QUADRATIC EIGENVALUE PROBLEMS: SOLVENTS AND INVERSE …

QUADRATIC EIGENVALUE PROBLEMS:SOLVENTS AND INVERSE PROBLEMS

Peter Lancaster, University of Calgaryand

Francoise Tisseur, University of Manchester

Spring, 2009.

Lancaster/Tisseur () QEVP Spring, 2009. 1 / 22

Early Gohberg/Lancaster/Rodman Theory

U. of Calgary, Dept of Math and Stat, Research report 419, 1979. (126pp)

Papers in LAA, Canadian J.Math., IEOT, SIMAX, LAMA. 1977-1982.

Gohberg, I., Lancaster, P., and Rodman, L., Spectral analysis of matrixpolynomials, Annals of Math, 1982.

Gohberg, I., Lancaster, P., and Rodman, L., Matrix Polynomials,Academic Press, 1982. (SIAM 2009 - to appear)


The quadratic problem

Let M, D, K ∈ Cn×n, all Hermitian, with M > 0 and

L(λ) := Mλ2 + Dλ + K .

We will assume that L is monic; M = I .

We will also consider the important special case of D and K real andsymmetric.


Factorization of L(λ)

In general, do there exist matrices A (a right divisor) and S (a left divisor)such that

Inλ2 + Dλ + K = (Inλ− S)(Inλ− A)

and D∗ = D, K ∗ = K?

YES! But, with eigenvalues “evenly divided” with respect to the real axis.

Can we take advantage of this factorization to solve an inverse problem asfollows?

Assign n eigenvalues and eigenvectors as those of a right divisor A.

Determine a class of compatible left divisors S .

Choose a suitable S


Factorization of L(λ)

In general, do there exist matrices A (a right divisor) and S (a left divisor)such that

Inλ2 + Dλ + K = (Inλ− S)(Inλ− A)

and D∗ = D, K ∗ = K?YES! But, with eigenvalues “evenly divided” with respect to the real axis.

Can we take advantage of this factorization to solve an inverse problem asfollows?

Assign n eigenvalues and eigenvectors as those of a right divisor A.

Determine a class of compatible left divisors S .

Choose a suitable S


Preliminaries

Suppose that there are 2r real evs and 2(n − r) non-real ev (in n − rconjugate pairs).

A right divisor immediately determines n of these ev’s and n associatedright eigenvectors, because

(Iλ− A)x = 0, ⇒ L(λ)x = (Inλ− S)(Inλ− A)x = 0.

THE TRIVIAL SOLUTION: We can solve the Hermitian problem bytaking S = A∗. We eschew this solution! Why?

Because real eigenvalues of (Inλ− A∗)(Inλ− A) are necessarily defectiveand

(Inλ− A∗)(Inλ− A) ≥ 0 for all λ ∈ R.


Preliminaries

Suppose that there are 2r real evs and 2(n − r) non-real ev (in n − rconjugate pairs).

A right divisor immediately determines n of these ev’s and n associatedright eigenvectors, because

(Iλ− A)x = 0, ⇒ L(λ)x = (Inλ− S)(Inλ− A)x = 0.

THE TRIVIAL SOLUTION: We can solve the Hermitian problem bytaking S = A∗. We eschew this solution! Why?

Because real eigenvalues of (Inλ− A∗)(Inλ− A) are necessarily defectiveand

(Inλ− A∗)(Inλ− A) ≥ 0 for all λ ∈ R.


The Hermitian case

Let S = SR + iSI where SR = 12(S + S∗) and SI = −i

2 (S − S∗) areHermitian.It is not difficult to see that, for a left solvent S ,

SRA− A∗SR =1

2(A2 − (A∗)2). (1)

Then we can prove:

Theorem

Given a matrix A ∈ Cn×n a matrix S ∈ Cn×n is such that both S + A andSA are Hermitian if and only if

S = SR −1

2(A− A∗) = SR − iAI ,

where SR is an Hermitian solution of (1).

When the proposition holds we have, of course,

D = −(S + A) = −(SR + AR), K = SA.


The Hermitian case

Let S = SR + iSI where SR = 12(S + S∗) and SI = −i

2 (S − S∗) areHermitian.It is not difficult to see that, for a left solvent S ,

SRA− A∗SR =1

2(A2 − (A∗)2). (1)

Then we can prove:

Theorem

Given a matrix A ∈ Cn×n a matrix S ∈ Cn×n is such that both S + A andSA are Hermitian if and only if

S = SR −1

2(A− A∗) = SR − iAI ,

where SR is an Hermitian solution of (1).

When the proposition holds we have, of course,

D = −(S + A) = −(SR + AR), K = SA.Lancaster/Tisseur () QEVP Spring, 2009. 6 / 22

The real-symmetric case

With A ∈ Rn×n write

A = A1 + A2 where A1 =1

2(A + AT ), A2 =

1

2(A− AT ),

and similarly for S .

Corollary

Given a matrix A ∈ Rn×n a matrix S ∈ Rn×n is such that both S + A andSA are real and symmetric if and only if S2 = −A2 and S1 is a realsymmetric solution of

S1A− ATS1 =1

2(A2 − (AT )2). (2)


Real systems in real arithmetic

A is now to be a real right divisor and we seek real left divisors S . For thespectrum of A write

σ(A) = {σ1, . . . , σ2r} ∪ {ρ1, . . . , ρs} (3)

where 2r + s = n and

σ1 = µ1 + iω1, σ2 = µ2 − iω2, . . . , σ2r = µ2r − iω2r , are distinctnon-real numbers.

ρ1, . . . , ρs are distinct real numbers.

Consider a real Jordan canonical form, JR , for A, assuming that alleigenvalues are simple:

JR = diag

[[µ1 ω1

−ω1 µ1

], . . .

[µr ωr

−ωr µr

], ρ1, . . . , ρs

], (4)

and there is real (e-vector) matrix B such that A = BJRB−1.



A is now to be a real right divisor and we seek real left divisors S . For thespectrum of A write

σ(A) = {σ1, . . . , σ2r} ∪ {ρ1, . . . , ρs} (3)

where 2r + s = n and

σ1 = µ1 + iω1, σ2 = µ2 − iω2, . . . , σ2r = µ2r − iω2r , are distinctnon-real numbers.


Consider a real Jordan canonical form, JR , for A, assuming that alleigenvalues are simple:

JR = diag

[[µ1 ω1

−ω1 µ1

], . . .

[µr ωr

−ωr µr

], ρ1, . . . , ρs

], (4)

and there is real (e-vector) matrix B such that A = BJRB−1.Lancaster/Tisseur () QEVP Spring, 2009. 8 / 22


Lemma

If A has n distinct eigenvalues (as in (3)) and B−1AB = JR , then S1 is areal symmetric solution of ZA− ATZ = 0 if and only if S1 = B−TFB−1

where

F = diag

[[a1 b1

b1 −a1

], . . .

[ar br

br −ar

], c1, . . . , cs

], (5)

and the parameters aj , bj , ck are real.

Notice that FT = F and FJR = JTR F = (FJR)T .



Theorem

Given a matrix A ∈ Rn×n and a real matrix B for which A = BJRB−1 and(4) holds, all real matrices S for which A + S and AS are real andsymmetric have the form S = S1 + S2 where

S1 = A1 + B−TFB−1, S2 = −A2 (6)

and F is an arbitrary real symmetric matrix with the structure of (5).



Real symmetric system coefficients generated in this way are:

D = −(S + A) = −2A1 − B−TFB−1, (7)

K = SA = ATA + B−T (FJR)B−1. (8)

Clearly, they are determined by the n real parameters defining F in (5).

In this construction the number of real (or non-real) eigenvalues in A canbe doubled (in L(λ)), and there is no further constraint on their positionsin the complex plane.



Example: Assign

A = BJRB−1 =

2 −2 1 −42 −3 1 −32 −2 −1 −24 −3 1 −5

with eigenvalues −1± i , −2 − 3.If we take

F = diag

[[−2 00 2

], −4, −4

],

we obtain a matching left divisor

S =

−4 4 1 −41 0 1 −50 0 −9 100 −2 13 −22

with ev -28.9022, 1.5426, -4.8432, -2.7971.


The real symmetric system coefficients obtained are

D =

6 −3 −1 4

0 −1 510 −11

26

, K =

7 −8 −3 13

8 2 −824 −41

99

.


The Hermitian case

As we are interested in the case of mixed, real and non-real eigenvalues,suppose that

σ(A) = {σ1, . . . , σr} ∪ {ρ1, . . . , ρs} (9)

where r + s = n and

σ1, . . . , σr are distinct non-real numbers,

σj 6= σ̄k for j , k = 1, . . . , r (no conjugate pairs),


∆ := diag(σ1 · · ·σr ), R := diag(ρ1 · · · ρs). (10)


The Hermitian case

Then A has the spectral decomposition

A =r∑

j=1

σjPj +s∑

k=1

ρkQk , (11)

where P1, . . . ,Pr ,Q1, . . . Qs form a mutually biorthogonal system ofrank-one projectors onto the eigenspaces of A corresponding to the σj , ρk ,respectively.(If r > 0 then these assumptions do not admit the choice of real matricesA with non-real eigenvalues.)


The Hermitian case

Theorem

Given a matrix A ∈ Cn×n for which (11) and (10) hold, all matrices S forwhich A + S and AS are Hermitian have the form S = SR + iSI where, forsome diagonal E ∈ Rs×s ,

SR =1

2(A + A∗) + T−∗

[0 00 E

]T−1, SI =

i

2(A− A∗). (12)

This shows that S is determined by A and the s real parameters definingE .

(If A has no real eigenvalues, then the last term in the equation for SR

does not appear and the trivial solution

S =1

2(A + A∗) + i .

i

2(A− A∗) = A∗

is unique.)Lancaster/Tisseur () QEVP Spring, 2009. 16 / 22

Sign charcteristics

In both cases (real/symm. and Hermitian) we can keep track of the signcharacteristics of real eigenvalues.


Hermitian systems with A ∈ Rn×n

Right divisor Iλ− A be defined by

A =

1 1 02 3 21 1 2

.

Ev of A are 1, 2, 3.(a)If we choose e1 = e2 = e3 = 1, then

S =

24 1 2 1−1 3 10 2 2

35 +3X

j=1

ejYj =

24 3 5/2 3−1/2 7/2 1

2 2 5

35 .

L(λ) has ev 1, 2, 3 with sign characteristic -1.

The ev of S (1.78.., 2.69.., 7.02..) interlace those of A and have signcharacteristic +1.


(b) If e1 = e2 = e3 = 10, then

S =

24 21 7 114 8 120 2 32

35with ev 42.83..,12.91.., 5.25.., all of +ve type.

This determines a hyperbolic system since all ev are real and, assuming theev’s are ordered, L(λ) has n consecutive ev’s of one type followed by nconsecutive ev’s of the other type with a gap between the nth and n + 1stev’s.

(c) If we let e1 = e2 = 1, e3 = −1, then

S =

24 1 3/2 1−3/2 3 0

0 1 3

35with evs 1.82.., and a conjugate pair: 2.58..± i(1.05..).


EPILOGUE: Self-adjoint Jordan triples

n × n Hermitian systems: L(λ) =∑l

j=0 Ajλj , Al > 0.

Spectral properties encapsulated in a self-adjoint Jordan triple:

(X , J, PX ∗).

X is n × ln complex (defined by eigenvectors):J is ln × ln complex Jordan:P is ln × ln real: all entries 0 or ±1.(Accounts for sign charcteristics of real eigenvalues.)

⇒ Hermitian moments Γj = X (J j)PX ∗, j = 0, 1, ...⇒ formulae for Hermitian coefficients Aj .


EPILOGUE: Self-adjoint Jordan triples

n × n Hermitian systems: L(λ) =∑l

j=0 Ajλj , Al > 0.

Spectral properties encapsulated in a self-adjoint Jordan triple:

(X , J, PX ∗).

X is n × ln complex (defined by eigenvectors):J is ln × ln complex Jordan:P is ln × ln real: all entries 0 or ±1.(Accounts for sign charcteristics of real eigenvalues.)

⇒ Hermitian moments Γj = X (J j)PX ∗, j = 0, 1, ...⇒ formulae for Hermitian coefficients Aj .


Real symmetric systems

Are Hermitian! But have more structure.Spectral properties encapsulated in a real self-adjoint Jordan triple:

(XR , JR , PRXTR ).

XR is n × ln real (defined by eigenvectors):JR is ln × ln real Jordan:PR is ln × ln real: all entries 0 or ±1.(Essentially unchanged.)

⇒ real symmetric moments Γj = X (J j)PX ∗, j = 0, 1, ...⇒ formulae for real symmetric coefficients Aj .


Real symmetric systems

Are Hermitian! But have more structure.Spectral properties encapsulated in a real self-adjoint Jordan triple:

(XR , JR , PRXTR ).

XR is n × ln real (defined by eigenvectors):JR is ln × ln real Jordan:PR is ln × ln real: all entries 0 or ±1.(Essentially unchanged.)

⇒ real symmetric moments Γj = X (J j)PX ∗, j = 0, 1, ...⇒ formulae for real symmetric coefficients Aj .


Time to go!


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QUADRATIC EIGENVALUE PROBLEMS: SOLVENTS AND INVERSE …