Discrete Riccati equations and block Toeplitz matrices

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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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 ) 1 0 0 André Ran and Leonid Lerer 9

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 ) 1 0 0 X = KXK KXB(X (n) 00 + B XB) 1 B XK André Ran and Leonid Lerer 9

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

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

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) 00 0 0 X 1 K, André Ran and Leonid Lerer 10

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) 00 0 0 X 1 K, where K is given by André Ran and Leonid Lerer 10

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

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 ) 1 0 0 n P(λ) = ( λ n j K j )R 1/2 j=0 André Ran and Leonid Lerer 12

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 ) 1 0 0 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

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

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

K = 5 0 1 0 0 5/6 0 1 6 0 0 0, 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 = 1 0 3 0 Condition ( ) is satisfied 0, x 2 = 3 0, x 3 = 1 André Ran and Leonid Lerer 14 1 0 2 0 0, x 4 = 2 0 1

Solution to Stein equation The block Toeplitz invertible solution to the Stein equation: Y = 7/120 0 5/120 0 0 21/10 0 3/2 5/120 0 7/120 0 0 3/2 0 21/10 André Ran and Leonid Lerer 15

Solution to Stein equation The block Toeplitz invertible solution to the Stein equation: Y = 7/120 0 5/120 0 0 21/10 0 3/2 5/120 0 7/120 0 0 3/2 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

Solution to Stein equation The block Toeplitz invertible solution to the Stein equation: Y = 7/120 0 5/120 0 0 21/10 0 3/2 5/120 0 7/120 0 0 3/2 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

Invertible and rank 3 solutions Subspace Solution 35 0 25 0 N = (0) Y 1 = 0 35/36 0 25/36 25 0 35 0 0 25/36 0 35/36 45/4 0 15/4 0 N = span {x 1 } 0 35/36 0 25/36 15/4 0 5/4 0 0 25/36 0 35/36 35 0 25 0 N = span {x 2 } 0 1/30 0 1/10 25 0 35 0 0 1/10 0 3/10 96/11 0 48/11 0 N = span {x 3 } 0 35/36 0 25/36 48/11 0 24/11 0 0 25/36 0 35/36 35 0 25 0 N = span {x 4 } 0 2/33 0 4/33 25 0 35 0 0 4/33 0 8/33 André Ran and Leonid Lerer 16

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 0 0 0 0 0 35/36 0 25/36 0 0 0 0 0 25/36 0 35/36 35 0 25 0 0 0 0 0 25 0 35 0 0 0 0 0 45/4 0 15/4 0 0 1/30 0 1/10 15/4 0 5/4 0 0 1/10 0 3/10 45/4 0 15/4 0 0 2/33 0 4/33 15/4 0 5/4 0 0 4/33 0 8/33 96/11 0 48/11 0 0 1/30 0 1/10 48/11 0 24/11 0 0 1/10 0 3/10 96/11 0 48/11 0 0 2/33 0 4/33 48/11 0 24/11 0 0 4/33 0 8/33 André Ran and Leonid Lerer 17

Rank 1 solutions Subspace Solution 0 0 0 0 N = span {x 1, x 2, x 3 } 0 1/30 0 1/10 0 0 0 0 0 1/10 0 3/10 45/4 0 15/4 0 N = span {x 1, x 2, x 4 } 0 0 0 0 15/4 0 5/4 0 0 0 0 0 0 0 0 0 N = span {x 1, x 3, x 4 } 0 2/33 0 4/33 0 0 0 0 0 4/33 0 8/33 96/11 0 48/11 0 N = span {x 2, x 3, x 4 } 0 0 0 0 48/11 0 24/11 0 0 0 0 0 N = C 4 0 4 4 André Ran and Leonid Lerer 18

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

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