A Homotopy Method for Computing All Isolated Solvents of ... · A Homotopy Method for Computing All Isolated Solvents of the Quadratic Matrix Equation AX 2 +BX +C = 0 Yongwen Hou

IntroductionHomotopy method for quadratic matrix equation

More results for special problemsNumerical results and conclusion

A Homotopy Method for Computing All IsolatedSolvents of the Quadratic Matrix Equation

AX 2 + BX + C = 0

Yongwen Hou Bo Yu

School of Mathematical Sciences, Dalian University of Technology

ASCM Beijing Oct. 26-28, 2012

Yongwen Hou, Bo Yu HM for Solving AX 2 + BX + C = 0



Outline

1 Introduction

2 Homotopy method for quadratic matrix equation

3 More results for special problems

4 Numerical results and conclusion




Quadratic matrix equation

The unilateral quadratic matrix equation

P(X ) = AX 2 + BX + C = 0, (QME)

is considered, whose coefficients A, B, and C ∈ Cn×n, and thematrix solution X is called a solvent. The corresponding quadraticeigenvalue problem is:

P(λ)v = (λ2A + λB + C )v = 0,

λ ∈ C and v 6= 0 ∈ Cn,

(QEP)

where λ and v are eigenvalue and eigenvector respectively.




Relations between solvents and eigenvalues.

Given a solvent X , satisfying

P(X ) = AX 2 + BX + C = 0,

then, P(λ) can be divided by the linear term X − λI on theright:

λ2A + λB + C = −(B + AX + λA)(X − λI ),

and thus, eigenvalues of P(λ) are those of X and those of ageneralized eigenvalue problem (B + AX )v = −λAv .




Relations between solvents and eigenvalues.

For n eigenpairs {λi , vi}ni=1 of P(λ), where the eigenvectorsv1, ..., vn are linearly independent, denoting

V = [v1, ..., vn] and Λ = diag(λ1, ..., λn),

thenAV Λ2 + BV Λ + CV = 0

is satisfied. Multiplied by V−1 on the right, we have

AV Λ2V−1 + BV ΛV−1 + C = 0,

which indicates V ΛV−1 is a solvent of P(X ).




Solvents of quadratic matrix equation

P(X ) can have no solvents, a finite positive number, or infinitelymany. For example, the equation

X 2 − Jn(λ),

where Jn(λ) (n > 1) is a Jordan block with the eigenvalue λ, hasno solvents when λ = 0, and precisely two solvents when λ 6= 0.And the equation

X 2 − In

has infinite many solvents, including two isolated solvents and2n − 2 families of solvents.




Existing algorithms

Newton’s method is attractive for its local quadratic convergence,if a good approximation Z0 of the desired solvent Z ∗ is given. Itgenerates a sequence of matrices converging to Z ∗:

Solve P ′Zk(Ek) = −P(Zk) for Ek

Update Zk+1 = Zk + Ek

}k = 0, 1, 2, ...

where P ′X (E ) : Cn×n → Cn×n is the Frechet derivative of P at Xin the direction E along X .




Existing algorithms

Bernoulli’s method, used to find the dominant or minimal solvent(if there is one) of P(X ), is generalized from the case of thequadratic scalar equations.

Dominant solvent:

(AZk + B)Zk−1 + C = 0, for k = 1, 2, ...

limk→∞

Zk = Zdom.

Minimal solvent:

(AZk−1 + B)Zk + C = 0, for k = 1, 2, ...

limk→∞

Zk = Zmin.




Existing algorithms

Definition

Suppose P(λ) has exactly 2n eigenvalues

|λ1| ≥ |λ2| ≥ ... ≥ |λ2n|,

and denote the set of eigenvalues of a matrix Z by λ(Z ), a solventZ1 of P(X ) is a dominant solvent if λ(Z1) = {λ1, ..., λn}, and asolvent Z2 is a minimal solvent if λ(Z1) = {λn+1, ..., λ2n}.




Eigenvalue technique, used to construct all diagonalizable solvents.

Theorem (Higham et al., 2001)

Suppose Q(λ) = Mλ2 + Lλ+ K has p distinct eigenvalues{λi}pi=1, with n ≤ p ≤ 2n, and that the corresponding set of peigenvectors {λi}pi=1 satisfies the Haar condition (that is, everysubset of n of them is linearly independent). Then there are atleast

(pn

)different solvents of Q(X ), and exactly this many if

p = 2n, which are given by

X = Wdiag(µi )W−1, W = [w1, ...,wn],

where the eigenpaires {µi ,wi}ni=1 are chosen among theeigenvpairs {λi , vi}pi=1 of Q.




Existing algorithms

N. J. Higham and H.-M. Kim, Numerical analysis ofquadratic matrix equation, IMA Journal of NumericalAnalysis, 20 (2000), pp. 499-519.

N. J. Higham and H.-M. Kim, Solving a quadratic matrixequation by Newton’s method with exact line searches, SIAMJournal on Matrix Analysis and Applications, 23 (2001), pp.303-316.




The aim–finding all isolated solvents

We focus on locating all isolated solvents of P(X ), of which thereexists a neighborhood not containing other solvents. It is necessaryto consider the element-wise form of P(X ), which is a polynomialsystem of n2 equations and n2 variables, and denoted by:

p(x) = 0.

Here, we set

x = vec(X ) = (x11, ..., xn1, ..., x1n, ..., xnn)T ,

andp = vec(P) = (p11, ..., pn1, ..., p1n, ..., pnn)T .




Example 1:

The element-wise form of the equation

X 2 +

[−1 −62 −9

]X +

[0 12−2 14

]= 0,

can be written as

p(x) =

x2

11 + x12x21 − x11 − 6x21

x11x21 + x12x22 − x12 − 6x22 + 12x21x11 + x22x21 + 2x11 − 9x21 − 2x21x12 + x2

22 + 2x12 − 9x22 + 14

= 0.




Homotopy method for solving polynomial systems

Given a systemf (z) : Cr → Cr ,

we design a start system g(z) = 0, and construct a homotopy:

h(z , t) = (1− t)g(z) + tf (z),

satisfying

Triviality: The solutions of start system g(x) = 0 are known.

Smoothness: The solution set of h(x , t) = 0 with t ∈ [0, 1)consists of a finite number of smooth paths emanating fromisolated solutions of g(x) = 0, each parameterized by t.

Accessibility: Every isolated solution of h(x , 1) = p(x) = 0 isreached by a path.




Homotopy method for solving polynomial systems




Step1. Given a start solution z0 at t0 = 0, a stopping criteriaε > 0, and an initial step h0.Step2. For i = 1, 2, ...

Predict: Predict solution (zi , ti ) such that:

ti = ti−1 + hi−1

zi = zi−1 − hz(zi−1, ti−)−1ht(zi−1, ti−1)hi−1

Correct: Hold t constant, and correct

zi = zi − h(zi , ti )−1h(zi , ti )

until ‖h(zi , ti )‖ < ε, then let zi = zi .

Adjust: Cut the step length in half on failure of the corrector,and double it if several corrections at the current step sizehave been successful.

Terminate At t = 1, refine z to high accuracy by Newtoniteration.




Example 1.

There are different classes of homotopy methods, depending ondifferent upper bounds estimation of the number of isolatedsolutions. To show the effect on Example 1, three of them havebeen applied, in which we need to track 10 or more solution paths,while P(X ) has exactly just 5 isolated solutions.

Homotopy Upper bound No.pathsClassical homotopy Bezout number 16

m-Homogeneous homotopy m-Homogeneous Bezout number 16Polyhedral homotopy BKK bound 10




Two questions

1. Can we implement without transforming to a large polynomialsystem?

2. Can solution paths be reduced more in the path-followingprocedure?




Homotopy method for quadratic matrix equation

We propose a homotopy method to solve P(X ) = 0 in matrix form:

H(X , t) = (1− t)γQ(X ) + tP(X ),

whereQ(X ) = MX 2 + LX + K = 0

is the start quadratic matrix equation.





Theorem

Let N (P) denote the number of isolated solvents of

P(X ) = AX 2 + BX + C ,

(1)N (P) is finite and it is the same, say N , for almost allA, B, and C ∈ Cn×n, where

N =

(2n

n

);

(2)For all A, B, and C ∈ Cn×n, we have

N (P) ≤ N .





Theorem

For the homotopy

H(X , t) = γ(1− t)Q(X ) + tP(X ), t ∈ [0, 1) and γ ∈ C

whereQ(X ) = MX 2 + LX + K = 0

has exactly N mutually different isolated solvents, then, for almostall γ ∈ C,(1) The homotopy generates N smooth solution paths startingfrom isolated solvents of Q(X ).(2) As t → 1, the limits of the solution paths include all isolatedsolvents of P(X ).





Both of the theorems are based on the parameter continuation ofisolated solutions of polynomial systems, that is, for a class ofsystems whose coefficients are continuous functions of someparameters, a continuous curve in the parameter space determinesa continuous solution path accordingly.

A. J. Sommese and C. W. Wampler, II, The numericalsolution of systems of polynomials, Arising in engineering andscience. World Scientific Publishing Co. Pte. Ltd.,Hackensack, NJ, 2005.




Construct the start equation

Step1. Give 2n eigenpairs

{λi , vi}2ni=1

in which λ′i s are different from each other, and the eigenvectorssatisfy the Haar condition.Step2. Construct a quadratic lambda polynomial

Q(λ) = Mλ2 + Lλ+ K

as well as Q(X ), satisfying

Q(λ)vi = λivi , for i = 1, ..., 2n.

Step3. Compute all start solvents

X = W diag(µi )W−1, W = [w1, ...,wn], (1)

where the eigenpaires (µi ,wi )ni=1 are chosen from among the

eigenvpairs {λi , vi}2ni=1 of Q(λ).Yongwen Hou, Bo Yu HM for Solving AX 2 + BX + C = 0




Our first approach is based on the method for solving a specialkind of quadratic inverse eigenvalue problem, give by Chu and Xuin 2009:

M. T. Chu and S.-F. Xu, Spectral decomposition of realsymmetric quadratic λ-matrices and its applications., Math.Comput., 78 (2009), pp. 293-313.




Corollary

Choose eigenvalues λ1 > λ2 > ... > λ2n in R, and the columns of

V := [In,Un]

as the eigenvectors, where In and Un are n × n identity matrix andrandom orthogonal matrix respectively. If we let

Λ := diag{λ1, ..., λ2n} and S = diag{In,−In},

then, the matrix coefficients M, L and K of Q(X ) can be given by

M = (V ΛSV T )−1,L = −MV Λ2SV TM,K = −MV Λ3SV TM + CM−1C .





Another method allows us to reconstruct a monic matrixpolynomial

Q(X ) = X 2 + LX + K ,

along with its block companion matrix T of order n2

T =

(0 In−K −L

)from the information of the associated eigenpairs.

E. Pereira, On solvents of matrix polynomials, AppliedNumerical Mathematics, 47 (2003), pp. 197-208.




Corollary

Randomly choose λ1, ..., λ2n ∈ C as the eigenvalues of Q(λ), andv1, ..., v2n ∈ Cn as the eigenvectors, then the 2n × 2n nonsingularmatrix

S =

(v1 · · · vn · · · v2n

λ1v1 · · · λnvn · · · λ2nv2n

)is a similarity matrix of a block companion matrix associated withQ(X ), that is,

T = Sdiag(λ1...λ2n)S−1.

Thus, partitioning T as a 2× 2 block matrix, we have

M = In, L = −T22,K = −T12




Frechet derivative of the homotopy equation

Consider the homotopy equation

H(X , t) = ((1−t)M +tA)X 2 +((1−t)L+tB)X +((1−t)K +tC )X

.= AtX

2 + BtX + CtX ,and the expansion

H(X + E , t) = H(X ) + AtEX + (AtX + Bt)E + AtE2,

the Frechet derivative of H at X in the direction E along X is

H ′X (X , t; E ) = AtEX + (AtX + Bt)E ,

and the partial derivative of H to t is easy to get:

H ′t(X , t) = (A−M)X 2 + (B − L)X + (C − K )X

.Yongwen Hou, Bo Yu HM for Solving AX 2 + BX + C = 0



Predictor-corrector scheme in matrix form

Step 1. Construct a start equation Q(X ) = 0, along with a givenstart solvent X0, satisfying Q(X0) = 0;Step 2. Choose a tolerance ε > 0, and a step length h0;Step 3. Loop: for k = 0, 1, 2, ...

Predict: Solve (Euler-Predictor)

Ati EXi + (Ati Xi + Bti )E = −H ′t(Xi , ti ),

for E , and let

X = Xi + hiE , and ti+1 = ti + hi





Correct: Solve (Newton-Corrector)

Ati+1EX + (Ati+1X + Bti+1)E = −H(X , ti+1),

for E and correct X in the subspace of t = ti+1 by

X = X + E

repeatedly until the condition ‖H(X , ti+1)‖ ≤ ε is satisfied,and let

Xi+1 = X .





Adjust: Double the step length if several successivecorrections at the current step length have been successful,and cut the step length in half on failure of the corrector.Refine: Terminate and refine the solution at t = 1 usingthe Newton iteration.Next: Choose another start solvent and go to step 2.




Generalized sylvester equation

The generalized sylvester equation:

AEX + (AX + B)E = F ,

involved in the algorithm can be solved by the generalized schurmethod. Apply generalized schur decomposition to A and AX + B,and schur decomposition to X , we have

W ∗AZ = T , W ∗(AX + B)Z = S , and U∗XU = R,

and a substitution leads to

T E R + SE = F , E = Z ∗EU and F = W ∗FU.

Here, W , Z , and U are orthogonal matrices, while T , S , and S areupper diagonal matrices.




Generalized sylvester equation

Compute E column by column:

(S + RkkT )Ek = Fk −k−1∑i=1

RikT Ei , k = 1, 2, ..., n

and the solutionE = Z E U∗

is obtained. Hence, the predictor-corrector method respecting tothe matrix form can be done.




Computing all isolated matrix square roots

Consider the equationX 2 − C = 0,

which has at most 2n isolated solvents for general n × n complexmatrix C , we present a homotopy map with a trivial start equation

X 2 − D = 0,

where D = diag(d1, · · · , dn) with nonzero diagonal elementsdifferent from each other, to find all isolated square roots of C ifthere exists.




Principal-preserving property

The principal square root, which is the unique square root forwhich every eigenvalue has nonnegative real part, can be locatedjust by following one path, since the principal root goes to theprincipal along the homotopy path.




Overdamped problems

Definition

The quadratic eigenvalue problem

P(λ) = Aλ2 + Bλ+ C

is overdamped if A and B are symmetric positive definite, C issymmetric positive semidefinite and

(xTBx)2 > 4(xTAx)(xTCx) for all x 6= 0.




Properties overdamped problems

The eigenvalues are real and nonpositive and there is a gapbetween the n largest and the n smallest;

There are n linearly independent eigenvectors associated withthe n largest eigenvalues and likewise for the n smallest ones;

P(X ) has at least two real solvents, having as theireigenvalues the n largest eigenvalues and for the n smallestones respectively.




Order-preserving for overdamped problems

For overdamped problems, our algorithm guarantees the dominantsolvent of the start equation goes to the dominant solvent of thetarget equation, and likewise for the minimal one, that means onlyone solution curve needs to be tracked for computing a dominantor minimal solvent.




Numerical results

The algorithm is implemented in Matlab language, usingMatlab’s qz to do the generalized Schur decomposition . Allprogrammes are running on an ordinary computer, configured toInter Core i5 2.76GHz processor, 8G RAM, Windows 7 Ultimate.




Example 2.

We present a comparison between our algorithm with brute-forcemethod (here, applying the polyhedral homotopy method on thethe polynomial system p(x) = 0 with n2 equation in n2 variables)for random matrix equations of the size n = 2, n = 3, and n = 4.

nHOMOTOPY-QME HOM4PS2 PHCno.path time (s) no.path time (s) no.path time (s)

2 6 0.417 12 0.047 12 0.078

3 20 1.906 266 3.760 266 19.207

4 70 12.1486 26284 3037 26284 > 3600




Example 3.

Consider the matrix 6 6 3 5 17 1 5 6 17 1 2 8 22 4 7 9 86 9 2 5 2

,

all of its 32 isolated square roots can be found in 0.1s if thesymmetry property is considered, since one of the roots X and −Xis enough.




Recomputing Example 1

Again, consider the Example 1

X 2 +

[−1 −62 −9

]X +

[0 12−2 14

]= 0,

(42

)= 6 solution paths are tracked in our homotopy, 5 of which

converged to its all isolated solvents:(1 00 2

),

(1 02 3

),

(3 10 2

),

(1 00 4

), and

(4 20 2

),

and the other one diverged (converging to infinity).




Conclusion

We have proposed an algorithm to locate all isolated solventsof the unilateral quadratic matrix equation, with the strategyfor choosing the start equation.

In particular, it can be used to find all isolated square roots ofa given matrix it there exists.

We can locate a dominant or minimal solution foroverdamped problems of great interest in applications bytracing only one solution curve, and thus can be applied tosolve larger problems.




Thanks for your attention!


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A Homotopy Method for Computing All Isolated Solvents of ... · A Homotopy Method for Computing All Isolated Solvents of the Quadratic Matrix Equation AX 2 +BX +C = 0 Yongwen Hou