 q1 are

linearly independent. Since B1 is Schur and ½q1;BT1 q1� is of full rank, it follows that a symmetric

positive definite matrix P exists which is the solution of the Stein equation

BT1 PB1 � P ¼ �q1qT1 ð14Þ

Furthermore, using B1 ¼ c1A1; we obtain

AT1 PA1 � P ¼ �ðc21 � 1ÞAT

1 PA1 � q1qT1 ¼ �Q1 ð15Þ



If A1 has either complex or non-zero real eigenvalues, then it is non-singular, and hence Q1 ispositive definite. If A1 is singular (and so it has at least one zero eigenvalue), it still follows thatQ1 is positive definite. This is shown below by demonstrating that the assumption Q150 resultsin a contradiction. If Q1 is singular, a non-zero vector, x1 2 R2; exists such that x1AT

1 PA1x1 ¼ 0(or x1BT

1 PB1x1 ¼ 0), and qT1 x1 ¼ 0: This implies that x0 is an eigenvector of B1 which results in acontradiction by Lemma 2(iii). Thus, Q1 > 0; and vðxÞ ¼ xTPx is a Lyapunov function for SA1

:Now we show that vðxÞ ¼ xTPx is also a Lyapunov function for SA2

: From (11), we havexT0Gða0;B1;B2ÞPGða0;B1;B2Þx0 ¼ 0: Expanding the left-hand side of this equation and using thefact that qT1 ðI þ B1Þ

�1x0 ¼ 0; we obtain xT0 ðI þ B2Þ�TðBT

2 PB2 � P ÞðI þ B2Þ�1x0 ¼ 0: Since ðI þ

B2Þ�1x0=0; it follows that

BT2 PB2 � P ¼ �q2qT2 ð16Þ

for some q2 2 R2: Furthermore, using B2 ¼ c2A2; we obtain

AT2 PA2 � P ¼ �ðc22 � 1ÞAT

2 PA2 � q2qT2 ¼ �Q2 ð17Þ

The matrix B2 is either Schur or has complex eigenvalues on the unit circle. If B2 has complexeigenvalues on the unit circle, then q2 ¼ ½0; 0�T in Equation (16). In this case, A2 is non-singularand from Equation (17), Q2 is positive definite. If B2 is Schur, q2=½0; 0�T: In this case, Q2 beingpositive definite can be shown using arguments similar to those used for A1; thus vðxÞ ¼ xTPx isalso a Lyapunov function for SA2

:Case 2: a0 ¼ 0 or a0 ¼ 1: In this case, either I � B1 ða0 ¼ 1Þ or I � B2 ða0 ¼ 0Þ is singular. It is

enough to consider the case a0 ¼ 1: From the proof of Lemma 3, it follows that the pencils

Gðb1;B1;B2Þ ¼ ½0:5I � ðI þ B1Þ�1� þ b1½0:5I � ðI þ B2Þ

�1�; b1 2 ½0;1Þ

Gðb2;B1;�B2Þ ¼ ½0:5I � ðI þ B1Þ�1� þ b2½0:5I � ðI � B2Þ

�1�; b2 2 ½0;1Þ

are singular only for b1 ¼ b2 ¼ 0; and are Hurwitz for any other b1; b250:I � B1 may be singular for two cases: (i) B1 may have two eigenvalues of unity. (ii) B1 may

have a single eigenvalue of unity. We deal with these cases separately.Case (2i): Assume that B1 has two eigenvalues of unity. Without loss of generality, we assume

that B1 and A2 are in the following form:

B1 ¼1 K

0 1

" #; A2 ¼

a11 a12

a21 a22

" #

Then, it is easy to show that Gðb1;B1;B2Þ and Gðb2;B1;�B2Þ are singular for

b1 ¼ 0; b1 ¼�c2c02a21K=4

ð0:5c02 � 1� c2a22Þð0:5c02 � 1� c2a11Þ � c22a12a21ð18Þ

b2 ¼ 0; b2 ¼c2c002K=4

ð0:5c002 � 1þ c2a22Þð0:5c002 � 1þ c2a11Þ � c22a12a21ð19Þ

where c02 and c002 are the determinants of the matrices I þ c2A2 and I � c2A2; respectively. Sincethe constant c2 is chosen so that B2 is Schur, we have

c02 > 0; ð0:5c02 � 1� c2a22Þð0:5c02 � 1� c2a11Þ � c22a12a21 > 0 ð20Þ

c002 > 0; ð0:5c002 � 1þ c2a22Þð0:5c002 � 1þ c2a11Þ � c22a12a21 > 0 ð21Þ



Since the pencils, Gðb1;B1;B2Þ and Gðb2;B1;�B2Þ; can only be singular for b1 ¼ b2 ¼ 0; fromEquations (18)–(21), we conclude that the product a21K has to be zero. If a21 ¼ 0; then both A1

and A2 are upper-triangular, and a diagonal P ¼ diag½1;p22� exists (see Lemma 1(ix)a), where

p22 > maxK2c21

ðc21 � 1Þ2; a212ð1� a211Þð1� a222Þ

" #

If K ¼ 0; then any P that satisfies Equation (10) also satisfies Equation (9), since AT1 PA1 � P ¼

�ð1� 1=c21ÞP50:Case (2ii): Assume that B1 has a single eigenvalue of unity. Then, we assume (with no loss of

generality) that B1 and A2 are in the following form:

B1 ¼1 0

0 b22

" #; A2 ¼

a11 a12

a21 a22

" #; jb22j51

In this case, Gðb1;B1;B2Þ and Gðb2;B1;�B2Þ are singular for

b1 ¼ 0; b1 ¼0:5c02ð0:5c

02 � 1� c2a22Þð1� b22Þ=ð1þ b22Þ

ð0:5c02 � 1� c2a22Þð0:5c02 � 1� c2a11Þ � c22a12a21

b2 ¼ 0; b2 ¼0:5c002ð0:5c

002 � 1þ c2a22Þð1� b22Þ=ð1þ b22Þ

ð0:5c002 � 1þ c2a22Þð0:5c002 � 1þ c2a11Þ � c22a12a21

Using arguments similar to those in case 2(i), we conclude that jc2ða11 � a22Þj51� c22 �ða11a22 � a12a21Þ: A Schur matrix B2 which also satisfies this relation has a diagonal P ¼diag½1; ja12=a21j� [12]. Since B2 ¼ c2A2; jc2j > 1; the same P also satisfies Equation (10).Furthermore, P is also a Stein solution for A1; since any diagonal positive definite matrix is aStein solution for a diagonal Schur matrix (see Lemma 1(ix)b). &

4. COMPUTATIONAL ASPECTS

Given two systems SA1and SA2

; the existence of a CQLF can be concluded from Theorem 1 bychecking the Schur stability of the two matrix pencils H ða;A1;A2Þ and H ða;A1;�A2Þ: As seenfrom the examples given in Section 5, this can be done by plotting the root-loci of the twopencils for a 2 ½0; 1�: If both root-loci lie entirely within the unit circle, it can be concluded that aCQLF exists. The CQLF can then be determined following the steps given in the proof ofTheorem 1.

While plots of the root-loci are attractive in that they provide a better understanding of thedistribution of eigenvalues, they are nevertheless computationally intensive, since the roots haveto be computed for a continuum of values of the parameter a: Hence, computationally lessdemanding methods would be attractive to merely check whether a CQLF exists. In whatfollows, we describe briefly an algebraic approach in which checking a finite number ofconditions yields the desired result.

Since the Schur stability of the matrix pencils, H ða;A1;A2Þ and H ða;A1;�A2Þ; is equivalent tothe Hurwitz stability of the matrix pencils, Gða;A1;A2Þ and Gða;A1;�A2Þ; we use the latter forconvenience in the sequel. Consider Gða;A1;A2Þ ¼ aM1 þ ð1� aÞM2 first, where M1 ¼ 0:5I �ðI þ A1Þ

�1 and M2 ¼ 0:5I � ðI þ A2Þ�1: Since A1 and A2 are Schur, both M1 and M2 are Hurtwitz



matrices. The characteristic equation of the matrix pencil Gða;A1;A2Þ is s2 � trace½Gða;A1;A2Þ�sþ det½Gða;A1;A2Þ� ¼ 0: For the matrix pencil Gða;A1;A2Þ to be Hurwitz, the trace ofGða;A1;A2Þ has to be negative and the determinant (det½Gða;A1;A2Þ�) should be positive forall a 2 ½0; 1�: Since trace½Gða;A1;A2Þ� ¼ a traceðM1Þ þ ð1� aÞtraceðM2Þ50; only the determinantcondition, det½Gða;A1;A2Þ�40; needs to be checked for all a 2 ½0; 1�: Denoting the matrices M1

and M2 as M1 ¼ ½mij� and M2 ¼ ½ %mmij�; i; j ¼ 1; 2; det½Gða;A1;A2Þ� can be expressed in the form,det½Gða;A1;A2Þ� ¼ aa2 þ baþ c; where

a ¼ detðM1Þ þ detðM2Þ � d;

b ¼ d � 2detðM2Þ;

c ¼ detðM2Þ;

d ¼ m11 %mm22 þ %mm11m22 � m12 %mm21 � %mm12m21:

9>>>>>=>>>>>;

ð22Þ

Since det½Gða;A1;A2Þ� is a quadratic function of a whose coefficients can be computed using (22),whether or not it satisfies the positivity condition in the interval can be easily checked. Inparticular, the matrix pencil Gða;A1;A2Þ is Hurwitz if and only if one of the following conditionsis satisfied: (i) a40; or (ii) a > 0; �b=2a =2 ½0; 1� or (iii) a > 0; �b=2a 2 ½0; 1�; and c� b2=4a > 0:Compared to the plotting of the root-locus, checking the above algebraic conditions involvessubstantially less computational effort.

To establish the existence of a CQLF, similar conditions have also to be checked for thesecond matrix pencil Gða;A1;�A2Þ:

5. EXAMPLES

In this section, we consider several sets of examples that illustrate various aspects of the mainresult of Section 3.

Example 1

Given two matrices A1 and A2 where

A1 ¼0:9 0

0 0:2

" #; A2 ¼

0:78 0:56

�0:84 �1:18

" #

the problem is to determine whether a CQLF exists. Following the algebraic approach given inSection 4, a1; b1; and c1 for the pencil H ða;A1;A2Þ and a2; b2; and c2 for the pencil H ða;A1;�A2Þcan be computed as a1 ¼ 1:8596; b1 ¼ �3:4342; c1 ¼ 1:5833; a2 ¼ �0:5429; b2 ¼ 0:5122; andc2 ¼ 0:0395: Since a1 > 0; �b1=2a1 ¼ 0:9233 2 ½0; 1�; and c1 � b21=4a1 ¼ �0:002250; it followsthat the pencil H ða;A1;A2Þ is not Schur. Thus, a CQLF does not exist for the two systems SA1

;and SA2

:

The same conclusion can also be drawn by plotting the root-loci for the pencils H ða;A1;A2Þand H ða;A1;�A2Þ for a 2 ½0; 1�; and checking whether the eigenvalues lie entirely within the unitcircle. The root-loci for the two pencils are shown in Figure 1. Since the pencil H ða;A1;A2Þ is notSchur, it follows that a CQLF does not exist.



Remark 1. An interesting feature of this example is that even though a CQLF does not exist,the time-varying system (6) is exponentially stable. The proof of stability is based on theproperties of the transition matrices and CQLFs for three or more systems, and will be discussedfurther in Section 6.

Example (Set) 2. This example ‘set’ includes six examples consisting six pairs of matrices A1

and A2: All the examples in the set correspond to case 1 discussed in Section 3 where a0 2 ð0; 1Þ:In each case we establish the existence of a CQLF and proceed to determine the Lyapunovfunction xTPx (or equivalently the matrix P ). Due to space limitations, we briefly describe theapproach used in all the cases and the results are tabulated in Table I. For all the pairs ofmatrices the following steps were used:

1. Determine constants c1 and c2; (jc1j; jc2j > 1), such that the pencil H ða; c1A1; c2A2Þ is Schur(or has complex eigenvalues on the unit circle) for all a 2 ½0; 1�; a=a0; and has aneigenvalue of unity for a ¼ a0 2 ð0; 1Þ:

2. Determine x0 2 R2; jjx0jj ¼ 1; the unit eigenvector of H ða0; c1A1; c2A2Þ corresponding to theeigenvalue unity.

3. Determine q0 ¼ ½q01; q02�T ¼ ðI þ c1A1Þ�1x0; and q1 ¼ q?0 ¼ ½q02;�q01�T:

4. Compute the matrix P which is the solution of the equation

ðc1A1ÞTP ðc1A1Þ � P ¼ �q1qT1

Six pairs of matrices are considered in this example set to include the different possible casesthat can arise. In the first two examples, A1 and A2 have real eigenvalues. The root-loci for thefirst example is depicted in Figure 2 and it can be noted from this figure that the matrix pencilsH ða;A1;A2Þ and H ða;A1;�A2Þ are Schur stable. Similar root-loci are obtained for the otherexamples. In the next two examples, A1 has real eigenvalues while the eigenvalues of A2 arecomplex. In the last two cases, both matrices have complex eigenvalues. In all cases, theresulting Lyapunov function, xTPx; is shown to be a CQLF for both SA1

and SA2:

Figure 1. Root-loci for Example 1.



As stated above, all the examples in the set correspond to case 1 where a0 2 ð0; 1Þ: In Example3, we consider the special case where a0 2 f0; 1g:

Example (Set) 3. Two pairs of matrices are considered in this example. For the first pair

A1 ¼0:8 0

0 0:8

" #; A2 ¼

0:35 �0:25

0:50 0:85

" #

Table I. Examples for the case where a0 2 ð0; 1Þ

A1 A2 c1=c2 a0 x0 P

0:75 0:75�1:5 �1:5

� �0:96 0:32�0:48 �0:16

� �1:17541:1754

0.04210:8030�0:5960

� �9:7287 5:38815:3881 4:1000

� �

�2 �14 2

� ��0:6 �0:21:8 0:6

� �2:5�2

0.50:4008�0:9162

� �0:4459 0:20330:2033 0:0951

� �

1:8 0:9�1:8 �0:9

� �0:55 �0:250:50 1:05

� �1:02811:0281

0.46280:7664�0:6424

� �0:4123 0:25580:2558 0:2061

� �

0:4 00 0:6

� �1:15 0:25�0:50 0:65

� �1:5354�1:0706

0.98670:2808�0:9598

� �0:4007 0:20020:2002 0:2001

� �

0:8 0:2�0:2 0:8

� �1:15 0:25�0:50 0:65

� �1:0032�1:0032

0.97870:1015�0:9948

� �0:6146 0:26640:2664 0:3486

� �

0:4 0:2�0:2 0:4

� �1:15 0:25�0:50 0:65

� ��1:38211:0706

0.04350:7950�0:6067

� �3:9018 1:95251:9525 1:9537

� �

Figure 2. Root-loci for A1 ¼0:75 0:75�1:5 �1:5

� �and A2 ¼

0:96 0:32�0:48 �0:16

� �in Example Set 2.



both H ða; 1:25A1; 1:25A2Þ and H ða; 1:25A1;�1:25A2Þ are Schur for all a 2 ½0; 1Þ; and both of themhave two eigenvalues of unity for a ¼ 1: In this case, since A1 ¼ 0:8I ; any positive definite matrixis a Stein solution for A1; therefore we determine the CQLF by solving (10) with Q2 ¼ I andobtain

P ¼2:2728 1:0486

1:0486 2:5096

" #

For the second pair of matrices

A1 ¼0:8 0

0 0:6

" #; A2 ¼

0:21 �0:18

0:27 0:84

" #

both H ða; 1:25A1; 1:1A2Þ and H ða; 1:25A1;�1:1A2Þ are Schur for all a 2 ½0; 1� except for a ¼ 1;where each pencil has a single eigenvalue of unity. In this case,

B2 ¼ 1:1A2 ¼0:231 �0:198

0:297 0:924

" #

is Schur, and has a diagonal Stein solution, P ¼ diag½1; 23�; which also satisfies Equations (9) and

(10) simultaneously.

6. EXTENSION TO MORE THAN TWO SYSTEMS

Despite the elegance of the conditions in Theorem 1, given a set of more than two systems(N > 2), it is not enough to check the pairwise existence conditions to determine whether aCQLF exists for the whole set. This fact is illustrated by the following example.

Example 4

Consider the stable second-order LTI systems SA1; SA2

and SA3where

A1 ¼0:8 0

�1 �0:8

" #; A2 ¼

0:8 0

1 �0:8

" #; A3 ¼

0:95 �0:08

0:1 0:95

" #

Each pair of systems satisfy the conditions in Theorem 1, thus a CQLF exists for every pair ofsystems. However, the matrix A1A2A2

3 has an eigenvalue of 1.0712, which implies that the time-varying system (6) is unstable for the periodic switching sequence AðkÞ with period 4 andAð1Þ ¼ A1; Að2Þ ¼ A2; Að3Þ ¼ Að4Þ ¼ A3: Hence, a Lyapunov function cannot exist for (6), whichin turn implies that the systems SAi ; i 2 N; do not have a CQLF.

6.1. CQLF for three or more systems

Example 4 illustrates the need to extend the results of Section 3 to more than two discrete-timesystems. In this section, we will provide the necessary and sufficient conditions for the existence



of a solution for three and more systems. Due to space limitations, we will not provide theproofs, but the interested reader is referred to [13] for further details.

Given the set of matrices A ¼ fA1;A2; . . . ;ANg; let %AA ¼ f %AA1; %AA2; . . . ; %AANg where %AAi ¼ TAiT�1;i 2 N; for a non-singular matrix T : In Reference [9], it is shown that there exists a non-singularmatrix T which decomposes the set %AA as %AA ¼ %AAI [ %AAII where the transformed matrices in thefirst set AI are multiples of the identity matrix, whereas the matrices in the second set AII are inthe form

%aa11 %aa12

%aa21 %aa22

" #; %aa21=0 ð23Þ

Since any vðxÞ ¼ xTPx; P ¼ PT > 0; is a CQLF for the set %AAI; it is therefore enough to check theexistence of a CQLF for the second set %AAII: In view of this remark, from this point on we willassume that the matrices in the original set A are of form (23). Then for any A 2 A; the set ofpoints (p12;p22) for which detðATPA� P Þ ¼ 0 describes an ellipse (from Lemma 1(viii)). In thesequel, we will denote this ellipse by EA: The interior of the ellipse EA defines an open convex setfor which detðATPA� P Þ > 0 and P > 0: In this paper, the set of P matrices characterized by thisconvex set is referred to as the Lyapunov set associated with A; and is denoted by LA: Theconditions for a CQLF to exist for all three systems can be derived by checking whether thecondition,

TNi¼1 LAi=1; holds or not. To check this condition algebraically, we first define the

sets

ZAiAj ¼ P ¼1 p12

p12 p22

" #: detðAT

l PAl � P Þ ¼ 0; l ¼ i; j

( ); 14i5j4N ð24Þ

Then a necessary and sufficient condition for a CQLF to exist for three systems SA1; SA2

and SA3

is that there exist at least two symmetric positive semidefinite matrices P1 and P2 belonging to theset ZA ¼4 ZA1A2

[ZA1A3[ZA2A3

such that ATi PjAi � Pj; i ¼ 1; 2; 3; j ¼ 1; 2; are simultaneously

either negative definite or negative semidefinite. In this case, vðxÞ ¼ xTPx where P ¼ aP1 þ ð1�aÞP2; a 2 ð0; 1Þ can be shown to be a set of CQLFs for the systems SA1

; SA2and SA3

:

Theorem 2

Let SAi ; i ¼ 1; 2; 3; satisfy the conditions of Theorem 1 pairwise and the matrices Ai be in form(23). Then a CQLF for SAi ; i ¼ 1; 2; 3; exists if and only if one of the following conditions hold:

(i) EAi \ EAj ¼ � for some i ¼ in 2 f1; 2; 3g; j ¼ jn 2 f1; 2; 3g; in=jn:(ii) EAi \ EAj ¼ � for all i; j 2 f1; 2; 3g; i=j; and there exist two distinct matrices P1; P2 2 ZA

such that

ATi PjAi � Pj40; i ¼ 1; 2; 3; j ¼ 1; 2 ð25Þ

Theorems 1 and 2 provide necessary and sufficient conditions for a CQLF to exist for two andthree systems, respectively. These results with Helly’s theorem [14] are adequate to derive theCQLF existence conditions for more than three systems.



Theorem 3

Let SAi ; i 2 N; be stable systems with Ai in form (23). Then a CQLF exists for the systems SAi ;i 2 N; if and only if every 3-tuple of systems in the set A satisfy the conditions of Theorem 2.

Example 1 (revisited)It has already been shown that the conditions of Theorem 1 are not satisfied, therefore a CQLFdoes not exist for the two systems. This fact can also be noted from the Lyapunov sets depictedin Figure 3 for %AAi ¼ TAiT�1; i ¼ 1; 2 where

T ¼1 �1

2 2

" #ð26Þ

Even though SA1and SA2

do not have a CQLF, the time-varying system (6) is still exponentiallystable. We will show this fact by using the properties of system (6) and the results in this section.

We first note that the stability properties of the time-varying system (6) are equivalent tothose of the time-varying system

xðk þ 1Þ ¼ BðkÞxðkÞ; BðkÞ 2 B¼4 fB1;B2;B3;B4g ð27Þ

where B1 ¼ A21; B2 ¼ A1A2; B3 ¼ A2A1 and B4 ¼ A2

2: We next show that the systems SBi ; i 2N¼4 f1; 2; 3; 4g; have a CQLF. Figures 4 and 5 depict the Lyapunov sets for %BBi ¼ TBiT�1; i 2 N;where T is given in (26). Each pair of matrices in the set %BB satisfies the conditions in Theorem 1,thus a CQLF exists for every pair of systems. The following sets of matrices are obtained by

Figure 3. Lyapunov sets for %AA1 and %AA2 in Example 1.



Figure 4. Lyapunov sets for %BBi; i 2 N in Example 1.

Figure 5. Lyapunov sets for %BBi; i 2 N in Example 1 (zoomed version).



solving the algebraic equations in (24).

Z %BB1%BB2¼

1 �0:5812

�0:5812 0:3579

" #;

1 0:6067

0:6067 0:6639

" #( )

Z %BB1%BB3¼

1 �0:4286

�0:4286 0:8892

" #;

1 0:5568

0:5568 0:3180

" #( )

Z %BB1%BB4¼

1 �0:6213

�0:6213 0:6108

" #;

1 0:6211

0:6211 0:4673

" #( )

Z %BB2%BB3¼

1 �1:3968

�1:3968 3:8235

" #;

1 0:0963

0:0963 0:2314

" #(;

1 1:1291

1:1291 1:5064

" #;

1 2:4227

2:4227 11:8643

" #)

Z %BB2%BB4¼

1 �1:0371

�1:0371 1:1841

" #;

1 �0:7550

�0:7550 9:4161

" #(;

1 0:1120

0:1120 0:2387

" #;

1 1:3175

1:3175 1:9154

" #)

Z %BB3%BB4¼

1 �1:9965

�1:9965 6:5829

" #;

1 0:0832

0:0832 0:2386

" #(;

1 0:8183

0:8183 0:6942

" #;

1 1:4955

1:4955 3:0793

" #)

It is straightforward to verify that the conditions in Theorem 3 are satisfied. Furthermore, wehave %BBT

i Pj %BBi � Pj40; i 2 N; j ¼ 1; 2; where

P1 ¼1 0:6067

0:6067 0:6639

" #; P2 ¼

1 �0:4286

�0:4286 0:8892

" #

Hence, vðxÞ ¼ xTPx with P ¼ aP1 þ ð1� aÞP2; a 2 ð0; 1Þ; is a set of CQLFs for S %BBi; i 2 N:

Therefore, the original time-varying system is exponentially stable.

7. CONCLUSIONS

In this paper, necessary and sufficient conditions for the existence of a common quadraticLyapunov function for second-order discrete-time systems are derived. These are expressed interms of the Schur stability of matrix pencils. Methods for determining whether the conditionsare satisfied can be carried out algebraically or by using root-loci. Work is currently in progresson the extension of the results to higher-order systems.



ACKNOWLEDGEMENTS

The research reported here was supported by the Office of Naval Research under contract number N00014-97-1-0948.

APPENDIX A: PROOFS OF LEMMAS 1–3

Lemma 1

(ii) Let P be a Stein solution for A; i.e., it satisfies (2) for some symmetric positive definitematrix, Q: We want to show that ðAkÞTPAk � P ¼ �Q0

k50; k 2 Zþ; by induction on k:Obviously, the argument is true for k ¼ 1: Assume that the hypothesis holds for k � 1;i.e., ðAk�1ÞTPAk�1 � P ¼ �Q0

k�150; then after pre- and post-multiplying this equation byAT and A; respectively, and using Equation (2), we obtain the desired result, Q0

k ¼Qþ ATQ0

k�1A > 0:(iii) Pre- and post-multiply Equation (9) by AT

2 and A2; and use Equation (10) to get

ðA1A2ÞTP ðA1A2Þ � P ¼ �½Q2 þ AT

2Q1A2�50

(vii) Let Equation (2) hold for A 2 R2�2 for some symmetric positive definite matrices, P andQ: We want to show that P is also a Stein solution for %AA; i.e., %AATP %AA� P ¼ � %QQ for some%QQ > 0: By writing the elements of each matrix, and carrying out the algebra, one can seethat detð %QQÞ ¼ detðQÞ > 0: This implies that %QQ cannot be indefinite; it has to be eithernegative or positive definite. Assume that it is negative definite. Then P > 0 and �Q > 0imply that %AA is unstable which is a contradiction. Therefore, %QQ must be positive definite.

(viii) The equation detðATPA� P Þ ¼ 0 can be expressed in the equivalent form

c1p222 þ c2p12p22 þ c3p2

12 þ c4p12 þ c5p22 þ c6 ¼ 0 ðA1Þ

where c1 ¼ �a221; c2 ¼ 2a21ða22 � a11Þ; c3 ¼ 4a11a12a21a22 � ða11a22 þ a12a21 � 1Þ2; c4 ¼2a12ða11 � a22Þ; c5 ¼ ða211 � 1Þða222 � 1Þ þ a212a

221 � 2a11a12a21a22; and c6 ¼ �a212: Equation

(A1) defines a parabola if and only if c22 � 4c1c3 ¼ 0 and an ellipse if and only if c22 �4c1c350: Straightforward manipulation yields c22 � 4c1c3 ¼ �4a221½1þ detðAÞ � traceðAÞ�½1þ detðAÞ þ traceðAÞ�: Since A is assumed to be Schur, the result follows.

(ix) The proof for part (a) is straightforward. For part (b), if A1 has the stated property, thenA1 has a diagonal Stein solution [12]. Finally, it is easy to see that this diagonal solutionalso satisfies Equation (10) for the diagonal Schur matrix, A2:

Lemma 2

(i) Only the equivalence of the stability of the matrix pencils H ða1;A1;A2Þ andH ða2;�A1;�A2Þ; a1; a2 2 ½0; 1�; is shown here. The equivalent results for the other pencilsfollow along the same lines. From relation (7), it is clear that H ða1;A1;A2Þ is Schur if andonly if Gða1;A1;A2Þ is Hurwitz. Therefore to show the equivalence of the Schur stabilityof the pencils, H ða1;A1;A2Þ and H ða2;�A1;�A2Þ; we will equivalently show the Hurwitz



stability of the pencils Gða1;A1;A2Þ and Gða2;�A1;�A2Þ: First, note that the matricesGð0;A1;A2Þ ¼ 0:5I � ðI þ A2Þ

�1 and Gð0;�A1;�A2Þ ¼ 0:5I � ðI � A2Þ�1 are both Hur-

witz, since A2 and �A2 are both Schur. Excluding this case, we have a1; a2 > 0; and wecan equivalently consider the pencils

Gðb1;A1;A2Þ ¼ ½0:5I � ðI þ A1Þ�1� þ b1½0:5I � ðI þ A2Þ

�1�; b1 2 ½0;1Þ

Gðb2;�A1;�A2Þ ¼ ½0:5I � ðI � A1Þ�1� þ b2½0:5I � ðI � A2Þ

�1�; b2 2 ½0;1Þ

By choosing b1detðI þ A1Þ=detðI þ A2Þ ¼ b2detðI � A1Þ=detðI � A2Þ; it can be shown thatdetðI þ A1ÞGðb1;A1;A2Þ and detðI � A1ÞGðb2;�A1;�A2Þ are classical adjoints of eachother, and therefore have the same eigenvalues. Since both A1 and A2 are Schur, we havedetðI þ A1Þ > 0; detðI � A1Þ > 0; detðI þ A2Þ > 0; and detðI � A2Þ > 0: Hence, we concludethat the pencil Gðb1;A1;A2Þ; b1 2 ½0;1Þ; is Hurwitz if and only if the pencilGðb2;�A1;�A2Þ; b2 2 ½0;1Þ is Hurwitz.

(ii) Given that A1;A2 2 R2�2 are Schur, we show that either one of the pencils Gða;A1;A2Þ orGða;A1;�A2Þ has real eigenvalues for some a 2 ½0; 1�: Note that if either A1 or A2 has realeigenvalues, the conclusion is immediate. Therefore, in the sequel, both A1 and A2 havecomplex eigenvalues. For convenience, define G1ða;A1;A2Þ ¼ 0:5I � Gða;A1;A2Þ: Clearly,the claim for Gða;A1;A2Þ is true if and only if it is true for G1ða;A1;A2Þ: Assume thatG1ða;A1;A2Þ has complex eigenvalues for all a 2 ½0; 1�: From Cayley–Hamilton theorem,we have

ðI þ A2Þ2 � trace½I þ A2�ðI þ A2Þ þ detðI þ A2ÞI ¼ 0 ðA2Þ

Multiplying the left-hand side of the above equation by ðI þ A2Þ�1; we obtain

ðI þ A2Þ � trace½I þ A2�I þ detðI þ A2ÞðI þ A2Þ�1 ¼ 0 ðA3Þ

Similarly,

ðI � A2Þ � trace½I � A2�I þ detðI � A2ÞðI � A2Þ�1 ¼ 0 ðA4Þ

From Equations (A3) and (A4) it follows that

ðI þ A2Þ�1 ¼

2

detðI þ A2ÞI �

detðI � A2ÞdetðI þ A2Þ

ðI � A2Þ�1 ðA5Þ

Since both A1 and A2 are Schur matrices, we consider the following cases separately:

(a) A non-zero vector x 2 R2 exists such that ðI þ A1Þ�1x=mðI þ A2Þ

�1x where m isa non-zero constant.

(b) or ðI þ A1Þ�1x ¼ mðI þ A2Þ

�1x for all x 2 R2 where m is a non-zero constant.

In case (a), ðI þ A1Þ�1x and ðI þ A2Þ

�1x span R2; and we can find non-zero constants m1; andm2 so that

m1ðI þ A1Þ�1xþ m2ðI þ A2Þ

�1x ¼ x ðA6Þ

By the assumption that G1ða;A1;A2Þ has complex eigenvalues for a 2 ½0; 1�; it follows that m2=m150: From Equations (A5) and (A6) we obtain

ðI þ A1Þ�1 �

m2

m1�detðI � A2ÞdetðI þ A2Þ

ðI þ A2Þ�1

� �x ¼

1

m1�

2m2

m1detðI þ A2Þ

� �x



Since A2 is Schur and m2=m150; it follows that G1ða;A1;�A2Þ (and therefore G ða;A1;�A2Þ) hasreal eigenvalues for a ¼ 1=ð1� ðm2=m1Þ � detðI � A2Þ=detðI þ A2ÞÞ: In case (b), we have

ðI þ A1Þ�1x� mðI þ A2Þ

�1x ¼ 0 ðA7Þ

Since the pencil G1ða;A1;A2Þ is assumed to have complex eigenvalues for a 2 ½0; 1�; it follows thatm > 0: Furthermore, using (A5) in (A7) yields

ðI þ A1Þ�1 þ m

detðI � A2ÞdetðI þ A2Þ

ðI þ A2Þ�1

� �x ¼

2mdetðI þ A2Þ

x

Hence G1ða;A1;�A2Þ has real eigenvalues for a ¼ 1=ð1þ m � detðI � A2Þ=detðI þ A2ÞÞ:(iii) By Lemma 2(iii), A1 and A2 cannot have x0 for an eigenvector if Gða0;A1;A2Þx0 ¼ 0 for

some a0 2 ð0; 1Þ: The proof is by contradiction. Assume (with no loss of generality) that x0 is aneigenvector of A1 (hence A1 has real eigenvalues), with a corresponding eigenvalue l; jlj51; i.e.,A1x0 ¼ lx0: Since x0 is also the eigenvector of Gða0;A1;A2Þ corresponding to the zero eigenvalue,we have a0ðI þ A1Þ

�1x0 þ ð1� a0ÞðI þ A2Þ�1x0 ¼ 0:5x0: Manipulating this equation using A1x0 ¼

lx0 yields that x0 is also an eigenvector of A2; i.e., A2x0 ¼ %llx0 with

%ll ¼1� a0½2l=ð1þ lÞ�1� a0½2=ð1þ lÞ�

ðA8Þ

Therefore, if A2 has complex eigenvalues, this leads to an immediate contradiction. If A2 has realeigenvalues, from (A8) it follows that j%llj > 1; which contradicts the fact that A2 is stable.

Lemma 3

Given A1;A2 2 R2�2; let H ða;A1;A2Þ and H ða;A1;�A2Þ be Schur pencils. We want to show thatconstants c1 and c2 (jc1j; jc2j > 1) exist so that the pencil H ða;B1;B2Þ with B1 ¼ c1A1 and B2 ¼c2A2 have the following properties: (i) H ða;B1;B2Þ is Schur or has complex eigenvalues on theunit circle for all a 2 ½0; 1�; a=a0: (ii) The pencil H ða;B1;B2Þ � I is singular for a single value ofa ¼ a0 2 ½0; 1�: From relation (7), H ða;B1;B2Þ has the above properties if and only if Gða;B1;B2Þhas the following properties: (i) Gða;B1;B2Þ is Hurwitz, or has complex eigenvalues on theimaginary axis for all a 2 ½0; 1�; a=a0: (ii) The pencil Gða;B1;B2Þ is singular for a single value ofa ¼ a0 2 ½0; 1�:

As long as c1 and c2 are chosen so that B1 and B2 are Schur or have complex eigenvalues onthe unit circle, we have

trace½Gða;B1;B2Þ� ¼ �að1� detðB1ÞÞdetðI þ B1Þ

� ð1� aÞð1� detðB2ÞÞdetðI þ B2Þ

40

which implies that the pencil Gða;B1;B2Þ cannot have right half plane complex eigenvalues. Thefollowing represent the three possible cases that can arise:

(a) Both A1 and A2 have real eigenvalues,(b) A1 has real eigenvalues and A2 has complex eigenvalues,(c) Both A1 and A2 have complex eigenvalues.



In the cases where we deal with matrices having real eigenvalues, we will assume (with no lossof generality) that the eigenvalue with the maximum norm is positive.

(a) Both A1 and A2 have real eigenvalues: Denote the maximum eigenvalues of A1 and A2 by m1and m2; respectively. Without loss of generality assume that m15m2: We have two cases:

(i) m1 ¼ m2 ¼ 0: In this special case, both matrices have two zero eigenvalues. Ifeither of the matrices is a null matrix, then any Stein solution for the othermatrix is also a Stein solution for the null matrix, and thus SA1

and SA2have a

CQLF. In the case that neither of the matrices is a null matrix, we can assumethat the matrices have the following form:

A1 ¼0 1

0 0

" #; A2 ¼

a11 a12

a21 a22

" #ðA9Þ

with a11 þ a22 ¼ a11a22 � a12a21 ¼ 0: Since the pencil

Gðb1;A1;A2Þ ¼ b1½0:5I � ðI þ A1Þ�1� þ ½0:5I � ðI þ A2Þ

�1�; b1 2 ½0;1Þ

is Hurwitz for all b150; we have det½Gðb1;A1;A2Þ� ¼ 0:25b21 þ 0:5ð1� 2a21Þb1 þ0:25 > 0 for all b150: Thus, det½Gðb1;A1;A2Þ� can be zero only for negative realor complex b1: For negative real b1; we have a2140; whereas for complex b1; wehave 1 > a21 > 0: Similarly, since the pencil

Gðb2;�A1;A2Þ ¼ b2½0:5I � ðI � A1Þ�1� þ ½0:5I � ðI þ A2Þ

�1�; b2 2 ½0;1Þ

is Hurwitz for all b250; we have a2150 for negative b2; and �15a2150 forcomplex b2: Thus, for Hurwitz stability of the pencils Gðb1;A1;A2Þ andGðb2;�A1;A2Þ for all b1; b250; we have

ja21j51 ðA10Þ

Now consider the pencil Gðb;B1;B2Þ; with B1 ¼ c1A1; and B2 ¼ c2A2: We havedet½Gðb;B1;B2Þ� ¼ 0:25b2 þ 0:5ð1� 2c1c2a21Þbþ 0:25: This is equal to zero for asingle b50 if and only if

c1c2a21 ¼ 1 ðA11Þ

From (A10), if a21 ¼ 0; then both A1 and A2 are upper-triangular, and thereforeSA1

and SA2have a CQLF. If a21=0; it is clear from (A10) and (A11) that there

exist constants c1 and c2; jc1j; jc2j > 1; so that the pencil Gðb;B1;B2Þ is singularfor the single value of b ¼ 1:

(ii) m1 > 0: Consider the pencil

Gða;A1;A2Þ ¼ a½0:5I � ðI þ A1Þ�1� þ ð1� aÞ½0:5I � ðI þ A2Þ

�1�; a 2 ½0; 1�

and define the function

f ða; c1; c2Þ ¼4det½Gða; c1A1; c2A2Þ� ðA12Þ

Note that the function f ða; c1; c2Þ is continuous in its arguments and it isquadratic in a: Since f ða; 1; 1Þ ¼ det½Gðb; c1A1; c2A2Þ�40; there exists c1 ¼ c2 ¼cn; cn 2 ð1; 1=m1� such that

f ða; c; cÞ > 0; 8c 2 ½1; cnÞ; 8a 2 ½0; 1� ðA13Þ



and

f ða0; cn; cnÞ ¼ 0; for some a0 2 ½0; 1� ðA14Þ

Now, if cn41=m151=m2; then there exists a unique a0 2 ½0; 1� such that (A14) issatisfied. This is due to the fact that m1=m2 implies that the function, f ða; c1; c2Þ;cannot be identically zero. If cn51=m1 ¼ 1=m2; the above argument still applies.If cn ¼ 1=m1 ¼ 1=m2; then the constants c1 ¼ 1=m1; and any c251=m2 satisfy thedesired requirements. In this case, a0 ¼ 1:

(b) A1 has real eigenvalues and A2 has complex eigenvalues: Denote the maximumeigenvalue of A1 by m1; and the norm of the eigenvalues of A2 by m2: We have two cases:

(i) m1 ¼ 0: In this special case, A1 has two zero eigenvalues. If it is a null matrix,then any Lyapunov function of SA2

is also a Lyapunov function of SA1: If A1 is

not a null matrix, we can assume that the matrices are in form (A.9). Then, fromthe Hurwitz stability of the pencil Gða;A1;A2Þ for all a 2 ½0; 1�; it follows that

a2150:5ð1� detðA2ÞÞ þ detðI þ A2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidet½0:5I � ðI þ A2Þ

�1�q

: Similarly, from the

Hurwitz stability of the pencil Gða;A1;�A2Þ for all a 2 ½0; 1�; it follows that

a21 > �0:5ð1� detðA2ÞÞ � detðI þ A2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidet½0:5I � ðI þ A2Þ

�1�q

: Thus, we have

ja21j50:5ð1� detðA2ÞÞ þ detðI þ A2Þffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidet½0:5I � ðI þ A2Þ

�1�q

ðA15Þ

From the pencil

Gðb; c1A1; c2A2Þ ¼ b½0:5I � ðI þ c1A1Þ�1� þ ½0:5I � ðI þ c2A2Þ

�1�; b 2 ½0;1Þ

we have

det½Gðb; c1A1; c2A2Þ� ¼ 0:25ðc02Þ2b2 � 0:5c02ð�1þ c22detðA2Þ þ 2c1c2a21Þb

þ ðc02Þ2det½0:5I � ðI þ c2A2Þ

�1� ðA16Þ

where c02 ¼ detðI þ c2A2Þ > 0: The expression for det½Gðb; c1A1; c2A2Þ� in (A.16)is equal to zero for a non-negative value of b if and only if the auxiliary function

f0ðc1; c2Þ ¼ c1c2a21 � 0:5ð1� c22detðA2ÞÞ � c02

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffidet½0:5I � ðI þ c2A2Þ

�1�q

has a zero for some c1 and c2: Using (A15), we can conclude that there exists ac1; jc1j > 1; so that f0ðc1; 1Þ is zero. Finally, since the auxiliary function f0ðc1; c2Þis continuous in its arguments, it achieves a zero value for some constants, c1and c2 with jc1j; jc2j > 1:

(ii) m1 > 0: Consider the function in (A12). The desired constants c1 and c2 can stillbe found by following the same arguments as in part (a), if cn41=m141=m2 orcn51=m251=m1: Otherwise, let c2 ¼ 1=m2: In this case, we can assume that A1

and B2 are in the following form:

A1 ¼a11 a12

a21 a22

" #; B2 ¼

a �ffiffiffiffiffiffiffiffiffiffiffiffiffi1� a2

pffiffiffiffiffiffiffiffiffiffiffiffiffi1� a2

pa

24

35; jaj51 ðA17Þ



Since the pencil, Gða;B1;B2Þ cannot be singular for a ¼ 0; in the sequel weconsider the pencil

Gðb; c1A1; c2A2Þ ¼ ½0:5I � ðI þ c1A1Þ�1� þ b½0:5I � ðI þ c2A2Þ

�1�; b50

and define the function

f1ðb; c1Þ ¼4det½c01Gðb; c1A1; ð1=m2ÞA2Þ�; b50 ðA18Þ

where c01 ¼ detðI þ c1A1Þ > 0: Expression (A18) for f1ðb; c1Þ can be expressed inthe equivalent form

f1ðb; c1Þ ¼ b2ðc01Þ2b2 � bc1c01ða12 � a21Þbþ ðc01Þ

2det½0:5I � ðI þ c1A1Þ�1�

where b ¼ 0:5ffiffiffiffiffiffiffiffiffiffiffiffiffi1� a2

p=ð1þ aÞ > 0: We look for conditions under which

f1ðb; c1Þ ¼ 0 for a single value of b50: If a12 � a2140; it is clear thatf1ðb; c1Þ ¼ 0 holds for c1 ¼ 1=m1; and b ¼ 0: If a12 � a21 > 0; then in order tohave a single non-negative b so that f1ðb; c1Þ ¼ 0; the discriminant function,Dðc1Þ ¼ b2ðc01Þ

2½c21ða12 � a21Þ2 � 4ðc01Þ

2det½0:5I � ðI þ c1A1Þ�1��; should have a

zero for some c1 2 ð1; 1=m1�: We first note that Dð1Þ50; since cn51=m2: Wealso have Dð1=m1Þ ¼ b2ðc01Þ

2c21ða12 � a21Þ2 > 0; since det½0:5I � ðI þ ð1=m1ÞA1Þ

�1�¼ 0: Thus, Dð1Þ50; and Dð1=m1Þ > 0 imply that there exists a cn1 2 ð1; 1=m1Þ; sothat Dðcn1Þ ¼ 0: This implies that the constants c1 ¼ cn1 and c2 ¼ 1=m2 are thedesired values.

(c) Both A1 and A2 have complex eigenvalues: Denote the norms of the eigenvalues of A1 andA2 by m1 and m2; respectively, and assume (without loss of generality) that m25m1:Consider the function in (A12). Since f ða; 1; 1Þ > 0; there exists a cn 2 ð1; 1=m2� such that(A13) holds. If, in addition, f ða0; cn; cnÞ is zero for some a0 2 ½0; 1�; then the desiredconstants are c1 ¼ c2 ¼ cn: Otherwise, choose c2 ¼ 1=m2: Let A1 and B2 be in form (A17),and consider the function in (A.18). In this case, we have

f1ðb; c1Þ ¼ b2ðc01Þ

2

ðc02Þ2b2 � bc1

c01c02

ða12 � a21Þbþ ðc01Þ2det½0:5I � ðI þ c1A1Þ

�1� ðA19Þ

where c01 ¼ detðI þ c1A1Þ; and c02 ¼ detðI þ c2A2Þ: A necessary and sufficient conditionfor f1ðb; c1Þ in (A.19) to be equal to zero for a single value of b; is that the discriminantfunction, Dðc1Þ ¼ b2ðc01Þ

2=ðc02Þ2½c21ða12 � a21Þ

2 � 4ðc01Þ2det½0:5I � ðI þ c1A1Þ

�1��; is zero forsome c1 2 ð1;m1�: Since the pencil Gða;A1; ð1=m2ÞA2Þ is either Hurwitz or has complexeigenvalues on the imaginary axis, we have Dð1Þ50: On the other hand,

Dð1=m1Þ ¼ b2ðc01Þ

2

ðc02Þ2m21

½ða12 þ a21Þ2 þ ða11 � a22Þ

2�50:

Note that Dð1=m1Þ ¼ 0 if and only if a21 ¼ �a12; and a11 ¼ a22; thus A1; and A2; are in thefollowing specific form:

A1 ¼a11 a12

�a12 a11

" #; A2 ¼

%aa11 %aa12

� %aa12 %aa11

" #;a211 þ a21251

%aa211 þ %aa21251



For these specific forms of A1 and A2; it is straightforward to see that P ¼ I is a Steinsolution (note that the desired constants in this case are c1 ¼ 1=m1 and c2 ¼ 1=m2).

In other cases Dð1Þ50; and Dð1=m1Þ > 0 imply that there exists a constant cn1 2ð1; 1=m1Þ; so that Dðcn1Þ ¼ 0: Thus, if a12 � a2150; we have the desired constants asc1 ¼ cn1 ; and c2 ¼ 1=m2:

If a12 � a2150; then arguments similar to those given above yield c1 ¼ cn1 2 ½�1=m1;�1Þ and c2 ¼ 1=m2:

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On the existence of common quadratic Lyapunov functions for second-order linear time-invariant discrete-time systems