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 016 <hal v3> HAL Id: hal Submitted on 7 May 017 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 Kfardebian Lebanon Abstract The main purpose of this paper is to englobe ( some ) new and known A X types of Hermitian block-matrices M = X satisfying or not the inequality M A+ for all symmetric norms For positive definite block-matrices another inequality is established and it is shown that it can be sharper (for some symmetric norms) than the following holding inequality M A + Keywords: Symmetric norm; Matrix Inequalities; Partial positive transpose matrix; Hermitian lock-matrices AMS-010 subj class: 15A60 15A4 47A30 1 Introduction and preliminaries The first section presents some known results relatd to the inequality together with some preliminaries we used in the second section to derive some new and generalization results Let M + n denote the set of positive and semi-definite part of the space of n n complex matrices For positive semi-definite ( ) block-matrix M we say that M is PSD and we write A X M = X M + n+m with A M+ n M + m The modulus of a matrix X stands for (X X) 1 and is denoted by X A norm over the space of matrices is a symmetric norm if UAV = A for all A and all unitaries U and V 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 A = F = A 0 for all symmetric norms 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 norm by k 1

3 A positive partial transpose( matrix denoted ) by PPT ( is a) PSD block A X A X matrix M such that both M = X and M = are positive X semi definite Let Im(X) := X X respectively Re(X) := X +X be i the imaginary part respectively the real part of a matrix X Lemma 11 [] For every matrix in M + n written in blocks of the same size we have the decomposition: ( ) ( A X A+ ) ( ) X = U +Im(X) 0 U 0 0 +V A+ V Im(X) for some unitaries UV M n Lemma 1 [] For every matrix in M + n written in blocks of the same size we have the decomposition: ( ) ( A X A+ ) ( ) X = U +Re(X) 0 U 0 0 +V A+ V Re(X) for some unitaries UV M n Remark 13 The proofs of Lemma 11 respectively Lemma 1 suggests that we have A+ (X X ) and A+ (X X ) respectively i i A+ (X + X ) and A+ (X + X ) since if we let hereafter M 1 := A+ + Im(X) M := A+ Im(X) N 1 := A+ + Re(X) and N := A+ Re(X) then N 1 N M 1 M are PSD as diagonal blocks of the PSD matrix JMJ for some unitary matrix J ( []) [ ] A Lemma 14 [3] Let M = be any square matrix written by blocks C D of same size if AC = CA then det(m) = det(ad C) Main results 1 Symmetric norm inequality ( ) It is well known that if M M + A X n+m with M = X then M A + (1)

4 for all symmetric norms (see [4]) Hereafter our block matrices are such their diagonal blocks are of equal size Lemma 1 [] Let M = X M + n if X is Hermitian or Skew- Hermitian then M A+ () for all symmetric norms See [6] for another proof of Lemma 1 (the case X is Hermitian) Lemma [5] Let M = X M + n be a positive partial transpose matrix then M A+ (3) for all symmetric norms Proposition 3 Let M = X M + n be a given positive semidefinite matrix If X commute with A or then M is unitarily congruent to a PPT matrix and M A+ for all symmetric norms In addition if X is normal then M is a positive partial transpose matrix Proof We will assume without loss of generality that X commute with A as the other case is similar Take the 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 ][ A X X ][ ] U 0 = 0 I n [ ] A X X consequently we have M A + for all symmetric norms and that completes the proof If X is normal then X = X The polar decomposition discussed above and the following known decomposition: X = U X = X U let us write [ U 0 0 I n ][ A X X which implies that M is a PPT matrix ][ ] [ U 0 A X = 0 I n X ] 3

5 Remark 4 It is easily seen that if X commute with the Hermitian matrix A so is X and conversely The following is a slight generalization result: Lemma 5 Let M = X be a positive semi definite matrix if Im(X) = ri n or Re(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 13 the matrices M 1 = A+ + Im(X) and M = A+ Im(X) are positive semi definite since Im(X) = ri n we have: k ( ) A+ k ( ) A+ k σ i +Im(X) + σ i Im(X) = σ i (A+) i=1 i=1 In other words by Lemma 11 M k M 1 k + M k = A+ k for all Ky-Fan k norms which from the Ky-Fan dominance theorem (see [1] -Sec 107-) completes the proof Using Lemma 1 the other case is similarly proven One can ask if Lemma gives in a certain way the sharpest inequality satisfied by PPT matrices and the answer is yes Example 6 Let where A = Lemma 7 Let [ ] 0 and = [ M = A X = X [ ] 1 0 Then we have 0 ] A+ s = 3 = M s < A s + s = 4 N = a a a n D 4 D b b b n i=1

6 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 Lemma 14 we get that the eigenvalues of N are the roots of det((a µi n )( µi n ) D D) = 0 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 (S) 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 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 one Since A+ = ( a1 +b a n+b n summing over indexes weseethat N k > A+ k fork = 1 nwhichyieldsto N > A+ for all symmetric norms Remark 8 The result of Lemma 7 still holds if the diagonal condition of D is replaced by: D commute with A or and the matrix D D is diagonal 5

7 It seems easy to construct examples of non positive semi definite matrices N such that N s > A + s let us have a look of such inequality for PSD matrices Example 9 Let 0 0 [ ] N y = 0 y 0 0 A X = X 0 0 [ ] [ ] 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 y () > A+ () = 4(3+y)+y +1 New Inequality Theorem 10 Let M = X 0 and let r 1 r be two nonnegative numbers if M 1 r 1 I n and M r I n or N 1 r 1 I n and N r I n then and A+ (A+) (r 1 +r )I n (4) M (A+) (r 1 +r )I n (5) for all symmetric norms In particular if M ri n for some r 0 then for all symmetric norms Proof If we have the case M (A+) ri n (6) A+ A+ +Im(X) r 1 I n (7) Im(X) r I n (8) 6

8 or the case A+ +Re(X) r 1 I n (9) A+ Re(X) r I n (10) thensummingbothequations ineach casegives A+ (r 1 +r )I n and(4) follows Given that M 1 k A+ r I n k and M k A+ r 1 I n k for all k n we derive the following inequality: M k M 1 k + M k = (A+) (r 1 +r )I n k for all Ky-Fan k norms y replacing M i by N i for i = 1 Lemma 1 gives the same inequality The particular case can be easily concluded since by the decompositions in Lemma 11 and Lemma 1 if M ri n then all of M 1 ri n N 1 ri n M ri n and N ri n are positive semi definite matrices Inequality (5) can be sharper than (1) as these examples show: Example 11 Let where A = 5 with M = [ ] 4 0 and = [ ] A X = X [ ] 0 M is positive semi-definite r = r = 8 = A s + s > (A+) (r 1 +r )I n s = 7 = M s > A+ s = 6 Example 1 Let M = = [ A X X [ ] [ ] where A = and = It can be verified that M is positive semi-definite r 1 = r = 0375 and we have M () = 15 > A+ () = with A () + () = for: > (A+) (r 1 +r )I n () = 315 > M () > A+ () ] 7

9 Acknowledgments Iwant tothankdrminghwa Linforofferingmereferences withmanyrecent related results and Dr Ajit Iqbal Singh for her helpful comments References [1] FZhang Matrix Theory asic Results and Techniques nd Edition (Universitext) Springer 013 [] J C ourin E Y Lee andm Lin On a decomposition lemma for positive semi-definite block-matrices Linear Algebra and its Applications 437 (01) pp [3] J R Silvester Determinant of lock Martrices Math Gazette 84 (000) pp [4] J C ourin E Y Lee Unitary otbits of Hermitian operators with convex and concave functions ulllondon Math Soc 44 (01) pp [5] M Lin H Wolkowicz Hiroshima s theorem and matrix norm inequalities Acta Sci Math (Szeged) 81 (015) pp [6] M Lin and H Wolkowicz An eigenvalue majorization inequality for positive semidefinite block matrices Linear Multilinear Algebra 60 (01) pp