PAMM · Proc. Appl. Math. Mech. 16, 873 – 874 (2016) / DOI 10.1002/pamm.201610425
On the parametric eigenvalue behavior of matrix pencils under rank one perturbations Hannes Gernandt1,∗ and Carsten Trunk1 1
Institute of Mathematics, TU Ilmenau, Weimarer Straße 25, 98693 Ilmenau, Germany.
We study the eigenvalues of rank one perturbations of regular matrix pencils depending linearly on a complex parameter. We prove properties of the corresponding eigenvalue sets including a convergence result as the parameter tends to infinity and an eigenvalue interlacing property for real valued pencils having real eigenvalues only. c 2016 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim
We consider matrix pencils A(s) = sE − A for s ∈ C with E, A ∈ Cn×n . Here the pencil A is assumed to be regular, which means that det(sE − A) is not the zero polynomial. Otherwise we say that A is singular. For A regular the finite eigenvalues are given by the zeros of the characteristic polynomial det(sE − A) and ∞ is said to be an eigenvalues of A if E is not invertible. The set of all eigenvalues of A is denoted by σ(A). For regular A and λ ∈ C we denote the geometric multiplicity of λ by gmA (λ) := dim ker(λE − A) and the algebraic multiplicity amA (λ) is the zero order of det A(s) at λ. For λ = ∞ one defines gmA (∞) := dim ker E and amA (∞) := n − deg det A(s). It is our aim to investigate the eigenvalues under parameter dependent rank one perturbations of the form τ P(s) := τ (αs − β)uv T ,
α, β ∈ C, u, v ∈ Cn ,
(1)
where τ is a parameter varying over C. Low-rank perturbation results for matrix pencils were obtained in [1, 4]. We follow the approach from [5], where generic results on the parametric eigenvalue behavior of matrices, i.e. E = In were obtained. For A(s) = sE − A regular there exists S, T ∈ Cn×n and r ∈ N such that SA(s)T is in Weierstraß canonical form, i.e. I 0 J 0 S(sE − A)T = s r − , J ∈ Cr×r , N ∈ C(n−r)×(n−r) (2) 0 N 0 In−r with J and N in Jordan canonical form and N nilpotent. For λ ∈ σ(A)\{∞} we denote by muv (λ) the order of the pole of s 7→ (αs−β)v T (sE −A)−1 u at λ and set muv (λ) := 0 if there is no pole. For ∞ ∈ σ(A) the number muv (∞) is the order of the pole of s 7→ (α − βs)v T (−sA + E)−1 u at 0. = sE − A be regular, let P be of the form (1) and consider the polynomials m(s) := Q Proposition 1 Let mA(s) uv (λ) (s − λ) and p(s) := (αs − β)v T m(s)(sE − A)−1 u. λ∈σ(A)\{∞} (a) Assume that there exists λ ∈ σ(A) with muv (λ) > 0. Then A + τ P is regular for all τ ∈ C. (i) For all µ ∈ σ(A) with muv (µ) > 0 we have amA+τ P (µ) = amA (µ) − muv (µ) for all τ ∈ C \ {0}. (ii) For µ ∈ σ(A) \ {∞} with muv (µ) = 0 and p(µ) = 0 we have amA+τ P (µ) = amA (µ) for all τ ∈ C. (iii) For µ ∈ σ(A) \ {∞} with muv (µ) = 0 and p(µ) 6= 0 we have amA+τ P (µ) = amA (µ) for all τ ∈ C \ {− m(µ) p(µ) }. (b) Assume that muv (λ) = 0 for all λ ∈ σ(A). Then m ≡ 1 and the polynomial p is constant. (i) If p ≡ 0 then det(A + τ P)(s) = det A(s) for all τ ∈ C. Hence amA+τ P (λ) = amA (λ) for all τ ∈ C and all λ ∈ σ(A). (ii) If p ≡ c for some c ∈ C \ {0} then amA+τ P (λ) = amA (λ) for all λ ∈ σ(A) and τ ∈ C \ {−1/c} and A − 1/cP is singular. P r o o f. From Sylvester’s determinant identity we conclude the following factorization det(A + τ P)(s) =
det A(s) det A(s) (m(s) + τ (αs − β)m(s)v T (sE − A)−1 u) = (m(s) + τ p(s)). m(s) m(s)
(3)
For muv (λ) > 0 and λ 6= ∞ one has m(λ) = 0 and p(λ) 6= 0 which implies that the second factor does not vanish identically. Since both factors are not the zero polynomial, A + τ P is regular for all τ ∈ C with amA+τ P (λ) = amA (λ) − muv (λ). For λ = ∞ one can proof the above by considering (3) for the dual pencil A0 + τ P 0 (s) := −As + E + τ (−βs + α)uv T and the fact amA (∞) = amA0 (0) (cf. [1, 3]). The remaining statements (ii) and (iii) of (a) are consequence of m(µ) 6= 0 for all µ ∈ σ(A) \ {∞} with muv (µ) = 0. The assertion (b) follows from the definition of m, muv (λ) and equation (3). ∗
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c 2016 Wiley-VCH Verlag GmbH & Co. KGaA, Weinheim
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Section 23: Applied operator theory
In the next theorem we describe the eigenvalue behavior as τ → ∞. Theorem 2 Let A be P regular and let P be of the form (1) such that there exists λ ∈ σ(A) with muv (λ) > 0. Then there exists τ0 ∈ C such that µ∈σ(A+τ P)\σ(A) amA+τ P (µ) = max{deg m, deg p} for all τ ≥ τ0 . Then deg p eigenvalues, counting with multiplicity, converge for τ → ∞ to the zero set of p and max{0, deg m − deg p} eigenvalues converge to ∞. P r o o f. For τ ∈ C \ {0} we write the last factor in (3) as τ −1 m + p and consider this polynomial in a neighborhood of the roots of p given by Cj (ε) := {λ ∈ C | |λ − µj | < ε} with j = 1, . . . , k. Here ε > 0 is chosen such that these discs are pairwise disjoint. As τ → ∞ the polynomial converges on ∪kj=1 Cj (ε) uniformly to p. By Rouche’s theorem the number of zeros of p and τ −1 m + p inside the discs coincide and hence these eigenvalues of A + τ P by (3) converge to the zeros of p. For deg p < deg m the convergence to ∞ follows from considering the dual pencil A0 + τ P 0 and applying the above argument to a disc Cj (ε) around µj = 0. For matrix pencils with σ(A) ⊆ R ∪ {∞} and semi-simple eigenvalues, i.e. amA (λ) = gmA (λ) for all λ ∈ σ(A), we describe the interlacing of the eigenvalues in the next theorem. Theorem 3 Let A(s) = sE −A be regular with E, A ∈ Rn×n and only real semi-simple eigenvalues λ0 < . . . < λm < ∞ and let P be given by (1) such that u, v ∈ Rn , α ∈ R \ {0}, β ∈ R and β/α ∈ / σ(A). Then there transformation matrices S, T ∈ Rn×n such that (2) holds with N = 0 and J diagonal and we decompose according to the eigenspaces Su = T T (uT0 , . . . , uTm , uT∞ )T , v T T = (v0T , . . . , vm , v∞ ) with ui , vi ∈ Rmi and u∞ , v∞ ∈ Rn−r . Denote by i1 < . . . < im0 < ∞ all T T indices with (β − αλik )vik uik 6= 0 for k = 1, . . . , m0 and we assume v∞ u∞ = 0 and m0 ≥ 2. (a) If viT ui 6= 0 then amA+τ P (λi ) = amA (λi ) − 1 for all τ ∈ C \ {0}. If viT ui = 0 then amA+τ P (λi ) = amA (λi ) for all i) τ ∈ C \ {− m(λ p(λi ) }. (b) If (β − αλik )viTk uik > 0 for all k = 1, . . . , m0 then for all τ ∈ (0, ∞) there exists λik (τ ) ∈ C with σ(A + τ P) ∩ (λik , λik+1 ) = {λik (τ )} ∪ (σ(A) ∩ (λik , λik+1 )) for all k = 1, . . . , m0 − 1 and σ(A + τ P) ∩ ((λm0 , ∞] ∪ (−∞, λi1 )) = {λim0 (τ )} ∪ (σ(A) ∩ ((λm0 , ∞] ∪ (−∞, λi1 ))), i.e. that all eigenvalues λik move to the right. (c) If (β − αλik )viTk uik < 0 for all k = 1, . . . , m0 then for all τ ∈ (0, ∞) there exists λik+1 (τ ) ∈ C with σ(A + τ P) ∩ (λik , λik+1 ) = {λik+1 (τ )}∪(σ(A)∩(λik , λik+1 )) for all k = 1, . . . , m0 −1 and σ(A+τ P)∩((λm0 , ∞]∪(−∞, λi1 )) = {λi1 (τ )} ∪ (σ(A) ∩ ((λm0 , ∞] ∪ (−∞, λi1 ))), i.e. that all eigenvalues λik move to the left. The assumptions of (b) and (c) imply that σ(A + τ P) ⊆ R ∪ {∞} holds for all τ ∈ R. P r o o f. First note that for E, A ∈ R and σ(A) ⊂ R ∪ {∞} there exist S, T ∈ Rn×n such that (2) holds with N = 0 and J diagonal (cf. [2]). The property (a) follows from Proposition 1 because viT ui 6= 0 implies muv (λi ) = 1 and viT ui = 0 implies muv (λi ) = 0. We continue with the proof of (b). For given τ > 0 and λ ∈ σ(A + τ P) \ σ(A) we obtain from (3) the equation m(λ) + τ p(λ) = 0. We show that there is a solution to this equation λik (τ ) ∈ (λik , λik+1 ). By the assumption Pm0 (β−αλ)viTk uik T . On the interval (λik , λik+1 ) the right hand side is v∞ u∞ = 0, the equation above reduces to τ −1 = k=1 λ−λik a continuous function that maps onto R in the case β/α ∈ / (λik , λik+1 ) and onto [0, ∞) for β/α ∈ (λik , λik+1 ). Hence we conclude that for every τ ∈ (0, ∞) there is a solution λik (τ ) ∈ (λik , λik+1 ) for all k = 1, . . . , m0 − 1. For the interval around ∞ given by (λim0 , ∞] ∪ (−∞, λi1 ) one considers the dual pencil which satisfies λ ∈ σ(A0 ) \ {0} if and only if λ−1 ∈ σ(A) \ {∞} as it can be derived from (2), to see that there is at least one eigenvalue in this set. Therefore we have shown that there are m0 eigenvalues in these disjoint intervals. Furthermore (a) and Proposition 1 imply that these are all eigenvalues in σ(A + τ P) \ σ(A). Hence each of these eigenvalues λik (τ ) is simple when there are not an element of σ(A) which proves (b) and we also have σ(A + τ P) ⊆ R ∪ {∞} for all τ ∈ R under the assumptions of (b). The proof of (c) can be carried out in the same way as the proof of (b). T Remark 4 The assertions of Theorem 3 remain true for m0 ≤ 1 and also for v∞ u∞ 6= 0. In the latter case one has to add n−r T in Theorem 3 (b) the assumption (−1) αv∞ u∞ < 0 which implies for all τ ∈ (0, ∞) the existence of λ∞ (τ ) ∈ C with T σ(A + τ P) ∩ (−∞, λi1 ) = {λ∞ (τ )}. In Theorem 3 (c) one needs to add the condition (−1)n−r αv∞ u∞ > 0 which implies for all τ ∈ (0, ∞) the existence of λ∞ (τ ) ∈ C with σ(A + τ P) ∩ (λim0 −1 , ∞) = {λ∞ (τ )}.
References [1] [2] [3] [4] [5]
F. Dopico, J. Moro and F. De Teran, SIAM J. Matrix Anal. Appl. 30, 538–547 (2008). F. Gantmacher, Theory of Matrices (Chelsea, New York, 1959). H. Gernandt and C. Trunk, submitted, arXiv:1604.06671 (2016). C. Mehl, V. Mehrmann and M. Wojtylak, Oper. Matrices 9, 733–772 (2015). A. Ran and M. Wojtylak, Linear Algebra Appl. 437, 589–600 (2012).
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