Shift techniques for Quasi-Birth-and-Death processes: canonical - PowerPoint PPT Presentation

Shift techniques for Quasi-Birth-and-Death processes: canonical factorizations and matrix equations Beatrice Meini Universit` a di Pisa Joint work with D.A. Bini and G. Latouche MAM9 - Budapest

QBD processes Let   A ′ A ′ 0 0 1   A − 1 A 0 A 1 P =   ... ... ... 0 be the transition matrix of a QBD with space state N × S , S = { 1 , . . . , n } . Here A − 1 , A 0 , A 1 are n × n nonnegative matrices such that A − 1 + A 0 + A 1 is stochastic and irreducible Define the matrix polynomial A ( z ) = A − 1 + z ( A 0 − I ) + z 2 A 1 We call eigenvalues of the matrix polynomial A ( z ) the roots of a ( z ) = det A ( z ) Remark: Since A (1) 1 = 0 then z = 1 is an eigenvalue of A ( z )

Quadratic matrix equations and canonical factorizations Let G , R , � G and � R be the minimal nonnegative solutions of the matrix equations A − 1 + A 0 X + A 1 X 2 = X X 2 A − 1 + XA 0 + A 1 = X A − 1 X 2 + A 0 X + A 1 = X X 2 A 1 + XA 0 + A − 1 = X . Then A ( z ) and the reversed matrix polynomial � A ( z ) = z 2 A − 1 + z ( A 0 − I ) + A 1 have the weak canonical factorizations A ( z ) = ( I − zR ) K ( zI − G ) � A ( z ) = ( I − z � R ) � K ( zI − � G ) with K = A 0 − I + A 1 G and � K = A 0 − I + A − 1 � G .

Roots of the matrix polynomial A ( z ) The roots ξ i , i = 1 , . . . , 2 n of a ( z ) are such that | ξ 1 | ≤ · · · ≤ | ξ n − 1 | ≤ ξ n ≤ 1 ≤ ξ n +1 ≤ | ξ n +2 | ≤ · · · ≤ | ξ 2 n | where we have introduced 2 n − deg a ( z ) roots at infinity if deg a ( z ) < 2 n More specifically, we have the following scenario: ◮ ξ n = 1 < ξ n +1 positive recurrent ◮ ξ n = 1 = ξ n +1 null recurrent ◮ ξ n < 1 = ξ n +1 transient If Gu = λ u then A ( λ ) u = 0; if v T R = µ v T then Remark. v T A ( µ − 1 ) = 0. That is, the eigenvalues of G and the reciprocals of the eigenvalues of R are eigenvalues of A ( z ). In particular: ◮ G has eigenvalues ξ 1 , . . . , ξ n ◮ R has eigenvalues ξ − 1 n +1 , . . . , ξ − 1 2 n Assumption 1 : the process is recurrent, i.e., ξ n = 1 Assumption 2 : | ξ n − 1 | < ξ n and ξ n +1 < | ξ n +2 |

Motivation of the shift ◮ There exist algorithms for computing the minimal nonnegative solution G ; their efficiency deteriorates as ξ n /ξ n +1 gets close to 1 ◮ In the null recurrent case where ξ n = ξ n +1 , the convergence speed turns from linear to sublinear, or from superlinear to linear, according to the used algorithm Here we provide a tool for getting rid of this drawback The idea is an elaboration of the Brauer theorem and of the shift technique for matrix polynomials [He, Meini, Rhee 2001] It relies on transforming the matrix polynomial A ( z ) into a new one � a ( z ) = det � A ( z ) in such a way that � A ( z ) has the same roots of a ( z ) except for ξ n = 1 which is shifted to 0, and/or ξ n +1 = 1 which is shifted to infinity

Brauer’s theorem on eigenvalues Theorem (Brauer 1956) Let A be an n × n matrix with eigenvalues λ 1 , . . . , λ n . Let x k be an eigenvector of A associated with the eigenvalue λ k , 1 ≤ k ≤ n, and let q be any n-dimensional vector. Then the matrix A + x k q T has eigenvalues λ 1 , . . . , λ k − 1 , λ k + q T x k , λ k +1 , . . . , λ n . Remark: if q is such that q T x k = − λ k , then A + x k q T has eigenvalues 0 , λ 1 , . . . , λ k − 1 , λ k +1 , . . . , λ n , i.e., the eigenvalue λ k is shifted to 0. Question: can we generalize this shifting to the eigenvalues of matrix polynomials?

Functional interpretation of Brauer’s theorem Let A be an n × n matrix and let Au = λ u , u � = 0. Choose any vector v such that v T u = 1 and define � A = A − λ uv T . Remark : � Au = Au − λ uv T u = λ u − λ u = 0. Functional interpretation : by direct inspection, one has � � λ � z − λ uv T A − zI = ( A − zI ) I + Taking determinants: z det( � A − zI ) = det( A − zI ) z − λ Therefore: ◮ � A has the same eigenvalues of A except for λ which is shifted to zero ◮ A and � A share the right eigenvector u and the left eigenvectors not corresponding to λ Question: can we do anything similar for A ( z )?

YES! Shift to the right Let u G � = 0 such that A ( ξ n ) u G = 0, and let v be any vector such that v T u G = 1. Define: � � ξ n � Q = u G v T A r ( z ) = A ( z ) I + Q , z − ξ n Remark: similarly to the matrix case, det � z A r ( z ) = det A ( z ) z − ξ n

YES! Shift to the right Let u G � = 0 such that A ( ξ n ) u G = 0, and let v be any vector such that v T u G = 1. Define: � � ξ n � Q = u G v T A r ( z ) = A ( z ) I + Q , z − ξ n Remark: similarly to the matrix case, det � z A r ( z ) = det A ( z ) z − ξ n Theorem The function � A r ( z ) coincides with the quadratic matrix polynomial � A r ( z ) = � A − 1 + z ( � A 0 − I ) + z 2 � A 1 with matrix coefficients � � � A − 1 = A − 1 ( I − Q ) , A 0 = A 0 + ξ n A 1 Q , A 1 = A 1 . Moreover, the eigenvalues of � A r ( z ) are 0 , ξ 1 , . . . , ξ n − 1 , ξ n +1 , . . . , ξ 2 n .

Shift to the left Let v R � = 0 such that v T R A ( ξ n +1 ) = 0, and let w be any vector such that w T v R = 1. Define � � z � S = wv T A ℓ ( z ) = I − A ( z ) , S R z − ξ n +1 Remark: similarly to the right shift, det � 1 A ℓ ( z ) = det A ( z ) z − ξ n

Shift to the left Let v R � = 0 such that v T R A ( ξ n +1 ) = 0, and let w be any vector such that w T v R = 1. Define � � z � S = wv T A ℓ ( z ) = I − A ( z ) , S R z − ξ n +1 Remark: similarly to the right shift, det � 1 A ℓ ( z ) = det A ( z ) z − ξ n Theorem The function � A ℓ ( z ) coincides with the quadratic matrix polynomial � A ℓ ( z ) = � A − 1 + z ( � A 0 − I ) + z 2 � A 1 with matrix coefficients � � � A 0 = A 0 + ξ − 1 A − 1 = A − 1 , n +1 SA − 1 , A 1 = ( I − S ) A 1 . Moreover, the eigenvalues of � A ℓ ( z ) are ξ 1 , . . . , ξ n , ξ n +2 , . . . , ξ 2 n , ∞ .

Double shift The right and left shifts can be combined together. Define the matrix function � � � � ξ n z � A d ( z ) = I − S A ( z ) I + Q . z − ξ n +1 z − ξ n Theorem The function � A d ( z ) coincides with the quadratic matrix polynomial � A d ( z ) = � A − 1 + z ( � A 0 − I ) + z 2 � A 1 with matrix coefficients � A − 1 = A − 1 ( I − Q ) , � A 0 = A 0 + ξ n A 1 Q + ξ − 1 n +1 SA − 1 − ξ − 1 n +1 SA − 1 Q , � A 1 = ( I − S ) A 1 . Moreover, the eigenvalues of � A d ( z ) are 0 , ξ 1 , . . . , ξ n − 1 , ξ n +2 , . . . , ξ 2 n , ∞ . In particular, � A d ( z ) is nonsingular on the unit circle and on the annulus | ξ n − 1 | < | z | < | ξ n +2 | .

Shifts and canonical factorizations Under which conditions both the polynomials � Question: A s ( z ) and z 2 � A s ( z − 1 ) for s ∈ { r , ℓ, d } obtained after applying the shift have a (weak) canonical factorization? In different words: Under which conditions there exist the four minimal Question: solutions to the matrix equations associated with the polynomial � A s ( z ) obtained after applying the shift? These matrix solutions will be denoted by G s , R s , � G s , � R s , with s ∈ { r , ℓ, d } . They are the analogous of the solutions G , R , � G , � R to the original equations. We examine the case of the shift to the right The shift to the left and the double shift can be treated similarly.

Right shift: the polynomial � A r ( z ) Recall that � � ξ n � A r ( z ) = A ( z ) I + Q = z − ξ n � � ξ n = ( I − zR ) K ( zI − G ) I + Q z − ξ n with Q = u G v T .

Right shift: the polynomial � A r ( z ) Recall that � � ξ n � A r ( z ) = A ( z ) I + Q = z − ξ n � � ξ n = ( I − zR ) K ( zI − G ) I + Q z − ξ n with Q = u G v T . By a direct computation we obtain � � ξ n ( zI − G ) I + = zI − G r Q z − ξ n with G r = G − ξ n Q . Therefore � A r ( z ) = ( I − zR ) K ( zI − G r )

Right shift: the polynomial � A r ( z ) Theorem ◮ The polynomial � A r ( z ) has the following factorization � A r ( z ) = ( I − zR ) K ( zI − G r ) , G r = G − ξ n Q This factorization is canonical in the positive recurrent case, and weak canonical otherwise. ◮ The eigenvalues of G r are those of G, except for the eigenvalue ξ n which is replaced by zero ◮ X = G r and Y = R are the solutions with minimal spectral radius of the equations A 1 X 2 + � � A 0 X + � Y 2 � A − 1 + Y � A 0 + � A − 1 = X , A 1 = Y

Right shift: the reversed polynomial z 2 � A r ( z − 1 ) Recall that � � z ξ n z 2 � A r ( z − 1 ) = z 2 A ( z − 1 ) I + Q = 1 − z ξ n � � z ξ n = ( I − z � R ) � K ( zI − � G ) I + Q 1 − z ξ n with Q = u G v T .

Right shift: the reversed polynomial z 2 � A r ( z − 1 ) Recall that � � z ξ n z 2 � A r ( z − 1 ) = z 2 A ( z − 1 ) I + Q = 1 − z ξ n � � z ξ n = ( I − z � R ) � K ( zI − � G ) I + Q 1 − z ξ n with Q = u G v T . By a direct computation we obtain � � z ξ n ( zI − � = z ( I − ξ n Q ) − � G ) I + Q G 1 − z ξ n and I − ξ n Q is singular! Things are more complicated. We need some preliminary results