A Note on Eigenvalues of Perturbed Hermitian Matrices

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1 A Note on Eigenvalues of Perturbed Hermitian Matrices Chi-Kwong Li Ren-Cang Li July 2004 Let ( H1 E A = E H 2 Abstract and à = ( H1 H 2 be Hermitian matrices with eigenvalues λ 1 λ k and λ 1 λ k, respectively. Denote by E the spectral norm of the matrix E, and η the spectral gap between the spectra of H 1 and H 2. It is shown that λ i λ i 2 E 2 η + η E 2, which improves all the existing results. Similar bounds are obtained for singular values of matrices under block perturbations. AMS Classifications: 15A42, 15A18, 65F15. Keywords: Hermitian matrix, eigenvalue, singular value. 1 Introduction Consider a partitioned Hermitian matrix A = ( m n m H 1 E n E H 2, (1.1 where E is E s complex conjugate transpose. At various situations (typically when E is small, one is interested in knowing the impact of removing E and E on the eigenvalues Department of Mathematics, The College of William and Mary, Williamsburg, Virginia (ckli@math.wm.edu. Research supported in part by the NSF grant DMS Department of Mathematics, University of Kentucky, Lexington, KY (rcli@ms.uky.edu. This work was supported in part by the National Science Foundation CAREER award under Grant No. CCR

2 of A. More specifically, one would like to obtain bounds for the differences between that eigenvalues of A and those of its perturbed matrix à = ( m n m H 1 n H 2. (1.2 Let λ(x be the spectrum of the square matrix X, and let Y be the spectral norm of a matrix Y, i.e., the largest singular value of Y. There are two kinds of bounds for the eigenvalues λ 1 λ m+n and λ 1 λ m+n of A and Ã, respectively: 1. [1, 7, 8] λ i λ i E. ( [1, 2, 3, 5, 7, 8] If the spectra of H 1 and H 2 are disjoint, then where and λ(h i is the spectrum of H i. λ i λ i E 2 /η, (1.4 η def = min µ 1 λ(h 1, µ 2 λ(h 2 µ 1 µ 2, The bounds of the first kind do not use information of the spectral distribution of the H 1 and H 2, which will give (much weaker bounds when η is not so small; while the bounds of the second kind may blow up whenever H 1 and H 2 have a common eigenvalue. Thus both kinds have their own drawbacks, and it would be advantageous to have bounds that are always no bigger than E, of ( E as η 0, and at the same time behave like ( E 2 /η for not so small η. To further motivate our study, let us look at the following 2 2 example. Example 1 Consider the 2 2 Hermitian matrix ( α ɛ A =. (1.5 ɛ β Interesting cases are when ɛ is small, and thus α and β are approximate eigenvalues of A. We shall analyze by how much the eigenvalues of A differ from α and β. Without loss of generality, assume α > β. The eigenvalues of A, denoted by λ ±, satisfy λ 2 (α + βλ + αβ ɛ 2 = 0; and thus λ ± = α + β ± (α + β 2 4(αβ ɛ 2 2 = α + β ± (α β 2 + 4ɛ

3 Now { λ+ α 0 < β λ } = (α β + (α β 2 + 4ɛ 2 2 = 2ɛ 2 (α β + (α β 2 + 4ɛ 2 (1.6 which provides a difference that enjoys the following properties: 2ɛ 2 ɛ always, (α β + ɛ as α β +, (α β 2 + 4ɛ 2 ɛ 2 /(α β. The purpose of this note is to extend this 2 2 example and obtain bounds which improve both (1.3 and (1.4. Such results are not only of theoretical interest but also important in the computations of eigenvalues of Hermitian matrices [4, 6, 9]. As an application, similar bounds are presented for the singular value problem. 2 Main Result Theorem 2 Let A = ( m n m H 1 E n E H 2 and à = ( m n m H 1 n H 2 be Hermitian matrices with eigenvalues λ 1 λ 2 λ m+n and λ1 λ 2 λ m+n, (2.1 respectively. Define η i η def = min λ i µ 2, if λ i λ(h 1, µ 2 λ(h 2 min λ i µ 1, if λ i λ(h 2, µ 1 λ(h 1 (2.2 def = min η i = min µ 1 µ 2. (2.3 1 i m+n µ 1 λ(h 1, µ 2 λ(h 2 Then for i = 1, 2,, m + n, we have λ i λ i 2 E 2 (2.4 η i + η 2i + 4 E 2 2 E 2 η + η E 2. (2.5 3

4 Proof. Suppose U H 1 U and V H 2 V are in diagonal form with diagonal entries arranged in descending order. We may assume that U = I m and V = I n. therwise, replace A by (U V A(U V. We may perturb the diagonal of A so that all entries are distinct, and apply continuity argument for the general case. We prove the result by induction on m + n. If m + n = 2, the result is clear (from our Example. Assume that m + n > 2, and the result is true for Hermitian matrices of size m + n 1. First, refining an argument of Mathias [5], we show that (2.4 holds for i = 1. Assume that the (1, 1 entry of H 1 equals λ 1. By the min-max principle [1, 7, 8], we have λ 1 e 1Ae 1 = λ 1, where e 1 is the first column of the identity matrix. Let ( I X = m 0 (H 2 µ 1 I n 1. E I n Then where ( X H1 (λ (A λ 1 IX = H 2 λ 1 I n, H 1 (λ 1 = H 1 λ 1 I m E (H 2 λ 1 I n 1 E. Since A and X AX have the same inertia, we see that H 1 (λ 1 has zero as the largest eigenvalue. Notice that the largest eigenvalue of H 1 λ 1 I is λ 1 λ 1 0. Thus, for δ 1 = λ 1 λ 1 = λ λ 1, we have (see [7, (10.9] λ 1 λ 1 + E 2 2/(δ 1 + η 1, and hence Consequently, δ 1 E 2 /(δ 1 + η 1. 2 E δ 1 η 1 + η E 2 as asserted. Similarly, we can prove the result if the (1, 1 entry of H 2 equals λ 1. In this case, we will apply the inertia arguments to A and Y AY with ( I Y = m 0 E(H 1 λ 1 I m 1. I n Applying the result of the last paragraph to A, we see that (2.2 holds for i = m + n. 4

5 Now, suppose 1 < i < m + n. The result trivially holds if λ i = λ i. Suppose λ i λ i. We may assume that λ i > λ i. therwise, replace (A, Ã, i by ( A, Ã, m + n i + 1. Delete the row and column of A that contain the diagonal entry λ n. Suppose the resulting matrix  has eigenvalues ν 1 ν m+n 1. By the interlacing inequalities [7, Section 10.1], we have λ i ν i and hence λi λ i λ i νi. (2.6 Note that λ i is the ith largest diagonal entries in Â. Let η i be the minimum distance between λ i and the diagonal entries in the diagonal block Ĥj in  not containing λ i ; here j {1, 2}. Then η i η i because Ĥj may have one fewer diagonal entries than H j. Let Ê be the off-diagonal block of Â. Then Ê E. Thus, λ i λ i = λ i λ i because λ i > λ i λ i ν i by (2.6 2 Ê 2 by induction assumption η i + η 2i + 4 Ê 2 2 Ê 2 because η i η i η i + η 2i + 4 Ê 2 = 1 ηi Ê 2 η i 1 ηi E 2 η i because Ê E = 2 E 2 η i + ηi E 2 as asserted. 3 Application to Singular Value Problem In this section, we apply the result in Section 2 to study singular values of matrices. For notational convenience in connection to our discussion, we define the sequence of singular values of a complex p q matrix X by σ(x = (σ 1 (X,..., σ k (X, where k = max{p, q} and σ 1 (X σ k (X are the nonnegative square roots of the eigenvalues of the matrix XX or X X depending on which one has a larger size. Note that the nonzero eigenvalues of XX and X X are the same, and they give rise to the nonzero singular values of X which are of importance. We have the following result concerning the nonzero singular values of perturbed matrices. 5

6 Theorem 3 Let B = ( k l m G 1 E 1 n E 2 G 2 ( m G and B = 1 n G 2 k l be complex matrices with singular values σ 1 σ 2 σ max{m+n,k+l} and σ 1 σ 2 σ max{m+n,k+l}, (3.1 respectively, so that G 1 and G 2 are non-trivial. Define ɛ = max{ E 1, E 2 }, and min σ i µ 2, if σ i σ(g 1, µ 2 σ(g 2 η i η def = min σ i µ 1, if σ i σ(g 2, µ 1 σ(g 1 (3.2 def = min η i = min µ 1 µ 2. (3.3 1 i m+n µ 1 σ(g 1, µ 2 σ(g 2 Then for i = 1, 2,, min{m + n, k + l}, we have σ i σ i 2ɛ 2 (3.4 η i + η 2i + 4ɛ2 2ɛ 2 η + η 2 + 4ɛ 2, (3.5 and σ i = σ i = 0 for i > min{m + n, k + l}. Proof: By Jordan-Wielandt Theorem [8, Theorem I.4.2], the eigenvalues of ( B B are ±σ i and possibly some zeros adding up to m+n+k+l eigenvalues. A similar statement holds for B. Permuting the rows and columns appropriately, we see that ( G 1 E 1 B B is similar to G 1 E2 E 2 G 2, E1 G 2 and ( B B is similar to G 1 G 1 G 2 G 2. 6

7 Applying Theorem 2 with ( Gi H i = G i ( E2 and E =, E 1 we get the result. ne can also apply the above proof to the degenerate cases when G 1 or G 2 in the matrix B is trivial, i.e., one of the parameters m, n, k, l is zero. These cases are useful in applications. We state one of them, and one can easily extend it to other cases. Theorem 4 Suppose B = ( G E and B = ( G are p q matrices with singular values σ 1 σ max{p,q} and σ 1... σ max{p,q}, respectively. Then for i = 1,..., min{p, q}, σ i σ i 2 E. 2 σ i + σ 2i + 4 E 2 Acknowledgment This paper evolved from an early note of the second author in which Theorem 2 was proven for special cases. The authors wish to thank Editor Volker Mehrmann for making this collaboration possible. 7

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