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Lyapunov Operator Let A ∈ Fn×n be given, and define a linear operator LA : Cn×n → Cn×n as LA (X) := A∗X + XA Suppose A is diagonalizable (what follows can be generalized even if this is not possible - the qualitative results still hold). Let {λ1 , λ2 , . . . , λn } be the eigenvalues of A. In general, these are complex numbers. Let V ∈ Cn×n be the matrix of eigenvectors for A∗ , so V = [v1 v2 · · · vn ] ∈ Cn×n is invertible, and for each i A∗vi = λ̄i vi Let wi ∈ Cn be defined so that V −1 = w1T w2T .. . wnT Since V −1V = In, we have that wiT vj = δij := For each 1 ≤ i, j ≤ n define 0 1 if i 6= j if i = j Xij := vivj∗ ∈ Cn×n 130 Eigenvalues Lyapunov Operator Lemma 63 The set {Xij }i=1,.,n;j=1,.,n is a linearly independent set, and this set is a full set of eigenvectors for the linear operator LA . Moreover, the o n eigenvalues of LA is the set of complex numbers λ̄i + λj i=1,.,n;j=1,.,n Proof: Suppose {αij }i=1,.,n;j=1,.,n are scalars and n X n X αij Xij = 0n×n (7.19) j=1 i=1 Premultiply this by wlT and postmultiply by w̄k 0 = wlT = = n n X X αij Xij w̄k j=1 i=1 n X n X αij wlT vivj∗ w̄k j=1 i=1 n n X X αij δliδkj j=1 i=1 = αlk which holds for any 1 ≤ k, l ≤ n. Hence, all of the α’s are 0, and the set {Xij }i=1,.,n;j=1,.,n is indeed a linearly independent set. Also, by definition of LA , we have LA (Xij ) = A∗Xij + Xij A = A∗vi vj∗ + vi vj∗A = λ̄i vivj∗ + vi λj vj∗ = = λ̄i + λj vi vj∗ λ̄i + λj Xij So, Xij is indeed an eigenvector of LA with eigenvalue λ̄i + λj . ♯ 131 Lyapunov Operator Alternate Invertibility Proof Do a Schur decomposition of A, namely A = QΛQ∗ , with Q unitary and Λ upper triangular. Then the equation A∗X + XA = R can be written as A∗X + XA = R ↔ QΛ∗Q∗ X + XQΛQ∗ = R ↔ Λ∗ Q∗XQ + Q∗ XQΛ = Q∗RQ ↔ Λ∗ X̃ + X̃Λ = R̃ Since Q is invertible, the redefinition of variables is an invertible transformation. Writing out all of the equations (we are trying to get that for every R̃, there is a unique X̃ satisfying the equation) yields a upper triangular system with n2 unknowns, and the values λi + λ¯j on the diagonal. Do it. ♯ This is an easy (and computationally motivated) proof that the Lyapunov operator LA is an invertible operator if and only if λi + λ̄j 6= 0 for all eigenvalues of A. 132 Lyapunov Operator Integral Formula If A is stable (all eigenvalues have negative real parts) then λ̄i + λj 6= 0 for all combinations. Hence LA is an invertible linear operator. We can explicitly solve the equation LA (X) = −Q for X using an integral formula. Theorem 64 Let A ∈ Fn×n be given, and suppose that A is stable. The LA is invertible, and for any Q ∈ Fn×n, the unique X ∈ Fn×n solving LA (X) = −Q is given by Z ∞ ∗ X= eA τ QeAτ dτ 0 Proof: For each t ≥ 0, define S(t) := Z t ∗ eA τ QeAτ dτ 0 Since A is stable, the integrand is made up of decaying exponentials, and S(t) has a well-defined limit as t → ∞, which we have already denoted X := lim S(t). For any t ≥ 0, t→∞ ∗ Z t ∗ A∗ τ Aτ A∗ τ A e Qe + e d A∗ τ Aτ = dτ e Qe 0 dτ ∗ = eA t QeAt − Q A S(t) + S(t)A = 0 Z t Taking limits on both sides gives A∗X + XA = −Q as desired. ♯ 133 Aτ Qe A dτ Lyapunov Operator More A few other useful facts Theorem 65 Suppose A, Q ∈ Fn×n are given. Assume that A is stable and Q = Q∗ ≥ 0. Then the pair (A, Q) is observable if and only if X := Z ∞ ∗ eA τ QeAτ dτ 0 >0 Proof: ⇐ Suppose not. Then there is an xo ∈ Fn, xo 6= 0 such that QeAt xo = 0 ∗ for all t ≥ 0. Consequently x∗o eA t QeAt xo = 0 for all t ≥ 0. Integrating gives R ∗ 0 = 0∞ x∗o eA t QeAt xo dt = x∗o R ∞ A∗ t At e Qe dt xo 0 = x∗o Xxo Since xo 6= 0, X is not positive definite. ⇒ Suppose that X is not positive definite (note by virtue of the definition X is at least positive semidefinite). Then there exists xo ∈ Fn, xo 6= 0 such that x∗o Xxo = 0. Using the integral form for X (since A is stable by assumption) and the fact tha Q ≥ 0 gives Z ∞ 1 Q 2 eAτ x dτ o 0 =0 The integrand is nonnegative and continuous, hence it must be 0 for all τ ≥ 0. Hence QeAτ xo = 0 for all τ ≥ 0. Since xo 6= 0, this implies that (A, Q) is not observable. ♯ 134 Lyapunov Operator More Theorem 66 Let (A, C) be detectable and suppose X is any solution to A∗X + XA = −C ∗C (there is no apriori assumption that LA is an invertible operator, hence there could be multiple solutions). Then X ≥ 0 if and only if A is stable. Proof ⇒ Suppose not, then there is a v ∈ Cn , v 6= 0, λ ∈ C, Re (λ) ≥ 0 with Av = λv (and v ∗ A∗ = λ̄v ∗). Since (A, C) is detectable and the eigenvalue in the closed right-half plane Cv 6= 0. Note that − kCvk2 = −v ∗ C ∗Cv = v ∗ (A∗ X + XA) v = λ̄ + λ v ∗ Xv = 2Re (λ) v ∗ Xv Since kCvk > 0 and Re (λ) ≥ 0, we must actually have Re (λ) > 0 and v ∗Xv < 0. Hence X is not positive semidefinite. ⇐ Since A is stable, there is only one solution to the equation A∗X + XA = −C ∗C, moreover it is given by the integral formula, X= Z ∞ ∗ eA τ C ∗CeAτ dτ 0 which is clearly positive semidefinite. ♯ 135 Lyap Ordering Finally, some simple results about the ordering of Lyapunov solutions. Let A be stable, so LA is guaranteed to be invertible. Use L−1 A (−Q) to ∗ denote the unique solution X to the equation A X + XA = −Q. Lemma 67 If Q∗1 = Q1 ≥ Q2 = Q∗2, then −1 L−1 A (−Q1 ) ≥ LA (−Q2 ) Lemma 68 If Q∗1 = Q1 > Q2 = Q∗2, then −1 L−1 A (−Q1 ) > LA (−Q2 ) Proofs Let X1 and X2 be the two solutions, obviously A∗ X1 + X1 A = −Q1 A∗ X2 + X2 A = −Q2 A∗ (X1 − X2 ) + (X1 − X2 ) A = − (Q1 − Q2) Since A is stable, the integral formula applies, and X1 − X2 = Z ∞ ∗ eA τ 0 (Q1 − Q2) eAτ dτ Obviously, if Q1 − Q2 ≥ 0, then X1 − X2 ≥ 0, which proves Lemma 5. If Q1 − Q2 > 0, then the pair (A, Q1 − Q2) is observable, and by Theorem 3, X1 − X2 > 0, proving Lemma 6. ♯ Warning: The converses of Lemma 5 and Lemma 6 are NOT true. 136 Algebraic Riccati Equation The algebraic Riccati equation (A.R.E.) is AT X + XA + XRX − Q = 0 (7.20) where A, R, Q ∈ Rn×n are given matrices, with R = RT , and Q = QT , and solutions X ∈ Cn×n are sought. We will primarily be interested in solutions X of (7.20), which satisfy A + RX is Hurwitz and are called stabilizing solutions of the Riccati equation. Associated with the data A, R, and Q, we define the Hamiltonian matrix H ∈ R2n×2n by A R (7.21) H := Q −AT 137 Invariant Subspace Definition Suppose A ∈ Cn×n. Then (via matrix-vector multiplication), we view A as a linear map from Cn → Cn . A subspace V ⊂ Cn is said to be “A-invariant” if for every vector v ∈ V, the vector Av ∈ V too. In terms of matrices, suppose that the columns of W ∈ Cn×m are a basis for V. In other words, they are linearly independent, and span the space V, so V = {W η : η ∈ Cm } Then V being A-invariant is equivalent to the existence of a matrix S ∈ Cm×m such that AW = W S The matrix S is the matrix representation of A restricted to V, referred to the basis W . Clearly, since W has linearly independent columns, the matrix S is unique. If (Q, Λ) are the Schur decomposition of A, then for any k with 1 ≤ k ≤ n, the first k columns of Q span a k-dimensional invariant subspace of A, as seen below Λ Λ 11 12 A [Q1 Q2 ] = [Q1 Q2] 0 Λ22 Clearly, AQ1 = Q1Λ11 . 138 Eigenvectors in Invariant subspaces Note that since S ∈ Cm×m has at least one eigenvector, say ξ 6= 0m, with associated eigenvalue λ, it follows that W ξ ∈ V is an eigenvector of A, A (W ξ) = AW ξ = W Sξ = λW ξ = λ(W ξ) Hence, every nontrivial invariant subspace of A contains an eigenvector of A. 139 Stable Invariant Subspace Definition If S is Hurwitz, then V is a “stable, A-invariant” subspace. Note that while the matrix S depends on the particular choice of the basis for V (ie., the specific columns of W ), the eigenvalues of S are only dependent on the subspace V, not the basis choice. To see this, suppose that the columns of W̃ ∈ Cn×m span the same space as the columns of W . Obviously, there is a matrix T ∈ Cm×m such that W = W̃ T . The linear independence of the columns of W (and W̃ ) imply that T is invertible. Also, −1 AW̃ = AW T −1 = W ST −1 = W̃ T {z } | ST S̃ Clearly, the eigenvalues of S and S̃ are the same. The linear operator A|V is the transformation from V → V defined as A, simply restricted to V. Since V is A-invariant, this is well-defined. Clearly, S is the matrix representation of this operator, using the columns of W as the basis choice. 140 AT X + XA + XRX − Q = 0 Invariant Subspace The first theorem gives a way of constructing solutions to (7.20) in terms of invariant subspaces of H. Theorem 69 Let V ⊂ C2n be a n-dimensional invariant subspace of H (H is a linear map from C2n to C2n ), and let X1 ∈ Cn×n, X2 ∈ Cn×n be two complex matrices such that V = span X1 X2 If X1 is invertible, then X := X2X1−1 is a solution to the algebraic Riccati equation, and spec (A + RX) = spec (H|V ). Proof: Since V is H invariant, there is a matrix Λ ∈ Cn×n with which gives A R Q −AT X1 X2 = X1 X2 Λ (7.22) AX1 + RX2 = X1 Λ (7.23) QX1 − AT X2 = X2Λ (7.24) and Pre and post multiplying equation (7.23) by X2X1−1 and X1−1 respectively gives X2 X1−1 A + X2X1−1 R X2 X1−1 = X2 ΛX1−1 (7.25) and postmultiplying 7.24 by X1−1 gives Q − AT X2 X1−1 = X2ΛX1−1 (7.26) Combining (7.25) and (7.26) yields AT (X2X1−1 ) + (X2X1−1 )A + (X2 X1−1)R(X2X1−1 ) − Q = 0 so X2 X1−1 is indeed a solution. Equation (7.23) implies that A+R X2 X1−1 = X1 ΛX1−1 , and hence they have the same eigenvalues. But, by definition, Λ is a matrix representation of the map H|V , so the theorem is proved. ♯ 141 AT X + XA + XRX − Q = 0 Invariant Subspace The next theorem shows that the specific choice of matrices X1 and X2 which span V is not important. Theorem 70 Let V, X1, and X2 be as above, and suppose there are two X̃1 also span other matrices X̃1 , X̃2 ∈ Cn×n such that the columns of X̃2 V. Then X1 is invertible if and only if X̃1 is invertible. Consequently, both X2 X1−1 and X̃2 X̃1−1 are solutions to (7.20), and in fact they are equal. Remark: Hence, the question of whether or not X1 is invertible is a property of the subspace, and not of the particular basis choice for the subspace. Therefore, we say that an n-dimensional subspace V has the invertibility property if in any basis, the top portion is invertible. In the literature, this is often called the complimentary property. Proof: Since both sets of columns span the same n-dimensional subspace, there is an invertible K ∈ Cn×n such that X1 X2 = X̃1 X̃2 K Obviously, X1 is invertible if and only if X̃1 is invertible, and X2 X1−1 = −1 X̃2 K X̃1 K = X̃2 X̃1−1 . ♯. 142 AT X + XA + XRX − Q = 0 Invariant Subspace As we would hope, the converse of theorem 69 is true. Theorem 71 If X ∈ Cn×n is a solution to the A.R.E., then there exist matrices X1, X2 ∈ Cn×n , with X1 invertible, such that X = X2 X1−1 and the X1 span an n-dimensional invariant subspace of H. columns of X2 Proof: Define Λ := A + RX ∈ Cn×n. Multiplying this by X gives XΛ = XA + XRX = Q − AT X, the second equality coming from the fact that X is a solution to the Riccati equation. We write these two relations as Hence, the columns of A R Q −A In T In X = In X Λ span an n-dimensional invariant subspace X of H, and defining X1 := In , and X2 := X completes the proof. ♯. Hence, there is a one-to-one correspondence between n-dimensional invariants subspaces of H with the invertibility property, and solutions of the Riccati equation. 143 AT X + XA + XRX − Q = 0 Eigenvalue distribution of H In the previous section, we showed that any solution X of the Riccati equation is intimately related to an invariant subspace, V, of H, and the matrix A + RX is similiar (ie. related by a similarity transformation) to the restriction of H on V, H|V . Hence, the eigenvalues of A + RX will always be a subset of the eigenvalues of H. Recall the particular structure of the Hamiltonian matrix H, A R H := T Q −A Since H is real, we know that the eigenvalues of H are symmetric about the real axis, including multiplicities. The added structure of H gives additional symmetry in the eigenvalues as follows. Lemma 72 The eigenvalues of H are symmetric about the origin (and hence also about the imaginary axis). Proof: Define J ∈ R2n×2n by J = 0 −I I 0 Note that JHJ −1 = −H T . Also, JH = −H T J = (JH)T . Let p (λ) be the characteristic polynomial of H. Then p(λ) := det (λI − H) = det λI − JHJ −1 = det λI + H T = det (λI + H) (7.27) = (−1)2ndet (−λI − H) = p(−λ) Hence the roots of p are symmetric about the origin, and the symmetry about the imaginary axis follows. ♯ 144 AT X + XA + XRX − Q = 0 Hermitian Solutions Lemma 73 If H has no eigenvalues on the imaginary axis, then H has n eigenvalues in the open-right-half plane, and n eigenvalues in the open-lefthalf plane. Theorem 74 Suppose V is a n-dimensional invariant subspace of H, and X1 , X2 ∈ Cn×n form a matrix whose colums span V. Let Λ ∈ Cn×n be a matrix representation of H|V . If λi + λj 6= 0 for all eigenvalues λi , λj ∈ spec(Λ), then X1∗ X2 = X2∗ X1 . Proof: Using the matrix J from the proof of Lemma 72 we have that [X1∗ X2∗ ] JH X1 X2 is Hermitian, and equals (−X1∗ X2 + X2∗ X1 ) Λ. Define W := −X1∗ X2 + X2∗ X1 . Then W = −W ∗, and W Λ + Λ∗ W = 0. By the eigenvalue assumption on Λ, the linear map LΛ : Cn×n → Cn×n defined by LΛ (X) = XΛ + Λ∗X is invertible, so W = 0. ♯ Corollary 75 Under the same assumptions as in theorem 74, if X1 is invertible, then X2 X1−1 is Hermitian. 145 AT X + XA + XRX − Q = 0 Real Solutions Real (rather than complex) solutions arise when a symmetry condition is imposed on the invariant subspace. Theorem 76 Let V be a n-dimensional invariant subspace of H with the invertibility property, and let X1, X2 ∈ Cn×n form a 2n × n matrix whose columns span V. Then V is conjugate symmetric ({v̄ : v ∈ V} = V) if and only if X2 X1−1 ∈ Rn×n . Proof: ← Define X := X2 X1−1 . By assumption, X ∈ Rn×n , and span In X =V so V is conjugate symmetric. → Since V is conjugate symmetric, there is an invertible matrix K ∈ Cn×n such that X1 X2 = X1 X2 K Therefore, (X2X1−1 ) = X2X1−1 = X2 K(X1K)−1 = X2 X1−1 as desired. ♯ 146 AT X + XA + XRX − Q = 0 Stabilizing Solutions Recall that for any square matrix M, there is a “largest” stable invariant subspace. That is, there is a stable invariant subspace VS∗ that contains every other stable invariant subspace of M. This largest stable invariant subspace is simply the span of all of the eigenvectors and generalized eigenvectors associated with the open-left-half plane eigenvalues. If the Schur form has the eigenvalues (on diagonal of Λ) ordered, then it is spanned by the “first” set of columns of Q Theorem 77 There is at most one solution X to the Riccati equation such that A + RX is Hurwitz, and if it does exist, then X is real, and symmetric. Proof: If X is such a stabilizing solution, then by theorem 71 it can be constructed from some n-dimensional invariant subspace of H, V. If A + RX is stable, then we must also have spec (H|V ). Recall that the spectrum of H is symmetric about the imaginary axis, so H has at ◦ most n eigenvalues in C− , and the only possible choice for V is the stable eigenspace of H. By theorem 70, every solution (if there are any) constructed from this subspace will be the same, hence there is at most 1 stabilizing solution. Associated with the stable eigenspace, any Λ as in equation (7.22) will be stable. Hence, if the stable eigenspace has the invertibility property, from corollary 75 we have X := X2X1−1 is Hermitian. Finally, since H is real, the stable eigenspace of H is indeed conjugate symmetric, so the associated solution X is in fact real, and symmetric. 147 AT X + XA + XRX − Q = 0 domS Ric If H is (7.21) has no imaginary axis eigenvalues, then H has a unique ndimensional stable invariant subspace VS ⊂ C2n, and since H is real, there exist matrices X1 , X2 ∈ Rn×n such that span X1 X2 = VS . More concretely, there exists a matrix Λ ∈ Rn×n such that H X1 X2 = X1 X2 Λ ◦ spec (Λ) ⊂ C− , This leads to the following definition Definition 78 (Definition of domS Ric) Consider H ∈ R2n×2n as defined in (7.21). If H has no imaginary axis eigenvalues, and the matrices X1 , X2 ∈ Rn×n for which span X1 X2 is the stable, n-dimensional invariant subspace of H satisfy det (X1 ) 6= 0, then H ∈ domS Ric. For H ∈ domS Ric, define RicS (H) := X2 X1−1 , which is the unique matrix X ∈ Rn×n such that A + RX is stable, and AT X + XA + XRX − Q = 0. Moreover, X = X T . 148 AT X + XA + XRX − Q = 0 Useful Lemma Theorem 79 Suppose H (as in (7.21)) does not have any imaginary axis eigenvalues, R is either positive semidefinite, or negative semidefinite, and (A, R) is a stabilizable pair. Then H ∈ domS Ric. X1 form a basis for the stable, n-dimensional invariant subX2 space of H (this exists, by virtue of the eigenvalue assumption about H). Then X X A R 1 Λ 1 = (7.28) X2 X2 Q −AT Proof: Let where Λ ∈ Cn×n is a stable matrix. Since Λ is stable, it must be that X1∗ X2 = X2∗ X1 . We simply need to show X1 is invertible. First we prove that KerX1 is Λ-invariant, that is, the implication x ∈ KerX1 → Λx ∈ KerX1 Let x ∈ KerX1 . The top row of (7.28) gives RX2 = X1Λ − AX1. Premultiply this by x∗X2∗ to get x∗X2∗ RX2 x = x∗X2∗ (X1 Λ − AX1 ) x = x∗X2∗ X1Λx = x∗X1∗ X2Λx since X2∗ X1 = X1∗ X2 . = 0 Since R ≥ 0 or R ≤ 0, it follows that RX2 x = 0. Now, using (7.28) again gives X1 Λx = 0, as desired. Hence KerX1 is a subspace that is Λ-invariant. Now, if KerX1 6= {0}, then there is a vector v 6= 0 and λ ∈ C, Re (λ) < 0 such that X1 v = 0 Λv = λv RX2 v = 0 149 We also have QX1 − AT X2 = X2 Λ, from the second row of (7.28). Mutiplying by v ∗ gives v ∗ X2∗ h A − −λ̄I i R =0 Since Re λ̄ < 0, and (A, R) is assumed stabilizable, it must be that X1 is full-column rank. Hence, v = X2 0, and KerX1 is indeed the trivial set {0}. This implies that X1 is invertible, and hence H ∈ domS Ric. ♯. X2 v = 0. But X1 v = 0, and 150 AT X + XA + XRX − Q = 0 Quadratic Regulator The next theorem is a main result of LQR theory. Theorem 80 Let A, B, C be given, with (A, B) stabilizable, and (A, C) detectable. Define T A −BB H := −C T C −AT Then H ∈ domS Ric. Moreover, 0 ≤ X := RicS (H), and KerX ⊂ C CA .. . CAn−1 Proof: Use stabilizablity of (A, B) and detectability of (A, C) to show that H has no imaginary axis eigenvalues. Specifically, suppose v1, v2 ∈ Cn and ω ∈ R such that A T −C C −BB T −A T v1 v2 = jω v1 v2 Premultiply top and bottom by v2∗ and v1∗, combine, and conclude v1 = T v2 = 0. Show that if (A, B) is stabilizable, then A, −BB is also stabilizable. At this point , apply Theorem 79 to conclude H ∈ domS Ric. Then, rearrange the Riccati equation into a “Lyapunov” equation and use Theorem 64 to show that X ≥ 0. Finally, show that KerX ⊂ KerC, and that KerX is A-invariant. That shows that KerX ⊂ KerCAj for any j ≥ 0. ♯. 151 Derivatives Linear Maps Recall that the derivative of f at x is defined to be the number lx such that f (x + δ) − f (x) − lx δ =0 δ→0 δ lim Suppose f : Rn → Rm is differentiable. Then f ′(x) is the unique matrix Lx ∈ Rm×n which satisfies 1 (f (x + δ) − f (x) − Lx δ) = 0 δ→0n kδk lim Hence, while f : Rn → Rm , the value of f ′ at a specific location x, f ′ (x), is a linear operator from Rn → Rm . Hence the function f ′ is a map from Rn → L (Rn , Rm ). Similarly, if f : Hn×n → Hn×n is differentiable, then the derivative of f at x is the unique linear operator LX : Hn×n → Hn×n such that satisfying 1 (f (X + ∆) − f (X) − LX ∆) = 0 ∆→0n×n k∆k lim 152 Derivative of Riccati Lyapunov f (X) := XDX − A∗ X − XA − C Note, f (X + ∆) = (X + ∆) D (X + ∆) − A∗ (X + ∆) − (X + ∆) A − C = XDX − A∗X − XA − C + ∆DX + XD∆ − A∗ ∆ − ∆A + ∆D∆ = f (X) − ∆ (A − DX) − (A − DX)∗ ∆ + ∆D∆ For any W ∈ Cn×n , let LW : Hn×n → Hn×n be the Lyapunov operator LW (X) := W ∗ X + XW . Using this notation, f (X + ∆) − f (X) − L−A+DX (∆) = ∆D∆ Therefore 1 1 [f (X + ∆) − f (X) − L−A+DX (∆)] = lim [∆D∆] = 0 ∆→0n×n k∆k ∆→0n×n k∆k lim This means that the Lyaponov operator L−A+DX is the derivative of the matrix-valued function XDX − A∗ X − XA − C at X. 153 Iterative Algorithms Roots Suppose f : R → R is differentiable. An iterative algorithm to find a root of f (x) = 0 is f (xk ) xk+1 = xk − ′ f (xk ) Suppose f : Rn → Rn is differentiable. Then to first order f (x + δ) = f (x) + Lx δ An iterative algorithm to find a “root” of f (x) = 0n is obtained by setting the first-order approximation expression to zero, and solving for δ, yielding the update law (with xk+1 = xk + δ) xk+1 = xk − L−1 xk [f (xk )] Implicitly, it is written as Lxk [xk+1 − xk ] = −f (xk ) (7.29) For the Riccati equation f (X) := XDX − A∗ X − XA − C we have derived that the derivative of f at X involves the Lyapunov operator. The iteration in (7.29) appears as L−A+DXk (Xk+1 − Xk ) = −Xk DXk + A∗Xk + Xk A + C In detail, this is − (A − DXk )∗ (Xk+1 − Xk )−(Xk+1 − Xk ) (A − DXk ) = −Xk DXk +A∗Xk +Xk A+C which simplifies (check) to (A − DXk )∗ Xk+1 + Xk+1 (A − DXk ) = −Xk DXk − C Of course, questions about the well-posedness and convergence of the iteration must be addressed in detail. 154 Comparisons Gohberg, Lancaster and Rodman Theorems 2.1-2.3 in “On Hermitian Solutions of the Symmetric Algebraic Riccati Equation,” by Gohberg, Lancaster and Rodman, SIAM Journal of Control and Optimization, vol. 24, no. 6, Nov. 1986, are the basis for comparing equality and inequality solutions of Riccati equations. The definitions, theorems and main ideas are below. Suppose that A ∈ Rn×n , C ∈ Rn×n , D ∈ Rn×n , with D = DT 0, C = C T and (A, D) stabilizable. Consider the matrix equation XDX − AT X − XA − C = 0 (7.30) Definition 81 A symmetric solution X+ = X+T of (9.39) is maximal if X+ X for any other symmetric solution X of (9.39). Check: Maximal solutions (if they exist) are unique. Theorem 82 If there is a symmetric solution to (9.39), then there is a maximal solution X+ . The maximal solution satisfies max Reλi (A − DX+ ) ≤ 0 i Theorem 83 If there is a symmetric solution to (9.39), then there is a T sequence {Xj }∞ j=1 such that for each j, Xj = Xj and 0 Xj DXj − AT Xj − Xj A − C Xj X for any X = X T solving 9.39 A − DXj is Hurwitz limj→∞ Xj = X+ Theorem 84 If there are symmetric solutions to (9.39), then for every symmetric C̃ C, there exists symmetric solutions (and hence a maximal solution X̃+) to XDX − AT X − XA − C̃ = 0 Moreover, the maximal solutions are related by X̃+ X+ . 155 Identities For any Hermitian matrices Y and Ŷ , it is trivial to verify Y (A − DY ) + (A − DY )∗Y + Y DY ∗ = Y (A − DŶ ) + (A − DŶ ) Y + Ŷ DŶ − Y − Ŷ D Y − Ŷ (7.31) Let X denote any Hermitian solution (if one exists) to the ARE, XDX − A∗ X − XA − C = 0 (7.32) This (the existence of a hermitian solution) is the departure point for all that will eventually follow. For any hermitian matrix Z −C = −XDX + A∗X + XA = X (A − DX) + (A − DX)∗ X + XDX = X (A − DZ) + (A − DZ)∗ X + ZDZ − (X − Z) D (X − Z) (7.33) This exploits that X solves the original Riccati equation, and uses the identity (7.31), with Y = X, Ŷ = Z. 156 Iteration Take any C̃ C (which could be C, for instance). Also, let X0 = X0∗ 0 be chosen so that A − DX0 is Hurwitz. Check: Why is this possible? Iterate (if possible, which we will show is...) as Xν+1 (A − DXν ) + (A − DXν )∗ Xν+1 = −Xν DXν − C̃ (7.34) Note that by the Hurwitz assumption on A − DX0, at least the first iteration is possible (ν = 0). Also, recall that the iteration is analogous to a first-order n) algorithm for root finding, xn+1 := xn − ff′(x (xn ) . Use Z := Xν in equation (7.33), and subtract from the iteration equation (7.34), leaving (Xν+1 − X) (A − DXν ) + (A − DXν )∗ (Xν+1 − X) = − (X − Xν ) D (X − Xν ) − C̃ − C which is valid whenever the iteration equation is valid. 157 (7.35) Main Induction Now for the induction: assume Xν = Xν∗ and A − DXν is Hurwitz. Note that this is true for ν = 0. Since A−DXν is Hurwitz, there is a unique solution Xν+1 to equation (7.34). Moreover, since A−DXν is Hurwitz, and the right-hand-side of (7.35) is 0, it follows that Xν+1 − X 0 (note that X0 − X is not necessarily positive semidefinite). Rewrite (7.31) with Y = Xν+1 and Ŷ = Xν . Xν+1 (A − DXν+1 ) + (A − DXν+1 )∗ Xν+1 + Xν+1 DXν+1 = Xν+1 (A − DXν ) + (A − DXν )∗ Xν+1 + Xν DXν − (Xν+1 − Xν ) D (Xν+1 − Xν ) | {z −C̃ Rewriting, } −C̃ = Xν+1 (A − DXν+1 ) + (A − DXν+1 )∗ Xν+1 + Xν+1 DXν+1 + (Xν+1 − Xν ) D (Xν+1 − Xν ) Writing (7.33) with Z = Xν+1 gives −C = X (A − DXν+1 ) + (A − DXν+1 )∗ X + Xν+1 DXν+1 − (X − Xν+1) D (X − Xν+1 ) Subtract these, yielding − C̃ − C = (Xν+1 − X) (A − DXν+1 ) + (A − DXν+1 )∗ (Xν+1 − X) + (Xν+1 − Xν ) D (Xν+1 − Xν ) + (X − Xν+1 ) D (X − Xν+1 ) (7.36) Now suppose that Re (λ) ≥ 0, and v ∈ Cn such that (A − DXν+1 ) v = λv. Pre and post-multiply (7.36) by v ∗ and v respectively. This gives −v ∗ C̃ − C v = 2Re(λ) v ∗ (Xν+1 − X) v + | {z ≤0 } | {z ≥0 } | {z ≥0 } | D1/2 (Xν+1 − Xν ) D1/2 (X − Xν+1 ) Hence, both sides are in fact equal to zero, and therefore v ∗ (Xν+1 − Xν ) D (Xν+1 − Xν ) = 0. 158 {z ≥0 2 v } Since D 0, it follows that D (Xν+1 − Xν ) v = 0, which means DXν+1 v = DXν v. Hence λv = (A − DXν+1 ) v = (A − DXν ) v But A − DXν is Hurwitz, and Re(λ) ≥ 0, so v = 0n as desired. This means A − DXν+1 is Hurwitz. Summarizing: • Suppose there exists a hermitian X = X ∗ solving XDX − A∗X − XA − C = 0 • Let X0 = X0∗ 0 be chosen so A − DX0 is Hurwitz • Let C̃ be a Hermitian matrix with C̃ C. • Iteratively define (if possible) a sequence {Xν }∞ ν=1 via Xν+1 (A − DXν ) + (A − DXν )∗ Xν+1 = −Xν DXν − C̃ The following is true: • Given the choice of X0 and C̃, the sequence {Xν }∞ ν=1 is well-defined, ∗ unique, and has Xν = Xν . • For each ν ≥ 0, A − DXν is Hurwitz • For each ν ≥ 1, Xν X. It remains to show that for ν ≥ 1, Xν+1 Xν (although it is not necessarily true that X1 X0 ). 159 Decreasing Proof For ν ≥ 1, apply (7.31) with Y = Xν and Ŷ = Xν−1 , and substitute the iteration (shifted back one step) equation (7.34) giving Xν (A − DXν ) + (A − DXν )∗ Xν + Xν DXν = −C̃ − (Xν − Xν−1) D (Xν − Xν−1 ) The iteration equation is Xν+1 (A − DXν ) + (A − DXν )∗ Xν+1 = −Xν DXν − C̃ Subtracting leaves (Xν − Xν+1 ) (A − DXν ) + (A − DXν )∗ (Xν − Xν+1 ) = (Xν − Xν−1 ) D (Xν − Xν−1 ) The fact that A − DXν is Hurwitz and D 0 implies that Xν − Xν+1 0. Hence, we have shown that for all ν ≥ 1, X Xν+1 Xν 160 Monotonic Matrix Sequences Convergence Every nonincreasing sequence of real numbers which is bounded below has a limit. Specifically, if {xk }∞ k=0 is a sequence of real numbers, and there is a real number γ such that xk ≥ γ and xk+1 ≤ xk for all k, then there is a real number x such that lim xk = x k→∞ A similar result holds for symmetric matrices: Specifically, if {Xk }∞ k=0 is a sequence of real, symmetric matrices, and there is a real symmetric matrix Γ such that Xk Γ and Xk+1 Xk for all k, then there is a real symmetric matrix X such that lim Xk = X k→∞ n o∞ This is proven by first considering the sequence eTi Xk ei k=0, which shows that all of the diagonal entries of the sequence have limits. Then look at n o∞ (ei + ej )T Xk (ei + ej ) k=0 to conclude convergence of off-diagonal terms. 161 Limit of Sequence Maximal Solution The sequence {Xν }∞ ν=0 , generated by (7.34), has a limit lim Xν =: X̃+ ν→∞ Remember that C̃ was any hermitian matrix C. Facts: • Since each A − DXν is Hurwitz, in passing to the limit we may get eigenvalues with zero real-parts, hence max Reλi (A − DX+ ) ≤ 0 i • Since each Xν = Xν∗ , it follows that X̃+∗ = X̃+ . • Since each Xν X (where X is any hermitian solution to the equality XDX − A∗ X − XA − C = 0, equation (7.32)), it follows by passing to the limit that X̃+ X. • The iteration equation is Xν+1 (A − DXν ) + (A − DXν )∗ Xν+1 = −Xν DXν − C̃ Taking limits yields ∗ X̃+ A − DX̃+ + A − DX̃+ X̃+ = −X̃+DX̃+ − C̃ which has a few cancellations, leaving X̃+ A + A∗ X̃+ − X̃+ DX̃+ + C̃ = 0 • Finally, if X+ denotes the above limit when C̃ = C, it follows that the above ideas hold, so X+ X, for all solutions X. • And really finally, since X+ is a hermitian solution to (7.32) with C, the properties of X̃+ imply that X̃+ X+ . 162 dx2 − 2ax − c = 0 Analogies with Scalar Case Matrix Statement Scalar Specialization D = D∗ 0 d ∈ R, with d ≥ 0 (A, D) stabilizable a < 0 or d > 0 existence of Hermitian solution a2 + cd ≥ 0 √ a+ a2 +cd for d d c for d = 0 −2a Maximal solution X+ x+ = Maximal solution X+ x+ = √ − a2 + cd for d > 0 A − DX+ A − DX+ >0 a for d = 0 A − DX0 Hurwitz x0 > A − DX Hurwitz x> a d for d > 0 A − DX0 Hurwitz a < 0 for d = 0 A − DX Hurwitz a < 0 for d = 0 a d for d > 0 If d = 0, then the graph of −2ax − c (with a < 0) is a straight line with a c . positive slope. There is one (and only one) real solution at x = − 2a If d > 0, then the graph of dx2 − 2ax − c is a upward directed parabola. The minimum occurs at ad . There are either 0, 1, or 2 real solutions to dx2 − 2ax − c = 0. • If a2 + cd < 0, then there are no solutions (the minimum is positive) • If a2 + cd = 0, then there is one solution, at x = ad , which is where the minimum occurs. a d 2 • If a + cd √> 0, then there are two solutions, at x− = − 2 x+ = ad + a d+cd which are equidistant from the minimum ad . 163 √ a2 +cd d and