Bound entanglement and distillability of multipartite quantum systems
aa r X i v : . [ qu a n t - ph ] S e p Bound entanglement and distillability of multipartite quantumsystems
Hui Zhao † Xin-yu Yu † Naihuan Jing ‡† College of Applied Science, Beijing University of Technology, Beijing 100124, China and ‡ Department of Mathematics, North Carolina State University, Raleigh, NC 27695, USA ‡ School of Mathematical Sciences, South ChinaUniversity of Technology, Guangzhou 510640, China (Dated: August 21, 2018)
Abstract
We construct a class of entangled states in H = H A ⊗ H B ⊗ H C quantum systems with dim H A = dim H B = dim H C = 2 and classify those states with respect to their distillability properties. Thestates are bound entanglement for the bipartite split( AB ) − C . The states are NPT entanglementand 1-copy undistillable for the bipartite splits A − ( BC ) and B − ( AC ). Moreover, we generalizethe results of 2 ⊗ ⊗ n ⊗ n ⊗ n systems. PACS numbers: 03.65.Ud, 03.67.MnKeywords: Bound entanglement; Distillability . INTRODUCTION Quantum entanglement is one of the most astonishing quantum phenomena. It plays animportant role in quantum information such as dense coding [1], quantum teleportation [2]and quantum cryptographic schemes [3–5].Namely we say that a state of composite systems is considered to be entangled if it cannot be written as a convex combination of product states [6]. Considerable efforts have beendevoted to analyze the separability and entanglement [7–12]. Indeed there are two kindsof entangled states. One is the free entangled state which is distillable, and the other isthe bound entangled state. A bound entangled state is one which is entangled and doesnot violate Peres condition [13]. For 2 ⊗ ⊗ ρ ⊗ n can be projected to obtain a two-qubit state with NPT (non-PPT)[15]. Therefore the bound entanglement can not be brought to the singlet form by means oflocal quantum operations and classical communication from many copies of a given state.Instead, is an NPT state distillable? It was proved that for two-qubit systems all entangledstates are distillable [16]. That means there is no NPT bound entangled state in 2 ⊗ ρ with NPT there exists anotherbound entangled state ρ such that the joint state ρ ⊗ ρ is no longer a bound entangledstate. Such a phenomenon is called superactivation.In this paper, we analyze a class of tripartite entangled states. The paper is organizedas follows. In Section 2, first we construct certain entangled states, then we give a detaileddescription about the entanglement with respect to different bipartite splits in 2 ⊗ ⊗ n ⊗ n ⊗ n systems. Finally, conclusionand discussion are given in Section 4. 2 I. ENTANGLEMENT OF ⊗ ⊗ QUANTUM SYSTEMS
In this section we consider the entanglement of mixed states for different bipartite splitsin 2 ⊗ ⊗ H = H A ⊗ H B ⊗ H C , dim H A = dim H B = dim H C = 2. Let P φ = | φ ih φ | , e i stand for orthonormal basis of C , i = 1 ,
2. We define thevectors Ψ = 1 √ e ⊗ e ⊗ e + e ⊗ e ⊗ e ) , Ψ = 1 √ e ⊗ e ⊗ e + e ⊗ e ⊗ e ) , Ψ = 1 √ e ⊗ e ⊗ e + e ⊗ e ⊗ e ) , Φ b = e ⊗ e ⊗ ( r b e + r − b e ) , b ∈ [0 , . (1)We construct a state as following σ insep = 27 X i =1 P Ψ i + 17 P e ⊗ e ⊗ e , (2)which is inseparable for all bipartite splits. It can be verified by using the partial transpo-sition criterion. Now we define the following state σ b = 7 b b + 1 σ insep + 17 b + 1 P Φ b . (3)Its matrix is of the form σ b = 17 b + 1 b b b b
00 0 b b b b √ − b b √ − b − b b b
00 0 b b . (4)Next we analyze the inseparability of σ b for all possible bipartite splits namely ( AB ) − C , A − ( BC ), B − ( AC ). 3 . Bipartite split ( AB ) − C For the bipartite split ( AB ) − C , we have σ T C b = 17 b + 1 b b b b b b b b b √ − b √ − b b b b b b . (5)It is easy to see that the state σ T C b is positive as σ T C b = I ⊗ I ⊗ U σ b I ⊗ I ⊗ U † , (6)where U = . (7)We now prove that σ b is an entangled state with respect to bipartite split ( AB ) − C by usingthe range criterion. Assume that b = 0 ,
1, then any vector belonging to the range of σ b canbe presented as u = ( A , A , A , A , xA , A + A , A , A ) , A i ∈ C , i = 1 , · · · , , (8)where x = q b − b . On the one hand, for x = 0 ,
1, if u is positive it must be of the form u prod = ( r, s, t, q ) ⊗ ( e A , e A ) = ( r e A , r e A , s e A , s e A , t e A , t e A , q e A , q e A ) , (9)where r, s, t, q, e A , e A ∈ C .Comparing the two forms of vector u , we consider the following cases.(i) If rs = 0, we can put r = 1 , s = 1, then A = A = e A , A = A = e A , q e A = A , q e A = A , and ( q − e A = 0. If q = 1, then e A = e A = 0, u = 0. If q = 1, we put q = 1,then e A = e A , xA = t e A , A + A = t e A , and t = xx − . We have u = A (1 , , xx − , ⊗ (1 , , A ∈ C . (10)4ii) If r = 0, s = 0, we put r = 1, then A = e A , A = e A , A = A = 0, q e A = A , q e A = 0. For the case q = 0, we have e A = e A , then u = 0. For q = 0, we get A = e A = 0, A = − A , we get u = A (1 , , − x, ⊗ (1 , , A ∈ C . (11)(iii) If r = 0, s = 0, we put s = 1, then A = A = q e A = 0, q e A = A = e A . For q = 0,we have e A = e A = 0, then u = 0. For q = 0, we get e A = 0, e A = A , A = t e A = 0, if t = 0, then e A = 0, u = 0. Then we have u = A (0 , , , ⊗ (0 , , A ∈ C . (12)(iiii) If r = 0 , s = 0, then q e A = A = 0, q e A = A = 0, t e A = xA , t e A = A . For q = 0,one has e A = e A = 0, u = 0. For q = 0, u prod = (0 , , , , t e A , t e A , , , (13)if t = 0, then u = 0, we put t = 1, then u = A (0 , , , ⊗ ( x, , A ∈ C . (14)All partial complex conjugations of vectors u , u , u , u are u ⋆ = A (1 , , xx − , ⊗ (1 , ,u ⋆ = A (1 , , − x, ⊗ (1 , ,u ⋆ = A (0 , , , ⊗ (0 , ,u ⋆ = A (0 , , , ⊗ ( x, . (15)On the other hand, any vector belongs to the range of σ T C b can be written as u ′ = ( A ′ , A ′ , A ′ , A ′ , A ′ + A ′ , xA ′ , A ′ , A ′ ) , A ′ i ∈ C , i = 1 , · · · , . (16)Let us check whether the vectors u ⋆ , u ⋆ , u ⋆ , u ⋆ can be written in the above form. For u ⋆ , we obtain that u ⋆ belongs to the rang of ( σ T C b ). For u ⋆ , assuming that it is of the form u ′ , we get A = A ′ = 0, then u ⋆ is the trivial zero vector. For u ⋆ , we have A = A ′ = 0,then u ⋆ is the trivial zero vector. For u ⋆ , considering A ′ = 0 , A ′ + A ′ = xA , xA ′ = A ,we obtain x = 1. This contradicts the fact that x = q b − b = 0 , b = 0 ,
1, the state σ b is a bound entangled state with respect tobipartite split ( AB ) − C . 5 . Bipartite split A − ( BC ) For the bipartite split A − ( BC ), we have σ T BC b = 17 b + 1 b b b b b b b b b √ − b b √ − b b b b
00 0 0 0 0 0 0 b (17)For any nonzero real vector X = ( x , x , · · · , x ) T , we have X T σ T BC b X = f ( x , x , . . . , x ) = bx + bx − bx + b ( x + x ) + b ( x + x ) + b ( x + x ) + ( r b x + r − b x ) . (18)Obviously, the positive index of inertia is 6, and the rank of σ T BC b is 7. Therefore σ b is anNPT state with respect to bipartite split A − ( BC ).Next we will show that the state σ b is 1-copy undistillable with respect to bipartite split A − ( BC ). We begin with the following Theorem 1.
A bipartite state ρ acting on a Hilbert space H A ⊗ H B is distillable if andonly if there exist a positive integer N ∈ N and a Schmidt rank-2 state vector | ψ [ N ]2 i in H ⊗ NA ⊗ H ⊗ NB such that [15] h ψ [ N ]2 | ( ρ ⊗ N ) T B | ψ [ N ]2 i = h ψ [ N ]2 | ( ρ T B ) ⊗ N | ψ [ N ]2 i < . (19)For N = 1, the Schmidt rank-2 state is of the form | ψ [1]2 i = X k,i =1 4 X j =1 c k u ( k ) i v ( k ) j | i i A ⊗ | j i BC , (20)where P k =1 c k = 1, P i =1 u ( k ) ∗ i u ( k ) i = δ k k , P j =1 v ( k ) ∗ j v ( k ) j = δ k k . So we have h ψ [1]2 | σ T BC b | ψ [1]2 i = X k ,k ,i =1 4 X j =1 b + 1 c ∗ k c k u ( k ) ∗ i ( M ( k ,k ) ) i,j u ( k ) i = 17 b + 1 Y † M Y (21)with Y = ( c u , c u , c u , c u ) T . We get the matrix M is positive. According to theTheorem 1, the state σ b is 1-copy undistillable with respect to bipartite split A − ( BC ).6 . Bipartite split B − ( AC ) For the bipartite split B − ( AC ), we can use the same method as above, for any nonzeroreal vector X = ( x , x , · · · , x ) T , we have X T σ T AC b X = f ( x , x , . . . , x ) = bx + bx − bx + b ( x + x ) + b ( x + x ) + b ( x + x ) + ( r b x + r − b x ) . (22)The positive index of inertia is 6, and the rank of σ T AC b is 7, then σ b is also a NPT statewith respect to bipartite split B − ( AC ).In the similar way, by direct calculation we have h ψ [1]2 | ( σ T AC b | ψ [1]2 i ≥ | ψ [1]2 i in H ⊗ B ⊗ H ⊗ AC . Therefore, σ b is 1-copy undistillable with respect tobipartite split B − ( AC ). III. ENTANGLEMENT OF n ⊗ n ⊗ n QUANTUM SYSTEMS
Consider the Hilbert space H = H A ⊗ H B ⊗ H C , dim H A = dim H B = dim H C = 2 n . Let P φ = | φ ih φ | , e i stand for orthonormal basis of C n , i = 1 , , · · · , n . We define the vectorsΨ ijk = 1 √ e i ⊗ e j ⊗ e k + e n + i ⊗ e j ⊗ e k +1 ) , Ψ ik = 1 √ e i ⊗ e k ⊗ e n + e n + i ⊗ e k +1 ⊗ e ) , Φ a = e n +1 ⊗ e ⊗ ( r a e + r − a e n ) , a ∈ [0 , . (23)where i = 1 , · · · , n , j = 1 , · · · , n , k = 1 , · · · , n −
1. Now we define the following state ρ insep = 28 n − n X i =1 2 n X j =1 2 n − X k =1 ( P Ψ ijk + P Ψ ik ) + 18 n − P e n ⊗ e n ⊗ e n . (24)This state is inseparable with respect to all bipartite splits as there always exist a minormatrix of order 2 of its partial transposition is negative. Mixing the states ρ insep and P Φ a ,we have ρ a = (8 n − a (8 n − a + 1 ρ insep + 1(8 n − a + 1 P Φ a . (25)Next we analyze the different types of entanglement of ρ a for all possible bipartite splits.7 . Bipartite split ( AB ) − C For the bipartite split ( AB ) − C , ρ T C a is a 4 n × n matrix ρ T C a = 1(8 n − a + 1 F · · · G t H t · · · F · · · G t · · · · · · ... ... ... · · · ...0 0 · · · F · · · G t G · · · K · · · H G · · · F · · · · · · ... ... ... · · · ...0 0 · · · G · · · F , (26)with F = a · · · a · · · · · · ...0 0 · · · a , G = a · · ·
00 0 a · · · · · · ...0 0 0 · · · a · · · ,H = · · · ...0 0 · · · a · · · , K = a · · · √ − a a · · · · · · ... ...0 0 · · · a √ − a · · · a , where F , G , H , K are all 2 n × n matrices and G t stand for transposition of G .For any nonzero real vector X = ( x , x , · · · , x n ) T , we get X T ρ T C a X = n − X k =0 2 n X i =2 a ( x i +2 nk + x n + i +2 nk − ) + n − X k =0 a ( x nk + x n +4 n +2 nk ) + ax n − n +1 + ( r − a x n +1 + r a x n +2 n ) . (27)Obviously, the positive index of inertia is 4 n + 1, and the rank of ρ T C a is 4 n + 1. We drivethat the state ρ T C a is a PPT state. 8ext, we will show that the state ρ a is entangled with respect to bipartite split ( AB ) − C .For any vector belongs to the range of ρ T C a can be presented as v = ( A , A , · · · , A n − , A n , A n +1 , · · · , A n − , A n , · · · , A n − n +1 , · · · , A n − , A n ,A + B, A , · · · , A n , yB, A n +2 , · · · , A n , A , · · · , A n − n +2 , · · · , A n , A n − n +1 ) , (28)where y = q a − a , A i , B ∈ C , i = 1 , , · · · , n .For y = 0 ,
1, if v is positive, it must be of the form v prod = ( s , s , · · · , s n ) ⊗ ( e A , e A , · · · , e A n ) , s i , e A j ∈ C , i = 1 , · · · , n , j = 1 , · · · , n. (29)Let us now consider the following cases, comparing the two forms of vector v .(i) While s = 0, we have s m = s m +2 n = 0 and s n = 0, m = 2 , , · · · , n −
1. The proof isin Appendix A. Hence if s n = 0, then s n +1 = 0, otherwise v = 0, we can put s n +1 = 1,then we get v = B (0 , , · · · , , , , · · · , ⊗ (1 , , · · · , , y ) . (30)If s n = 0, combine with s n ( e A , · · · , e A n − ) = s n ( e A , · · · , e A n ) and s n x e A n = s n +1 e A one has s n +1 = 0, we put s n = 1, so we get v = A n − n +1 (0 , , · · · , , , , · · · , ⊗ (1 , , · · · , , . (31)(ii) While s = 0, we put s = 1, then e A i = A i , i = 1 , , · · · , n . According to therelation A k = s n +1 A k − , 3 ≤ k ≤ n , we have that if for some k , A k = 0, 2 ≤ k ≤ n ,then A , · · · , A n are not zero and s n +1 = 0, if for some k , A k = 0, 2 ≤ k ≤ n , then A = · · · = A n = 0, s n +1 = 0.If A = 0, from A = s n +2 A n , then s n +2 = 0, A n = 0, otherwise v = 0, according tothe conclusion of Appendix A and s n A n = s n A n − , one has s n = 0, therefore v = A (1 , , · · · , , s n +1 , , · · · , ⊗ (0 , , s n +1 , s n +1 , · · · , s n − n +1 ) . (32)If A = 0, we put s n +1 = 1, then A + B = A , yB = A n and A = , · · · , = A n . From A = s n +2 A n , we obtain s n +2 = y +1 y . Since s m A = s n + m A , 2 ≤ m ≤ n and s m A = s n + m +1 A n , 2 ≤ m ≤ n −
1, then s m = ( y +1 y ) m − , s n + m = ( y +1 y ) m − , we have v = A (1 , ( y + 1 y ) , ( y + 1 y ) , · · · , ( y + 1 y ) n − , , y + 1 y , · · · , ( y + 1 y ) n − ) ⊗ ( y + 1 y , , , · · · , . (33)9ll partial complex conjugations of vectors v , v , v , v are v ⋆ = B (0 , , · · · , , , , · · · , ⊗ (1 , , · · · , , y ) ,v ⋆ = A n − n +1 (0 , , · · · , , , , · · · , ⊗ (1 , , · · · , , ,v ⋆ = A (1 , , · · · , , s n +1 , , · · · , ⊗ (0 , , s n +1 ∗ , s n +1 ∗ , · · · , s n − n +1 ∗ ) , s n +1 = 0 ,v ⋆ = A (1 , ( y + 1 y ) , ( y + 1 y ) , · · · , ( y + 1 y ) n − , , y + 1 y , · · · , ( y + 1 y ) n − ) ⊗ ( y + 1 y , , , · · · , . (34)On the other hand, any vector belongs to the range of ρ a can be written as v ′ = ( A ′ , A ′ , · · · , A ′ n , · · · , A ′ n − n +1 , A ′ n − n +2 , · · · , A ′ n , yB ′ , A ′ , · · · , A ′ n − ,B ′ + A ′ n − , A ′ n , A ′ n +1 , · · · , A ′ n − , · · · , A ′ n − n , A ′ n − n +1 , · · · , A ′ n − ) , (35)Now we check whether vectors v ⋆ , v ⋆ , v ⋆ , v ⋆ can be written in the above form.For v ⋆ , assume it can be written as the form of v ′ , we get B = yB ′ , yB = B ′ , then y = 1, which contradicts the fact that y = 0 ,
1. For v ⋆ , we certainly have A n − n +1 = A ′ n − n +1 = 0, then v ⋆ is the zero vector.For v ⋆ , it must be hold s n +1 A = A ′ = 0, then s n +1 = 0. This contradicts the fact that s n +1 = 0. For v ⋆ , considering A ′ = y +1 y A and A ′ = A , we get A = 0, then v ⋆ is also the zero vector.Therefore, it leads to the conclusion that none of vectors v ⋆ , v ⋆ , v ⋆ , v ⋆ belongs to therange of ρ a . For any a = 0 ,
1, the state ρ a is a bound entangled state with respect to bipartitesplit ( AB ) − C . B. Bipartite split A − ( BC ) For the bipartite split A − ( BC ), we have ρ T BC a is a 2 n × n matrix ρ T BC a = 1(8 n − a + 1 F · · · G ′ H ′ · · · F · · · G ′ H ′ · · · · · · ... ... ... ... · · · ...0 0 0 · · · F · · · G ′ G · · · K · · · H G · · · F · · · · · · ... ... ... ... · · · ...0 0 0 · · · G · · · F (36)10ith F = a · · · a · · · · · · ...0 0 · · · a , G = a · · ·
00 0 a · · · · · · ...0 0 0 · · · a · · · , H = · · · ...0 0 · · · a · · · , and K = a · · · √ − a · · · a · · · · · · · · · ... ... ... · · · ...0 0 · · · a · · · √ − a · · · a · · ·
00 0 · · · a · · · · · · ... ... ... · · · ...0 0 · · · · · · a , which are 4 n × n matrices.For any nonzero real vector X = ( x , x , · · · , x n ) T , we get X T ρ T BC a X = n − X k =0 4 n X i =2 a ( x i +4 kn + x n +4 kn + i − ) + n − X k =0 a ( x kn + x n +(8+4 k ) n ) +( r − a x n +2 n + r a x n +1 ) + ax n − n +1 + ax n +4 n − ax n +1 . (37)Obviously, the state ρ T BC a is not positive, so ρ a is a NPT state with respect to the bipartitesplit A − ( BC ).Now we prove ρ a is 1-copy undistillable with respect to the bipartite split A − ( BC ) byusing Theorem 1.For N = 1, the Schmidt rank-2 state is of the form | ϕ [1]2 i = X k =1 2 n X i =1 4 n X j =1 c k u ( k ) i v ( k ) j | i i A ⊗ | j i BC , (38)where P k =1 c k = 1, P ni =1 u ( k ) ∗ i u ( k ) i = δ k k , P n j =1 v ( k ) ∗ j v ( k ) j = δ k k . Then we have h ϕ [1]2 | ρ T BC a | ϕ [1]2 i = X k ,k =1 2 n X i =1 4 n X j =1 n − a + 1 c ∗ k c k u ( k ) ∗ i ( M ( k ,k ) ) i,j u ( k ) i = 1(8 n − a + 1 Y † M Y (39)11ith Y = ( c u , c u , · · · , c u n , c u , c u , · · · , c u n ) T , and the matrix M are positive, thatis h ϕ [1]2 | ρ T BC a | ϕ [1]2 i ≥ | ϕ [1]2 i in H ⊗ A ⊗ H ⊗ BC . Therefore ρ a is 1-copy undistillable with respect to the bipartite split A − ( BC ). C. Bipartite split B − ( AC ) We can use the same method to analyze the case of bipartite split B − ( AC ). For anynonzero real vector X = ( x , x , · · · , x n ) T , we get X T ρ T AC a X = n − X k =0 2 n − X i =1 a ( x i +2 kn + x n +2 kn + i − ) + n − X k =0 a ( x n (2+ k ) + x n +2 kn +1 ) +( r − a x n +2 n + r a x n +1 ) + ax n +2 n +1 + ax n − ax n +1 , (40)then ρ T AC a is not positive, ρ a is a NPT state with respect to the bipartite split B − ( AC ).By direct calculation h ϕ [1]2 | ρ T AC a | ϕ [1]2 i is positive, where ϕ [1]2 ∈ H ⊗ B ⊗ H ⊗ AC .Therefore ρ a is 1-copy undistillable with respect to bipartite split B − ( AC ). IV. CONCLUSION AND DISCUSSION
In summary, we have constructed a class of tripartite entangled states, then presented adetailed description about the entanglement with respect to all possible bipartite splits in2 ⊗ ⊗ AB ) − C , the state is bound entanglement, foranother two bipartite splits, it is a NPT state and 1-copy undistillable. Finally, we havegeneralized the results to the case of 2 n ⊗ n ⊗ n systems.In order to avoid complicated calculations, we can also use the following method to prove1-copy undistillation. According to the Ref. [25], a bipartite state ρ acting on a Hilbert space H A ⊗ H B is distillable if and only if there exist a positive integer K and two 2-dimensionalprojectors P : ( H A ) ⊗ K −→ C and Q : ( H B ) ⊗ K −→ C such that (( P ⊗ Q ) ρ ⊗ K ( P ⊗ Q )) T B is not positive. For example, in 2 ⊗ ⊗ {| i , | i} and {| i , | i , | i , | i} beorthonormal bases of H A and H BC respectively, we take K = 1, considering the followingtwo-dimensional projectors P = | ih | + | ih | and Q = | ih | + | ih | . Then the nonzeroeigenvalues of matrix (( P ⊗ Q ) σ b ( P ⊗ Q )) T BC are b b +1 , b +1 ( b ± √ b − b +12 + ), whichare positive for b ∈ (0 , Q i of H BC ,12 = 2 , , · · · ,
6, we also get the matrix (( P ⊗ Q i ) σ b ( P ⊗ Q i )) T BC is positive by calculatingthe eigenvalues, then σ b is 1-copy undistillable with respect to the bipartite split A − ( BC ).Using the same method, it is also easy to get σ a is 1-copy undistillable with respect to thebipartite splits B − ( AC ).In 2 n ⊗ n ⊗ n systems, let K = 1, according to the form of matrix ρ T BC a , after takingevery possible two-dimensional projectors P and Q of H A and H BC respectively, the nonzerorows and columns of matrix ( P ⊗ Q ) ρ T BC a ( P ⊗ Q ) constituting a new matrix J only has fivekinds of form as following a a a
00 0 0 a , a a a
00 0 0 a , a a a a a a
00 0 0 a , a a a a a
00 0 0 a , a a a √ − a √ − a a . (41)Obviously, the nonzero eigenvalues of matrix ( P ⊗ Q ) ρ T BC a ( P ⊗ Q ) are equal to the oneof matrix J . It is easy to check that all eigenvalues of J are positive for a ∈ (0 , P ⊗ Q ) ρ T BC a ( P ⊗ Q ) is positive for all two-dimensional projectors P and Q . Therefore, ρ a is 1-copy undistillable with respect to the bipartite split A − ( BC ). Using the same methodto analyze the case of bipartite split B − ( AC ), we get ρ a is also 1-copy undistillable.We also hope that our results will help further investigations of multipartite quantumsystems. Appendix A
Comparing the two forms of v , we have s m ( e A , e A , · · · , e A n ) = s n + m ( e A , e A , · · · , e A n − ) , (A1) s m − e A = s n + m e A n , (A2)where m = 2 , , · · · , n . In fact, we can obtain the two relations from (A1) and (A2),(i) if s m − = 0, then s n + m = 0, m = 2 , , · · · , n ,(ii) if s n + m = 0, then s m = 0, m = 2 , , · · · , n − s m − = 0 and s n + m = 0, m = 2 , , · · · , n . From(A2), we have e A n = 0, then s m e A = s n + m +1 e A n = 0. Here, if s m = 0, the e A = 0,13ccording to (A1), we get e A = 0, e A = 0, · · · , e A n = 0, v is a zero vector, so s m = 0. From(A1) and s n + m = 0. e A = 0, e A = 0, · · · , e A n − = 0 must hold , v is also a zero vector, so s n + m = 0.For the second one, if s n + m = 0, and s m = 0, m = 2 , , · · · , n −
1, then e A = 0, e A = 0, · · · , e A n = 0. Since s m e A = s n + m +1 e A n , we get e A = 0, then v = 0. Therefore if s n + m = 0, then s m = 0. ACKNOWLEDGMENTS
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