Polynomial Jacobi Davidson Method for Large/Sparse Eigenvalue Problems

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Polynomial Jacobi Davidson Method for Large/Sparse Eigenvalue Problems

Tsung-Ming Huang

Department of Mathematics National Taiwan Normal University, Taiwan

April 28, 2011

T.M. Huang (Taiwan Normal Univ.) Poly. JD method for PEP April 28, 2011 1 / 42

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Outline

1 Jacobi’s orthogonal component correction

2 Davidson’s method

3 Jacobi-Davidson method

4 Polynomial Jacobi-Davidson method

5 Non-equivalence deflation

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References

SIAM J. Matrix Anal. Vol. 17, No. 2, PP. 401-425, 1996, Van der Vorst etc.

BIT. Vol. 36, No. 3, PP. 595-633, 1996, Van der Vorst etc.

T.-M. Hwang, W.-W. Lin and V. Mehrmann, SIAM J. Sci. Comput.

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Some basic theorems

Theorem 1

Let X be an eigenspace of A and let X be a basis for X . Then there is a unique matrix L such that

AX = XL.

The matrix L is given by

L = XÎAX, where XÎ is a matrix satisfying XÎX = I.

If (λ, x) is an eigenpair of A with x ∈ X , then (λ, X^Ix)is an eigenpair of L. Conversely, if (λ, u) is an eigenpair of L, then (λ, Xu) is an eigenpair of A.

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Proof: Let

X = [x₁· · · x_k] and Y = AX = [y₁· · · y_k] .

Since y_i∈ X and X is a basis for X , there is a unique vector `_i such that

yi = X`i. If we set L = [`1· · · `_k], then AX = XL and

L = X^IXL = X^IAX.

Now let (λ, x) be an eigenpair of A with x ∈ X . Then there is a unique vector u such that x = Xu. However, u = X^Ix. Hence

λx = Ax = AXu = XLu ⇒ λu = λX^Ix = Lu.

Conversely, if Lu = λu, then

A(Xu) = (AX)u = (XL)u = X(Lu) = λ(Xu), so that (λ, Xu) is an eigenpair of A.

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Theorem 2 (Optimal residuals) Let [X X⊥]be unitary. Let

R = AX − XL and S^H = X^HA − LX^H. Then kRk and kSk are minimized when

L = X^HAX, in which case

(a) kRk = kX_⊥^HAXk, (b) kSk = kX^HAX⊥k, (c) X^HR = 0.

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Proof: Set

X^H X_⊥^H

A

X X_⊥ =

Lˆ H

G M

. Then

X^H X_⊥^H

R =

Lˆ H

G M

X^H X_⊥^H

X −

X^H X_⊥^H

XL =

L − Lˆ G

. It implies that

kRk =

X^H X_⊥^H

R

=

L − Lˆ G

, which is minimized when L = ˆL = X^HAXand

min kRk = kGk = kX_⊥^HAXk.

The proof for S is similar. If L = X^HAX, then

X^HR = X^HAX − X^HXL = X^HAX − L = 0.

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Definition 3

Let X be of full column rank and let X^I be a left inverse of X. Then X^IAX is a Rayleigh quotient of A.

Theorem 4

Let X be orthonormal, A be Hermitian and R = AX − XL.

If `₁, . . . , `_kare the eigenvalues of L, then there are eigenvalues λj1, . . . , λjk of A such that

|`_i− λ_j_i| ≤ kRk₂ and v u u t

k

X

i=1

(`i− λ_j_i)²≤√

2kRkF.

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Jacobi’s orthogonal component correction, 1846

Consider the eigenvalue problem A

1 z

≡

α c^T b F

1 z

= λ

1 z

, (1)

where A is diagonal dominant and α is the largest diagonal element.

(1) is equivalent to

λ = α + c^Tz, (F − λI)z = −b.

Jacobi iteration: (with z₁ = 0)

θk= α + c^Tzk,

(D − θ_kI)z_k+1 = (D − F )z_k− b (2) where D = diag(F ).

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Davidson’s method (1975)

Algorithm 1 (Davidson’s method) Given unit vector v, set V = [v]

Iterate until convergence

Compute desired eigenpair (θ, s) ofV^TAV. Computeu = V sandr = Au − θu.

If (k r k₂< ε), stop.

Solve(D_A− θI)t = r.

Orthog. t ⊥ V → v, V = [V, v]

end

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Let u_k= (1, z_k^T)^T. Then

rk = (A − θkI)uk=

α − θ_k+ c^Tz_k (F − θ_kI)z_k+ b

Substituting the residual vector r_kinto linear systems (DA− θ_kI)t_k= −r_k, where DA=

α 0

0 D

, we get

(D − θkI)yk = −(F − θ_kI)zk− b

= (D − F )zk− (D − θ_kI)zk− b From (2) and above equality, we see that

(D − θ_kI)(z_k+ y_k) = (D − F )z_k− b = (D − θ_kI)z_k+1.

This implies that z_k+1= z_k+ y_kas one step of JOCC starting with z_k.

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Jacobi-Davidson method (1996)

(θk, uk): approx. eigenpair of A, θ_k≈ λ, with

u_k= V_ks_k, V_k^TAV_ks_k= θ_ks_k and k s_kk₂= 1.

Definition 5

(θk, uk)is called a Ritz pair of A. θ_kis called a Ritz value and u_kis a Ritz vector.

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Jacobi-Davidson method (1996)

(θk, uk): approx. eigenpair of A, θ_k≈ λ, with

u_k= V_ks_k, V_k^TAV_ks_k= θ_ks_k and k s_kk₂= 1.

Then

u^T_kr_k ≡ u^T_k(A − θ_kI)u_k = s^T_kV_k^TAV_ks_k− θ_ks^T_kV_k^TV_ks_k = 0 ⇒ r_k⊥ u_k Find the correctiont ⊥ uksuch that

A(u_k+ t) = λ(u_k+ t).

That is

(A − λI)t = λu_k− Au_k= −r_k+ (λ − θ_k)u_k.

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Since (I − u_ku^T_k)rk= rk, we have

I − u_ku^T_k (A − λI)t = −r_k.

Correction equation

(I − uku^T_k)(A − θkI) I − uku^T_k t = −r_k and t ⊥ u_k,

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Solving correction vector t

Correction equation:

I − uku^T_k (A − θ_kI)(I − uku^T_k)t = −rk, t ⊥ uk. (3) Scheme S_OneLS:

Use preconditioning iterative approximations, e.g., GMRES, to solve (3).

Use a preconditioner

M_p ≡ I − u_ku^T_k M I − u_ku^T_k , where M is an approximation of A − θ_kI. In each of the iterative steps, it needs to solve

M_py = b, y ⊥ u_k (4)

for a given b.

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Since y ⊥ u_k, Eq. (4) can be rewritten as

I − uku^T_k My = b ⇒ My = u^T_kMy u_k+ b ≡ ηkuk+ b.

Hence

y = M⁻¹b + η_kM⁻¹u_k, where

ηk= − u^T_kM⁻¹b u^T_kM⁻¹uk

.

SSOR preconditioner: Let A − θ_kI = L + D + U. Then M = (D + ωL)D⁻¹(D + ωU ).

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Scheme S_{T woLS}: Since t ⊥ u_k, Eq. (3) can be rewritten as

(A − θ_kI)t = u^T_k(A − θ_kI)t u_k− r_k≡ εu_k− r_k. (5) Let t₁and t₂ be approximated solutions of the following linear systems:

(A − θ_kI)t = −r_k and (A − θ_kI)t = u_k, respectively. Then the approximated solution ˜tfor (5) is

˜t = t1+ εt2 for ε = −u^T_kt₁ u^T_kt2

. Scheme S_OneStep: The approximated solution ˜tfor (5) is

˜t = −M⁻¹r_k+ εM⁻¹u_k for ε = u^>_kM⁻¹r_k u^>_kM⁻¹u_k, where M is an approximation of A − θ_kI.

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Algorithm 2 (Jacobi-Davidson Method) Choose an n-by-m orthonormal matrix V0

Do k = 0, 1, 2, · · ·

Compute all the eigenpairs ofV_k^TAVks = λs.

Select the desired (target) eigenpair (θ_k, s_k)with ks_kk₂ = 1.

Computeu_k = V_ks_kandr_k = (A − θ_kI)u_k. If (kr_kk₂< ε), λ = θ_k, x = u_k, Stop

Solve (approximately) at_k⊥ u_k from

(I − u_ku^T_k)(A − θ_kI)(I − u_ku^T_k)t = −r_k. Orthogonalize t_k⊥ V_k→ v_k+1, V_k+1= [V_k, v_k+1]

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Locking:

V_kwith V_k^∗V = I_kare convergentSchur vectors, i.e., AVk = VkTk

for someupper triangularT_k. SetV = [V_k, V_q]with V^∗V = I_k+q in k + 1-th iteration of Jacobi-Davidson Algorithm. Then

V^∗AV =

V_k^∗AV_k V_k^∗AV_q V_q^∗AV_k V_q^∗AVq

=

T_k V_k^∗AV_q V_q^∗V_kT_k V_q^∗AVq

=

T_k V_k^∗AV_q 0 V_q^∗AVq

. Restarting:

Keep the locked Schur vectors as well as the Schur vectors of interest in the subspace and throw away those we are not interested.

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Remark 1

If ε = 0, we obtain Davidson method with t₁ = −(D_A− θ_kI)⁻¹r.

( ˜tis NOT orthogonal to u_k )

If the linear systems in (6) are exactly solved, then the vector t becomes

t = ε(A − θ_kI)⁻¹u_k− u_k. (6) Since t is made orthogonal to u_k, (6) is equivalent to

t = (A − θkI)⁻¹uk

which is equivalent to shift-invert power iteration. Hence it is quadratic convergence.

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Consider Ax = λx and assume that λ is simple.

Lemma 6

Consider w with w^Tx 6= 0. Then the map Fp ≡

I −xw^T w^Tx

(A − λI)

I −xw^T w^Tx

is a bijection from w^⊥ to w^⊥.

Proof: Suppose y⊥w and Fpy = 0. That is

I − xw^T w^Tx

(A − λI)

I −xw^T w^Tx

y = 0.

Then it holds that

(A − λI)y = εx.

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Therefore, both y and x belong to the kernel of (A − λI)². The

simplicity of λ implies that y is a scale multiple of x. The fact that y⊥w and x^Tw 6= 0implies y = 0, which proves the injectivity of Fp. An obvious dimension argument implies bijectivity.

Extension F_pt ≡

I − uu^T u^Tu

(A − θI)

I − uu^T u^Tu

t = −r, t ⊥ u, r ⊥ u.

Then

t ∈ u^⊥→

Fp

r ∈ u^⊥.

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Theorem 7

Assume that the correction equation

I − uu^T (A − θI) I − uu^T t = −r, t ⊥ u (7) is solved exactly in each step of Jacobi-Davidson Algorithm. Assume uk= u → xand u^T_kxhas non-trivial limit. Then if u_kis sufficiently chosen to x, then u_k→ x with locally quadratical convergence and

θk= u^T_kAuk → λ.

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Proof: Suppose Ax = λx with x such that x = u + z for z ⊥ u. Then (A − θI) z = − (A − θI) u + (λ − θ) x = −r + (λ − θ) x. (8) Consider the exact solution z₁ ⊥ u of (7):

(I − P ) (A − θI) z₁ = − (I − P ) r, (9) where P = uu^T. Note that (I − P ) r = r since u⊥r. Since

x − (u + z₁) = z − z₁ and z = x − u, for quadratic convergence, it suffices to show that

kx − (u + z₁) k = kz − z1k = O kzk² . (10) Multiplying (8) by (I − P ) and subtracting the result from (9) yields

(I − P ) (A − θI) (z − z₁) = (λ − θ) (I − P ) z + (λ − θ) (I − P ) u. (11)

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Multiplying (8) by u^T and using r ⊥ u leads to λ − θ = u^T(A − θI) z

u^Tx . (12)

Since u^T_kxhas non-trivial limit, we obtain k (λ − θ) (I − P ) zk =

u^T (A − θI) z

u^Tx (I − P ) z

. (13)

From (11), Lemma 6 and (I − P ) u = 0, we have kz − z₁k =

h

(I − P ) (A − θkI) |_u^⊥

k

i−1

(λ − θ) (I − P ) z

= h

(I − P ) (A − θkI) |_u^⊥

k

i−1 u^T_k (A − θ_kI) z

u^T_kx (I − P ) z

= O kzk² .

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Jacobi Davidson method as on accelerated Newton Scheme

Consider Ax = λx and assume that λ is simple. Choose w^Tx = 1.

Consider nonlinear equation

F (u) = Au − θ(u)u = 0 with θ (u) = w^TAu w^Tu ,

where kuk = 1 or w^Tu = 1. Then F : {u|w^Tu = 1} → w^⊥. In particular, r ≡ F (u) = Au − θ (u) u ⊥ w.

Suppose u_k≈ x and the next Newton approximation uk+1: u_k+1 = u_k− ∂F

∂u u=uk

!−1

F (u_k) is given by u_k+1= u_k+ t, i.e., t satisfies that

∂F

∂u u=u_k

!

t = F (u_k) = −r.

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Since 1 = u^T_k+1w = (u_k+ t)^Tw = 1 + t^Tw, it implies that w^Tt = 0. By the definition of F , we have

∂F

∂u = A − θ (u) I −− w^TAu uw^T + w^Tu uw^TA (w^Tu)²

= A − θI + w^TAu

(w^Tu)²uw^T −uw^TA w^Tu =

I −u_kw^T w^Tu_k

(A − θ_kI) . Consequently, the Jacobian of F acts on w^⊥and is given by

∂F

∂u u=uk

! t =

I − u_kw^T w^Tu_k

(A − θ_kI) t, t ⊥ w.

Hence the correction equation of Newton method read as t ⊥ w,

I − ukw^T w^Tu_k

(A − θkI) t = −r,

which is the correction equation of Jacobi-Davidson method in (7) with w = u.

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Polynomial Jacobi-Davidson method

(θ_k, u_k): approx. eigenpair of A(λ) ≡Pτ

k=0λ^kA_k, θ_k≈ λ, with u_k= V_ks_k, V_k^>A(λ)V_ks_k= 0 and k s_kk₂= 1.

Let

r_k= A(θ_k)u_k. Then

u^>_krk= u^>_kA(θk)uk= s^>_kV_k^>A(θk)Vksk= 0 ⇒ rk ⊥ u_k Find the correctiontsuch that

A(λ)(u_k+ t) = 0.

That is

A(λ)t = −A(λ)u_k = −r_k+ (A(θ_k) − A(λ))u_k.

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Let

p_k = A⁰(θ_k)u_k≡

τ

X

i=1

iθⁱ⁻¹_k Ai

! u_k.

A(λ) = A − λI: p_k= −u_k,

(A(θk) − A(λ))uk = (λ − θk)uk= (θk− λ_k)pk

A(λ) = A − λB:

p_k= −Bu_k,

(A(θ_k) − A(λ))u_k= (λ − θ_k)Bu_k= (θ_k− λ)p_k A(λ) =Pτ

i=0λⁱAiwith τ ≥ 2:

(A(θ_k) − A(λ))u_k =

(θ_k− λ)A⁰(θ_k) −1

2(θ_k− λ)²A⁰⁰(ξ_k)

u_k

= (θk− λ)p_k−1

2(θk− λ)²A⁰⁰(ξk)uk

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Hence

A(λ)t = −r_k+ (θ_k− λ)p_k−1

2(θ_k− λ)²A⁰⁰(ξ_k)u_k Since r_k⊥ u_k, we have

I − p_ku^>_k u^>_kpk

A(λ)t = −rk−1

2(θk− λ)²

I − p_ku^>_k u^>_kpk

A⁰⁰(ξk)uk. Correction equation:

I − pku^>_k u^>_kp_k

A(θ_k)(I − u_ku^>_k)t = −r_k and t ⊥ u_k,

or

I −pku^>_k u^>_kp_k

(A − θ_kB)

I −ukp^>_k p^>_ku_k

t = −r_k and t ⊥_Bu_k, with symmetric positive definite matrix B.

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Algorithm 3 (Jacobi-Davidson Algorithm for solving A(λ)x = 0) Choose an n-by-m orthonormal matrix V₀

Do k = 0, 1, 2, · · ·

Compute all the eigenpairs ofV_k^>A(λ)V_k = 0.

Select the desired (target) eigenpair (θ_k, s_k)with ks_kk₂ = 1.

Computeuk = Vksk,rk= A(θk)ukandpk= A⁰(θk)uk. If (kr_kk₂< ε), λ = θ_k, x = u_k, Stop

Solve (approximately) at_k⊥ u_k from (I −^p_u^kT^u^T^k

kpk)A(θ_k)(I − u_ku^T_k)t = −r_k. Orthogonalize t_k⊥ V_k→ v_k+1, V_k+1= [Vk, vk+1]

pk = A⁰(θk)uk≡

τ

X

i=1

iθⁱ⁻¹_k Ai

! uk.

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Non-equivalence deflation of quadratic eigenproblems

Let λ1 be a real eigenvalue of Q(λ) ≡ λ²M + λC + K and x1, z1be the associated right and left eigenvectors, respectively, with z^T₁Kx₁= 1.

Let

θ1 = (z₁^TM x1)⁻¹. We introduce a deflated quadratic eigenproblem

Q(λ)x ≡e h

λ²M + λ ef C + eK i

x = 0, where

Mf = M − θ1M x1z₁^TM, Ce = C + θ₁

λ₁(M x₁z₁^TK + Kx₁z₁^TM ), Ke = K − θ₁

λ²₁Kx₁z^T₁K.

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Complex deflation

Let λ1 = α1+ iβ1be a complex eigenvalue of Q(λ) and

x₁= x_1R+ ix_1I, z₁ = z_1R+ iz_1I be the associated right and left eigenvectors, respectively, such that

Z₁^TKX1= I2, where X1 = [x1R, x1I]and Z1= [z1R, z1I]. Let

Θ₁= (Z₁^TM X₁)⁻¹.

Then we introduce a deflated quadratic eigenproblem with Mf = M − M X₁Θ₁Z₁^TM,

Ce = C + M X₁Θ₁Λ^−T₁ Z₁^TK + KX₁Λ⁻¹₁ Θ^T₁Z₁^TM, Ke = K − KX1Λ⁻¹₁ Θ1Λ^−T₁ Z₁^TK

in which Λ₁ =

α1 β1

−β₁ α₁

.

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Theorem 8

(i) Let λ₁ be a simple real eigenvalue of Q(λ). Then the spectrum of Q(λ)e is given by

(σ(Q(λ)){λ1}) ∪ {∞}

provided that λ²₁6= θ₁.

(ii) Let λ₁ be a simple complex eigenvalue of Q(λ). Then the spectrum of eQ(λ)is given by

σ(Q(λ)){λ1, ¯λ₁} ∪ {∞, ∞}

provided that Λ1Λ^T₁ 6= Θ₁.

Furthermore, in both cases (i) and (ii), if λ₂6= λ₁ and (λ₂, x₂)is an eigenpair of Q(λ) then the pair (λ₂, x₂)is also an eigenpair of eQ(λ).

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Suppose that M, C, K are symmetric. Given an eigenmatrix pair (Λ1, X1) ∈ R^r×r× R^n×r of Q(λ), where Λ₁ is nonsingular and X₁ satisfies

X₁^TKX1= Ir, Θ1 := (X₁^TM X1)⁻¹. We define ˜Q(λ) := λ²M + λ ˜˜ C + ˜K, where

M˜ := M − M X₁Θ₁X₁^TM,

C˜ := C + M X₁Θ₁Λ^−T₁ X₁^TK + KX₁Λ⁻¹₁ Θ₁X₁^TM, K˜ := K − KX1Λ⁻¹₁ Θ1Λ^−T₁ X₁^TK.

Theorem 9

Suppose that Θ₁− Λ₁Λ^T₁ is nonsingular. Then the eigenvalues of the real symmetric quadratic pencil ˜Q(λ)are the same as those of Q(λ) except that the eigenvalues of Λ₁, which are closed under complex conjugation, are replaced by r infinities.

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Proof: Since (Λ1, X1)is an eigenmatrix pair of Q(λ), i.e., M X1Λ²₁+ CX1Λ1+ KX1 = 0,

we have

Q(λ)˜ = Q(λ) + [M X1(λIr+ Λ1) + CX1] Θ1Λ^−T₁ (X₁^TK − λΛ^T₁X₁^TM )

= Q(λ) + Q(λ)X₁(λI_r− Λ₁)⁻¹Θ₁Λ^−T₁ (X₁^TK − λΛ^T₁X₁^TM ).

By using the identity

det(I_n+ RS) = det(I_m+ SR), where R, S^T ∈ R^n×m, we have

det[ ˜Q(λ)]

= det[Q(λ)]det[I + X1(λIr− Λ₁)⁻¹Θ1Λ^−T₁ (X₁^TK − λΛ^T₁X₁^TM )]

= det[Q(λ)]det[I_r+ (λI_r− Λ₁)⁻¹Θ₁Λ^−T₁ (I_r− λΛ^T₁Θ⁻¹₁ )]

= det[Q(λ)]

det(λIr− Λ₁)det(Θ₁Λ^−T₁ − Λ₁).

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Since (Θ₁− Λ₁Λ^T₁) ∈ R^r×ris nonsingular, we have det(Θ₁Λ^−T₁ − Λ₁) 6= 0.

Therefore, ˜Q(λ)has the same eigenvalues as Q(λ) except that r eigenvalues of Λ₁ are replaced by r infinities.

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Non-equiv. deflation for cubic poly. eigenproblems

Let (Λ, Vu) ∈ R^r×r× R^n×r be an eigenmatrix pair of

A(λ) ≡ λ³A₃+ λ²A₂+ λA₁+ A₀ (14) with V_u^TV_u= I_rand 0 /∈ σ(Λ), i.e.,

A3VuΛ³+ A2VuΛ²+ A1VuΛ + A0Vu = 0. (15) Define a new deflated cubic eigenvalue problem by

A(λ)u = (λe ³A˜3+ λ²A˜2+ λ ˜A1+ ˜A0)u = 0, (16) where











A˜0 = A0,

A˜₁ = A₁− (A₁V_uV_u^T + A₂V_uΛV_u^T + A₃V_uΛ²V_u^T), A˜₂ = A₂− (A₂V_uV_u^T + A₃V_uΛV_u^T),

A˜₃ = A₃− A₃V_uV_u^T.

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Lemma 10

Let A(λ) and eA(λ)be cubic pencils given by (14) and (16), respectively. Then it holds

A(λ) = A(λ) Ie n− λV_u(λIr− Λ)⁻¹V_u^T . (18) Theorem 11

Let (Λ, V_u)be an eigenmatrix pair of A(λ) with V_u^TV_u = I_r. Then (i) (σ(A(λ))σ(Λ)) ∪ {∞} = σ( eA(λ)).

(ii) Let (µ, z) be an eigenpair of A(λ) with k z k2= 1and µ /∈ σ(Λ).

Define

˜

z = (In− µV_uΛ⁻¹V_u^T)z ≡ T (µ)z. (19) Then (µ, ˜z)is an eigenpair of eA(λ).

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Proof of Lemma: Using (17) and (15), and the fundamental matrix calculation, we have

A(λ)e = A(λ) − λ λ²A3VFV_F^T + λA2VFV_F^T + λA3VFΛV_F^T +A₁V_FV_F^T + A₂V_FΛV_F^T + A₃V_FΛ²V_F^T

= A(λ) − λ A3VF(λIr− Λ)³(λIr− Λ)⁻¹V_F^T +3A3VFΛ(λIr− Λ)²(λIr− Λ)⁻¹V_F^T +3A₃V_FΛ²(λI_r− Λ)(λI_r− Λ)⁻¹V_F^T +A₂V_F(λI_r− Λ)²(λI_r− Λ)⁻¹V_F^T +2A2V_FΛ(λIr− Λ)(λI_r− Λ)⁻¹V_F^T

+A1VF(λIr− Λ)(λI_r− Λ)⁻¹V_F^T

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A(λ)e = A(λ) − λA₃V_F(λ³Ir− Λ³) + A2V_F(λ²Ir− Λ²) +A1VF(λIr− Λ) + A₀VF − A₀VF] (λIr− Λ)⁻¹V_F^T

= A(λ) − λA(λ)V_F(λI_r− Λ)⁻¹V_F^T

= A(λ)I_n− λV_F(λIr− Λ)⁻¹V_F^T .

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Proof of Theorem: (i) Using the identity

det(I_n+ RS) = det(I_m+ SR) and Lemma 10, we have

det( eA(λ)) = det(A(λ)) det In− λV_F(λIr− Λ)⁻¹V_F^T

= det(A(λ)) det In− λ(λI_r− Λ)⁻¹

= det(A(λ)) det(λIr− Λ)⁻¹det(−Λ).

Since 0 /∈ σ(Λ), det(−Λ) 6= 0. Thus, eA(λ)and A(λ) have the same finite spectrum except the eigenvalues in σ(Λ). Furthermore, dividing Eq. (16) by λ³ and using the fact that

Ae3VF = (A3− A₃VFV_F^T)VF = 0,

we see that (diag_r{∞, · · · , ∞}, V_F)is an eigenmatrix pair of eA(λ) corresponding to infinite eigenvalues.

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(ii) Since µ /∈ σ(Λ), the matrix T (µ) = (I − µV_FΛ⁻¹V_F^T)in (19) is invertible with the inverse

T (µ)⁻¹= I_n− µV_F(µI_r− Λ)⁻¹V_F^T. (20) From Lemma 10, we have

A(µ)˜e z = A(µ)I_n− µV_F(µIr− Λ)⁻¹V_F^T I_n− µV_FΛ⁻¹V_F^T z = 0.

This completes the proof.