Discrete Riccati equations and block Toeplitz matrices

Transcription

1 Discrete Riccati equations and block Toeplitz matrices André Ran Vrije Universiteit Amsterdam Leonid Lerer Technion-Israel Institute of Technology Haifa André Ran and Leonid Lerer 1

2 Discrete algebraic Riccati equation X = AXA + Q AXB(R + B XB) 1 B XA Standing assumptions: A is invertible, R is Hermitian and invertible, Q is Hermitian No definitiness conditions on Q and R André Ran and Leonid Lerer 2

3 Discrete algebraic Riccati equation X = AXA + Q AXB(R + B XB) 1 B XA Standing assumptions: A is invertible, R is Hermitian and invertible, Q is Hermitian No definitiness conditions on Q and R A solution is a Hermitian matrix X for which R + B XB is invertible, and the is satisfied André Ran and Leonid Lerer 2

4 Discrete algebraic Riccati equation X = AXA + Q AXB(R + B XB) 1 B XA Standing assumptions: A is invertible, R is Hermitian and invertible, Q is Hermitian No definitiness conditions on Q and R A solution is a Hermitian matrix X for which R + B XB is invertible, and the is satisfied Goal: describe the set of all solutions in terms of two given solutions X 1 and X 0, when X 1 X 0 is invertible Second goal: to apply this to an interesting special case André Ran and Leonid Lerer 2

5 Reduction to the case Q = 0 Let X 0 be a fixed solution and let X be any other solution Denote by A 0 = A AX 0 B(R + B X 0 B) 1 B, and R 0 = R + B X 0 B The latter matrix is invertible by assumption, since X 0 is a solution Then Y = X X 0 satisfies the equation Y = A 0 Y A 0 A 0 Y B(R 0 + B Y B) 1 B Y A 0 So, without loss of generality Q = 0, supposing we know a solution André Ran and Leonid Lerer 3

6 Reduction to Lyapunov equation Assume that A is invertible, and let X be an invertible solution of : X = AXA AXB(R + B XB) 1 B XA Then Y = X 1 solves the Stein equation Y A Y A = BR 1 B and conversely, if Y is an invertible solution of the Stein equation, then X = Y 1 solves the discrete algebraic Riccati equation above André Ran and Leonid Lerer 4

7 Reduction to Lyapunov equation Assume that A is invertible, and let X be an invertible solution of : X = AXA AXB(R + B XB) 1 B XA Then Y = X 1 solves the Stein equation Y A Y A = BR 1 B and conversely, if Y is an invertible solution of the Stein equation, then X = Y 1 solves the discrete algebraic Riccati equation above Note: the zero matrix is an obvious solution of the algebraic Riccati equation André Ran and Leonid Lerer 4

8 Non-invertible solutions X = AXA AXB(R + B XB) 1 B XA X + an invertible solution, A invertible André Ran and Leonid Lerer 5

9 Non-invertible solutions X = AXA AXB(R + B XB) 1 B XA X + an invertible solution, A invertible Theorem Let N be an A -invariant subspace which is X + -nondegenerate, that is, C r = (X + N) +N Let P N be the projection onto (X + N) along N, and put X N = X + P N Then X N is a Hermitian solution of the algebraic Riccati equation for which kerx N = N André Ran and Leonid Lerer 5

10 Non-invertible solutions X = AXA AXB(R + B XB) 1 B XA X + an invertible solution, A invertible Theorem Let N be an A -invariant subspace which is X + -nondegenerate, that is, C r = (X + N) +N Let P N be the projection onto (X + N) along N, and put X N = X + P N Then X N is a Hermitian solution of the algebraic Riccati equation for which kerx N = N For converse we need an additional assumption André Ran and Leonid Lerer 5

11 Description of all solutions A is invertible, so X = AXA AXB(R + B XB) 1 B XA X(A ) 1 = ( A AXB(R + B XB) 1 B ) X Consequently, for any solution X the subspace N = kerx is invariant under A André Ran and Leonid Lerer 6

12 Description of all solutions A is invertible, so X = AXA AXB(R + B XB) 1 B XA X(A ) 1 = ( A AXB(R + B XB) 1 B ) X Consequently, for any solution X the subspace N = kerx is invariant under A Would like to conclude that for any solution X its kernel is X + -nondegenerate This is true under an extra assumption on A André Ran and Leonid Lerer 6

13 Main theorem I Theorem Assume that A and (A ) 1 have no common eigenvalues Then, if X solves ker X is X + -nondegenerate There is a one-to-one correspondence between solutions of and subspaces N that are A -invariant and X + -nondegenerate, which is given by X = X + P N, where P N is the projection onto (X + N) along N In particular, there can be only one invertible solution of André Ran and Leonid Lerer 7

14 Main theorem I Theorem Assume that A and (A ) 1 have no common eigenvalues Then, if X solves ker X is X + -nondegenerate There is a one-to-one correspondence between solutions of and subspaces N that are A -invariant and X + -nondegenerate, which is given by X = X + P N, where P N is the projection onto (X + N) along N In particular, there can be only one invertible solution of Observe that the uniqueness of the solution of the Stein equation is in fact equivalent to A and A 1 having no common eigenvalues André Ran and Leonid Lerer 7

15 Main theorem II Assume that A and (A ) 1 have no common eigenvalues, and let Y be the unique solution of the Stein equation Y A Y A = BR 1 B Theorem There is a one-to-one correspondence between solutions of and subspaces N that are A -invariant and Y -nondegenerate which is given by X = Y 1 P N, where P N is the projection onto Y N along N André Ran and Leonid Lerer 8

16 Special case: companion type A = K = K 1 I 0 0 K 2 0 I, B = I K n 0 0 I 0 0 Define X (n) 00 Assume K is invertible = R, X(n) j0 = K jr André Ran and Leonid Lerer 9

17 Special case: companion type A = K = K 1 I 0 0 K 2 0 I, B = I K n 0 0 I 0 0 Define X (n) 00 Assume K is invertible Stein equation: = R, X(n) j0 = K jr Y K Y K = diag ( ) (X (n) 00 ) André Ran and Leonid Lerer 9

18 Special case: companion type A = K = K 1 I 0 0 K 2 0 I, B = I K n 0 0 I 0 0 Define X (n) 00 Assume K is invertible Stein equation: : = R, X(n) j0 = K jr Y K Y K = diag ( ) (X (n) 00 ) X = KXK KXB(X (n) 00 + B XB) 1 B XK André Ran and Leonid Lerer 9

19 Connection to block Toeplitz matrices The invertible solution X of the Riccati equation X = KXK KXB(X (n) 00 + B XB) 1 B XK is congruent to the inverse of a block Toeplitz matrix André Ran and Leonid Lerer 10

20 Connection to block Toeplitz matrices The invertible solution X of the Riccati equation X = KXK KXB(X (n) 00 + B XB) 1 B XK is congruent to the inverse of a block Toeplitz matrix Use Gohberg-Lerer result: André Ran and Leonid Lerer 10

21 Connection to block Toeplitz matrices The invertible solution X of the Riccati equation X = KXK KXB(X (n) 00 + B XB) 1 B XK is congruent to the inverse of a block Toeplitz matrix Use Gohberg-Lerer result: Y = X 1 = T n 1, where T n is the block Toeplitz matrix given by ( ) T n = K X (n) X 1 K, André Ran and Leonid Lerer 10

22 Connection to block Toeplitz matrices The invertible solution X of the Riccati equation X = KXK KXB(X (n) 00 + B XB) 1 B XK is congruent to the inverse of a block Toeplitz matrix Use Gohberg-Lerer result: Y = X 1 = T n 1, where T n is the block Toeplitz matrix given by ( ) T n = K X (n) X 1 K, where K is given by André Ran and Leonid Lerer 10

23 Connection to block Toeplitz matrices K = 00 ) K 1 I 0 0 K 2 0 I I 0 K n 0 0 I (X (n) André Ran and Leonid Lerer 11

24 Discussion of the extra assumption Gohberg-Lerer: If R = X 00 > 0, K and (K ) 1 are even allowed to have common eigenvalues!!! To ensure existence of an invertible solution to the Stein equation Y K Y K = diag ( ) (X (n) 00 ) n P(λ) = ( λ n j K j )R 1/2 j=0 André Ran and Leonid Lerer 12

25 Discussion of the extra assumption Gohberg-Lerer: If R = X 00 > 0, K and (K ) 1 are even allowed to have common eigenvalues!!! To ensure existence of an invertible solution to the Stein equation Y K Y K = diag ( ) (X (n) 00 ) P(λ) = ( n λ n j K j )R 1/2 André Ran and Leonid Lerer 12 j=0 Theorem[Gohberg-Lerer] The Stein equation has an invertible solution if and only if for every symmetric pair of eigenvalues λ 0, λ 1 0 and for any right Jordan chains x 1,, x k and y 1,, y l of P corresponding to λ 0 and λ 1 0, respectively, there holds, with ν = max{k, l}: ( ) ν x ν j, y j = 0 j=1

26 Putting it all together K, B and R as above, assume K invertible, R > 0 If ( ) holds whenever λ 0, λ 1 0 are both eigenvalues of K, then: there is an invertible block Toeplitz Hermitian matrix Y = T n 1 solving the Stein equation, André Ran and Leonid Lerer 13

27 Putting it all together K, B and R as above, assume K invertible, R > 0 If ( ) holds whenever λ 0, λ 1 0 are both eigenvalues of K, then: there is an invertible block Toeplitz Hermitian matrix Y = T n 1 solving the Stein equation, the solutions of are in 1-1 correspondence to K-invariant subspaces N which are Y -nondegenerate: X N = P NY 1 P N where P N is the projection onto Y N along N André Ran and Leonid Lerer 13

28 K = / , R = I 2 0 1/6 0 0 σ(k) = {2, 1/2, 3, 1/3} There are pairs of eigenvalues symmetrically placed with respect to the unit circle The corresponding eigenvectors are, respectively, x 1 = Condition ( ) is satisfied 0, x 2 = 3 0, x 3 = 1 André Ran and Leonid Lerer , x 4 = 2 0 1

29 Solution to Stein equation The block Toeplitz invertible solution to the Stein equation: Y = 7/ / /10 0 3/2 5/ / /2 0 21/10 André Ran and Leonid Lerer 15

30 Solution to Stein equation The block Toeplitz invertible solution to the Stein equation: Y = 7/ / /10 0 3/2 5/ / /2 0 21/10 14 nontrivial K-invariant subspaces: four one-dimensional subspaces, six two-dimensional ones, and again four three-dimensional ones All of them are Y -nondegenerate André Ran and Leonid Lerer 15

31 Solution to Stein equation The block Toeplitz invertible solution to the Stein equation: Y = 7/ / /10 0 3/2 5/ / /2 0 21/10 14 nontrivial K-invariant subspaces: four one-dimensional subspaces, six two-dimensional ones, and again four three-dimensional ones All of them are Y -nondegenerate So 14 solutions in total André Ran and Leonid Lerer 15

32 Invertible and rank 3 solutions Subspace Solution N = (0) Y 1 = 0 35/ / / /36 45/4 0 15/4 0 N = span {x 1 } 0 35/ /36 15/4 0 5/ / / N = span {x 2 } 0 1/30 0 1/ /10 0 3/10 96/ /11 0 N = span {x 3 } 0 35/ /36 48/ / / / N = span {x 4 } 0 2/33 0 4/ /33 0 8/33 André Ran and Leonid Lerer 16

33 Rank 2 solutions Subspace N = span {x 1, x 3 } N = span {x 2, x 4 } N = span {x 1, x 2 } N = span {x 1, x 4 } N = span {x 2, x 3 } N = span {x 3, x 4 } Solution / / / / /4 0 15/ /30 0 1/10 15/4 0 5/ /10 0 3/10 45/4 0 15/ /33 0 4/33 15/4 0 5/ /33 0 8/33 96/ / /30 0 1/10 48/ / /10 0 3/10 96/ / /33 0 4/33 48/ / /33 0 8/33 André Ran and Leonid Lerer 17

34 Rank 1 solutions Subspace Solution N = span {x 1, x 2, x 3 } 0 1/30 0 1/ /10 0 3/10 45/4 0 15/4 0 N = span {x 1, x 2, x 4 } /4 0 5/ N = span {x 1, x 3, x 4 } 0 2/33 0 4/ /33 0 8/33 96/ /11 0 N = span {x 2, x 3, x 4 } / / N = C André Ran and Leonid Lerer 18

35 Thank you for your attention! André Ran and Leonid Lerer 19