ELE7029: Optimal Digital Control System Lecture III Linear Matrix Inequalities Goals of this lecture · LMIs · LMI Problems · Solving LMIPs S.-H. Lee ([email protected]) Elec. Engr., Hanyang University Spring 2022 Outline 1. Linear Matrix Inequalities 2. Solving LMIPs S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 1 Linear Matrix Inequalities An inequality of the form F(x) = F0 + x1F1 + · · ·+ xnFn = F0 + n∑ i=1 xiFi ≥ 0, (3.1) is called linear matrix inequality (LMI) Here x = (x1, · · · ,xn) ∈ Rn a vector of unknown scalars (decision, optimization variable) Fi = FTi ∈ Rm×m, i= 0, · · · ,n, are given symmetric matrices F : Rn→ Rm×m is an affine fcn. of the variable x ∈ Rn S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 2 The canonical form is very inefficient from a storage point of view. In most applications, LMIs arise rather in the form L (X1, . . . ,Xm)where L(·) and R(·) are affine functions of some structured matrix variables X1, . . . ,Xm. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 3 e.g. An affine function of x ∈ R2 F(x) = x1 + x2 x2 + 1 x2 + 1 x3 = F0 + x1F1 + x2F2 + x3F3 ≥ 0 where F0 = 0 1 1 0 , F1 = 1 0 0 0 , F2 = 1 1 1 0 , F3 = 0 0 0 1 The LMI F(x)≥ 0 is equivalent to a set of nonlinear inequalities: x1 + x2 ≥ 0, x3 ≥ 0, (x1 + x2)x3 (x2 + 1)2 ≥ 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 4 e.g. Lyapunov inequality Consider a system x˙ = Ax with A ∈ Rn×n. Suppose the Lyapunov inequality ATY + YA+Q≤ 0 is affine inequality of the variable Y. Y satisfies the Lyapunov inequality iff the quadratic form V(x) = xTYx satisfies V˙(x)≤ xTQx for the system x˙ = Ax. The inequality F(Y) = ATY + YA+Q≤ 0 is an LMI in Y. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 5 e.g. Bounded Real Lemma Consider a system x˙ = Ax+ Bw z= Cx with A ∈ Rn×n, B ∈ Rn×m, C ∈ Rp×n, and γ > 0. The BR lemma LMIs are ATY + YA+ CTC YB BTY γI ≤ 0, Y ≥ 0 with a variable Y. By this we mean if Y satisfies the LMIs, then the quadratic Lyapunov fcn. V(x) = xTYx proves the RMS gain of the system is no more than γ. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 6 e.g. A time-varying linear dynamic system x˙(t) = A(t)x(t) with A(t) ∈ {A1, . . . ,Aq}. Want to find a quadratic Lyapunov fcn. V(x) = xTYx: Y > 0 and V˙(x)≤ 0 for all x and all possible values of A(t) Y > 0, ATi Y + YAi ≤ 0, i= 1, . . . ,q We can write as LMIs Y ≥ I, ATi Y + YAi ≤ 0, i= 1, . . . ,q If such Y exists, it proves boundedness of trajectories of the system x˙ = A(t)x with A(t) = θ1(t)A1 + · · ·+ θq(t)Aq where θi(t)≥ 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 7 ■ S-procedure: Let F0 = FT0 , F1 = F T 1 ∈ Rn×n Assume xTF1x > 0 When is it true that for all x satisfying xTF1x ≥ 0 xTF0x ≥ 0 The implication xTF1x ≥ 0 = xTF0x ≥ 0 holds iff there exists τ≥ 0 s.t. F0 ≥ τF1. Note: For a given F1 ≥ 0, find Fo ≥ 0 and τ≥ 0 satisfying Fo ≥ τF1. Note: The S-procedure provides conditions under which a particular quadratic inequality is a consequence of another quadratic inequality. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 8 Let F1 = FT1 , F0 = F T 0 ∈ Rn×n, v1, v0 ∈ Rn, and α1, α0 ∈ R. Assume that there is some x s.t. xTF1x+ 2v T 1x+α1 > 0 holds. Then the implication xTF1x+ 2v T 1x+α1 ≥ 0 = xTF0x+ 2vT0x+α0 ≥ 0 holds iff if there exists τ≥ 0 s.t. F0 v0 vT0 α0 ≥ τ F1 v1 vT1 α1 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 9 e.g. Let Li(·) be quadratic fcns Li(ζ) = ζ TTiζ, Ti = T T i ∈ Rn×n, i= 0, . . . ,p When is it true that ζTTiζ≥ 0, i= 1, . . . ,p = ζTToζ > 0 (hard to find) Holds if there exist τi ≥ 0, i= 1, . . . ,p s.t. To > p∑ i=1 τiTi (LMIs in To and τi’s) S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 10 e.g. Stability of a system x˙ = Ax+ d(x), ∥d(x)∥ ≤ γ∥x∥ Let us use a quadratic Lyapunov fcn V(x) = xTPx to prove stability of the system We need P> 0 and V˙(x)≤ αV(x) for all x (α > 0 given): V˙(x) +αV(x) = 2xTP(Ax+ d(x)) +αxTPx = xT(ATP+ PA)x+ xTPd(x) + d(x)TPx+αxTPx = xT(ATP+ PA+αP)x+ d(x)TPx+ xTPd(x) = x d(x) T ATP+ PA+αP P P 0 x d(x) ≤ 0 whenever d(x)Td(x)≤ γ2xTx S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 11 So we need P> 0 and x d(x) ATP+ PA+αP P P 0 x d(x) ≥ 0 whenever x d(x) T γ2I 0 0 I x d(x) ≥ 0 (hard to find) By S-procedure, this happens iff ATP+ PA+αP P P 0 ≥ τ γ2I 0 0 I for some τ≥ 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 12 Thus, necessary and sufficient conditions for the existence of quadratic Lyapunov fcn. can be expressed as LMI P> 0, ATP+ PA+αP+τγ2I P P τI ≤ 0 in variables P and τ. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 13 e.g. Find P> 0 satisfying the constraints on P: For all ξ = 0 and P satisfying ξ η T ATP+ PA PB BTP 0 ξ η < 0, whenever ξ η T CTC 0 0 I ξ η ≤ 0. By S-procedure, this happens iff ATP+ PA PB BTP 0 < τ CTC 0 0 I ≤ 0 for some τ≥ 0. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 14 What if we have a matrix inequality such as X22 > 0 and X11 X12X 122 XT12 > 0 (nonlinear) Find its linear equivalent form. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 15 Good to Note: Matrix partition Let A= A11 A12 A21 A22 , then A= A11 A12 A21 A22 = I 0 A21A 1 11 I A11 0 0 I A 111A12 0 I (3.3a) where = A22 A21A 111A12, A11 nonsingular: (Schur complement of A11 in A) Or A= A11 A12 A21 A22 = I A12A 1 22 0 I 0 0 A22 I 0 A 122A21 I (3.3b) where = A11 A12A 122A21, A22 nonsingular: (Schur complement of A22 in A) S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 16 Let X = X ≥ 0 be partitioned as X = X11 X12 X 12 X22 If X+22 is the pseudo-inverse of X22, X11 X12 X 12 X22 = I X12X + 22 0 I X11 X12X+22X 12 0 0 X22 I 0 X+22X 12 I Find X11 X12 X 12 X22 ≥ 0 X11 X12X+22X 12 0 0 X22 ≥ 0 X11 X12X+22X 12 ≥ 0 X22 ≥ 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 17 ■ Schur complement: For matrices Q(x) = Q(x)T, R(x) = R(x)T, and S(x) depend affinely on x: R(x)> 0, Q(x) S(x)R(x) 1S(x)T > 0 Q(x) S(x) S(x)T R(x) > 0 Note: Q S ST R = I S 1R 0 I Q SR 1ST 0 0 R I 0 R 1S I If Q S ST R > 0, then so is the Schur complement of R in Q S ST R . Nonlinear (convex) inequalities can be converted to an LMI form using the Schur complement. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 18 Schur complement conditions: Let A> 0. Then A B BT C > 0 A> 0, C BTA 1B> 0. Let C > 0. Then A B BT C > 0 C > 0, A BC 1BT > 0. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 19 Proof: Consider the quadratic program inf (u,v) u v T A B BT C u v = uTAu+ 2uTBv+ vTCv The solution of the minimization problem satisfies: u v = 0 (and attained uniquely at 0) A B BT C > 0. First, solve min u u v T A B BT C u v = min u uTAu+ 2uTBv + vTCv The optimum u is d du = 2Au+ 2Bv= 0 u = A 1Bv S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 20 Observe uTAu= uTATu= ( A 1Bv)TAT( A 1Bv) = vTBTA 1Bv and uTBv= vTBTu= vTBTA 1Bv which leads to uTAu+ 2uTBv + vTCv= vT C BTA 1B v Now if we minimize over v, then the minimum is achieved and attained uniquely at 0 iff C BTA 1B> 0. □ S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 21 Good to Note: A block matrix A B BT C < 0 is equivalent to C < 0, A BC 1BT < 0 A B BT C > 0 or A< 0, C BTA 1B< 0 C BT B A < 0 C BT B A > 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 22 Extensions to the Schur complement: A BC 1BT Q QT R < 0 A B QBT C 0 QT 0 R < 0 C BT 0B A Q 0 QT R < 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 23 P Q QT A BC 1BT < 0 P Q 0QT A B 0 BT C < 0 P 0 Q0 C BT QT B A < 0 Exercise: Proof S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 24 ■ Redundant (Slack) Variables: Parametrization, Dilated LMI conditions Assume there exists X > 0 with I A X 0 0 X I AT = AXAT X < 0 Equivalent reformulation: There exist X and G with X AG GTAT X G GT < 0 New parametrization independent of the original variables Dilated synthesis approaches always encompass the corresponding standard-LMI-based ones S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 25 Equivalent reformulation I A X 0 0 X I AT + I A A I G 0 I I AT + I A 0 I GT AT I I AT = I A X AG GTAT X G GT I AT = I A X 0 0 X I AT = AXAT X S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 26 Exercise: Equivalent reformulation i) P> 0 s.t. ΦTPΦ P< 0 ii) P> 0 s.t. P ΦTP PΦ P < 0 iii)
P> 0 and G s.t. P ΦTGT GΦ P (G+GT) < 0 iv) P> 0 and G and F s.t. P ΦTF FTΦ FT +ΦTGT F+GΦ P (G+GT) < 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 27 Proof: ii) → iii) Apply Schur complement by choosing G= GT = P> 0 ΦTPΦ P< 0 = P ΦTP PΦ P < 0 = P ΦTGT GΦ P (G+GT) < 0 iii) → ii) I Φ T P ΦTGT GΦ P (G+GT) I Φ = ΦTPΦ P< 0 ii) → iv) P> 0 s.t. ΦTPΦ P< 0 = F s.t. ΦTPΦ P+ FP 1F < 0. ΦTPΦ P+ FP 1F = P ΦTF FTΦ+ (F+ PΦ)T P 1 (F+ PΦ) S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 28 Apply Schur complements by choosing G= GT = P> 0 as in the proof of ii) → iii) P ΦTF FTΦ+ (F+ PΦ)T P 1 (F+ PΦ)< 0 = P ΦTF FTΦ FT +ΦTGT F+GΦ P G+GT < 0 iv) → ii) I Φ T P ΦTF FTΦ FT +ΦTGT F+GΦ P (G+GT) I Φ = ΦTPΦ P< 0 □ S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 29 ■ Elimination of matrix variables: Lemma 3.1 (Projection Lemma) Given a symmetric matrix Ψ and two matrices U, V, there exist M such that Ψ +UMVT + VMTUT < 0 (3.4) iff the projection inequalities U TΨU < 0, V TΨV < 0 (3.5) hold, where U and V are orthogonal complements (bases of the null spaces) of U and V, respectively, i.e., U TU = 0 and V TV = 0. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 30 Proof: Using U TU = 0 and V TV = 0, we find U T Ψ +UMVT + VMTUT U = U TΨU + U TUMVTU + U TVMTUTU = U TΨU < 0 and V T Ψ +UMVT + VMTUT V = V TΨV + V TUMVTV + V TVMTUTV = V TΨV < 0 □ S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 31 ■ Generic LMI Problems: Feasibility problem (FEASP): Given an LMI F(x)> 0, find xfeas s.t. F(xfeas)> 0 or determine that the LMI is feasible. e.g. Given Ai ∈ Rn×n, i= 1, . . . ,n, find X satisfying the LMI X > 0, ATi X + XAi < 0, i= 1, . . . ,n or determine that no such X exists. Given Ai ∈ Rn×n, i= 1, . . . ,n, and γ > 0, find X satisfying ATi X + XAi + C TC XB BTX γI < 0, i= 1, . . . ,n, X > 0 or determine that no such X exists. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 32 Find X satisfying the LMI: AXAT X < 0, X > 0 Find a soln. X to N TL(X)N This feasibility problem is solved by the auxiliary problem: minimize t subject to: L(X) R(X)≤ t I Feasible iff tmin ≤ 0. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 33 Linear objective minimization problem (MINCX): Given an LMI F(x)> 0, minimize a linear objective with F(xmincx)> 0 being satisfied. e.g. minimize cTx subject to F(x)> 0 minimize tr (X) subject to F(X)> 0 minimize t subject to F(X)< t I (with t as a DV) S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 34 Generalized eigenvalue problem (GEVP): Given an LMI F(x,λ)> 0, minimize λ with F(xgevp,λ)> 0 being satisfied. e.g. minimize λ subject to λB(x) A(x)> 0, B(x)> 0, C(x)> 0 maximize α subject to ATX + XA+ 2αX < 0, X > 0 minimize γ subject to ATX + XA+ CTC XB BTX γI < 0, X > 0 minimize γ subject to L(X)< γR(X), L(X)> 0, R(X)> 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 35 Solving LMIPs Develop an LMI system: S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 36 Assign a name to the LMI system e.g. lmisys Describe the matrix variables and LMIs Write the commands as an m file. e.g. lmimodel.m S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 37 % Load an LMI model: lmimodel % in LMIEDIT, write commands as an m file lminbr(lmisys) % # LMIs matnbr(lmisys) % # matrix variables lmiinfo(lmisys) % info. about variables and term content % Choose a solver for generic LMIP: feasp/mincx/gevp % <==== Call feasp/mincx/gevp here % Solution matrix from the decision vector(s) Xopt=dec2mat(lmisys,xopt,X) % Evaluate LMIs to check if the DV(s) indeed feasible evlmi=evallmi(lmisys,xopt); [lhs,rhs]=showlmi(evlmi,1) % LHS and RHS S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 38 e.g. Choosing a solver Feasibility problem: feasp [tmin,xfeas]=feasp(lmisys); % feasible iff tmin<=0 Note: feasp solves minimize tmin for L(·) R(·)≤ tminI Linear objective problem: mincx ndc=decnbr(lmisys); % total # decision variables c=zeros(ndc,1); c(1:q)=nu; % for nu*x(1:q) c=zeros(ndc,1); c(diag(sX))=1; % for trace(X) [cxopt,xopt]=mincx(lmisys,c); Generalized eigenvalue problem: gevp [mu,xopt]=gevp(lmisys,nlfc); % last nlfc LMIs involving obj. Note: gevp solves minimize γ for L(·)< γR(·) with R(·)> 0. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 39 Note: “Solve” means that we can find DVs that satisfy the LMIP, or determine that no solns. exist. Note: The dimension of the variable Y = YT ∈ Rn×n is n(n+ 1)/2: Y11 Y12 Y22 Y13 … Ynn ∈ R n(n+1)/2 or Y11 Y21 Y22 Y31 … Ynn ∈ R n(n+1)/2 Note: The dimension of the variable Y ∈ Rn×n is n2. S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 40 e.g. >> n=3; setlmis([]) >> [X,nX,sX]=lmivar(1,[n 1]) X = 1 nX = 6 sX = 1 2 4 2 3 5 4 5 6 >> [Y,nY,sY]=lmivar(2,[n n]) Y = 2 nY = 15 sY = 7 8 9 10 11 12 13 14 15 >> S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 41 e.g. Linear objective Given A1 = 1 2 1 3 , A2 = 0.8 1.5 1.3 2.7 Find Y > I to minimize trY satisfying I < Y AT1Y + YA1 < Y AT2Y + YA2 < Y S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 42 A1=[-1 2; 1 -3]; A2=[-1.4 0.9; 0.7 -2]; setlmis([]); [Y,nY,sY]=lmivar(1,[2 1]); lmiterm([1 1 1 0],1); % LMI #1: I lmiterm([-1 1 1 Y],1,1); % LMI #1: Y lmiterm([2 1 1 Y],A1,1); % LMI #2: A1*Y lmiterm([2 1 1 Y],1,A1’); % LMI #2: Y*A1’ lmiterm([-2 1 1 Y],1,1); % LMI #2: Y lmiterm([3 1 1 Y],A2,1,’s’); % LMI #3: A2*Y+Y*A2’ (with FLAG=’s’) lmiterm([-3 1 1 Y],1,1); % LMI #3: Y lmis=getlmis; S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 43 >> ndc=decnbr(lmis); >> c=zeros(ndc,1); c(diag(sY))=1; >> [cyopt,yopt]=mincx(lmis,c); >> cyopt cyopt = 2.0012 >> Yopt=dec2mat(lmis,yopt,Y) Yopt = 1.0005 0.0000 0.0000 1.0007 >> evlmi=evallmi(lmis,yopt); >> [lhs,rhs]=showlmi(evlmi,3) lhs = -2.8013 1.6008 1.6008 -4.0027 rhs = 1.0005 0.0000 0.0000 1.0007 >> eig(lhs-rhs) ans = -6.1123 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 44 -2.6928 >> eig(A2*Yopt+Yopt*A2’-Yopt) ans = -6.1123 -2.6928 >> >> [lhs,rhs]=showlmi(evlmi,2) lhs = -2.0008 3.0017 3.0017 -6.0040 rhs = 1.0005 0.0000 0.0000 1.0007 >> eig(lhs-rhs) ans = -8.6108 -1.3951 >> eig(A1*Yopt+Yopt*A1’-Yopt) ans = -8.6108 -1.3951 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 45 Note: To solve a FEASP, we need to write LMI conditions L(X)The FEASP solver finds tmin ≤ 0 to guarantee L(X) R(X)< tminI ≤ 0 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 46 e.g. Feasibility Given A1 = 1 2 1 3 , A2 = 0.8 1.5 1.3 2.7 Find Y > I satisfying I < Y AT1Y + YA1 < Y AT2Y + YA2 < Y S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 47 >> [tmin,yopt]=feasp(lmis); >> tmin tmin = -2.9180 >> Yopt=dec2mat(lmis,yopt,Y) Yopt = 99.9899 27.8674 27.8674 54.9808 >> evlmi=evallmi(lmis,yopt); >> [lhs,rhs]=showlmi(evlmi,3) lhs = -229.8104 24.7265 24.7265 -180.9089 rhs = 99.9899 27.8674 27.8674 54.9808 >> eig(lhs-rhs) ans = -329.9052 -235.7848 >> eig(A2*Yopt+Yopt*A2’-Yopt) S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 48 ans = -329.9052 -235.7848 >> >> [lhs,rhs]=showlmi(evlmi,2) lhs = -88.5102 98.4820 98.4820 -274.1501 rhs = 99.9899 27.8674 27.8674 54.9808 >> eig(lhs-rhs) ans = -358.4683 -159.1628 >> eig(A1*Yopt+Yopt*A1’-Yopt) ans = -358.4683 -159.1628 >> S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 49 Note: To solve a GEVP, we need to write LMI conditions L(X) 0. The GEVP solver finds λmin to guaaraantee L(X)< λminR(X). S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 50 e.g. Generalized eigenvalue problem Given A1 = 1 2 1 3 , A2 = 0.8 1.5 1.3 2.7 Find Y > I to minimize λ: I < Y AT1Y + YA1 < λY AT2Y + YA2 < λY S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III - 51 >> [lambda,yopt]=gevp(lmis,2); % the last 2 lmis are LFTs >> lambda lambda = -0.5359 >> Yopt=dec2mat(lmis,yopt,Y) Yopt = 493.1196 -671.1353 -671.1353 917.6951 >> evlmi=evallmi(lmis,yopt); >> [lhs,rhs]=showlmi(evlmi,3) lhs = 1.0e+03 * -2.5888 3.4530 3.4530 -4.6104 rhs = 493.1196 -671.1353 -671.1353 917.6951 >> eig(lhs-lambda*rhs) ans = 1.0e+03 * -6.4423 S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 52 -0.0008 >> eig(A2*Yopt+Yopt*A2’-lambda*Yopt) ans = 1.0e+03 * -6.4423 -0.0008 >> [lhs,rhs]=showlmi(evlmi,2) lhs = 1.0e+03 * -3.6708 5.0131 5.0131 -6.8484 rhs = 493.1196 -671.1353 -671.1353 917.6951 >> eig(lhs-lambda*rhs) ans = 1.0e+03 * -9.7632 -0.0000 >> eig(A1*Yopt+Yopt*A1’-lambda*Yopt) ans = S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 53 1.0e+03 * -9.7632 -0.0000 >> S.-H. Lee : ODC / LMIs Elec. Engr., Hanyang University III – 54