Symmetric Norm Inequalities And Positive Semi-Definite Block-Matrices

Symmetric Norm Inequalities And Positive Semi-Definite lock-matrices Antoine Mhanna To cite this version: Antoine Mhanna Symmetric Norm Inequalities And Positive Semi-Definite lock-matrices 15 <hal-11844v5> HAL Id: hal-11844 https://halinriafr/hal-11844v5 Submitted on 14 Sep 15 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

Symmetric Norm Inequalities And Positive Semi-Definite lock-matrices Antoine Mhanna 1 1 Dept of Mathematics, Lebanese University, Hadath, eirut, Lebanon tmhanat@yahoocom Abstract For positive semi-definite block-matrix M, we say that M is PSD and we write M = X M + n+m, with A M+ n, M + m The focus is on studying the consequences of a decomposition lemma due to C ourrin and the main result is extending the class of PSD matrices M written by blocks of same size that satisfies the inequality: M A+ for all symmetric norms Keywords : Matrix Analysis, Hermitian matrices, symmetric norms 1 Introduction 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, 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 M + n denote the 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 = X M + n+m, with A M+ n, M+ m 1

Decomposition of block-matrices Lemma 1 For every matrix M in M + n+m written in blocks, we have the decomposition: A X A X = U U +V V for some unitaries U,V M n+m Proof Factorize the positive matrix as a square of positive matrices: A X C Y C Y X = Y D Y D we verify that the right hand side can be written as T T +S S so : C Y C Y C C Y Y Y D Y = D Y + D Y D }{{}}{{}}{{}}{{} T T S S Since TT = CC +YY = A, SS = Y Y +DD AA is unitarily congruent to A A for any square matrix A, the lemma follows = and Remark 1 As a consequence of this lemma we have: M A + for all symmetric norms Equations involving unitary matrices are called unitary orbits representations Recall that if A M n, RA = A+A and IA = A A i Corollary 1 For every matrix in M + n written in blocks of the same size, we have the decomposition: A X A+ X = U RX U +V A+ V +RX for some unitaries U,V M n Proof Let J = 1 I I where I is the identity of M I I n, J is a unitary matrix, and we have: A+ A A X J X J RX + X X = A X X A+ +RX }{{} N

Now we factorize N as a square of positive matrices: A X X = J L M L M M F M J F and let: A direct computation shows that: δψ = 1 L M δ = J M = L+M 1 M +F F M L F M L M ψ = M J = L+M M L 1 F F +M F M L+M L+M+M+FF+M L+M M L+M+FF M M LL+M+F MF+M M LM L+F MF M = Γ Γ+Φ Φ 1 where: Γ = 1 L+M M L, and Φ = 1 F +M F M to finish notice that for any square matrix A, A A is unitarily congruent to AA and, ΓΓ, ΦΦ have the required form The previous corollary implies that A+ RX and A+ RX Corollary For every matrix in M + n written in blocks of the same size, we have the decomposition: X = U A+ +IX U +V A+ V IX for some unitaries U,V M n+m A X A Proof The proof is similar to Corollary 1, we have: J 1 X J 1 = ix I where J 1 =, and ii K = JJ 1 X J 1J = A+ +IX A+ IX ix here * means an unspecified entry, the proof is similar to that in Corollary 1 but for reader s convenience we give the main headlines: first factorize K as a square of positive matrices; that is, M = X = J 1 J L JJ 1 next decompose L as in Lemma 1 to obtain M = J 1J T T +S SJJ 1 = J 1J T TJJ 1 +J 1J S SJJ 1 3

A+ where TT = +IX property completes the proof and SS = A+ finally the congruence IX The existence of unitaries U and V in the decomposition process need not to be unique as one can take the special case; that is, M any diagonal matrix with diagonal k k entries equals a nonnegative number k, explicitly M = ki = U I U +V I V for any U and V unitaries Remark Notice that from the Courant-Fischer theorem if A, M + n, then the eigenvalues of each matrix are the same as the singular values and A = A k k, for all k = 1,,n, also A < = A k < k, for all k = 1,,n Corollary 3 For every matrix in M + n written in blocks of the same size, we have: X 1 { U for some unitaries U,V M n+m A+ + X X } U +V V A+ + X X Proof This a consequence of the fact that IX IX 3 Symmetric Norms and Inequalities In [1] they found that if X is hermitian then M A+ for all symmetric norms It has been given counter-examples showing that this does not necessarily holds if X is a normal but not Hermitian matrix, the main idea of this section is to give examples and counter-examples in a general way and to extend the previous inequality to a larger class of PSD matrices written by blocks satisfying Theorem 31 If A and are positive definite matrices of same size Then X > A X 1 X 4

A X I X 1 A X Proof Write X = 1 X I I X 1 I the identity matrix, and that complete the proof since for any matrix A, where I is A X AX, X A Theorem 3 Let M = be any square matrix written by blocks of same size, C D if AC = CA then detm = detad C Proof Suppose first that A is invertible, let us write M as M = Z I E V I F 3 upon calculation we find that: Z = A, V = C, E = A 1, F = D CA 1 taking the determinant on each side of 3 we get: detm = detad CA 1 = detad C the result follows by a continuity argument since the Determinant function is a continuous function Given thematrix M = X amatrix inm + n written byblocks of samesize, we know that it M is not PSD, to see this notice that all the extracted principle submatrices of M are PSD if and only if X = and A is positive semi-definite Even if a proof of this exists and would take two lines, it is quite nice to see a different constructive proof, a direct consequence of Lemma 1 Theorem 33 Given X a matrix in M + n written in blocks of same size: 1 If X is positive semi-definite, IX > or IX <, then there exist a A Y matrix Y such that M = Y is positive semi-definite and: A Y Y > A 4 for all symmetric norms 5

X If X is positive semi-definite, IX > or IX < then there exist a Y matrix Y such that M = Y is positive semi-definite and: The same result holds if we replaced IX by RX because congruent to X Y Y > 5 A ix ix is unitarily Proof Without loss of generality we can consider IX > cause X and A X X are unitarily congruent, we will show the first statement as the second one has a similar proof, from Corollary we have: X U A U +U IX U +V A V A X A lx Since X is congruent to L = lx for any l C, L is PSD A is a fixed A matrix, we have U k U +V A V = β A k for some β 1 finally we set Y = lx where l R is large enough to have M k > A k, k thus M > A for all symmetric norms Notice that there exist a permutation matrix P such that P X P 1 = X andsinceix > ifandonlyifix X A <,thetwoassertions oftheorem33 are equivalent up to a permutation similarity Corollary 31 If M = X, A a positive semi-definite matrix, and we have one of the following conditions: 1 RX > RX < 3 IX > 4 IX < 6

Then M can t be positive semi-definite Proof y Remark 1 any positive semi-definite matrix M written in blocks must satisfy M A + for all symmetric norms which is not the case of the matrix M constructed in Theorem 33 Finally we get: Theorem 34 If X and =, A, the matrix M = positive semi-definite X cannot be Proof Suppose the converse, so M = X is positive semi-definite, without loss of generality the only case we need to discuss is when RX has positive and negative eigenvalues, by Corollary 1 we can write: M = U A RX U +V A +RX V for some unitaries U,V M n Now if RX has α the smallest negative eigenvalue RX+α+ǫI > consequently the matrix A H = U RX U +V A V 6 +α+ǫi +RX+α+ǫI A+α+ǫI X +α+ǫi = X +α+ǫi 7 is positive semi-definite with RY >, where Y = X +α+ǫi, by Corollary 31 this is a contradiction A natural question would be how many are the nontrivial PSDmatrices written by blocks? The following lemma will show us how to construct some of them Lemma 31 Let A and be any n n positive definite matrices, then there exist an ta X integer t 1 such that the matrix F t = X is positive definite t Proof Recall from Theorem 31 that F 1 is positive definite if and only if A > X 1 X, which is equivalent to x Ax > x X 1 X x for all x C n Set fx := x Ax and gx := x X 1 X x and let us suppose, to the contrary, that there exist a vector z such 7

that fz gz since fx and gx are homogeneous functions of degre d = over R if fx gx for all x such that x s = 1 then fx gx for any x C n So let us set K = max gx, and L = min fx since gx and fx are continuous functions x s =1 x s =1 and {x; x s = 1} is compact, there exist a vector w respectively v such that K = gw, respectively L = fv Now choose t 1 such that tfv > gw, to obtain t x tax v tav > w Xt 1 X w x Xt 1 X x for all x such that x s = 1, thus x tax > x Xt 1 X x for any x C n which completes the proof Theorem 35 Let A = diagλ 1,,λ n, = diagν 1,,ν n and M = X a given positive semi-definite matrix If X commute with A and X X equals a diagonal matrix, then M A+ for all symmetric norms The same inequality holds if X commute with and XX is diagonal Proof It suffices to prove the inequality for the Ky Fan k norms k = 1,,n, let P = In X A X wherei I n n istheidentity matrixofordern, since = P X A X P 1 and X have same singular values, we will discuss only the first case; that is, when X commute with Aand X X is diagonal, as the second case will follows Let D := X X = d 1 d d n that the eigenvalues of, as X commute with A, from Theorem 3 we conclude X are the roots of deta µi n µi n D = 8

Equivalently the eigenvalues are all the solutions of the n equations: 1 λ 1 µν 1 µ d 1 = λ µν µ d = 3 λ 3 µν 3 µ d 3 = i λ i µν i µ d i = n λ n µν n µ d n = Each equation is of nd degree, if we denote by a i and b i the two solutions of the i th equation we deduce that: a 1 +b 1 = λ 1 +ν 1 a +b = λ +ν a n +b n = λ n +ν n ut λ 1 +ν 1 λ +ν A+ = λ n +ν n and each diagonal entry of A+ is equal the sum of two nonegative eigenvalues of M, thus we have necessarily: M k A + k for all k = 1,,n which completes the proof Example 31 Let i x 99 i M x = 1 i 99 1 i 1 3 If 1 x 1, M x is positive definite and we have: for all symmetric norms, where A = M x A+ 8 x 99 99 and = 1 1 1 definite for x = 3 1 then M x is PD for all x > 3 The eigenvalues of M 3 1 1 If M x is positive which are 9

the same as the singular values of M 3 are: 1 λ 1 = 149 141 + λ = 19 + λ 3 = 149 λ 4 = 19 And the 8 inequality follows from Theorem 35 131 9 14761 15 1 141 188 11 14761 375 1 Let us study the commutation condition in Theorem 35 First notice that any square matrix X = x ij M n will commute with A = diaga 1,,a n if and only if : x 1,1 a 1 x 1, a x 1,n a n x 1,1 a 1 x 1, a 1 x 1,n a 1 Y x,1 a 1 x, a x,n a n = = x,1 a x, a x,n a = Y x n,1 a 1 x n, a x n,n a n x n,1 a n x n, a n x n,n a n An i,j entry of Y is equal to that of Y if and only if x i,j a j = x i,j a i, ie either a i = a j or x i,j = Corollary 3 Let A = diagλ 1,,λ n, = diagν 1,,ν n and M = X a given positive semi-definite matrix If X commute with A, or X commute with, then M A+ for all symmetric norms Proof As in Theorem 35, we will assume without loss of generality that X commute with A, as theother case is similar If X is diagonal the resultfollows from Theorem 35, suppose there is an off diagonal entry x i,j of X different from, from the commutation condition we have a i = a j and the same goes for all such entries, of course if AX = XA then PAXP 1 = PXAP 1 = PAP 1 PXP 1 = PXP 1 PAP 1 = PXAP 1 Take P to be the permutation matrix that will order the same diagonal entries of A in a one diagonal block and keeps the matrix the same, since M is Hermitian so is PMP 1 1

because we can consider the permutation matrix as a product of transposition matrices P 1,,P n wich are orthogonal; in other words PMP 1 = P 1 P P n MP T n PT PT 1 Consequently P T = P 1 for any permutation matrix and M = PMP T for all symmetric norms If H = PMP T, D := PX and X i is some i i extracted submatrix of X, we will have the block written matrix PAP T PX H = = X P T ai i O j O s O i bi j O s O i O j ri s X i O j O s O i X j O s O i O j X s X i O i O i O j X j O j O s O s X s ν 1 ν ν n where we denoted the diagonal matrix of order i whose diagonal entries are equal to a by ai i and the zero block of order i by O i Let us calculate the roots of the characteristic polynomial of H; that is, the roots of det a λi i O j O s O i b λi j O s O i O j r λi s ν 1 λ ν λ ν n λ D D = we translate this to a system of blocks, while each eigenvalue of H, which is the same as its singular value, will verify one of the following equations: 1 det a λi i ν 1 λ X i X i ν i λ det b λi j ν i+1 λ X j X j c det ν i+j λ r λi s ν n s λ ν n λ X sx s = = = where c is the number of diagonal blocks we have Let us have a closer look to any of the equations above, without loss of generality we will take the first one, the same will hold for the others, notice that all eigenvalues λ are nonnegative and we have ai i X i ν1 C1 X M 1 = X i = i Xi K 1 ν i 11 T

is positive semi-definite because it s eigenvalues are a subset of those of M The key idea is that for this matrix C 1 +K 1 = C 1 + K 1 for all symmetric norms where C 1 = ai i and K = ν1 ν i Now back to the system T we associate like we did to M 1 each equation whose number is i to a positive semi-definite matrix M i to obtain by Remark 1 M 1 k M k M c k ai i + bi j + ri s + = ai i k + k = bi j k + k ν1 ν i νi+1 ν i+j νn s = ri s k + k ν n ν1 ν i k νi+1 ν i+j νn s ν n k k for all k, but the order of the entries of are arbitrary chosen, thus from Theorem 35 M k A+ k for all k = 1,,n and that completes the proof Corollary 33 Let M = X be a positive semi-definite matrix written by blocks There exist a unitary V and a unitary U such that for all symmetric norms A X X UAU +VV := A + Proof Let U and V be two unitary matrix such that UAU = D o and VV = G o where D o and G o are two diagonal matrices having the same ordering o, of eigenvalues with respect to their indexes ie, if λ n λ 1 are the diagonal entries of D o, and ν n ν 1 are those of G o, then if λ i is in the j,j position then ν i will be also Consequently UAU +VV = D o +G o = D o + G o = A +, for all the Ky-Fan k norms and thus for all symmetric norms To complete the proof notice that U if T = UXV and Q is the unitary matrix, by Remark 1 V A X X = A X Q X Q = Do T T D G o + G o 13 o for all symmetric norms 1

Theorem 36 Let M = X, if X is normal, X commute with A and X commute with, then we have M A+ for all symmetric norms Proof We consider first that the normal matrix X has all of its eigenvalues distinct, by Theorem?? and the normality condition, there exist a unitary matrix U such that U AU and U X U are both diagonal A direct computation shows that: U U A X X U = U U AU U XU U X U U = G U Now U XU also commute with U U, since U XU is diagonal and all of its diagonal entries are distinct by Remark?? U U must be also diagonal, applying Theorem 35 to the matrix G yields to: A X X = G U AU +U U = A+, for all symmetric norms The inequality holds for any X normal by a continuity argument Lemma 3 Let N = a 1 a a n D D b 1 b b n where a 1,,a n respectively b 1,,b n are nonnegative respectively negative real numbers, A = a 1 a a n, = b 1 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 and a i b i d i < 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 3 we get that the eigenvalues of N are the roots of deta µi n µi n D D = 13

Equivalently the eigenvalues are all the solutions of the n equations: 1 a 1 µb 1 µ d 1 = a µb µ d = 3 a 3 µb 3 µ d 3 = i a i µb i µ d n = n a n µb n µ d n = 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 x +y = a +b x n +y n = a n +b n x 1 y 1 = a 1 b 1 d 1 < x y = a b d < x n y n = a n b n d n < 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 1 a n+b n, summing over indexes we see that N k > A + k for k = 1,,n which yields to N > A+ for all symmetric norms It seems easy to construct examples of non PSD matrices N written in blocks such that N s > A+ s, let us have a look of such inequality for PSD matrices Example 3 Let 4 1 1 3 1 1 C = 5 A X 3 = 1 X 1 1 5 4 where A = 3 3, = Since the eigenvalues of C are all positive with 1 λ 1 38, λ 17, λ 3 9, λ 4 89, C is positive definite and we verify that 38 C s > A+ s = 3 Example 33 Let N y = y A X 1 = X 14

where A = y and = 1 The eigenvalues of N y are the numbers: λ 1 = 4, λ = 1, λ 3 = y, λ 4 =, thus if y, N y is positive semi-definite and for all y such that y < 1 we have 1 4 = N y s > A+ s = 3 16+y +1 = N > A+ = 43+y+y +1 15

References [1] J C ourin, E Y Lee, and M Lin, On a decomposition lemma for positive semidefinite block-matrices, Linear Algebra and its Applications 437, pp196 191, 1 [] J C ourin, F Hiai, Norm and anti-norm inequalities for positive semi-definite matrices, InternatJMath63, pp111-1138, 11 16