On Symmetric Norm Inequalities And Hermitian Block-Matrices

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1 On Symmetric Norm Inequalities And Hermitian lock-matrices Antoine Mhanna To cite this version: Antoine Mhanna On Symmetric Norm Inequalities And Hermitian lock-matrices 015 <hal v1> HAL Id: hal Submitted on 1 Nov 015 (v1, last revised 7 May 017 (v3 HAL is a multi-disciplinary open access archive for the deposit and dissemination of scientific research documents, whether they are published or not The documents may come from teaching and research institutions in France or abroad, or from public or private research centers L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés

2 ON SYMMETRIC NORM INEQUALITIES AND HERMITIAN LOCK-MATRICES ANTOINE MHANNA 1 Abstract The mainpurposeofthis paperis ( toenglobe somenew andknown types of Hermitian block-matrices M = X satisfying or not the inequality M A+ for all symmetric norms 1 Introduction and preliminaries The first section will deal with some known results on the inequality and some preliminaries we used in the second section to derive some new generalization results For positive( semi-definite block-matrix M, we say that M is PSD and we write M = + n+m, with A M + n, M + m Let A be an n n matrix and F an m m matrix, (m > n written by blocks such that A is a diagonal block and all entries other than those of A are zeros, then the two matrices have the same singular values and for all unitarily invariant norms A = F = A 0, we say then that the symmetric norm on M m induces a symmetric norm on M n, so for square matrices we may assume that our norms are defined on all spaces M n, n 1 The spectral norm is denoted by s, the Frobenius norm by (, and the Ky Fan k norms by k Let Im(X := X X be the imaginary part of a matrix X and let M + n denote the i set of positive and semi-definite part of the space of n n complex [ matrices ] and M be any positive semi-definite block-matrices; that is, M = + n+m, with A M + n, M+ m Lemma 11 [1] For every matrix in M + n written in blocks of the same size, we have the decomposition: ( ( A+ ( +Im(X 0 X = U 0 0 U 0 0 +V 0 A+ V Im(X for some unitaries U,V M n Remark 1 TheproofofLemma11suggeststhatwehaveA+ (X X i and A+ (X X i Date: 1 November 015 Corresponding author Key words and phrases Symmetric norms, Hermitian lock-matrices 1

3 AMHANNA A Lemma 13 [3] Let M = be any square matrix written by blocks of C D same size, if AC = CA then det(m = det(ad C Main results 1 Symmetric norm inequality Hereafter our block matrices are such their diagonal blocks are of equal size Lemma 1 [1] Let M = + n, if X is hermitian then for all symmetric norms M A+ (1 The fact is that there exist PSD matrices with non Hermitian off-diagonal blocks satisfying (1 Definition A block matrix N = X is said to be a Hermitio matrix A Y if it is unitarly congruent to a matrix M = with Hermitian off diagonal Y blocks Clearly if N is PSD it satisfies N A+ for all symmetric norms (by Lemma 1 Proposition 3 Let M = + n be a given positive semi-definite matrix If X commute with A, or X commute with, then M A+ for all symmetric norms Proof We will assume without loss of generality that X commute with A, as the other case is similar We show that such M is a Hermitio matrix Take the right polar decomposition of X so X = U X and X = X U Since U is unitary and X commute with A, X and X commute with A thus AU = U A If I n is the identity matrix of order n, a direct computation shows that [ U 0 0 I n ][ X ] U 0 = 0 I n [ A X X consequently by Lemma 11, M A+ for all symmetric norms and that completes the proof Here s a counter example: Example 4 Let 3 i T = 0 5 i 0 0 i 5 0 = X i ],

4 where X = [ 0 i 11 i 0 ], A = SYMMETRIC NINEQ 3 [ 3 ] and = 3 As seen X does not 10 commute with A The eigenvalues of T are the positive numbers: λ 1 = 6, λ = 4, λ 3 039, λ 4 01, T 0, but 6 = T s > A+ s = Remark 5 It is easily seen that if X commute with the Hermitian matrix A so is X and conversely The following is a slight generalization of Lemma 1 unless one proves that this is a case of a Hermitio matrix Theorem 6 Let M = X be a positive semi definite matrix, if Im(X = ri n for some r, then M A+ for all symmetric norms Proof Let σ i (H denote the singular values of a matrix H ordered in decreasing order, by Remark 1 the matrices M 1 = A+ +Im(X and M = A+ Im(X are positive semi definite since Im(X = ri n we have: k i=1 σ i ( A+ +Im(X + k i=1 σ i ( A+ Im(X = k σ i (A+ In other words M k M 1 k + M k = A + k for all Ky-Fan k norms which completes the proof i=1 Lemma 7 Let N = ( a a a n D D b b b n where a 1,,a n respectively b 1,,b n are nonnegative respectively negative real numbers, A = ( a a a n, = b b b n and D is any diagonal matrix, then nor N neither N is positive semi-definite Set (d 1,,d n as the diagonal entries of D D, if a i +b i 0 and a i b i d i < 0 for all i n, then N > A+ for all symmetric norms Proof The diagonal of N has negative and positive numbers, thus nor N neither N is positive semi-definite, now any two diagonal matrices will commute, in particular D and A, by applying Theorem 13 we get that the eigenvalues of N are the roots of det((a µi n ( µi n D D = 0

5 4 AMHANNA Equivalently the eigenvalues are all the solutions of the n equations: 1 (a 1 µ(b 1 µ d 1 = 0 (a µ(b µ d = 0 3 (a 3 µ(b 3 µ d 3 = 0 i (a i µ(b i µ d n = 0 n (a n µ(b n µ d n = 0 Let us denote by x i and y i the two solutions of the i th equation then: x 1 +y 1 = a 1 +b 1 0 x +y = a +b 0 x n +y n = a n +b n 0 (S x 1 y 1 = a 1 b 1 d 1 < 0 x y = a b d < 0 x n y n = a n b n d n < 0 This implies that each equation of (S has one negative and one positive solution, their sum is positive, ( thus the positive root is bigger or equal than the negative a1 +b 1 0 one Since A+ =, summing over indexes we see that N k > 0 a n+b n A + k for k = 1,,n which yields to N > A + for all symmetric norms Itseems easytoconstruct examplesofnonpsdmatricesn, suchthat N s > A+ s, let us have a look of such inequality for PSD matrices Example 8 Let 0 0 N y = 0 y = X where A = and = The eigenvalues of N 0 y 0 y are the numbers: λ 1 = 4, λ = 1, λ 3 = y, λ 4 = 0, thus if y 0, N y is positive semi-definite and for all y such that 0 y < 1 we have (1 4 = N y s > A+ s = 3 ( 16+y +1 = N ( > A+ ( = 4(3+y+y +1 References 1 J C ourin, E Y Lee, and M Lin, On a decomposition lemma for positive semi-definite block-matrices, Linear Algebra and its Applications 437, pp , (01 FZhang, Matrix Theory asic Results and Techniques, nd Edition (Universitext Springer, J R Silvester, Determinant of lock Martrices, Math Gazette, 84 (000, pp

6 SYMMETRIC NINEQ 5 1 Department of Mathematics, Lebanese University Hadath, eirut, Lebanon address: tmhanat@yahoocom