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Hamiltonian Matrices and the Algebraic Riccati Equation Seminar Presentation Hamiltonian Matrices and the Algebraic Riccati Equation Seminar presentation by Piyapong Yuantong 7th December 2009 Technische Universit¨atChemnitz Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation 1 Hamiltonian matrices We start by introducing the square matrix J 2 R 2n×2n defined by " # O I J = n n ; (1) −In On where n×n On 2 R − zero matrix n×n In 2 R − identity matrix: Remark It is not very complicated to prove the following properties of the matrix J: i. J T = −J ii. J −1 = J T T iii. J J = I2n T T iv. J J = −I2n 2 v. J = −I2n vi. det J = 1 Definition 1 A matrix A 2 R 2n×2n is called Hamiltonian if JA is symmetric, so JA = (JA) T ) A T J + JA = 0 where J 2 R 2n×2n is from (1). We will denote { } Hn = A 2 R 2n×2n j A T J + JA = 0 the set of 2n × 2n Hamiltonian matrices. 1 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation Proposition 1.1 The following are equivalent: a) A is a Hamiltionian matrix b) A = JS, where S = S T c) (JA) T = JA Proof a ! b A = JJ −1 A ! A = J(−J)A 2Hn A ! (J(−JA)) T J + JA = 0 ! (−JA) T J T J = −JA T J J=!I2n (−JA) T = −JA ! J(−JA) T = A If − JA = S ! A = J(−JA) = J(−JA) T ! A = JS = JS T =) S = S T a ! c A T J + JA = 0 ! A T J = −JA ()T ! (A T J) T = (−JA) T ! J T A = −(JA) T ! −JA = −(JA) T ! (JA) T = JA 2 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation Proposition 1.2 Let A; B 2 Hn. The following are true: a) A + B 2 Hn b) αA 2 Hn, α 2 R def: c) [A; B] 2 Hn, where [A; B] = AB − BA Proof a) Because A and B are Hamiltonian matrices, it results that A T J + JA = 0, and respectively B T J + JB = 0. By adding those two relations we obtain: (A T + B T )J + J(A + B) = 0 () (A + B) T J + J(A + B) = 0 ! A + B 2 H b) A T J + JA = 0 ! A T Jα + JAα = 0 ! A T αJ + J(Aα) = 0 ! (Aα) T J + J(Aα) = 0 ! αA 2 H c) We will prove that [A; B] = JM, where M = M T . We know that A = JS and B = JR, where S = S T and R = R T . [A; B] = AB − BA = JSJR − JRJS = J(SJR − RJS) from where, using the notation SJR − RJS = M, we obtain [A; B] = JM. Now we will show that M = M T : M T = (SJR − RJS) T = (SJR) T − (RJS) T = RT J T ST − ST J T RT = −RJS + SJR = SJR − RJS = M 3 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation Consequence (H; [·; ·]) is a Lie algebra. Proof We will prove the necessary properties of the bracket [·; ·] in terms of bilin- earity, antisymmetry and Jacobi's relation. i. [αA + βB; C] = α[A; C] + β[B; C] - evidently =) the operation is bilinear. ii. [A; B] = AB − BA = −(BA − AB) = −[B; A] =) the operation is antisymmetric. iii. Jacobi's relation is satisfied: [[A; B];C] + [[C; A];B] + [[B; C];A] = [AB − BA; C] + [CA − AC; B] + [BC − CB; A] == ABC − BAC − (CAB − CBA) + CAB − ACB −(BCA − BAC) + BCA − CBA − (ABC − ACB)) = 0 Proposition 1.3 n Let A 2 H and pA(x) - the characteristic polynomial of the matrix A. Then: a) pA(x) = pA(−x) b) if pA(c) = 0, then pA(−c) = pA(¯c) = pA(−c¯) = 0, where c 2 R. 4 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation Proof a) T pA(x) = det(A − xI2n); but A = JA J =) T −1 pA(x) = det(JA J − xI2n) A = −JA J = det(JAT J + JxJ) T = det(J(A + xI2n)J) T = detJ det(A + xI2n)detJ T T T = det(A + xI2n) = det(A + xI2n) T = det(A + xI2n) = det(A + xI2n) = det(A − (−x)I2n) = pA(−x) b) a) pA(c) = 0 =) pA(−c) = 0 a) pA(x) is a real coefficients polynomial =) pA(¯c) = 0 =) pA(−c¯) = 0 2 The Algebraic Riccati Equation Definition 2.1 The linear space ν spanned by the vectors v1; : : : ; vk is called an invariant subspace of the matrix A (or A-invariant) if for every v 2 ν, Av 2 ν. This is equivalent to Avi 2 ν for i = 1; : : : ; k. We consider the general Algebraic Riccati Equation (ARE) 0 = R(x) = F + A T X + XA − XGX (2) where A; F; G; X 2 Rn×n and F; G are symmetric matrices (F = F T , G = GT ). Equation (2) is for a symmetric unknown X. First, we define the following 2n × 2n matrix " # AG H = (3) F −AT 5 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation Let the columns of [U T ;V T ]T , U; V; 2 Rn×n, span a H-invariant, n-dimensional subspace, i.e., " #" # " # AG U U = Z; Z 2 Rn×n; λ(Z) ⊂ λ(H) (4) F −AT V V Assuming that U is nonsingular, we obtain from the first row of (4) AU + GV = UZ ! U −1AU + U −1GV = Z Inserting this into the equation resulting from evaluating the second row of (4) yields FU − AT V = VZ = VU −1AU + VU −1GV The above equation is equivalent to 0 = F − AT VU −1 − VU −1A − VU −1GV U −1 setting X := −VU −1 we see that X solves (2). Hence," # from an H-invariant subspace of dimension n, given as the range U of with U nonsingular, we obtain a solution of the Algebraic Riccati V Equation (ARE). What remains is the problem of how to choose U; V such that U is nonsin- gular, VU −1 is symmetric and X is stabilizing. Before we can solve this problem, we need some properties of the matrix H in (3). Remark It is easy to see that any Hamiltonian matrix must have the block repre- sentation as shown in (3). Moreover, it is easy to verify that the matrix H defined in (3) is Hamiltonian according to (JH)T = JH. By using the similarity transformation J −1HJ = −JHJ = −H T (5) it can be shown that if λ is an eigenvalue of H, then −λ¯ is also an eigenvalue of H. The subspace χ is the subspace spanned by the eigenvectors of H corre- sponding to the negative eigenvalues. This subspace is invariant under the transformation H; i.e., if x 2 χ then Hx 2 χ . 6 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation Theorem Consider the Hamiltonian matrix H with no eigenvalue on the imaginary axis and the invariant subspace χ ⊂ R 2n×n as follows, " # X χ = Im 1 ; X2 which is spanned by eigenvectors associated with stable eigenvalues. If X1 − −1 is invertible then X = X2X1 is a (stabilizing and symmetric) solution of the corresponding Riccati equation and A − GX is stable. Proof Because χ is an invariant subset, there exists a stable matrix H~ such that " # " # " #" # " # X X AG X X H 1 = 1 H~ =) 1 = 1 H~ T X2 X2 F −A X2 X2 " #" # " # − − AG I I −1 =) = X1HX~ F −AT X X 1 " #" # h i AG −I =) XI = 0 F −AT X " # h i −I =) XA + FXG − AT = 0 X =) −XA − F + XGX − AT X = 0 =) F + AT X + XA − XGX = 0 From the other side we have − − ~ −1 ! − ~ −1 GX A = X1HX1 A GX = X1HX1 which proves stability. To show that X is symmetric, we use equation (5) " #T " # " #T " # X X X X 1 JH 1 = 1 J 1 H~ X2 X2 X2 X2 " #T " #" # X F −AT X ) 1 1 − T T ~ = = ( X2 X1 + X1 X2)H X2 −A −G X2 7 Piyapong Yuantong Hamiltonian Matrices and the Algebraic Riccati Equation The left-hand side of this equation is symmetric and therefore the right-hand side is also symmetric, − T T ~ ~ T − T T ( X2 X1 + X1 X2)H = H ( X1 X2 + X2 X1) − ~ T − T T = H ( X2 X1 + X1 X2) ) − T T ~ ~ T − T T = ( X2 X1 + X1 X2)H + H ( X2 X1 + X1 X2) = 0: This is a Lyapunov equation with the stable matrix H~ , and the only solution should be zero, − T T ) − T T X2 X1 + X1 X2 = 0 = X2 X1 = X1 X2; and therefore − −1 − −T T −1 X = X2X1 = X1 X1 X2X1 − −T T −1 = X1 X2 X1X1 − −T T = X1 X2 is also symmetric. Remark − −1 As the matrix H is real, it can be shown that the solution X = X2X1 is also real. Furthermore, this solution does not depend on the bases chosen and is the unique solution determined by H. 8.
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