Inequalities on generalized matrix functions
aa r X i v : . [ m a t h . F A ] A ug Inequalities on generalized matrix functions
Shaowu Huang ∗ , Chi-Kwong Li † , Yiu-Tung Poon ‡ , Qing-Wen Wang § In memory of Marvin Marcus.
Abstract
We prove inequalities on non-integer powers of products of generalized matrices functions onthe sum of positive semi-definite matrices. For example, for any real number r ∈ { } ∪ [2 , ∞ ),positive semi-definite matrices A i , B i , C i ∈ M n i , i = 1 ,
2, and generalized matrix functions d χ , d ξ such as the determinant and permanent, etc., we have( d χ ( A + B + C ) d ξ ( A + B + C )) r + ( d χ ( A ) d ξ ( A )) r + ( d χ ( B ) d ξ ( B )) r + ( d χ ( C ) d ξ ( C )) r ≥ ( d χ ( A + B ) d ξ ( A + B )) r + ( d χ ( A + C ) d ξ ( A + C )) r + ( d χ ( B + C ) d ξ ( B + C )) r . A general scheme is introduced to prove more general inequalities involving m positive semi-definite matrices for m ≥ AMS Classifcations. 15A15, 15A45, 15A63, 15B57.Keywords. Positive semi-definite matrices, generalized matrix functions, majorization.
Let G be a subgroup of the symmetric group S n of degree n , and let χ be a linear character of G .The generalized matrix function associated with G and χ (also known as the G -immanant) of amatrix A = ( a ij ) ∈ M n is defined by d Gχ ( A ) = X σ ∈ G χ ( σ )Π ni =1 a i σ ( i ) . For simplicity, we write d ( A ) = d Gχ ( A ) if the function d Gχ is understood in the context.Denote by X ⊗ Y the tensor (Kronecker) product of two matrices X and Y . It is known thatthere is a decomposable tensor v χ ∈ C n such that d ( A ) = v ∗ χ ( ⊗ n A ) v χ ; see [4, 6, 7]. So, one canuse the theory of tensor products and quadratic forms to study inequalities on generalized matrix ∗ Department of Mathematics, Shanghai University, Shanghai, China. [email protected] † Department of Mathematics, College of William and Mary, Williamsburg, Virginia, USA. [email protected] ‡ Department of Mathematics, Iowa State University, Ames, Iowa, USA. [email protected] § Corresponding author. Department of Mathematics, Shanghai University, Shanghai, China. [email protected]@yahoo.com
A, B ∈ M n and any positive integer k , we have ⊗ k ( A + B ) ≥ ⊗ k A + ⊗ k B. Letting k = nℓ with ℓ ∈ N , and using ⊗ ℓ v χ , we see that d ( A + B ) ℓ ≥ d ( A ) ℓ + d ( B ) ℓ . (1.1)Using similar techniques, one can obtain inequalities on the integer powers of d ( A ) , . . . , d ( A m ) , d ( A + A ) , . . . , d ( A m − + A m ) , etc.for positive semi-definite matrices A , . . . , A m ∈ M n .In this paper, we are interested in inequalities related to non-integer powers of generalizedmatrix functions. In some situations, such inequalities can be obtained by using the theory ofmajorization and Schur convex functions. Recall that for real vectors u, v ∈ R n , we say that u isweakly majorized by v , denoted by u ≺ w v , if the sum of the k largest entries of u is not larger thanthat of v for k = 1 , . . . , n . Furthermore, if the sums of the entries of u and v are the same, we saythat u is majorized by v , denoted by u ≺ v . A function f : R n → R is Schur convex if f ( u ) ≤ f ( v )whenever u ≺ v . It is known that if f is Schur convex and if f is increasing in each coordinate,then f ( u ) ≤ f ( v ) whenever u ≺ w v . One may see [2, 8] for the background on majorization andSchur convex functions.For two positive semi-definite matrices A and B , we have the weak majorization relation( d ( A ) , d ( B )) ≺ w ( d ( A + B ) , . Thus, we have f ( d ( A + B ) , ≥ f ( d ( A ) , d ( B ))for any Schur convex and increasing function f ; see [2, Chapter II] and [8, Chatper 3]. For example,for any p ≥
1, the ( x , x ) x p + x p is a Schur convex function. So, we have d ( A + B ) p ≥ d ( A ) p + d ( B ) p . Actually, for any two positive semi-definite matrices
A, B ∈ M n , it is known that (see [2, SectionIX.8.16]), ⊗ n ( A + B ) /n ≥ ⊗ n A /n + ⊗ n B /n . Hence, d (( A + B ) /n ) ≥ d ( A /n ) + d ( B /n ) . As a result, we have the weak majorization relation( d ( A /n ) , d ( B /n )) ≺ w ( d (( A + B ) /n ) , . f ( d (( A + B ) /n ) , ≥ f ( d ( A /n ) , d ( B /n ))for any Schur convex and increasing function f . For example, for any p ≥
1, the ( x , x ) x p + x p is a Schur convex function. So, we have d (( A + B ) /n ) p ≥ d ( A /n ) p + d ( B /n ) p . Applying this to the determinant and permanent functions, we havedet( A + B ) q ≥ det( A ) q + det( B ) q , for all q ≥ /n, per(( A + B ) /n ) p ≥ per( A /n ) p + per( B /n ) p for all p ≥ . However, for three or more positive semi-definite matrices, one may or may not be able to applythese arguments. For example, it is known that (see [1, 5]) d ( A + B + C ) + d ( A ) ≥ d ( A + B ) + d ( A + C ) . (1.2)Because ( d ( A + B ) , d ( A + C )) ≺ w ( d ( A + B + C ) , d ( A )), it follows that f ( d ( A + B + C ) , d ( A )) ≥ f ( d ( A + B ) , d ( A + C ))for any Schur convex and increasing functions f . In particular, for any p ≥ d ( A + B + C ) p + d ( A ) p ≥ d ( A + B ) p + d ( A + C ) p . On the other hand, it is also known that d ( A + B + C ) + d ( A ) + d ( B ) + d ( C ) ≥ d ( A + B ) + d ( A + C ) + d ( B + C ) . However, in general,( d ( A + B ) , d ( A + C ) , d ( B + C ) , w ( d ( A + B + C ) , d ( A ) , d ( B ) , d ( C )) . For example, if A = B = C = I , then ( d ( A + B ) , d ( A + C ) , d ( B + C ) ,
0) = (2 n , n , n , d ( A + B + C ) , d ( A ) , d ( B ) , d ( C )) = (3 n , , , f ( d ( A + B + C ) , d ( A ) , d ( B ) , d ( C )) ≥ f ( d ( A + B ) , d ( A + C ) , d ( B + C ) ,
0) for a Schur convexfunction. We will give examples in the next section showing that there is p > d ( A + B + C ) p + d ( A ) p + d ( B ) p + d ( C ) p ≥ d ( A + B ) p + d ( A + C ) p + d ( B + C ) p (1.3)is not valid even though one can use the tensor product and quadratic form techniques to showthat (1.3) holds for all positive integer p .In this paper, we will develop a general scheme to prove inequalities involving the (non-integer)powers of generalized matrix functions. For example, we will prove in Section 2 that (1.3) is validfor any p ∈ { } ∪ [2 , ∞ ). A general scheme and more results will be described in Section 3. Furtherextensions of our techniques will be mentioned in Section 4.3 Results on three matrices
Suppose we have three positive semi-definite matrices
A, B, C ∈ M n . It was proved in [1] that ⊗ n ( A + B + C ) + ⊗ n A + ⊗ n B + ⊗ n C ≥ ⊗ n ( A + B ) + ⊗ n ( A + C ) + ⊗ n ( B + C ) . (2.1)Applying the quadratic form using v χ on both sides, we have d ( A + B + C ) + d ( A ) + d ( B ) + d ( C ) ≥ d ( A + B ) + d ( A + C ) + d ( B + C ) . (2.2)For any ℓ ∈ N , ⊗ ℓ ( ⊗ n ( A + B + C )) + ⊗ ℓ ( ⊗ n A ) + ⊗ ℓ ( ⊗ n B ) + ⊗ ℓ ( ⊗ n C ) ≥ ⊗ ℓ ( ⊗ n ( A + B )) + ⊗ ℓ ( ⊗ n ( A + C )) + ⊗ ℓ ( ⊗ n ( B + C )) . Applying the quadratic form using ⊗ ℓ ( v χ ) on both sides, we have d ( A + B + C ) ℓ + d ( A ) ℓ + d ( B ) ℓ + d ( C ) ℓ ≥ d ( A + B ) ℓ + d ( A + C ) ℓ + d ( B + C ) ℓ . (2.3)As mentioned in Section 1, in general,( d ( A + B ) , d ( A + C ) , d ( B + C ) , w ( d ( A + B + C ) , d ( A ) , d ( B ) , d ( C )) . Thus, we cannot apply the Schur convex function result to conclude that d ( A + B + C ) r + d ( A ) r + d ( B ) r + d ( C ) r ≥ d ( A + B ) r + d ( A + C ) r + d ( B + C ) r for r ≥
1. Nevertheless, we have the following.
Theorem 2.1
Suppose
A, B, C ∈ M n are positive semi-definite matrices, and r ∈ { } ∪ [2 , ∞ ) .Then for any generalized matrix function d ( X ) , we have d ( A + B + C ) r + d ( A ) r + d ( B ) r + d ( C ) r ≥ d ( A + B ) r + d ( A + C ) r + d ( B + C ) r . (2.4) Proof.
To prove (2.4), let d ( A ) = x , d ( B ) = x , d ( C ) = x ,d ( A + B ) = x + x + x , d ( A + C ) = x + x + x , d ( B + C ) = x + x + x . Then x , x , x ≥
0. Moreover, x = d ( A + B ) − ( d ( A ) + d ( B )) ≥
0. Similarly, x , x ≥ d ( A + B + C ) + d ( A ) + d ( B ) + d ( C ) ≥ d ( A + B ) + d ( B + C ) + d ( A + C ) , so that d ( A + B + C ) ≥ ( x + x + x ) + ( x + x + x ) + ( x + x + x ) − ( x + x + x )= ( x + x + x + x + x + x ) . X = { ( x , x , x , x , x , x ) : x i , x jk ≥ ≤ i ≤ , ≤ j < k ≤ } . For r ≥ f : X → R by f ( x , x , x , x , x , x ) = ( x + x + x + x + x + x ) r + X i =1 x ri − X ≤ j 3. Also, ∂f∂x = r (cid:0) ( x + x + x + x + x + x ) r − + x r − − (cid:0) ( x + x + x ) r − + ( x + x + x ) r − (cid:1)(cid:1) ≥ X because(( x + x + x ) , ( x + x + x )) ≺ w (( x + x + x + x + x + x ) , x ) and r ≥ ∂f∂x i ≥ X for all 1 ≤ i ≤ 3. Therefore, for every x ∈ X , we have f ( x ) ≥ f ( ) = 0 for all x ∈ X. The following example shows that the bound the region for r in Theorem 2.1 is best possiblefor n = 1. Example 2.2 Let A = B = C = [1] . Then det( A + B + C ) r + det( A ) r + det( B ) r + det( C ) r − (det( A + B ) r + det( A + C ) r + det( B + C ) r )= 3 r + 3 − r ) . Let f ( r ) = 3 r + 3 − r ) . Then f (1) = f (2) = 0 and f ′′ ( r ) = 3 r (ln(3)) − r ) (ln(2)) > for r ≥ . Therefore, f ( r ) < for < r < . The following example shows that even for n > r in Theorem 2.1 cannot beextended to [1 , ∞ ). Example 2.3 Let A = B = (cid:18) (cid:19) and C ( x ) = x (cid:18) − − (cid:19) . Then per( A ) = per( B ) = 2 , per( C ( x )) = 2 x , per( A + B ) = 8 , per( A + C ( x )) = per( B + C ( x )) = (1 + x ) + (1 − x ) , per( A + B + C ) = (2 + x ) + (2 − x ) . Let x = 0 . and r = 1 . . Then direct calculation shows that per( A + B + C ( x )) r + per( A ) r + per( B ) r + per( C ( x )) r − (per( A + B ) r + per( A + C ( x )) r + per( B + C ( x )) r ) < − . . A general scheme and additional results The following observations capture the main idea in the proof of Theorem 2.1. Proposition 3.1 Let X = { ( x , . . . , x N ) : x i ≥ for all ≤ i ≤ N } and f : X → R be afunction with continuous partial derivatives.1. If ∂f∂x j ≥ on X for all ≤ j ≤ N . Then f ( x ) ≥ f ( ) .2. If ( a , . . . , a N ) , ( b , . . . , b N ) ∈ X are such that ( a , . . . , a N ) ≺ w ( b , . . . , b N ) and p ≥ , then a p + · · · + a pN ≤ b p + · · · + b pN . Suppose A , . . . , A m ∈ M n are positive semi-definite matrices with m ≥ d a generalizedmatrix function on M n . For any subsequence J of the sequence K = { , . . . , m } , denoted by J ≤ K ,let A J = P j ∈ J A j and | J | be the number of terms in J .The following two generalizations of (2.2) are given in [1, Corollary 3.4, Theorem 4.3]. m X j =1 ( − m − j X J ≤ K, | J | = j d ( A J ) ≥ , (3.1) d ( A + · · · + A m ) + ( m − m X j =1 d ( A j ) ≥ X i Suppose d ( A ) is a generalized matrix function on M n and A , . . . , A m ∈ M n arepositive semi-definite matrices. Let x i = d ( A i ) ≥ for ≤ i ≤ m , and x J = d ( A J ) − X L ≤ J,L = J x L for J ≤ K with | J | > . Then x J ≥ for every J ≤ K .Proof. Let x J satisfy the hypothesis of the lemma. We may relabel A , . . . , A m , and assumethat J = { , . . . , | J |} .We prove the result by induction on | J | . The case for | J | = 1 is trivial as d ( A ) = x ≥ | J | = 2 follows from x = d ( A + A ) − x − x = d ( A + A ) − d ( A ) − d ( A ) ≥ q matrices chosen from { A , . . . , A | J | } with q < | J | . Then we seethat x I ≥ I ≤ J with | I | < | J | . It remains to show that x J ≥ 0. By (3.1), | J | X i =1 ( − | J |− i X I ≤ J, | I | = i d ( A I ) ≥ ⇒ d ( A J ) ≥ − | J |− X i =1 ( − | J |− i X I ≤ J, | I | = i d ( A I ) . (3.3)6eplace d ( A I ) by P L ≤ I x L for each term on the right hand side. After the replacement, let usdetermine the coefficient of x L on the right side of (3.3) for each L ≤ J with 1 ≤ | L | < | J | . Notethat x L is a summand of d ( A I ) = P L ≤ I x L if and only if L ≤ I , and there are (cid:0) | J |−| L || I |−| L | (cid:1) so many d ( A I ) with L ≤ I . Thus, the coefficient of x L on the right side of (3.3) is − | J |− X i = | L | ( − | J |− i (cid:18) | J | − | L | i − | L | (cid:19) = − | J |−| L |− X j =0 ( − | J |−| L |− j (cid:18) | J | − | L | j (cid:19) = − (cid:16) (1 − | J |−| L | − (cid:17) = 1 . Therefore, we can choose x J = d ( A J ) − X L ≤ J,L = J x L ≥ . The following theorem extends [1, Corollary 3.5], which corresponds to the case when r = 1. Theorem 3.3 Let A , . . . , A m ∈ M n be positive semi-definite matrices, and r ∈ { , . . . , m − } ∪ [ m − , ∞ ) . Then for any generalized matrix function d ( X ) , we have m X j =1 ( − m − j X J ≤ K, | J | = j d ( A J ) r ≥ , where K = { , . . . , m } .Proof. By Theorem 3.3 in [1], for all positive integer p , we have m X j =1 ( − m − j X J ≤ K, | J | = j ⊗ p A J ≥ . (3.4)For a positive integer r , let p = nr and v ∈ C n such that d ( A ) = v ∗ ( ⊗ n A ) v for all A ∈ M n . Thenthe result follows by applying the quadratic form using ⊗ r v on both sides of (3.4).Suppose r ≥ m − 1. Let X = { ( x J ) : x J ≥ , J ≤ K } , K j = { J ≤ K : | J | = j } for 1 ≤ j ≤ m ,and s J = P I ≤ J x I for J ∈ K j . Suppose f ( x ) = m X j =1 ( − m − j X J ∈ K j s rJ . (3.5)We will prove that ∂f∂x I ≥ I ≤ K , so that f ( x ) ≥ f ( ) = 0 for x ∈ X . It turns out thatwe need to use the nonnegativity of the higher order partial derivatives of f to prove that the firstpartial derivatives of f are nonnegative, which corresponds to the case when t = 1 in the followingclaim. 7 laim: Let 1 ≤ t ≤ m − I , . . . , I t ≤ K such that I j \ ∪ j − i =1 I i = ∅ for 2 ≤ j ≤ t . Then ∂ t f∂x I . . . ∂x I t ≥ . Note that for I, J ≤ K , ∂s rJ ∂x I = rs r − J if I ≤ J and 0 otherwise. More generally, for I , I , . . . , I t , J ≤ K , ∂ t s rJ ∂x I . . . ∂x I t = r ( r − . . . ( r − t + 1) s r − tJ if ∪ tj =1 I j ≤ J , and 0 otherwise.For I ≤ K , let X I = { x ∈ X : x J = 0 for all J I } . Then for every x ∈ X I and I ≤ J ≤ K ,we have s J ( x ) = s I ( x ). Let I = ∪ ti =1 I i and x ∈ X I . Then if ∪ tj =1 I j ≤ J , we have ∂ t s rJ ∂x I . . . ∂x I t ( x ) = r ( r − . . . ( r − t + 1) s r − tJ ( x ) = r ( r − . . . ( r − t + 1) s r − tI ( x ) . Let | I | = p . We have ∂ t f∂x I . . . ∂x I t ( x )= r ( r − . . . ( r − t + 1) (cid:16)P mj = p ( − m − j P J ∈ K j , I ≤ J s r − tJ (cid:17) ( x )= r ( r − . . . ( r − t + 1) (cid:16)P mj = p ( − m − j P J ∈ K j , I ≤ J s r − tI (cid:17) ( x )= r ( r − . . . ( r − t + 1) (cid:18)P mj = p ( − m − j (cid:18) m − pj − p (cid:19)(cid:19) s r − tI ( x )= ( − m − p r ( r − . . . ( r − t + 1) (cid:18)P m − pj =0 ( − j (cid:18) m − pj (cid:19)(cid:19) s r − tI ( x )= 0 . (3.6)We will prove the claim by (backward) induction on t . For t = m − 1, by the condition on I i , wehave | I | = m − m . If | I | = m − 1, then ∂ t f∂x I . . . ∂x I t = r ( r − . . . ( r − m + 2) (cid:0) s r − m +1 K − s r − m +1 I (cid:1) ≥ r ≥ m − 1. If | I | = m , then ∂ t f∂x I . . . ∂x I t = r ( r − . . . ( r − m + 2) s r − m +1 K ≥ . Suppose the result holds for some 1 < t ≤ m − 1. Let I , . . . I t − ≤ K such that I j \ ∪ j − i =1 I i = ∅ for2 ≤ j ≤ t − 1. Let I = ∪ t − i =1 I i . For each I t ≤ K , with I t I , we have8 ∂x I t (cid:18) ∂ t − f∂x I . . . ∂x I t − (cid:19) = ∂ t f∂x I . . . ∂x I t ≥ . (3.7)For x = ( x J ) ∈ X , let x I = ( x ′ J ) ∈ X I , where x ′ J = x J if J ≤ I and 0 otherwise. By (3.7), wehave ∂ t − f∂x I . . . ∂x I t − ( x ) ≥ ∂ t − f∂x I . . . ∂x I t − ( x I ) = 0by (3.6). Example 3.4 Suppose m ≥ and A i = [1] for ≤ i ≤ m . Let r > f ( m, r ) = m X j =1 ( − m − j X J ≤ K, | J | = j d ( A J ) r = m X j =1 ( − m − j (cid:18) mj (cid:19) j r . Then f ( m, r ) is the m th finite difference (with step size = 1) of the function g ( x ) = x r at . Bythe mean value theorem of finite difference [10], f ( m, r ) has the same sign as g ( m ) ( x ) for x > .Therefore, f ( m, r ) = 0 for r = 1 , , . . . , m − and f ( m, r ) < for m − i < r < m − i + 1 with ≤ i ≤ [ m/ and f ( m, r ) > for m − i − < r < m − i with ≤ i ≤ [( m − / . Hence, thecondition on r ≥ m − in Theorem 3.3 is necessary. The next result extend the inequality in [1, Theorem 4.8] to non-integer powers. Theorem 3.5 Suppose A , . . . , A m ∈ M n are positive semi-definite matrices, and r ∈ { }∪ [2 , ∞ ) .Let K = { , . . . , m } and K j = { J ≤ K : | J | = j } for ≤ j ≤ m . For each J ≤ K , let A J = P j ∈ J A j . Then for every r ≥ , ≤ k < ℓ < p ≤ m and any generalized matrix function d ( A ) , we have ℓ − kp (cid:18) mp (cid:19) X J ∈ K p d ( A J ) r + p − ℓk (cid:18) mk (cid:19) X J ∈ K k d ( A J ) r ≥ p − kℓ (cid:18) mℓ (cid:19) X J ∈ K ℓ d ( A J ) r . (3.8)To prove the theorem, one needs only consider the case when m = p for the following reason.If the special case when m = p is proved, then for any ˆ J ∈ K p , we have ℓ − kp d (cid:0) A ˆ J (cid:1) r + p − ℓk (cid:18) pk (cid:19) X J ∈ K k ,J ≤ ˆ J d ( A J ) r ≥ p − kℓ (cid:18) pℓ (cid:19) X J ∈ K ℓ ,J ≤ ˆ J d ( A J ) r . (3.9)Note that every J ∈ K k will appear in (cid:0) m − kp − k (cid:1) different ˆ J ∈ K p , and every J ∈ K ℓ will appearin (cid:0) m − ℓp − ℓ (cid:1) many ˆ J ∈ J p . Hence, summing the inequalities (3.9) for different ˆ J , and dividing theresulting inequality by (cid:0) mp (cid:1) , we have ℓ − kp (cid:0) mp (cid:1) X J ∈ K p d ( A J ) r + ( p − ℓ ) (cid:0) m − kp − k (cid:1) k (cid:0) mp (cid:1)(cid:0) pk (cid:1) X J ∈ K k d ( A J ) r ≥ ( p − k ) (cid:0) m − ℓp − ℓ (cid:1) ℓ (cid:0) mp (cid:1)(cid:0) pℓ (cid:1) X J ∈ K ℓ d ( A J ) r , ℓ = p − k = p − Lemma 3.6 Let X = { ( x J ) : x J ≥ , J ≤ K } . For ≤ p ≤ m , define f on X by f ( x ) = ( m − p )!( p − X J ∈ K p s rJ + ( m − p + 2)!( p − X J ∈ K p − s rJ − m − p + 1)!( p − X J ∈ K p − s rJ . Then f ( x ) ≥ for all x ∈ X .Proof. By the comment after Theorem 3.5, it suffices to prove the case when p = m . We needto show that for all x ∈ X ,( m − s rK + 2!( m − P J ∈ K m − s rJ − m − P J ∈ K m − s rJ ≥ ⇔ g ( x ) = ( m − m − s rK + 2 P J ∈ K m − s rJ − m − P J ∈ K m − s rJ ≥ . For 1 ≤ q ≤ m , let K q = { J ≤ K : | J | = q } . Let I ∈ K q . Then ∂g∂x I = r ( m − m − s r − K + 2 X J ∈ K m − , I ≤ J s r − J − m − X J ∈ K m − , I ≤ J s r − J . There are ( m − q ) J ∈ K m − such that I ≤ J . For each such J , we have s K ≥ s J . If( m − m − ≥ m − m − q ) ⇔ q ≥ m + 12 , then we have ∂g∂x I ≥ 0. Let X = { x ∈ X : x I = 0 for all I with | I | ≥ m + 12 } . For x ∈ X , let x be the projection of x to X . Then we have g ( x ) ≥ g ( x ). It suffices to provethat ∂g∂x I ( x ) ≥ x ∈ X , and I ≤ K with | I | < m + 12 . (3.10)Then we have g ( x ) ≥ g ( x ) ≥ g ( ) = 0.Note that for x ∈ X , s J ( x ) = P I ≤ J, | I | < m +12 x I . Without loss of generality, we may assumethat I = { m − q + 1 , . . . , m } with q < m + 12 .For 1 ≤ i ≤ m − q , (1 ≤ i < j ≤ m − q ), let K ( i ) = K \ { i } ( K ( i, j ) = K \ { i, j } ). Let s ( i ) = s K ( i ) and s ( i, j ) = s K ( i,j ) . We may assume that s (1) ≥ s (2) ≥ · · · ≥ s ( m − q ). Let a = ( a i )and b = ( b i ) ∈ R M , where M = ( m − m − 2) + 2 (cid:0) m − q (cid:1) = ( m − m − 2) + ( m − q )( m − q − a = ( s K , . . . , s K | {z } ( m − m − terms , s (1 , , s (1 , , s (1 , , s (1 , , s (2 , , s (2 , , · · ·· · · , s (1 , m − q ) , . . . , s ( m − q − , m − q )) , b = ( s (1) , . . . , s (1) | {z } m − terms , s (2) , . . . , s (2) | {z } m − terms , . . . , s ( m − q ) , . . . , s ( m − q ) | {z } m − terms , , . . . , | {z } ( q − q − terms ) . M − m − m − q ) = ( q − q − b ≺ w a in the following. For each 1 ≤ k ≤ M , P ki =1 a i = P J ≤ K n ( a, J ) x J and P ki =1 b i = P J ≤ K n ( b, J ) x J for some non-negative integers n ( a, J ) , n ( b, J ). Since x J ≥ J ≤ K , it suffices to show that n ( a, J ) ≥ n ( b, J ) for all J ≤ K .For each J ≤ K with | J | < m + 12 , note that every s K contains a copy of x J and each s ( i )(respectively, s ( i, j )) contains a copy of s J if and only if i J (respectively, both i, j J ). Considerthe following cases: Case 1: ≤ k ≤ ( m − m − n ( a, J ) = k ≥ n ( b, J ). Case 2: ( m − m − < k ≤ M . We may assume that k ≤ m − m − q ). Choose t > t − t − < k − ( m − m − ≤ ( t − t and t ≥ m − t − < k ≤ m − t . Let k = k − ( m − m − − ( t − t − P ki =1 a i consists of ( m − m − 2) copies of s K and 2 copies each of s ( i, j ), where 1 ≤ i < j ≤ t − s ( i, t ), where 1 ≤ i ≤ (cid:20) k (cid:21) and a copy of s ( k + 12 , t ) if k is odd. On the otherhand, P ki =1 b i consists of 2( m − 2) copies of s ( i ) for 1 ≤ i ≤ t − k − m − t − 1) copiesof s ( t ).For every integer m, t , we have( m − m − 2) + t ( t − 1) = 2( m − t + ( m − ( t + 1))( m − ( t + 2)) ≥ m − t . (3.11)In particular, we have ( m − m − 2) + t ( t − ≥ m − t ⇒ t ( t − − ( k − ( m − m − ≥ m − t − k ≥ ⇒ t ( t − ≥ k − ( m − m − . This shows that t ≤ t .If J ≤ I , then n ( a, J ) = k = n ( b, J ).Suppose J I . Let J \ I = { j , . . . , j u } with j < j < · · · < j v ≤ t − < t ≤ j v +1 < · · · < j u .Then n ( b, J ) = k − m − v if j v +1 = t m − t − − v ) if j v +1 = t . On the other hand, n ( a, J ) = k − number of copies of s ( i, j ) in k X i =1 a i such that either i or j ∈ J \ I. Consider the following subcases: 11 ase 2a: t < t or t < j v +1 . Then t < j v +1 . We have n ( a, J ) ≥ k − v ( t − ≥ k − v ( m − q − ≥ k − v ( m − ≥ n ( b, J ) . Case 2b: t = t and t = j v +1 . Then by (3.11), k = k − ( m − m − − ( t − t − ≤ k − m − t − 1) = k − m − t − . So we have n ( a, J ) ≥ k − v ( t − − k ≥ k − v ( m − − ( k − m − t − m − t − − v )= n ( b, J ) . Proof of Theorem 3.5. For 1 ≤ q ≤ m , let t q = 1 q (cid:0) mq (cid:1) P J ∈ K q d ( A J ) r . Then (3.8) is equivalent to( ℓ − k )( t p − t ℓ ) ≥ ( p − ℓ )( t ℓ − t k ) (3.12)for all 1 ≤ k < ℓ < p ≤ m .By Lemma 3.6, (3.12) holds for ℓ = p − k = p − 2. Thus, for every p ≥ q > ℓ ≥ q ′ > k ,we have t q − t q − ≥ t q − − t q − ≥ · · · ≥ t q ′ − t q ′ − . Therefore, we have ( ℓ − k )( t q − t q − ) ≥ P ℓq ′ = k +1 ( t q ′ − t q ′ − ) = t ℓ − t k ⇒ ( ℓ − k )( t p − t ℓ ) = ( ℓ − k ) P pq = ℓ +1 ( t q − t q − ) ≥ ( p − ℓ )( t ℓ − t k ) . Remark 3.7 When m = p = 3 , ℓ = 2 , k = 1 , (3.8) reduces to (2.4). Therefore, Example 2.2shows that the condition r ≥ in Theorem 3.5 is best possible. Additional results and techniques For any partition { I , . . . , I k } of { , . . . , m } , and for any Schur Convex function f : R k → R , wehave f ( d ( A + · · · + A m ) , , . . . , ≥ f ( d ( A I , · · · , d ( A I k ))) . More generally, suppose ( I , . . . , I k ) and ( J , . . . , J k ) are two collections of muti-subsets { , . . . , m } ,we can define ( I , . . . , I k ) ⊳ ( J , . . . , J k ) if the union of ℓ subsets in the family ( I , . . . , I k ) is alwayscontained in the union of ℓ subsets in the family ( J , . . . , J k ) for ℓ = 1 , . . . , k . We have the following. Theorem 4.1 Let A , . . . , A m ∈ M m be positive semi-definite matrices. Suppose ( I , . . . , I k ) and ( J , . . . , J k ) are two collections of multi-subsets { , . . . , m } such that ( I , . . . , I k ) ⊳ ( J , . . . , J k ) . Thenfor any Schur Convex function f : R k → R , we have f ( d ( A J ) + · · · + d ( A J k )) ≥ f ( d ( A I , · · · d ( A I k )) . One can also take partial sum of the positive semi-definite matrices A , . . . , A m in Theorem 3.5,and obtain the following result that removes the restriction r ≥ Theorem 4.2 Suppose A , . . . , A m ∈ M n are positive semi-definite matrices, and Φ is a convexfunction on [0 , ∞ ) . Let K = { , . . . , m } and K j = { J ≤ K : | J | = j } for ≤ j ≤ m . Then forevery ≤ k < ℓ < p ≤ m and any generalized matrix function d ( X ) , we have ℓ − k (cid:18) mp (cid:19) X J ∈ K p Φ ( d ( A J )) + p − ℓ (cid:18) mk (cid:19) X J ∈ K k Φ ( d ( A J )) ≥ p − k (cid:18) mℓ (cid:19) X J ∈ K ℓ Φ ( d ( A J )) . (4.1) Proof. Suppose J ∈ K p . Then for every I ∈ K p − with I ≤ J , let J \ I = { j , j } and J i = I ∪{ j i } for i = 1 , 2. Then J , J are the only ˆ J ∈ K p − such that I ≤ ˆ J ≤ J . By (1.2), we have( d ( A J ) , d ( A J )) ≺ w ( d ( A J ) , d ( A I )), we have Φ ( d ( A J )) + Φ ( d ( A I )) ≥ Φ ( d ( A J )) + Φ ( d ( A J )).Summing over all I ∈ K p − , with I ≤ J we have (cid:18) pp − (cid:19) Φ ( d ( A J )) + X I ∈ K p − , I ≤ J Φ ( d ( A I )) ≥ (cid:0) pp − (cid:1) p X ˆ J ∈ K p − , ˆ J ≤ J Φ (cid:0) d (cid:0) A ˆ J (cid:1)(cid:1) . Then summing over all J ∈ K p , we have (cid:0) pp − (cid:1) P J ∈ K p Φ ( d ( A J )) + (cid:0) m − ( p − (cid:1) P I ∈ K p − Φ ( d ( A I )) ≥ ( p − m − ( p − P ˆ J ∈ K p − Φ (cid:0) d (cid:0) A ˆ J (cid:1)(cid:1) . (4.2)For 1 ≤ j ≤ m , let t j = 1 (cid:0) mj (cid:1) P J ∈ K j Φ ( d ( A J )), then (4.2) is equivalent to t p + t p − ≥ t p − ⇔ t p − t p − ≥ t p − − t p − . ℓ − k )( t p − t ℓ ) ≥ ( p − ℓ )( t ℓ − t k ) , which is equivalent to (4.1).Let P k be the set of functions Φ on [0 , ∞ ) such that Φ ( i ) ( x ) ≥ ≤ i ≤ k and x ≥ d ( A J ) r in Theorem 2.1 (respectively, Theorem 3.3 and Theorem 3.5) can be replacedby Φ( d ( A J )) for all Φ ∈ P (respectively, P m − and P ).Finally, we point out that following the same proof in [1], (2.1) can be generalized to thefollowing: Proposition 4.3 Suppose A i , B i , C i ∈ M n i are positive semi-definite matrices for ≤ i ≤ k . Then ⊗ ki =1 ( A i + B i + C i ) + ⊗ ki =1 A i + ⊗ ki =1 B i + ⊗ ki =1 C i ≥ ⊗ ki =1 ( A i + B i ) + ⊗ ki =1 ( A i + C i ) + ⊗ ki =1 ( B i + C i ) . Consequently, we have ⊗ ki =1 ( ⊗ n i ( A i + B i + C i )) + ⊗ ki =1 ( ⊗ n i A i ) + ⊗ ki =1 ( ⊗ n i B i ) + ⊗ ki =1 ( ⊗ n i C i ) ≥ ⊗ ki =1 ( ⊗ n i ( A i + B i )) + ⊗ ki =1 ( ⊗ n i ( A i + C i )) + ⊗ ki =1 ( ⊗ n i ( B i + C i )) . (4.3)For given generalized matrix functions d i on M n i , we can choose unit vectors v i ∈ C n i such that d i ( X ) = v ∗ i Xv i for X ∈ M n i . Let v = ⊗ ki =1 v i and apply the quadratic form using v on both sidesof (4.3). We have ⊗ ki =1 d i ( A i + B i + C i ) + ⊗ ki =1 d i ( A i ) + ⊗ ki =1 d i ( B i ) + ⊗ ki =1 d i ( C i ) ≥ ⊗ ki =1 d i ( A i + B i ) + ⊗ ki =1 d i ( A i + C i ) + ⊗ ki =1 d i ( B i + C i ) . Thus we can replace d ( X ) in Theorem 2.1 by a product of generalized matrix functions d ( X , . . . , X k ) = d ( X ) · · · d k ( X k ). For example, if we set d ( X , X ) = det( X )per( X ) , then for any positive semi-definite matrices A , B , C ∈ M n , A , B , C ∈ M n and r ∈ { } ∪ [2 , ∞ ), we have[det( A + B + C )per( A + B + C )] r +[det( A )per( A )] r + [det( B )per( B )] r + [det( C )per( C )] r ≥ [det( A + B )per( A + B )] r + [det( A + C )per( A + C )] r + [det( B + C )per( B + C )] r . Theorem 3.3, Theorem 3.5 and Theorem 4.2 can also be generalized in a similar way. Acknowledgment This research was supported by the grant from the National Natural Science Foundation ofChina 11571220. Li is an honorary professor of the University of Hong Kong and the ShanghaiUniversity. His research was also supported by USA NSF grant DMS 1331021, Simons FoundationGrant 351047. 14 eferenceseferences