# Linear Transformations between Multipartite Quantum Systems That Map the Set of Tensor Product of Idempotent Matrices into Idempotent Matrix Set

Authors: Jinli Xu; Baodong Zheng; Hongmei Yao
Journal: Journal of Function Spaces and Applications
 (2013)

---

## Abstract

LetHn be the set of n×n complex Hermitian matrices and 𝒫n (resp., 𝒯n) be the set of all idempotent (resp., tripotent) matrices in Hn. In l-partite quantum system Hm1Tml=⊗1lHmi, ⊗1l𝒫mi (resp., ⊗1l𝒯mi) denotes the set of all decomposable elements ⊗1lAi such that Ai∈𝒫mi (resp., Ai∈𝒯mi). In this paper, linear maps ϕ from Hm1⋯ml to Hn with n≤m1⋯ml such that ϕ⊗1l𝒫mi∈𝒫n are characterized. As its application, the structure of linear maps ϕ from Hm1⋯ml to Hn with n≤m1⋯ml such that ϕ⊗1l𝒯mi∈𝒯n is also obtained.


---

## Body

## 1. Introduction

LetMn be the vector space of n×n complex matrices and Hn⊆Mn be the vector space of n×n complex Hermitian matrices. In quantum information theory, a l-partite system can be represented as the tensor product space Hm1⋯ml≡⊗1lHmi, where ⊗ is the usual Kronecker product of matrices; a quantum state is represented as a positive semidefinite with trace one in Hn, see [1]. In quantum information science and quantum computing, it is important to understand, characterize, and construct different classes of maps on quantum states [2]. For example, entanglement is one of the main concepts in quantum information theory, and entangled state involves at least bipartite system or multipartite system [3]; to study entangled states, one should construct entanglement witnesses, which are special types of positive maps, see [4]. On this background, the research on the characterizations of maps leaving invariant, some important subsets or quantum properties attracted more and more researchers’ attention, see ([1, 3, 5]). Especially, in [5], Lim characterized the linear and additive maps on tensor products of spaces of Hermitian matrices that carry the set of tensor product of rank one matrices into itself.Preserver problem is a hot area in Banach algebra; there are many results about this area, see ([6–11]). Specially, the idempotent preservers and the rank one preservers play an important role; therefore, it is meaningful to study the two preservers. Chan et al. [12] first characterized linear transformations on Mn preserving idempotent matrices. In [13], the authors obtained the following result: linear map T:Hn→Hm with n≤m satisfies T(𝒫n)⊂𝒫m if and only if T=0 or
(1)T(X)=QXQ-1orT(X)=QXTQ-1,∀X∈Hn,
where Q is unitary matrix.Since the study of linear transformations, preserving idempotents is important in many aspects of mathematics and physics, see [14–17], and inspired by the above, the purpose of this paper is to study linear maps from l-partite system to the space of Hermitian matrices that carry the set of tensor product of idempotent matrices into the set of idempotent matrices in the space of Hermitian matrices, that is, ϕ:Hm1⋯ml→Hn with n≤m1⋯ml satisfying(2)ϕ(⊗1l𝒫mi)⊂𝒫n,
where 𝒫n is the set of all idempotent matrices in Hn, that is, 𝒫n={A∈Hn|A2=A} and ⊗1l𝒫mi denotes the set of all decomposable elements ⊗1lAi such that Ai∈𝒫mi. As application, the forms of linear maps from l-partite system to the space of Hermitian matrices that carry the set of tensor product of tripotent matrices into the set of tripotent matrices in the space of Hermitian matrices are obtained, that is, ϕ:Hm1⋯ml→Hn with n≤m1⋯ml satisfying
(3)ϕ(⊗1l𝒯mi)⊂𝒯n,
where 𝒯n is the set of all tripotent matrices in Hn, that is, 𝒯n={A∈Hn|A3=A} and ⊗1l𝒯mi denotes the set of all decomposable elements ⊗1lAi such that Ai∈𝒯mi.Throughout this paper, we always assume integersl≥1 and n,m1, …, ml≥2. Let Ik be the k×k identity matrix, 0 be the zero matrix which order is omitted in different matrices just for simplicity, and XT (resp., X*, rankX) be the transpose (resp., conjugate transpose, rank) of X. Eij(n) (1≤i,j≤n) stands for the n×n matrix with 1 at the (i,j)th entry and 0 otherwise. Dij(n)=Eij(n)+Eji(n) and Kij(n)=-1Eij(n)--1Eji(n), where 1≤i≠j≤n. For real numbers a and b with a≤b, let [a,b] be the set of all integers between a and b. ⊕ is the usual direct sum of matrices. A linear map π=⊗1lπi:Hm1⋯ml→Hm1⋯ml is canonical if π satisfies π(⊗1lAi)=⊗1lπi(Ai), where πi:Hmi→Hmi satisfies πi(Ai)=Ai or πi(Ai)=AiT for all Ai∈Hmi.


## 2. Main Results

Lemma 1 (see [18]).
SupposeP1,…,Pk∈𝒫n such that Pi+Pj∈𝒫n, ∀1≤i<j≤k. Let ri=rankPi. Then, there exists a unitary matrix U∈Mn such that
(4)Pi=Udiag(0,…,0,1,…,1,0,…,0)U*,
where diag(0,…,0,1,…,1,0,…,0) is the diagonal matrix in which all diagonal entries are zero except those in the (r1+⋯+ri-1+1)st to the (r1+⋯+ri)th rows.Lemma 2.
LetP∈𝒫r. Suppose X, 0s⊕P⊕0n-r-s-X∈𝒫n. Then, there exists Xr∈𝒫r such that X=0s⊕Xr⊕0n-r-s, where 0≤r,s≤n.Proof.
LetT∈Mr be a unitary matrix satisfying P=T(It⊕0)T*. Set
(5)[Ir000T*000In-r-s]X[Ir000T000In-r-s]=[X11X12X13X21X22X23X31X32X33],
where X11∈Hs,X22∈Ht. Then,
(6)[X11X12X13X21X22X23X31X32X33]∈𝒫n,[0r000It0000]-[X11X12X13X21X22X23X31X32X33]∈𝒫n.
A straightforward computation shows that X=0s⊕[T(X22⊕0r-t)T*]⊕0n-r-s; therefore, the result holds.Lemma 3.
LetA=Pr⊕0s⊕0n-r-s and B=0r⊕Ps⊕0n-r-s with Pr∈𝒫r, Ps∈𝒫s. If
(7)αA+βB+γX∈𝒫nforanyrealγwith0≠|γ|≤12,
where X∈Mn, α and β are all roots of x2-x+γ2=0, then, X2=AB, and X is of the form
(8)[0rX12X210s]⊕0n-r-s,
where X12∈Mr×s, X21∈Ms×r.Proof.
It is clear thatA,B∈𝒫n and AB=BA=0. By a direct computation, one can obtain that
(9)X=α(AX+XA)+β(BX+XB)+γ(X2-A-B).
Replacing γ by -γ, we have X2=A+B and
(10)X=α(AX+XA)+β(BX+XB).
Choosing γ=2/5, then α=1/5,β=4/5 or α=4/5,β=1/5, this yields that
(11)X=15(AX+XA)+45(BX+XB),X=45(AX+XA)+15(BX+XB).
Thus, X=AX+XA and X=BX+XB. This, together with the form of A and B, implies that X has the form [0rX12X210s]⊕0n-r-s, where X12∈Mr×s, X21∈Ms×r.Lemma 4.
LetJ=Pp⊕-Pq⊕0n-p-q with Pp∈𝒫p and Pq∈𝒫q. Suppose X∈𝒯n satisfying J-X∈𝒯n and J-2X∈𝒯n. Then, X=Xp⊕-Xq⊕0n-p-q with Xp∈𝒫p and Xq∈𝒫q.Proof.
By a direct computation, we have(12)3X=J2X+JXJ+XJ2,(13)3X=JX2+XJX+X2J.
Let U∈Mp and V∈Mq be unitary matrices satisfying Pp=U(Ir⊕0p-r)U* and Pq=V(Is⊕0q-s)V*, respectively. Then,
(14)[U*000V*000In-p-q]J[U000V000In-p-q]=[Ir⊕0p-r000-Is⊕0q-s0000n-p-q].
By (12), one can assume that
(15)[U*000V*000In-p-q]X[U000V000In-p-q]=[Xr⊕0p-r000-Xs⊕0q-s0000n-p-q],
where Xr∈Hr and Xs∈Hs. This, together with (13), implies that Xr∈𝒫r and Xs∈𝒫s. Thus,
(16)X=[U(Xr⊕0p-r)U*]⊕-[V(Xs⊕0q-s)V*]⊕0n-p-q,
as desired.Lemma 5.
Letψ1, ψ2 be canonical maps on Hm1⋯ml, and U, V are invertible matrices of order m1⋯ml. If ψ1(X)U=Vψ2(X) for all X∈Hm1⋯ml, then ψ1=ψ2 and U=V=λIm1⋯ml for some λ≠0.Proof.
Let us first point out a simple observation which will be used in our proof. If a matrixU∈Mmn commutes with Im⊗S for all real symmetric S∈Mn, then U has the form W⊗In with W∈Mm.
Sinceψ1,ψ2 are canonical maps on Hm1⋯ml, we assume that
(17)ψ1=τ1⊗⋯⊗τl,ψ2=η1⊗⋯⊗ηl,
where τi and ηi are the identity maps X↦X or the transposition maps X↦XT for i∈[1,l].
Clearly,ψi(Im1⋯ml)=Im1⋯ml, i=1,2, hence U=V.
We proveU=λIm1⋯ml by induction on l. The case of l=1 is the well-known fact. We assume that the statement holds true for l-1. We will prove it for l. For any real symmetric S∈Mml, since ψi(Im1⋯ml-1⊗S)=Im1⋯ml-1⊗S, i=1,2, it follows that U=W⊗Iml for some matrix W∈Mm1⋯ml-1. We define linear maps
(18)φ1=τ1⊗⋯⊗τl-1,φ2=η1⊗⋯⊗ηl-1.
It is easy to see that φ1 and φ2 are canonical maps on Hm1⋯ml-1. For any Y∈Hm1⋯ml-1, since
(19)(φ1(Y)W)⊗Iml=ψ1(Y⊗Iml)U=Uψ2(Y⊗Iml)=(Wφ2(Y))⊗Iml,
we have φ1(Y)W=Wφ2(Y), thus, by induction hypothesis, W=λIm1⋯ml-1. Since U is invertible, we have λ≠0. Thus, U=λIm1⋯ml and ψ1(X)=(1/λ)ψ1(X)U=(1/λ)Uψ2(X)=ψ2(X) for all X∈Hm1⋯ml.Theorem 6.
Supposeϕ is a linear map from Hm1⋯ml to Hn with n≤m1⋯ml. Then, ϕ(⊗1l𝒫mi)⊂𝒫n if and only if either ϕ=0 or n=m1⋯ml, there exists a unitary matrix U∈Mn and a canonical map π on Hm1⋯ml such that
(20)ϕ(X)=Uπ(X)U*,∀X∈Hm1⋯ml.Proof.
The sufficiency part is obvious. We prove the necessity part by induction onl. When l=1, it is the result in [13], which has been given in the above introduction. We assume now that the result holds true for l-1 and give the proof of the case l by the following five steps.
Step
1. Suppose P∈𝒫m1 and ϕ(P⊗Im2⋯ml)=0s⊕Ir⊕0n-r-s.Then,
(21)ϕ(P⊗X)=0s⊕ψ(X)⊕0n-r-s,∀X∈Hm2⋯ml,
where ψ is a linear map from Hm2⋯ml to Hr satisfying ψ(⊗2l𝒫mi)⊂𝒫r.
Proof of Step
1. Set
(22)Γ0={Im2⊗⋯⊗Iml},Γ1={P2⊗Im3⋯ml:P2∈𝒫m2},Γ2={P2⊗P3⊗Im4⋯ml:Pi∈𝒫mi,i∈[2,3]},⋯Γl-1={P2⊗⋯⊗Pml:Pi∈𝒫mi,i∈[2,l]}.
We prove by induction onk that ϕ(P⊗X)=0s⊕ψ(X)⊕0n-r-s,∀X∈Γk,k∈[0,l-1]. Assume for a moment that we have already proved this. Then, when ϕ is linear, we can complete the proof of Step 1.
Now, we prove the assertion. The case ofk=0 is just the assumption. Then, we assume that our statement holds true for k-1, and we consider the image of P⊗P2⊗⋯⊗Pk+1⊗Imk+2⋯ml for Pi∈𝒫mi, i∈[2,k+1]. Because
(23)P⊗P2⊗⋯⊗Pk⊗Imk+1⊗Imk+2⋯ml∈⊗1l𝒫mi,P⊗P2⊗⋯⊗Pk⊗Pk+1⊗Imk+2⋯ml∈⊗1l𝒫mi,P⊗P2⊗⋯⊗Pk⊗(Imk+1-Pk+1)⊗Imk+2⋯ml∈⊗1l𝒫mi,
we obtain using the property of ϕ that
(24)ϕ(P⊗P2⊗⋯⊗Pk⊗Imk+1⊗Imk+2⋯ml)∈𝒫n,ϕ(P⊗P2⊗⋯⊗Pk⊗Pk+1⊗Imk+2⋯ml)∈𝒫n,ϕ(P⊗P2⊗⋯⊗Pk⊗Imk+1⊗Imk+2⋯ml),-ϕ(P⊗P2⊗⋯⊗Pk+1⊗Imk+2⋯ml)∈𝒫n.
We obtain by induction hypothesis and Lemma2 that ϕ(P⊗P2⊗⋯⊗Pk+1⊗Imk+2⋯ml)=0s⊕Xr⊕0n-r-s, where Xr∈𝒫r, as desired.
Step
2. If m1⋯ml>n. Then, ϕ=0.
Proof of Step
2. Since Eii(m1)⊗Im2⋯ml∈⊗1l𝒫mi,∀i∈[1,m1] and (Eii(m1)+Ejj(m1))⊗Im2⋯ml∈⊗1l𝒫mi,∀i≠j∈[1,m1], we obtain using the property of ϕ that
(25)ϕ(Eii(m1)⊗Im2⋯ml)∈𝒫n,∀i∈[1,m1],ϕ(Eii(m1)⊗Im2⋯ml)+ϕ(Ejj(m1)⊗Im2⋯ml)∈𝒫n,∀i≠j∈[1,m1].
Denote by ri=rankϕ(Eii(m1)⊗Im2⋯ml), ∀i∈[1,m1]. Using Lemma 1 and composing ϕ by a similarity transformation, we may obtain that
(26)ϕ(Eii(m1)⊗Im2⋯ml)=diag(0,…,0,1,…,1,0,…,0),
where diag(0,…,0,1,…,1,0,…,0) is the diagonal matrix in which all diagonal entries are zero except those in the (r1+⋯+ri-1+1)st to the (r1+⋯+ri)th rows.
Sincen<m1⋯ml, there exists some ri<m2⋯ml. Without loss of generality, we may assume r1<m2⋯ml. Thus, ϕ(E11(m1)⊗Im2⋯ml)=Ir1⊕0n-r1. By Step 1, we see that there exists a linear map ψ from Hm2⋯ml to Hr1 satisfying ψ(⊗2l𝒫mi)⊂𝒫r1 such that ϕ(E11(m1)⊗X)=ψ(X)⊕0n-r1 for all X∈Hm2⋯ml. We obtain by induction hypothesis ψ=0. Thus,
(27)ϕ(E11(m1)⊗X)=0,∀X∈Hm2⋯ml.
For any j∈[2,m1], P∈⊗2l𝒫mi and real γ with 0≠|γ|≤1/2, we have (αE11(m1)+βEjj(m1)+γD1j(m1))⊗P∈⊗1l𝒫mi, where α and β are all roots of x2-x+γ2=0. We obtain using the property of ϕ, (27), and Lemma 3 that ϕ(Ejj(m1)⊗P)=0 and ϕ(D1j(m1)⊗P)=0. By using the similar argument, we see that ϕ(Dij(m1)⊗P)=0, ϕ(Kij(m1)⊗P)=0, ∀i≠j∈[1,m1]. Thus,
(28)ϕ(A⊗P)=0,∀A∈Hm1.
By the arbitrariness of P∈⊗2l𝒫mi, we have
(29)ϕ(A⊗X)=0,∀A∈Hm1,X∈Hm2⋯ml.
Therefore, ϕ=0.
We next always assumem1⋯ml=n and ϕ≠0.
Step
3. Suppose F1,…,Fm1∈𝒫m1 are of rank one with Fi+Fj∈𝒫m1, ∀i≠j∈[1,m1]. Then, there exists a unitary matrix U of order m1⋯ml such that
(30)ϕ(Fk⊗X)=U[Ekk(m1)⊗πk(X)]U*,∀X∈Hm2⋯ml,k∈[1,m1],
where πk is a canonical map on Hm2⋯ml,∀k∈[1,m1].
Proof of Step
3. Using the similar approach as in the proof of Step 2, we may show that rankϕ(Fk⊗Im2⋯ml)=m2⋯ml, ∀k∈[1,m1]. In fact, if rankϕ(Fk⊗Im2⋯ml)>m2⋯ml for some k, then, by Lemma 1, there exists rankϕ(Fi⊗Im2⋯ml)<m2⋯ml for some i. Then, using the similar approach as in the proof of Step 2, we have ϕ=0, which is a contradiction, and there exist a unitary matrix Q such that ϕ(Fk⊗Im2⋯ml)=Q[Ekk(m1)⊗Im2⋯ml]Q*. Thus, applying Step 1, we obtain that
(31)ϕ(Fk⊗X)=Q[Ekk(m1)⊗ψk(X)]Q*,∀X∈Hm2⋯ml,
where linear map ψk:Hm2⋯ml→Hm2⋯ml satisfying ψk(⊗2l𝒫mi)⊂𝒫m2⋯ml, for any k∈[1,m1]. We obtain by induction hypothesis that there exists unitary matrices Uk∈ℳm2⋯ml and canonical maps πk on Hm2⋯ml such that
(32)ψk(X)=Ukπk(X)Uk*,∀X∈Hm2⋯ml,k∈[1,m1].
Set U=Qdiag(U1,…,Um1), we complete the proof of Step 3.
By Step 3, we may assume that(33)ϕ(Ekk(m1)⊗X)=Ekk(m1)⊗πk(X),∀X∈Hm2⋯ml,k∈[1,m1],
where πk is canonical map on Hm2⋯ml,∀k∈[1,m1].
Step
4. For ∀i≠j∈[1,m1], πi=πj and there exists λij, μij with |λij|=|μij|=1 such that
(34)ϕ(Dij(m1)⊗X)=(λijEij(m1)+λ¯ijEji(m1))⊗π1(X),∀X∈Hm2⋯ml,ϕ(Kij(m1)⊗X)=(μijEij(m1)+μ¯ijEji(m1))⊗π1(X),∀X∈Hm2⋯ml.
Proof of Step
4. Without loss of generality, we only prove the case of i=1, j=2.
SetF1=(1/2)E11(m1)+(1/2)E22(m1)+(1/2)D12(m1), F2=(1/2)E11(m1)+(1/2)E22(m1)-(1/2)D12(m1), Fk=Ekk(m1) for k∈[3,m1] (if m1≥3). By Step 3, there exists a unitary matrix V and canonical maps ηk on Hm2⋯ml such that
(35)ϕ(Fk⊗X)=V[Ekk(m1)⊗ηk(X)]V*,∀X∈Hm2⋯ml,k∈[1,m1].
This, together with (33), implies that
(36)Ekk(m1)⊗Im2⋯ml=V[Ekk(m1)⊗Im2⋯ml]V*,∀k∈[3,m1].
Thus, we may set
(37)V=[V11V12V21V22]⊕diag(V33,…,Vm1m1),
where Vkk∈Mm2⋯ml,k∈[1,m1].
On one hand, noting that(38)D12(m1)=2F1-E11(m1)-E22(m1),D12(m1)=E11(m1)+E22(m1)-2F2,
we obtain using (33) and (35) that for all X∈Hm2⋯ml,
(39)ϕ(D12(m1)⊗X)=[2V11η1(X)V11*-π1(X)2V11η1(X)V21*2V21η1(X)V11*2V21η1(X)V21*-π2(X)]⊕0,(40)ϕ(D12(m1)⊗X)=[π1(X)-2V12η2(X)V12*-2V12η2(X)V22*-2V22η2(X)V12*π2(X)-2V22η2(X)V22*]⊕0.
On the other hand, for anyP∈⊗2l𝒫mi and real γ with 0≠|γ|≤1/2, we have (αE11(m1)+βE22(m1)+γD12(m1))⊗P∈⊗1l𝒫mi, where α and β are all roots of x2-x+γ2=0. We obtain using the property of ϕ, (33), and Lemma 3 that
(41)[ϕ(D12(m1)⊗Im2⋯ml)]2=ϕ(E11(m1)⊗Im2⋯ml)+ϕ(E22(m1)⊗Im2⋯ml),(42)ϕ(D12(m1)⊗X)=[0m1Y12Y12*0m1]⊕0,∀X∈Hm2⋯ml,
where Y12∈Mm1. Choosing X=Im2⋯ml, we get using (39), (40), and (42) that 2V11V11*=2V22V22*=2V21V21*=2V12V12*=Im2⋯ml. Hence, by (39), (40), and (42) again, one can obtain that
(43)π1(X)(V11*)-1V21*=2V11η1(X)V21*=-2V12η2(X)V22*=-V12V22-1π2(X).
By Lemma 5, we have π1=π2 and (V11*)-1V21*=λ12Im2⋯ml with λ12≠0. Hence,
(44)ϕ(D12(m1)⊗X)=(λ12E12(m1)+λ¯12E21(m1))⊗π1(X),∀X∈Hm2⋯ml,λ12≠0.
This, together with (41), implies that |λ12|=1. Similarly,
(45)ϕ(K12(m1)⊗X)=(μ12E12(m1)+μ¯12E21(m1))⊗π1(X),∀X∈Hm2⋯ml,|μ12|=1.
This completes the proof of Step 4.
Step
5. There exists ɛ∈{-1,1} such that μij=ɛ-1λij, ∀i≠j∈[1,m1]. If m1≥3, then λijλjk=λik, for distinct i,j,k∈[1,m1].
Proof of Step
5. Since
(46)(12Eii(m1)+12Ejj(m1)+24Dij(m1)+24Kij(m1))⊗Im2⋯ml∈⊗1l𝒫mi,
we obtain using the property of ϕ, (33), and Step 4 that
(47)[1224(λij+μij)24(λ¯ij+μ¯ij)12]∈𝒫2.
Hence, (λij+μij)(λ¯ij+μ¯ij)=2. It follows from |λij|=|μij|=1 that λij2=-μij2. Thus, there exists ɛij∈{-1,1} such that μij=ɛij-1λij.
Ifm1≥3, since
(48)13(Eii(m1)+Ejj(m1)+Ekk(m1)+Dij(m1)+Dik(m1)+Djk(m1))⊗Im2⋯ml∈⊗1l𝒫mi,13(Eii(m1)+Ejj(m1)+Ekk(m1)+Kij(m1)+Kik(m1)+Djk(m1))⊗Im2⋯ml∈⊗1l𝒫mi,13(Eii(m1)+Ejj(m1)+Ekk(m1)+Dij(m1)+Kik(m1)+Kjk(m1))⊗Im2⋯ml∈⊗1l𝒫mi,
we obtain using the property of ϕ, (33), and Step 4 that
(49)13[1λijλikλ¯ij1λjkλ¯ikλ¯jk1]∈𝒫3,(50)13[1ɛijλij-1ɛikλik-1-ɛijλ¯ij-11λjk-ɛikλ¯ik-1λ¯jk1]∈𝒫3,(51)13[1λijɛikλik-1λ¯ij1ɛjkλjk-1-ɛikλ¯ik-1-ɛjkλ¯jk-11]∈𝒫3.
It follows from (49) that λijλjk=λik. This, together with (50), implies that ɛij=ɛik. Similarly, (51) implies that ɛik=ɛjk. Thus, all numbers ɛij have to be the same for any i≠j∈[1,m1]. This completes the proof of this step.
By Steps3–5, we set Q=diag(1,λ12,…,λ1m1), then Q is unitary such that
(52)ϕ(Ekk(m1)⊗X)=[QEkk(m1)Q*]⊗π1(X),∀X∈Hm2⋯ml,k∈[1,m1],ϕ(Dij(m1)⊗X)=[QDij(m1)Q*]⊗π1(X),∀X∈Hm2⋯ml,i≠j∈[1,m1],ϕ(Kij(m1)⊗X)=ɛ[QKij(m1)Q*]⊗π1(X),∀X∈Hm2⋯ml,i≠j∈[1,m1].
Thus, set U=Q⊗Im2⋯ml, we have
(53)ϕ(A⊗X)=U[A⊗π1(X)]U*,∀A∈Hm1,X∈Hm2⋯ml
or
(54)ϕ(A⊗X)=U[AT⊗π1(X)]U*,∀A∈Hm1,X∈Hm2⋯ml.
This completes the proof of Theorem 6.Remark 7.
Whenl≥2, the linear transformation that maps the set of tensor product of idempotent matrices into idempotent matrix set does not necessarily preserve idempotent matrices. For example, let
(55)ϕ(A1⊗A2⊗⋯⊗Al)=A1T⊗A2⊗⋯⊗Al,∀Ai∈ℋmi,i∈[1,l],C=12(E11(m1)⊗E11(m2)+E22(m1)⊗E22(m2)+E12(m1)⊗E12(m2)+E21(m1)⊗E21(m2))⊗Im3⋯ml.
Then ϕ(⊗1l𝒫mi)⊂𝒫m1⋯ml and C∈𝒫m1⋯ml, but
(56)ϕ(C)=12(E11(m1)⊗E11(m2)+E22(m1)⊗E22(m2)+E21(m1)⊗E12(m2)+E12(m1)⊗E21(m2))⊗Im3⋯ml∉𝒫m1⋯ml.
As the application of Theorem6, we give the following theorem.Theorem 8.
Supposeϕ is a linear map from Hm1⋯ml to Hn with n≤m1⋯ml. Then, ϕ(⊗1l𝒯mi)⊂𝒯n if and only if either ϕ=0 or n=m1⋯ml, there exist a unitary matrix U∈Mn, ɛ∈{-1,1} and a canonical mapπ on Hm1⋯ml such that
(57)ϕ(X)=ɛUπ(X)U*,∀X∈Hm1⋯ml.Proof.
The sufficiency part is clear. We give the proof of the necessity part.
Set(58)Γ0={Im1⊗⋯⊗Iml},Γ1={P1⊗Im2⋯ml:P1∈𝒫m1},Γ2={P1⊗P2⊗Im3⋯ml:Pi∈𝒫mi,i∈[1,2]},⋯Γl={P1⊗⋯⊗Pml:Pi∈𝒫mi,i∈[1,l]}.
It is obvious that Γl=⊗1l𝒫mi and Γk⊂⊗1l𝒯mi, ∀k∈[0,l].
SinceIm1⋯ml∈⊗1l𝒯mi, we have ϕ(Im1⋯ml)∈𝒯n. Without loss of generality, we may assume that ϕ(Im1⋯ml)=Ip⊕-Iq⊕0n-p-q. We next prove by induction on k that
(59)ϕ(X)=ψp(X)⊕-ψq(X)⊕0n-p-q,∀X∈Γk,
where ψp(X)∈𝒫p and ψq(X)∈𝒫q.
We assume that our statement holds true fork-1 and prove it for k. For any X=P1⊗⋯⊗Pk⊗Imk+1⋯ml∈Γk, choosing Y=P1⊗⋯⊗Pk-1⊗Imk⊗Imk+1⋯ml∈Γk-1, we have by induction hypothesis that
(60)ϕ(Y)=ψp(Y)⊕-ψq(Y)⊕0n-p-q,
with ψp(Y)∈𝒫p and ψq(Y)∈𝒫q. Since X∈⊗1l𝒯mi, Y-X∈⊗1l𝒯mi and Y-2X∈⊗1l𝒯mi. Using the property of ϕ, we have ϕ(X)∈𝒯n, ϕ(Y)-ϕ(X)∈𝒯n and ϕ(Y)-2ϕ(X)∈𝒯n. By Lemma 4, we have
(61)ϕ(X)=ψp(X)⊕-ψq(X)⊕0n-p-q,
with ψp(X)∈𝒫p and ψq(X)∈𝒫q. This implies that (59) holds.
Asϕ is linear, we can expend ψp to be a linear map from Hm1⋯ml to Hp and ψq to be a linear map from Hm1⋯ml to Hq, then
(62)ϕ(X)=ψp(X)⊕-ψq(X)⊕0n-p-q,∀X∈Hm1⋯ml,
with ψp(⊗1l𝒫mi)⊂𝒫p and ψq(⊗1l𝒫mi)⊂𝒫q, we obtain using Theorem 6 that(a)
ifn<m1⋯ml, then p,q<m1⋯ml, ψp=0, ψq=0, thus, ϕ=0, and(b)
ifϕ≠0, then n=m1⋯ml and p=n or q=n, thus ϕ=ψp or ϕ=-ψq.
This completes the proof of Theorem8.



---

*Source: 182569-2013-08-12.xml*