Matrix limit theorems of Kato type related to positive linear maps and operator means

Transcription

1 Matrix it theorems of Kato type related to positive linear maps and operator means Fumio Hiai 1 arxiv: v1 [math.fa] 1 Oct Tohoku University (Emeritus), Hakusan , Abiko , Japan Abstract We obtain it theorems for Φ(A p ) 1/p and (A p σb) 1/p as p for positive matrices A,B, where Φ is a positive linear map between matrix algebras (in particular, Φ(A) = KAK ) and σ is an operator mean (in particular, the weighted geometric mean), which are considered as certain reciprocal Lie-Trotter formulas and also a generalization of Kato s it to the supremum A B with respect to the spectral order. 010 Mathematics Subject Classification: Primary 15A45, 15A4, 47A64 Key words and phrases: positive semidefinite matrix, Lie-Trotter formula, positive linear map, operator mean, operator monotone function, geometric mean, antisymmetric tensor power, Rényi relative entropy 1 Introduction For any matrices X and Y, the well-known Lie-Trotter formula is the convergence n (ex/n e Y/n ) n = e X+Y. The symmetric form with a continuous parameter is also well-known for positive semidefinite matrices A,B 0 as pց0 (Ap/ B p A p/ ) 1/p = P 0 exp(loga +logb), (1.1) where P 0 is the orthogonal projection onto the intersection of the supports of A,B and loga +logb is defined as P 0 (loga)p 0 +P 0 (logb)p 0. When σ is an operator mean [13] corresponding to an operator monotone function f on (0, ) such that α := f (1) is in (0,1), the operator mean version of the Lie-Trotter formula is also known to hold as pց0 (Ap σb p ) 1/p = P 0 exp((1 α)loga +αlogb) (1.) 1 address: hiai.fumio@gmail.com 1

2 for matrices A,B 0. In particular, when σ is the geometric mean A#B (introduced first in [15] and further discussed in [13]) corresponding to the operator monotone function f(x) = x 1/, (1.) yields pց0 (Ap #B p ) /p = P 0 exp(loga +logb), (1.3) which has the same right-hand side as (1.1). Due to the Araki-Lieb-Thirring inequality and the Ando-Hiai log-majorization[4, 1], it is worthwhile to note that (A p/ B p A p/ ) 1/p and (A p #B p ) /p both tend to P 0 exp(loga +logb) as p ց 0, with the former decreasing and the latter increasing in the log-majorization order (see [1] for details on log-majorization for matrices). In the previous paper [6], under the name reciprocal Lie-Trotter formula, we considered the question complementary to (1.1) and (1.3), that is, about what happens to the its of (A p/ B p A p/ ) 1/p and (A p #B p ) /p as p tends to instead of 0. If A and B are commuting, then (A p/ B p A p/ ) 1/p = (A p #B p ) /p = AB, independently of p > 0. However, if A and B are not commuting, then the question is rather complicated. Indeed, although we can prove the existence of the it (A p/ B p A p/ ) 1/p, the description of the it has a rather complicated combinatorial nature. Moreover, it is unknown so far whether the it of (A p #B p ) /p as p exists or not. In the present paper, we consider a similar (but seemingly a bit simpler) question about what happens to the its of (BA p B) 1/p and (A p #B) 1/p as p tends to, the case where B is fixed without the p-power, in certain more general settings. The rest of the paper is organized as follows. In Section, we first prove the existence of the it of (KA p K ) 1/p as p and give the description of the it in terms of the diagonalization (eigenvalues and eigenvectors) data of A and the images of the eigenvectors by K. We then extend the result to the it of Φ(A p ) 1/p as p for a positive linear map Φ between matrix algebras. For instance, this it is applied to the map Φ(A B) := (A + B)/ to reformulate Kato s it theorem ((A p + B p )/) 1/p A B in [1]. Another application is given to find the it formula as α ց 0 of the sandwiched α-rényi divergence [14, 18], a new relative entropy relevant to quantum information theory. In Section 3, we discuss the it behavior of (A p σb) 1/p as p for operator means σ. To do this, we may assume without loss of generality that B is an orthogonal projection E. Under a certain condition on σ, we prove that (A p σe) 1/p is decreasing as 1 p, so that the it as p exists. Furthermore, when σ is the the weighted geometric mean, we obtain an explicit description of the it in terms of E and the spectral projections of A. It is worth noting that a it formula in the same vein as those in [1] and this paper was formerly given in [, 3] for the spectral shorting operation.

3 Φ(A p ) 1/p for positive linear maps Φ For each n N we write M n for the n n complex matrix algebra and M + n for the set of positive semidefinite matrices in M n. When A M n is positive definite, we write A > 0. We denote by Tr the usual trace functional on M n. For A M + n, λ 1 (A) λ n (A) are the eigenvalues of A in decreasing order with multiplicities, and rana is the range of A. Let A M + n be given, whose diagonalization is A = Vdiag(a 1,...,a n )V = a i v i v i (.1) with the eigenvalues a 1 a n and a unitary matrix V = [v 1 v n ] so that Av i = a i v i for 1 i d. Let K M n and assume that K 0 (our problem below is trivial when K = 0). Consider the sequence of vectors Kv 1,...,Kv n in C n. Let 1 l 1 < l < < l m be chosen so that if l k 1 < i < l k for 1 k m, then Kv i is in span{kv l1,...,kv lk 1 } (this means, in particular, Kv i = 0 if i < l 1 ). Then {Kv l1,...,kv lm } is a linearly independent subset of {Kv 1,...,Kv n }, so we perform the Gram-Schmidt orthogonalization to obtain an orthonormal vectors u 1,...,u m from Kv l1,...,kv lm. In particular, u 1 = Kv l1 / Kv l1. The next theorem is our first it theorem. This implicitly says that the right-hand side of (.) is independent of the expression of (.1) (note that v i s are not unique for degenerate eigenvalues a i ). Theorem.1. We have and in particular, (KAp K ) 1/p = λ k((ka p K ) 1/p ) = a lk u k u k, (.) { a lk, 1 k m, 0, m < k n. (.3) Proof. Write Z p := (KA p K ) 1/p and λ i (p) := λ i (Z p ) for p > 0 and 1 i n. First we prove (.3). Note that Z p p = KAp K = KVdiag(a p 1,...,ap n )V K = [ a p 1 Kv 1 a p Kv a p n Kv n][ Kv1 Kv Kv n ]. (.4) Since λ 1 (p) p TrZp p = Tr[Kv ] [ 1 Kv n a p 1 Kv 1 a p n Kv ] n = a p i Kv i,kv i a p l 1 Kv i, 3

4 we have Moreover, since supλ 1 (p) a l1. nλ 1 (p) p TrZ p p = a p i Kv i,kv i a p l 1 Kv l1, we have Therefore, (.3) holds for k = 1. inf λ 1(p) a l1. To prove (.3) for k > 1, we consider the antisymmetric tensor powers A k and K k for each k = 1,...,n. Note that Z k p = ( (K k )(A k ) p (K k ) ) 1/p (.5) and A k = 1 i 1 < <i k d The above case applied to A k and K k yields that λ ( ) 1(p)λ (p) λ k (p) = λ 1 Z k p a i1 a ik v i1 v ik v i1 v ik. = max { a i1 a ik : 1 i 1 < < i k d, K k (v i1 v ik ) 0 }. (.6) Since K k (v i1 v ik ) = Kv i1 Kv ik is non-zero if and only if {Kv i1,...,kv ik } is linearly independent, it is easy to see that (.6) is equal to a l1 a lk if k m and equal to 0 if k > m. Therefore, which implies (.3). λ 1(p)λ (p) λ k (p) = { a l1 a lk, 1 k m, 0, m < k n, Now, for p > 0 choose an orthonormal basis {u 1 (p),...,u n (p)} of C n for which Z p u i (p) = λ i (p)u i (p) for 1 i n. To prove (.), write ã k := a lk for 1 k m. If ã 1 = 0 then it is obvious that Z p = 0. So assume that ã 1 > 0 and furthermore ã 1 > ã, i.e., λ 1 (p) > λ (p) at the moment. From (.4) we have Zp p = a p i Kv i Kv i l 1 = a p i Kv i Kv i + i=l 1 4 i=l a p i Kv i Kv i

5 so that ( Zp ã 1 ) p l 1 = i=l 1 ( ai l 1 = a p i Kv i u 1 u 1 + i=l 1 ã 1 ) p Kv i u 1 u 1 + i=l i=l a p i Kv i Kv i ( ai ã 1 ) p Kv i Kv i α u 1 u 1 as p for some α > 0, since a i /ã 1 ã /ã 1 < 1 for i l. Hence, for any p > 0 sufficiently large, the largest eigenvalue of (Z p /ã 1 ) p is simple and the corresponding eigen projection converges to u 1 u 1 as p. Since the eigen projection E 1 (p) of Z p corresponding to the largest eigenvalue λ 1 (p) (simple for any large p > 0) is the same as that of (Z p /ã 1 ) p, we have E 1 (p) = u 1 (p) u 1 (p) u 1 u 1 as p. In the general situation, we assume that ã 1 ã k > ã k+1 with 1 k m, where ã k+1 = 0 if k = m. From (.3) note that λ 1(Zp k ) = ã 1 ã k 1 ã k > ã 1 ã k 1 ã k+1 = λ (Zp k ). Hence, for any sufficiently large p > 0, the largest eigenvalue λ 1 (Zp k) = λ 1(p) λ k (p) of Zp k is simple, and from the above case applied to (.5) it follows that u 1 (p) u k (p) u 1 (p) u k (p) u 1 u k u 1 u k as p, since the vector in the present situation corresponding to u 1 = Kv l1 / Kv l1 is K k (v l1 v lk ) K k (v l1 v lk ) = Kv l 1 Kv lk Kv l1 Kv lk = u 1 u k. (The last identity follows from the fact that, for linearly independent w 1,...,w k, w 1 w k / w 1 w k = w 1 w k if w 1,...,w k are the Gram-Schmidt orthogonalization of w 1,...,w k.) By [6, Lemma.4] we see that the orthogonal projection E k (p) onto span{u 1 (p),...,u k (p)} converges to the orthogonal projection E k of span{u 1,...,u k }. Finally, let 0 = k 0 < k 1 < < k s 1 < k s = m be such that ã 1 = = ã k1 > ã k1 +1 = = ã k > > ã ks 1 +1 = = ã ks. The above argument says that, for every r = 1,...,s 1, the orthogonal projection E kr (p) onto span{u 1 (p),...,u kr (p)} converges to the orthogonal projection E kr onto span{u 1,...,u kr }. When ã ks > 0, this holds for r = s as well. Therefore, when ã ks > 0, we have Z p = λ i (p) u i (p) u i (p) 5

6 = = s k r r=1 i=k r 1 +1 s k r r=1 i=k r i=k s+1 λ i (p) u i (p) u i (p) + i=k s+1 (λ i (p) ã i ) u i (p) u i (p) + λ i (p) u i (p) u i (p) s ã kr (E kr E kr 1 ) = r=1 λ i (p) u i (p) u i (p) s ã kr (E kr (p) E kr 1 (p)) r=1 ã i u i u i, where E 0 (p) = E 0 = 0. When ã ks = 0, we may modify the above estimate as Z p = s 1 k r r=1 i=k r 1 +1 λ i (p) u i (p) u i (p) + i=k s 1 +1 s 1 k s 1 ã kr (E kr E kr 1 ) = ã i u i u i = r=1 λ i (p) u i (p) u i (p) ã i u i u i. The following corollary of Theorem.1 is an improvement of [7, Theorem 1.]. Corollary.. Let A M n be positive definite. We have λ i ((KA p K ) 1/p ) = a i for all i = 1,...,n if and only if {Kv 1,...,Kv n } is linearly independent. Remark.3. Note that Theorem.1 can easily extend to the case where K is a rectangle[ n ] n matrix. In fact, when n < n we may apply Theorem.1 to n n K matrices and A, and when n O > n we may apply to n n matrices [ K O ] and A O n n. A linear map Φ : M n M n is said to be positive if Φ(A) M + n for all A M + n, which is further said to be strictly positive if Φ(I n ) > 0, that is, Φ(A) > 0 for all A M n, A > 0. The following is an extended and refined version of Theorem.1. Theorem.4. Let Φ : M n M n be a positive linear map. Let A M + n be given as A = n a i v i v i with a 1 a n and an orthonormal basis {v 1,...,v n } of C n. Then Φ(A p ) 1/p exists and where M 1 := ranφ( v 1 v 1 ), Φ(Ap ) 1/p = 6 a i P Mi,

7 M i := i ranφ( v j v j ) j=1 i 1 j=1 ranφ( v j v j ), i n, and P Mi is the orthogonal projection onto M i for 1 i n. Proof. Let C (I,A) be the commutative C -subalgebra of M n generated by I,A. We can consider the composition of the conditional expectation from M n onto C (I,A) with respect to Tr and Φ C (I,A) : C (I,A) M n instead of Φ, so we may assume that Φ is completely positive. By the Stinespring representation there are a ν N, a -homomorphism π : M n M nν and a linear map K : C nν C n such that Φ(X) = Kπ(X)K for all X M n. Moreover, since π : M n M nν is represented, under a suitable change of an orthonormal basis of C nν, as Φ(X) = I ν X for all X M n under identification M nν = M ν M n, we can assume that Φ is given (with a change of K) as Φ(X) = K(I ν X)K, X M n. We then write I ν A = (I ν V)diag ( ) a 1,...,a }{{} 1,a,...,a }{{},...,a n,...,a n (Iν V) }{{} ν ν ν = a i ( e 1 v i e 1 v i + + e ν v i e ν v i ). Now, we consider the following sequence of nν vectors in C n : K(e 1 v 1 ),...,K(e ν v 1 ),K(e 1 v ),...,K(e ν v ),...,K(e 1 v n ),...,K(e ν v n ), and if K(e j v i ) is a linear combination of the vectors in the sequence preceding it, then we remove it from the sequence. We write the resulting linearly independent subsequence as K(e j v l1 ) (j J 1 ), K(e j v l ) (j J ),..., K(e j v lm ) (j J m ), where 1 l 1 < l < < l m n and J 1,...,J m {1,...,ν}. Furthermore, by performing the Gram-Schmidt orthogonalization to this subsequence, we end up making an orthonormal sequence of vectors in C n as follows: u (l 1) j (j J 1 ), u (l ) j (j J ),..., u (lm) j (j J m ). Since Φ(A p ) 1/p = K((I ν A) p ) 1/p K, Theorem.1 and Remark.3 imply that Φ(A p ) 1/p exists and ( ) Φ(Ap ) 1/p = a lk u (l k ) (l j u k ) j, j J k 7

8 where u (l k ) (l j J k j u k ) = 0 if Jk =. j The next step of the proof is to find what is u (l k ) j J k j this we first note that ( ν ν ) K(e j v i ) K(e j v i ) = K e j v i e j v i j=1 From Lemma.5 below this implies that j=1 (l u k ) j for 1 k m. For K = K(I ν v i v i )K = Φ( v i v i ). R i := ranφ( v i v i ) = span{k(e j v i ) : 1 j ν}. Through the procedure of the Gram-Schmidt diagonalization we see that R i = 0, 1 i < l 1, R l1 = span { u (l 1) j : j J 1 }, R i R l1, l 1 < i < l, (R l1 R l ) R l1 = span { u (l ) j : j J }, R i R l1 R l, l < i < l 3, (R l1 R l R l3 ) (R l1 R l ) = span { u (l 3) j : j J 3 },. R i R l1 R lm 1, l m 1 < i < l m, (R l1 R lm ) (R l1 R lm 1 ) = span { u (lm) j : j J m }, R i R l1 R lm, l m < i n. Now, let P Mi be the orthogonal projections, respectively, onto the subspaces M 1 := R 1, M i := (R 1 R i ) (R 1 R i 1 ), i n, so that P Mi = 0 if i {l 1,...,l m } and P Mlk is the orthogonal projection onto span { u (l } k) j : j J k for 1 k m. Therefore, we have ( ) Φ(Ap ) 1/p = a lk u (l k ) (l j u k ) j = a lk P Mlk = a i P Mi. j J k Lemma.5. For any finite set {w 1,...,w k } in C n, span{w 1,...,w k } is equal to the range of w 1 w w k w k. More generally, for every B 1,...,B k M + n, k j=1 ranb j is equal to the range of B 1 + +B k. 8

9 Proof. Let Q := w 1 w w k w k. Since Qx = w 1,x w w k,x w k span{w 1,...,w k } for all x C n, we have ranq span{w 1,...,w k }. Since w i w i Q, we have w i ran w i w i ranq, 1 i k. Hence we have span{w 1,...,w k } ranq. The proof of the latter assertion is similar. Thanks to the lemma we can restate Theorem.4 as follows: Theorem.6. Let Φ : M n M n be a positive linear map. Let A M + n be given with the spectral decomposition A = m a kp k, where a 1 > a > > a m > 0. Define Then M 1 := ranφ(p 1 ), M k := ranφ(p 1 + +P k ) ranφ(p 1 + +P k 1 ), k m. Φ(Ap ) 1/p = a k P Mk. Example.7. Consider a linear map Φ : M n M n given by ([ ]) X11 X Φ 1 := X 11 +X, X X 1 X ij M n. Clearly, Φ is completely positive. For any A,B M + n and p > 0 we have Thus, it is well-known [1] that Φ ([ ] p ) 1/p ( ) A 0 A p +B p 1/p Φ =. 0 B ([ A 0 0 B ] p ) 1/p = ( ) A p +B p 1/p = A B, (.7) where A B is the supremum of A,B in the spectral order. Here let us show (.7) from Theorem.6. The spectral decompositions of A,B are given as A = m a i P i, B = b j Q j, j=1 where a 1 > > a m 0, b 1 > > b m 0 and m P i = m j=1 Q j = I. Then A B = l c k R k, 9

10 where {c k } l = {a i} m {b j} m j=1 with c 1 > > c l and P i Q j if a i = b j = c k, R k = P i 0 if a i = c k and b j c k for all j, 0 Q j if b j = c k and a i c k for all i. Note that Φ(R 1 + +R k ) = 1 ( i:a i c k P i + j:b j c k Q j so that by Lemma.5 the support projection F k (i.e., the orthogonal projection onto the range) of Φ(R 1 + +R k ) is ( ) ( ) F k =. Theorem.6 implies that i:a i c k P i ([ ] p ) 1/p A 0 Φ = C := 0 B j:b j c k Q j ) l c k (F k F k 1 ). For every x R we denote by E [x, ) (A) the spectral projection of A corresponding to the interval [x, ), i.e., E [x, ) (A) := P i, i:a i x and similarly for E [x, ) (B) and E [x, ) (C). If c k x > c k+1 for some 1 k < l, then we have E [x, ) (C) = F k = E [x, ) (A) E [x, ) (B). This holds also when x > c 1 and x c l. Indeed, when x > c 1, E [x, ) (C) = 0 = E [x, ) (A) E [x, ) (B). When x c l, E [x, ) (C) = I = E [x, ) (A) E [x, ) (B). This description of C is the same as A B in [1], so we have C = A B. Example.8. The example here is relevant to quantum information. For density matrices ρ,σ M n (i.e., ρ,σ M + n with Trρ = Trσ = 1) and for a parameter α (0, )\{1}, the traditional Rényi relative entropy is { 1 D α (ρ σ) := log[ Trρ α σ 1 α] if ρ 0 σ 0 or 0 < α < 1, α 1 + otherwise, where ρ 0 denotes the support projection of ρ. On the other hand, the new concept recently introduced and called the sandwiched Rényi relative entropy [14, 18] is { 1 α 1 D α (ρ σ) := log[ Tr ( ) σ 1 α α ρσ 1 α α ] α if ρ 0 σ 0 or 0 < α < 1, + otherwise. 10

11 By taking the it we also consider D 0 (ρ σ) := D α (ρ σ) = logtr(ρ 0 σ), αց0 [ D 0 (ρ σ) := Dα (ρ σ) = log αց0 Tr( σ 1 α αց0 1 α α ρσ α ) α ]. (We remark that the notations D α and D α are interchanged from those in [8].) Here, note that Tr( σ 1 α αց0 1 α α ρσ α ) α = Tr(σ p/ ρσ p/ ) 1/p = Tr(ρ 0 σ p ρ 0 ) 1/p, where the existence of Tr(ρ 0 σ p ρ 0 ) 1/p follows from the Araki-Lieb-Thirring inequality [4] (also [1]), and the latter equality above follows since λρ 0 ρ µρ 0 for some λ,µ > 0 and λ 1/p Tr(σ p/ ρ 0 σ p/ ) 1/p Tr(σ p/ ρσ p/ ) 1/p µ 1/p Tr(σ p/ ρ 0 σ p/ ) 1/p. It was proved in [8] that D 0 (ρ σ) D 0 (ρ σ) (.8) and equality holds in (.8) if ρ 0 = σ 0. Let us here prove the following: (1) D 0 (ρ σ) = log Q 0 (ρ σ), where Q 0 (ρ σ) := max { Tr(Pσ) : P an orthogonal projection, [P,σ] = 0, (Pρ 0 P) 0 = P }. () D 0 (ρ σ) = D 0 (ρ σ) holds if and only if [ρ 0,σ] = 0. (Obviously, [ρ 0,σ] = 0 if ρ 0 = σ 0.) Indeed, to prove (1), first note that (Pρ 0 P) 0 = P means that the dimension of ranρ 0 P is equal to that of P, that is, ρ 0 v 1,...,ρ 0 v d are linearly independent when {v 1,...,v d } is an orthonormal basis of ranp. Choose 1 l 1 < l < < l m as in the first paragraph of this section (before Theorem.1) for A = σ and K = ρ 0. Let P 0 be the orthogonal projection onto span{v l1,...,v lm }. Then [P 0,σ] = 0, (P 0 ρ 0 P 0 ) 0 = P 0, and Theorem.1 gives Tr(ρ0 σ p ρ 0 ) 1/p = a lk = Tr(P 0 σ). Ontheother hand, let P beanorthogonalprojectionwith[p,σ] = 0and(Pρ 0 P) 0 = P. From[P,σ] = 0wemayassumethatP = d v i k v ik forsome1 i 1 < < i d n 11

12 (after, if necessary, changing v i for degenerate eigenvalues a i ). Since (Pρ 0 P) 0 = P implies that ρ 0 v i1,...,ρ 0 v id are linearly independent, we have d m and Tr(Pσ) = d a ik a lk = Tr(P 0 σ). Next, to prove (), note that Tr(ρ 0 σ p ρ 0 ) 1/p is increasing in p > 0 by the Araki-Lieb- Thirring inequality mentioned above, which shows that Tr(ρ 0 σ) Tr(ρ 0 σ p ρ 0 ) 1/p. This means inequality (.8), and equality holds in (.8) if and only if Tr(ρ 0 σ p ρ 0 ) 1/p is constant for p 1. By [10, Theorem.1] this is equivalent to the commutativity ρ 0 σ = σρ 0. Finally, we consider the complementary convergence of Φ(A p ) 1/p as p, or Φ(A p ) 1/p as p. Here, the expression Φ(A p ) 1/p for p > 0 is defined in such a way that the ( p)-power of A is restricted to the support of A, i.e., defined in the sense of the generalized inverse, and the ( 1/p)-power of Φ(A p ) is also in this sense. The next theorem is the complementary counterpart of Theorem.6. Theorem.9. Let Φ : M n M n be a positive linear map. Let A M + n be given with the spectral decomposition A = m a kp k, where a 1 > a > > a m > 0. Define M k := ranφ(p k + +P m ) ranφ(p k+1 + +P m ), 1 i m 1, M m := ranφ(p m ). Then Φ(A p ) 1/p = a k P M k. Proof. The proof is just a simple adaptation of Theorem.6. We can write for any p > 0 Φ(A p ) 1/p = { Φ((A 1 ) p ) 1/p} 1, where A 1 and { } 1 are defined in the sense of the generalized inverse so that A 1 = a 1 k P k = a 1 m+1 k P m+1 k with a 1 m > > a 1 1 > 0. By Theorem.6 we have Φ((A 1 ) p ) 1/p = a 1 m+1 k P M m+1 k = 1 a 1 k P M k,

13 where M m := ranφ(p m+1 1 ) = ranφ(p m ), M m+1 k := ranφ(p m P m+1 k ) ranφ(p m P m+1 (k 1) ) = ranφ(p m+1 k + +P m ) ranφ(p m+ k + +P m ), k m. According to the proofs of Theorems.1 and.4, we see that the ith eigenvalue λ i (p) of Φ((A 1 ) p ) 1/p converges to a positive real as p, or otherwise, λ i (p) = 0 for all p > 0. That is, λ i (p) 0 as p occurs only when λ i (p) = 0 for all p > 0. This implies that { } 1 m Φ(A p ) 1/p = Φ((A 1 ) p ) 1/p = a k P Mk. Remark.10. AssumethatΦ : M n M n isaunital positivelinearmap. LetA M n bepositivedefiniteand1 p < q. Sincex p/q andx 1/p areoperatormonotoneon[0, ), we have Φ(A q ) p/q Φ(A p ) and so Φ(A q ) 1/q Φ(A p ) 1/p. Hence Φ(A p ) 1/p increases as 1 p ր. Similarly, Φ(A q ) p/q Φ(A p ) and so Φ(A q ) 1/q Φ(A p ) 1/p since x 1/p is operator monotone decreasing on (0, ). Hence Φ(A p ) 1/p decreases as 1 p ր. Moreover, since x 1 is operator convex on (0, ), we have Φ(A 1 ) 1 Φ(A). (See [5, Theorem.1] for more details.) Combining altogether, when A is positive definite, we have and in particular, Φ(A p ) 1/p Φ(A q ) 1/q, p,q 1, (.9) Φ(A p ) 1/p Φ(A p ) 1/p. However, the latter inequality does not hold unless Φ is unital. For example, let [ ] [ ] [ ] [ ] / 1/ 1/ 1/ P 1 :=, P 0 0 :=, Q :=, Q 1/ 1/ :=, 1/ 1/ and consider Φ : M M given by ([ ]) a11 a Φ 1 := a a 1 a 11 P 1 +a Q 1, and A := ap 1 + bp where a > b > 0. Since P ranφ(p1 +P ) = I, P ranφ(p1 ) = P 1 and P ranφ(p ) = Q 1, Theorems.6 and.9 give Φ(Ap ) 1/p = ap 1 +b(i P 1 ) = ap 1 +bp, Φ(A p ) 1/p = a(i Q 1 )+bq 1 = aq +bq 1. 13

14 We compute which is not positive semidefinite. (ap 1 +bp ) (aq +bq 1 ) = [ a b a b a b b a ], Remark.11. We may always assume that Φ : M n M n is strictly positive. Indeed, we may consider Φ as Φ : M n Q 0 M n Q 0 = Mn, where Q 0 is the support projection of Φ(I n ). Under this convention, another reasonable definition of Φ(A p ) 1/p for p 1 is Φ(A p ) 1/p := εց0 Φ((A+εI n ) p ) 1/p, which is well defined since Φ((A + εi) p ) is increasing so that Φ((A + εi) p ) 1/p is decreasing as ε ց 0. But this definition is different from the above definition of Φ(A p ) 1/p. For example, let Φ : M M be given by Φ(A) := KAK with an invertible K = inverse) so that and so [ a b c d On the other hand, is equal to ], and let A = P = KA p K = [ ]. Then A p = P (in the generalized [ ] a ac ac c (KA p K ) 1/p 1 = ( a + c ) 1 1 p εց0 (K(A+εI) p K ) 1/p = (K 1 (A+εI) p K 1 ) 1/p εց0 1 ad bc /p ( b + d ) 1 1 p [ ] a ac ac c. (.10) = (K 1 A p K 1 ) 1/p = (K 1 PK 1 ) 1/p [ ] d bd bd b. (.11) Hence we find that (.10) and (.11) are very different, even after taking the its as p. Here is a simpler [ example. ] Let ϕ : M[ C ] = M 1 be a state (hence, unital) with 1/ 1/ 1 0 density matrix, and let A =. For the first definition we have 1/ 1/ 0 0 ϕ(a p ) 1/p = 1/p = 1. For the second definition, { } (1+ε) εց0 ϕ((a+εi) p ) 1/p p +ε p 1/p = = 0 εց0 14

15 for all p > 0. Moreover, since ϕ(a p ) 1/p = 1/p for p > 0, this example says also that (.9) does not hold for general positive semidefinite A. Problem.1. It is also interesting to consider the it of (A p BA p ) 1/p as p for A,B M + n, a version different from the it treated in Theorem.1. To consider (A p BA p ) 1/p, we may assume without loss of generality that B is an orthogonal projectione (see theargument around(3.) below). Since (A p EA p ) 1/p = (A p E p A p ) 1/p converges as p by [6, Theorem.5], the existence of the it (A p BA p ) 1/p follows. But it seems that the description of the it is a combinatorial problem much more complicated than that in Theorem.1. 3 (A p σb) 1/p for operator means σ In theory of operator means due to Kubo and Ando [13], a main result says that each operator mean σ is associated with a non-negative operator monotone function f on [0, ) with f(1) = 1 in such a way that AσB = A 1/ f(a 1/ BA 1/ )A 1/ for A,B M + n with A > 0, which is further extended to general A,B M + n as AσB = εց0 (A+εI)σ(B +εi). We write σ f for the operator mean associated with f as above. For 0 α 1, the operator mean corresponding to the function x α (x 0) is the weighted geometric mean # α, i.e., A# α B = A 1/ (A 1/ BA 1/ ) α A 1/ for A,B M + n with A > 0. In particular, # = # 1/ is the so-called geometric mean first introduced by Pusz and Woronowicz [15]. The transpose of f above is given by f(x) := xf(x 1 ), x > 0, which is again an operator monotone function on [0, ) (after extending to [0, ) by continuity) corresponding to the transposed operator mean of σ f, i.e., Aσ fb = Bσ f A. We also write f(x) := { f(x 1 ) = f(x)/x if x > 0, 0 if x = 0. (3.1) In the rest of the section, let f be such an operator monotone function as above and σ f be the corresponding operator mean. We are concerned with the existence and 15

16 the description of the it (A p σ f B) 1/p, in particular, (A p # α B) 1/p for A,B M + n. For this, we may assume without loss of generality that B is an orthogonal projection. Indeed, let E be the support projection of B. Then we can choose λ,µ > 0 with λ < 1 < µ such that λe B µe. Thanks to monotonicity and positive homogeneity of σ f, we have λ(a p σ f E) = (λa p )σ f (λe) A p σ f B (µa p )σ f (µe) = µ(a p σ f E). Hence, for every p 1, since x 1/p (x 0) is operator monotone, λ 1/p (A p σ f E) 1/p (A p σ f B) 1/p µ 1/p (A p σ f B), (3.) so that (A p σ f B) 1/p exists if and only if (A p σ f E) 1/p does, and in this case, both its are equal. In particular, when B > 0, since (A p σ f I) 1/p = f(a p ) 1/p, we note that (Ap σ f B) 1/p = f ( ) (A) whenever f ( ) (x) := f(x p ) 1/p exists for all x 0. For instance, if f(x) = 1 α + αx where 0 < α < 1, then σ f = α, the α-arithmetic mean A α B := (1 α)a+αb, and f ( ) (x) = max{x,1}, if f(x) = x α where 0 α 1, then σ f = # α and f ( ) (x) = f(x) = x 1 α, if f(x) = x/((1 α)x+α) where 0 < α < 1, then σ f =! α, the α-harmonic mean A! α B := (A 1 α B) 1, and f ( ) (x) = min{x,1}. But it is unknown to us that, for any operator monotone function f on [0, ), the it f(x p ) 1/p exists for all x 0, while it seems so. When E is an orthogonal projection, the next proposition gives a nice expression for Aσ f E. This was shown in [11, Lemma 4.7], while the proof is given here for the convenience of the reader. Lemma 3.1. Assume that f(0) = 0. If A M n is positive definite and E M n is an orthogonal projection, then where f is given in (3.1). Proof. For every m = 1,,... we have Aσ f E = f(ea 1 E), (3.3) A 1/ (EA 1 E) m A 1/ = (A 1/ EA 1/ ) m+1. (3.4) Note that the eigenvalues of EA 1 E and those of A 1/ EA 1/ are the same including multiplicities. Choose a δ > 0 such that the positive eigenvalues of EA 1 E and 16

17 A 1/ EA 1/ are included in [δ,δ 1 ]. Then, since f(x) is continuous on [δ,δ 1 ], one can choose a sequence of polynomials p k (x) with p k (0) = 0 such that p k (x) f(x) uniformly on [δ,δ 1 ] as n. By (3.4) we have A 1/ p k (EA 1 E)A 1/ = A 1/ EA 1/ p k (A 1/ EA 1/ ) for every k. Since f(0) = 0 by definition, we have and p k (EA 1 E) f(ea 1 E) A 1/ EA 1/ p k (A 1/ EA 1/ ) A 1/ EA 1/ f(a 1/ EA 1/ ) as k. Since f(0) = 0 by assumption, we have f(x) = x f(x) for all x [0, ). This implies that Therefore, A 1/ EA 1/ f(a 1/ EA 1/ ) = f(a 1/ EA 1/ ). A 1/ f(ea 1 E)A 1/ = f(a 1/ EA 1/ ) so that we have f(ea 1 E) = A 1/ f(a 1/ EA 1/ )A 1/ = Aσ f E, as asserted. Formula (3.3) can equivalently be written as Aσ f E = f((ea 1 E) 1 ), (3.5) where (EA 1 E) 1 is the inverse restricted to rane (in the sense of the generalized inverse) and f((ea 1 E) 1 ) is also restricted to rane. In particular, if f is symmetric (i.e., f = f) with f(0) = 0, then Aσ f E = f((ea 1 E) 1 ). Example 3.. Assume that 0 < α 1 and A,E are as in Lemma 3.1. (1) When f(x) = x α and σ f = # α, f(x) = x α 1 for x > 0 so that A# α E = (EA 1 E) α 1, where the (α 1)-power in the right-hand side is defined with restriction to ran E. () When f(x) = x/((1 α)x+α) and σ f =! α, f(x) = (1 α+αx) 1 for x > 0 so that A! α E = {(1 α)e +αea 1 E} 1 = {E((1 α)i +αa 1 )E} 1, where the inverse of E((1 α)i +αa 1 )E in the right-hand side is restricted to rane. 17

18 (3) When f(x) = (x 1)/logx and so σ f is the logarithmic mean, f(x) = (1 x 1 )/logx for x > 0 so that Aσ f E = (E (EA 1 E) 1 )(logea 1 E) 1, where the right-hand side is defined with restriction to ran E. Theorem 3.3. Assume thatf(0) = 0 and f(x r ) f(x) r for all x > 0 and all r (0,1). Let A M + n and E M n be an orthogonal projection. Then (A p σ f E) 1/p (A q σ f E) 1/q if 1 p < q. (3.6) Proof. First, note that f(x r ) = x r f(x r ) x r f(x 1 ) r = f(x) r for all x > 0, r (0,1). By replacing A with A+εI and taking the it as ε ց 0, we may assume that A is positive definite. Let 1 p < q and r := p/q (0,1). By (3.5) we have (A q σ f E) r = f((ea q E) 1 ) r f((ea q E) r ). (3.7) Since x r is operator monotone on [0, ), we have by Hansen s inequality [9] (EA q E) r EA qr E = EA p E so that (EA q E) r (EA p E) 1. Since f(x) is operator monotone on [0, ), we have Combining (3.7) and (3.8) gives f((ea q E) r ) f((ea p E) 1 ) = A p σ f E. (3.8) (A q σ f E) r A p σ f E. Since x 1/p is operator monotone on [0, ), we finally have (A q σ f E) 1/q (A p σ f E) 1/p. Corollary 3.4. Assume that f(0) = 0 and f(x r ) f(x) r for all x > 0, r (0,1). Then for every A,B M + n, the it exists. (Ap σ f B) 1/p Proof. From the argument around (3.) we may assume that B is an orthogonal projection E. Then Theorem 3.3 implies that (A p σ f E) 1/p converges as p. 18

19 Remark 3.5. Following [17], an operator monotone function f on [0, ) is said to be power monotone increasing (p.m.i. for short) if f(x r ) f(x) r for all x > 0, r > 1 (equivalently, f(x r ) f(x r ) for all x > 0, r (0,1)), and power monotone decreasing (p.m.d.) if f(x r ) f(x) r for all x > 0, r > 1. These conditions play a role to characterize the operator means σ f satisfying Ando-Hiai s inequality [1], see [17, Lemmas.1,.]. For instance, the p.m.d. condition is satisfied for f in (1) and () of Example 3., while f in Example 3.(3) does the p.m.i. condition. Hence, for any α [0,1], (A p # α E) 1/p and (A p! α E) 1/p converge decreasingly as 1 p ր. In fact, for the harmonic mean, we have the it A B := (A p!b p ) 1/p, the decreasing it as 1 p ր for any A,B 0, which is the infimum counterpart of A B in [1] (see also Example.7). The reader might be wondering if the opposite inequality to (3.6) holds (i.e., (A p σ f E) 1/p is increasing as 1 p ր ) when f satisfies the p.m.i. condition. Although this is the case when σ = α the weighted arithmetic mean, it is not the case in general. In fact, if it were true, (A p # α E) 1/p must be constant for p 1 since x α satisfies both p.m.i. and p.m.d. conditions, that is impossible. Finally, for the weighted geometric mean # α we obtain the explicit description of (A p # α E) 1/p for any A M + n. For the trivial cases α = 0,1 note that (A p # 0 E) 1/p = A and (A p # 1 E) 1/p = E for all p > 0. Theorem 3.6. Assume that 0 < α < 1. Let A M + n be given with the spectral decomposition A = m a kp k where a 1 > > a m > 0, and E M n be an orthogonal projection. Then where Q 1 := P 1 E, (Ap # α E) 1/p = a 1 α k Q k, (3.9) Q k := (P 1 + +P k ) E (P 1 + +P k 1 ) E, k m. Proof. First, assume that A is positive definite so that P P m = I. When f(x) = x α with 0 < α < 1, formula (3.5) is given as Since A# α E = (EA 1 E) (1 α). (Ap # α E) 1/p = (EA p E) (1 α)/p = (E(A 1 α ) p E) 1/p, it follows from Theorem.9 that (Ap # α E) 1/p = 19 a 1 α k P M k,

20 where M k := rane(p k + +P m )E rane(p k+1 + +P m )E, 1 k m 1, M m := ranep m E. From Lemma 3.7 below we have and for k m, M 1 = rane ranep 1 E = ranp 1 E, M k = rane(p 1 + +P k 1 ) E rane(p 1 + +P k ) E = [ rane rane(p 1 + +P k ) E ] [ rane rane(p 1 + +P k 1 ) E ] = ran(p 1 + +P k ) E ran(p 1 + +P k 1 ) E = ran [ (P 1 + +P k ) E (P 1 + +P k 1 ) E ]. Therefore, (3.9) is established when A is positive definite. Next, when A is not positive definite, let P m+1 := (P P m ). For any ε (0,a m ) define A ε := A+εP m+1. Then the above case implies that where (A ε# α E) 1/p = a 1 α k Q+ε 1 α Q m+1, Q m+1 := E (P 1 + +P m ) E. Assume that 0 < ε < ε < a m. For every p 1, since A p ε A p ε, we have A p ε# α E A p ε # α E andhence (A p ε # αe) 1/p (A p ε # α E) 1/p. Furthermore, since A p ε # αe A p # α E as a m > ε ց 0, we have (A p # α E) 1/p = a m>εց0 (Ap ε# α E) 1/p decreasingly. (3.10) Now, we can perform a calculation of its as follows: 1 (Ap # α E) 1/p = 1 a (Ap ε # αe) 1/p m>εց0 = a m>εց0 = a m>εց0 = 1 (Ap ε# α E) 1/p ( m a 1 α k Q k. a 1 α k Q+ε 1 α Q m+1 ) In the above, the second equality (the exchange of two its) is confirmed as follows. Let X p,ε := (A p ε# α E) 1/p for p 1 and 0 < ε < a m. Then X p,ε is decreasing as 0

21 1 p ր by Theorem 3.3 and also decreasing as a m > ε ց 0 as seen in (3.10). Let X p := ε X p,ε (= (A p # α E) 1/p ), X ε := p X p,ε, and X := p X p. Since X p,ε X p, we have X ε X and hence ε X ε X. On the other hand, since X p,ε X ε, we have X p ε X ε and hence X ε X ε. Therefore, X = ε X ε, which gives the assertion. Inparticular, whena = P isanorthogonalprojection, we have(a p # α E) 1/p = P E for all p > 0 (see [13, Theorem 3.7]) so that both sides of (3.9) are certainly equal to P E. Lemma 3.7. For every orthogonal projections E and P, ranep E = ran(e P E), or equivalently, ranp E = rane ranep E. Proof. According to the well-known representation of two projections (see [16, pp ]), we write [ ] I 0 E = I I 0 0, 0 0 [ ] C CS P = I 0 I CS S 0, where 0 < C,S < I with C +S = I. We have P E = I Since we also have whose range is that of P = 0 I 0 EP E = 0 I 0 0 I 0 [ ] S CS CS C I, [ ] S 0 0, 0 0 [ ] I 0 0 = E P E, 0 0 which yields the conclusion. Acknowledgments This work was supported by JSPS KAKENHI Grant Number JP17K

22 References [1] T. Ando and F. Hiai, Log majorization and complementary Golden-Thompson type inequalities, Linear Algebra Appl. 197/198 (1994), [] J. Antezana, G. Corach and D. Stojanoff, Spectral shorted matrices, Linear Algebra Appl. 381 (004), [3] J. Antezana, G. Corach and D. Stojanoff, Spectral shorted operators, Integral Equations Operator Theory 55 (006), [4] H. Araki, On an inequality of Lieb and Thirring, Lett. Math. Phys. 19 (1990), [5] K. M. R. Audenaert and F. Hiai, On matrix inequalities between the power means: counterexamples, Linear Algebra Appl. 439 (013), [6] K. M. R. Audenaert and F. Hiai, Reciprocal Lie-Trotter formula, Linear and Multilinear Algebra 64 (016), [7] J.-C. Bourin, Convexity or concavity inequalities for Hermitian operators, Math. Ineq. Appl. 7 (004), [8] N. Datta and F. Leditzky, A it of the quantum Rényi divergence, J. Phys. A: Math. Theor. 47 (014), [9] F. Hansen, An operator inequality, Math. Ann. 46 (1980), [10] F. Hiai, Equality cases in matrix norm inequalities of Golden-Thompson type, Linear and Multilinear Algebra 36 (1994), [11] F. Hiai, A generalization of Araki s log-majorization, Linear Algebra Appl. 501 (016), [1] T. Kato, Spectral order and a matrix it theorem, Linear and Multilinear Algebra 8 (1979), [13] F. Kubo and T. Ando, Means of positive linear operators, Math. Ann. 46 (1980), [14] M. Müller-Lennert, F. Dupuis, O. Szehr, S. Fehr and M. Tomamichel, On quantum Rényi entropies: A new generalization and some properties, J. Math. Phys. 54 (013), 103. [15] W. Pusz and S. L. Woronowicz, Functional calculus for sesquilinear forms and the purification map, Rep. Math. Phys. 8 (1975),

23 [16] M. Takesaki, Theory of Operator Algebras I, Encyclopaedia of Mathematical Sciences, Vol. 14, Springer-Verlag, Berlin, 00. [17] S. Wada, Some ways of constructing Furuta-type inequalities, Linear Algebra Appl. 457 (014), [18] M. M. Wilde, A. Winter and D. Yang, Strong converse for the classical capacity of entanglement-breaking and Hadamard channels via a sandwiched Rényi relative Entropy, Comm. Math. Phys. 331 (014),