www.elsevier.com/locate/laa
On completely copositive and decomposable
linear transformations
David A. Yopp
a ,∗,1, Richard D. Hill
baCollege of William and Mary, Williamsburg, VA 23185, USA bIdaho State University, Pocatello, ID 83209, USA Received 13 April 1999; accepted 6 December 1999
Submitted by G.P. Barker
Abstract
Characterization theorems and other results for the cone of completely copositive linear transformations are given. Further results including the cone of decomposable linear transfor-mations and its dual follow. © 2000 Published by Elsevier Science Inc. All rights reserved.
Keywords: Completely copositive; Completely positive decomposable; Positive semidefinite
1. Introduction
Let Mn denote the space of all complex matrices over the complex numbers, let
HPm,ndenote the real vector space of all complex linear transformations which
pre-serve the hermitian matrices, and let CPm,ndenote the set of all completely positive
maps from Mmto Mn. The completely postive maps have been shown to be an
im-portant subcone of the linear transformations that preserve the positive semidefinite matrices (denoted π(PSDm, PSDn)) [2,3,9]. Poluikis and Hill [9, Theorem 1] give
characterizations of CPm,n. To give these results we must introduce some notation.
Let S = {(i, j) : i = 1, . . . , q; j = 1, . . . n}. The usual ordering on S is called the lexicographical ordering; viz.,
(i, j ) < (r, s) ⇐⇒ i < r or (i = r and j < s).
∗ Corresponding author.
E-mail addresses: [email protected] (D.A. Yopp), [email protected] (R.D. Hill). 1 This is part of Yopp’s D.A. thesis written at Idaho State University under the direction of Hill.
0024-3795/00/$ - see front matter2000 Published by Elsevier Science Inc. All rights reserved. PII: S 0 0 2 4 - 3 7 9 5 ( 0 0 ) 0 0 0 1 2 - 4
A second ordering, the antilexicographical ordering, is given by
(i, j ) < (r, s) ⇐⇒ j < s or (j = s and i < r).
Corresponding to each of these orderings we define bijections from S onto{1, . . . ,
qn} by [i, j] = (i − 1)n + j and hi, ji=(j − 1)q + i, respectively. Letting Mn(Mq)
denote the set of n× n block matrices whose elements are from Mq, we define
mappings C : Mq2,n2 → Mn(Mq) and W : Mq2,n2 → Mn(Mq) by C(T)ij rs = t[i,j],[r,s]= tj sir, i, j = 1, . . . , q, r, s = 1, . . . , n and W(T)ij rs= thr,si,hi,ji= trisj, i, j = 1, . . . , n, r, s = 1, . . . , q.
Here trsijdenotes the (r, s)th element of the (i, j )th block.
IfT is a linear transformation from Mmto Mn,hTi will denote its matrix
repre-sentation with respect to the unit matrices Eij ordered antilexicographically.
With this notation Poluikis and Hill [9, Theorem 1] have shown thatT ∈ CPm,n
if and only if either of the following hold: (a)C(hTi) is positive semidefinite; or (b)
W(hTi) is positive semidefinite.
The concept of completely copositive linear transformations has also been ob-served in the study of π(PSDm, PSDn) [4,7,11,12]. A mapping is said to be
com-pletely copositive (in coCPm,n) if it is the composition of an element of CPm,nwith
the transpose mapping, T(A)= Atr. Since T is an element of π(PSDm), coCPm,n⊂ π(PSDm, PSDn). Since T itself is not in CPn, coCPm,nis not a subset of CPm,n. For
example, if T∈ L(M2), then hTi = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 . Thus, C(hTi) = 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 .
SinceC(hTi) is not an element of PSD4, [9, Theorem 1] gives us that T /∈ CP2. In Theorem 3.4, we shall show that CPm,nis not a subset of coCPm,n. Specifically, we
shall show that if A∈ Mn,m, then A⊗KA represents an element of coCPm,nif and
only if rank A= 1. Thus, CPm,nand coCPm,nare distinct sets.
Choi [3] defined a positive linear map (an element of π(PSDm, PSDn)) to be
of coCPn. Woronowics [16] showed that all elements of π(PSD2, PSDn) where n6 3 (and those of π(PSD3, PSD2)) are decomposable. Thus, π(PSD2, PSDn)=
CP2,n+ coCP2,n where n6 3 and π(PSD3, PSD2)= CP3,2+ coCP3,2 (the sum here is definitely not a direct sum). However, Choi [3] and Tang [11] have shown that π(PSDm, PSDn) is not decomposable outside of the above-mentioned cases.
Choi [3, p. 287] stated that “since there are no elegant expressions [in general] for positive (positive-semidefinite-preserving) linear maps, we may be convinced that completely positive linear maps, rather than positive linear maps, deserve the adjective positive”. We shall show that coCPm,nis as rich a topic as CPm,n.
2. The cone coCPm,n
In order to characterize coCPm,n, we must introduce new matrix reorderings
which play a similar role to the mapsC and W used in [15]. We shall use two (more) of the Oxenrider–Hill (matrix) reorderings [8]
N(T)ij
rs = t[i,j],hr,si= tj ris
and
K(T)ij
rs = thr,si,[i,j]= trjsi.
Intuitively,C rearranges the q2rows ofT ∈ Mq2,n2 lexicographically into n× n
blocks ordered lexicographically, whereasN rearranges the q2 rows ofT antilex-icographically into n× n blocks ordered lexicographically. Also, W rearranges the
n2columns of T antilexicographically into q× q blocks ordered antilexicographi-cally, whereasK rearranges the n2columns ofT antilexicographically into q × q blocks ordered lexicographically. Thus, we observe the following natural relations for Aij ∈ Mn:
C(T) = (Aij)16i,j6q ⇐⇒ N(T) = (Atrij)16i,j6q
and
W(T) = (Aij)16i,j6n ⇐⇒ K(T) = (Aj i)16i,j6n.
To find a natural reordering for coCPm,nwe define the following: Let Aij ∈ Mq
and define gPSDnq to be the set of all matrices (Aij)16i,j6n such that (Aj i)16i,j6n
is in PSDnq. Then we see that T ∈ CPn,q if and only if N(hTi) (K(hTi)) is in g
PSDnq.
Consider again T∈ L(M2). Observe that N(hTi) = 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 1 ∈ PSD4.
We shall show thatN and K are the natural reorderings for coCPmn.
Lemma 2.1. For everyT ∈ L(Mn, Mq),K(hTi) = N
hTitr.
Proof. From the definitions above observe that for U = hTitr
N(U)ij
rs= uisj r = trjsi= K(hTi) ij rs.
Lemma 2.2. For everyT ∈ L(Mn, Mq),
(T(Ej i))16i,j6n= K(hTi) = W(hT ◦ Ti).
Proof. LettinghTi = (tij), we have T(Ej i)16i,j6n= q X r,s=1 thr,si,hj,iiErs = q X r,s=1 thr,si,[i,j]Ers = q X r,s=1 K hTiij rsErs =K hTiij.
Finally, from [9, Lemma 2], we have that
WhT ◦ Ti=T ◦ T(Eij) 16i,j6n = T(Ej i) 16i,j6n.
Lemma 2.3. Let E[i,j]= Eij ∈ Mqn, i= 1, . . . , q, j = 1, . . . , n, and T ∈ L(Mn, Mq). Then hT ◦ Ti = qn X l,m=1 (N(hTi))lmEl⊗KEm.
Proof. Letting l= [i, j] and m = [r, s], from [9, Lemma 3] we have that
hTi= qn X l,m=1 (C(hTi)lmEl⊗KEm = qn X [i,j],[r,s]=1 (C(hTi)[i,j],[r,s]E[i,j]⊗KE[r,s]
= q X i,r=1 n X j,s=1 C(hTi)ir j sEij⊗KErs.
Hill and Bernet [6] have shown that T=Pnu,v=1Euv⊗KEvu. Thus, hT ◦ Ti= Xq i,r=1 n X j,s=1 C(hTi)ir j sEij⊗KErs Xn u,v=1 Euv⊗KEvu = q X i,r=1 n X j,s=1 C(hTi)ir j s X u,v EijEuv⊗KErsEvu.
However, EijEuv= 0 if j /= u, and EijEj q= Eiq. Also, ErsEqp= 0 if q /= s,
and ErsEsp= Erp. Thus, we have that hT ◦ Ti = q X i,r=1 n X j,s=1 (C(hTi))irj sEis⊗KErj.
Now observe thatC(hTi)irj s= N(hTi)irsj. Thus,
hT ◦ Ti= q X i,r=1 n X j,s=1 N(hTi)ir sjEis⊗KErj = q X i,r=1 n X j,s=1 N(hTi)[i,s],[r,j]Eis⊗KErj = qn X l0,m0=1 N(hTi)l0,m0En0⊗KEm0.
We shall now give a set of characterization theorems for coCPm,nwhich are the
analogues to those given for CPm,nin [9, Theorem 1]. For notational convenience,
we shall also use T to denotehTi.
Theorem 2.1. LetT : Mm→ Mnbe a linear transformation. Then the following
are equivalent:
(i) T is completely copositive.
(ii) T = S ◦ T, where S is completely positive.
(iii) There exist A1, . . . , As ∈ Mn,m such that T(C) = Ps
i=1AiCtrA∗i, or
equivalently,hTi =Psi=1Ai⊗K Ai◦ T.
(iv) There exist AP 1, . . . , As ∈ Mn,m and (di,j)∈ PSDs such that hTi = s
i,j=1di,jAi⊗KAj◦ T. (v) The block matrix T(Ej,i)
(vi)N hTiis positive semidefinite.
(vii) N hTitris positive semidefinite.
(viii) K hTiis positive semidefinite.
(ix)K hTitris positive semidefinite.
(x)T∗is completely copositive.
Proof. Since T is its own inverse, we have thatT is completely copositive if and
only ifT ◦ T is completely positive. Thus, (i) and (ii) are equivalent. Statements (iii) and (iv) are equivalent to (ii), by [9, Theorem 1].
That (ii) is equivalent to (v) follows from Lemma 2.2 and [9, Theorem 1]. Also, (viii) and (ix) are equivalent to (v) by the same lemma and theorem. By Lemma 2.1 we see the equivalence of (vii) and (viii).
Note that Lemma 2.1 can be reformulated asK(hTitr)= (N(hTi)), and thus, we
see that (vi) and (ix) are equivalent. Finally, since
K(hT∗i) = K(hTi∗)= N(hTi) = N(hTi),
we have that (i) and (x) are equivalent.
Poluikis and Hill [9, Theorem 3] have shown that if A1, . . . , As are linearly
in-dependent in Mn,m, then Ps
i,j=1di,jAi⊗KAj represents an element of CPm,n if
and only if (di,j) is positive semidefinite. We would like to prove a similar result
for coCPm,n. By Lemma 2.3 we have that
Pmn
l,m=1dl,mEl⊗KEm(where the El’s
are as in Lemma 2.3) represents an element of coCPm,n if and only if (dl,m) is in g
PSDmn. Unfortunately, this result does not generalize to the case where the
linear-ly independent set in Mn,m is chosen to be other that the standard basis ordered
lexicographically.
There are two reasons why the coCP case differs from the CP case. First, principal submatrices of ]PSDnare not necessarily elements of a lower dimensional ]PSDr.
Sec-ondly, the key to the proof of [9, Theorem 3] is that PSDnis closed under
∗-congru-ence transformations, whereas gPSD is not. Now gPSD does give us a characterization
of coCPm,nin the standard HPm,nsetting [9, Theorem, 2] as follows:
Theorem 2.2. Let A1, . . . As be linearly independent in Mn,m. If Ps
i,j=1di,jAi⊗KAjrepresents a completely copositive linear transformation, then
there exists a nonsingular F ∈ Mnm such that F∗((di,j)⊕ 0mn−s)F ∈ gPSDmn.
Conversely, suppose that F∗(di,j)F ∈ gPSDmn where F is nonsingular. Then there
exist linearly independent A1, . . . Amnin Mn,msuch that
Pmn
i,j=1di,jAi⊗K Aj
Proof. To prove the first statement, extend{A1, . . . As} to a basis {A1. . . Amn}.
There exists some nonsingular F = (fi,j)∈ Mmnsuch that Ai=Pnmk=1fk,iEkwhere
the Ek’s are as in Lemma 2.3. Let G= (gi,j)= F∗((dij)⊕ 0mn−s)F . Let (clk)= ((dij)⊕ 0mn−s). Thus, by an argument similar to one made in [9, Theorem 3],
we havePmnl,k=1clkAl⊗KAk =
Pmn
l,k=1glkEl⊗KEk, and we must have that G= (gi,j)= F∗(dij⊕ 0mn−s)F . By [9, Theorem 3] and Lemma 2.3, we have that G∈
g
PSDmn. The converse follows from a similar argument.
Theorem 2.3. The cones CPm,nand coCPm,nare isometrically isomorphic.
Proof. Define ψ: L(Mm, Mn)→ L(Mm, Mn) by
ψ: T 7→ T ◦ T.
Since T ◦ T ∈ L(Mm, Mn), ψ is well-defined. Since (T + S) ◦ T = T ◦ T + S ◦ T, we have that ψ is linear. To show that ψ is onto, let S ∈ L(Mm, Mn). Then ψ(S ◦ T) = S ◦ T ◦ T = S. Also, ψ is one-to-one since
ψ(T) = 0 ⇐⇒ T ◦ T = 0 ⇐⇒ T = 0.
Since ψ(CPm,n)= coCPm,n, these cones are isomorphic.
Finally, ψ is an isometry sincehT ◦ T, S ◦ Ti = hT, Si.
In [2], it is shown that CPm,nis a closed, pointed, full (i.e. proper), self-dual cone
in HPm,n. In [15], it is shown that all the faces of CPm,nare exposed. Thus, we have
the following corollary to the above theorem:
Corollary 2.1. The set coCPm,n forms a proper, self-dual cone in HPm,n.
More-over, all of its faces are exposed.
Since the mapping ψ gives a one-to-one correspondence between the extremals of CPm,nand coCPm,n, the following result is immediate:
Corollary 2.2. LetT ∈ coCPm,n. ThenT is an extremal of coCPm,nif and only if
hTi = A ⊗KA◦ T for some A ∈ Mn.
3. The cone of decomposable transformations
An element of π(PSDm, PSDn) is said to be decomposable if it is the sum of an
element of CPm,nand an element of coCPm,n[4]. The set of all decomposable maps
shall be denoted Dm,n.
Proof. Let CP0m,n= {T ∈ CPm,n | kTk 6 1} and coCP 0 m,n= {T ∈ coCPm,n | kTk 6 1}. Observe that Dm,n= {αT | T ∈ CP 0 m,n+ coCP 0 m,nand α> 0}. Since
the sum of two convex compact sets is also a convex compact set, the set of all nonnegative multiples of CP0m,n+ coCP0m,n is a closed convex cone. The fact that Dm,nis a pointed and full is obvious.
In order to characterize the extremals of Dm,n, we first determine if an extremal
of CPm,nor coCPm,nlies in a higher dimensional face of Dm,n; and, secondly, we
determine whether an extremal of CPm,n(coCPm,n) belongs to coCPm,n(CPm,n). In
[14] we have shown that the extremals of both CPm,nand coCPm,nare extremals of π(PSDm, PSDn). Since Dm,n⊆ π(PSDm, PSDn), these elements are also extremals
of Dm,n. In the following we describe the relationship between coCPm,n and the
extremals of CPm,nas well as that between CPm,nand the extremals of coCPm,n.
Theorem 3.2. If A∈ Mn,mand rank A= 1, then A ⊗KA◦ T represents an
extre-mal of CPm,n.
Proof. Since T=Pmi,j=1Eij⊗KEj i,
A⊗KA◦ T =
m X i,j=1
AEij⊗KAEj i.
Now AEij is the n× m matrix with the ith column of A in its jth column and
zeros elsewhere. Since A is rank 1, there exists a column of A, say A(k), such that
A(i)= αiA(k)for i= 1, . . . , m, where each αi ∈ C. Thus, AEij = αiAEkjfor every
i. Therefore, m X i,i=1 AEij⊗KAEj i = n X i,j=1 αiαjAEkj ⊗KAEki= n X j,i=1 αjαiBj⊗KBi,
where Bi = AEki. Since (αjαi) is an extremal of PSDn, by [15, Theorem 4.6], we
have that A⊗KA◦ T is an extremal of CPn.
Our next result will show that if rankA > 1, then A⊗KA◦ T is not an element
of CPn.
Theorem 3.3. The nonzero extremal of coCPm,n, A⊗KA◦ T, is an element of
CPm,niff rank(A)= 1.
Proof. We have already seen that if rank(A)= 1, then A ⊗KA◦ T is an
extre-mal of CPn. Suppose that rank(A)= m > 1. Then the columns of A are linearly
independent. Thus, the set{AEij} is linearly independent. Ordering the AEij’s
A⊗KA◦ T = m2 X k,l=1
dklAEk⊗KAEl,
where the AE1, . . . , AEm2 are linearly independent and
dkl=
1 if l= (j − 1)m + i when k = (i − 1)m + j,
0 otherwise.
By [9, Theorem 3], it suffices to show that (dkl) is not positive semi-definite.
To see this observe that
k= l ⇐⇒ (j − i)m + i = (i − 1)m + j
⇐⇒ (j − i)m + i − j = 0 ⇐⇒ (j − i)(m − 1) = 0 ⇐⇒ j = i whenever m > 1.
If i /= j then dk,l= d(i−1)m+j,(j−1)m+i= 1 = dl,k. However, dk,k= 0 and dl,l=
0. Thus (dkl) has a negative principal minor, and hence, it is not positive semidefinite.
Now suppose that rank(A)= r where 1 < r < m. We may assume that the first
r columns of A are linearly independent, since a similar argument would hold if we
rearranged the AEij’s. Thus, for any q > r
A(q)= r X i=1 αiA(i) or, equivalently, AEqj = r X i=1 αiAEij.
Since{AEij:1 6 i 6 r, 1 6 j 6 m} is a linearly independent set, using the above
expression for the AEqj’s where q > m and ordering the AEijlexicographically, we
have that A⊗KA◦ T = rm X k,l=1 dklAEk⊗KAEl,
where E1, . . . , Emnare linearly independent. Now if we choose i and j such that 16 i, j 6 m and i /= j, then dk,l= d(i−1)m+j,(j−1)m+i = 1 = dl,k. However, dk,k= 0
and dl,l= 0. Then (dkl) has a negative principal minor, and hence, it is not
posi-tive semidefinite. By [9, Theorem 1], A⊗KA◦ T does not represent a completely
positive map.
Theorem 3.4. Let A∈ Mn,m. Then A⊗KA represents an element of coCPm,n if
Proof. (⇒) Suppose that A ⊗KA is an element of coCPm,n. Since T is its own
inverse, A⊗KA◦ T is an element of CPm,n. By Theorem 3.3, A must be of rank 1.
(⇐) Suppose that rank A = 1. Then by Theorem 3.3 A ⊗KA◦ T is an element
of CPm,n. Since T is its own inverse, A⊗KA is an element of coCPm,n.
The proof of the following theorem is immediate from the definition Dm,n=
CPm,n+ coCPm,nby using the fact that the dual of the sum is the intersection of the
duals and the facts that both CPm,nand coCPm,nare self-dual.
Theorem 3.5. The dual cone Dm,n∗ = CPm,n∩ coCPm,n.
Corollary 3.1. LetT ∈ L(Mm, Mn). ThenT ∈ Dm,niffh(T(Eij)16i,j6m), Ai >
0 for all A=(Aij)16i,j6m∈ Mm(Mn) with the property (Aij)∈ PSDmnand (Aj i)∈
PSDmn.
Proof. Barker et al. [2] show thatW ◦ h·i also gives an isometric isomorphism
be-tween CPm,nand PSDmn. By Theorem 2.1 we have thatK ◦ h·i gives an isomorphism
between coCPm,nand PSDmn. Again observe that for Aij ∈ Mn W(hTi) = (Aij)16i,j6m ⇐⇒ K(hTi) = (Aj i)16i,j6m.
Thus,W ◦ h·i gives an isometric isomorphism between D∗m,nand the set of ma-trices A= (Aij)16i,j6m∈ Mm(Mn) with the property (Aij)∈ PSDmnand (Aj i)∈
PSDmn. By [9, Lemma 2], we know that (T(Eij))16i,j6m= W(hTi). Thus, hT, (W ◦ h·i)−1(A)i = hW(hTi), Ai = h(T(Eij))16i,j6m, Ai.
Since Dm,n∗∗ = Dm,nwe have the result.
Kye [7] has studied the generalized Choi map defined as follows: LetH : M3→ M3be a linear map defined by
H[a; c1, c2, c3](A) = 1[a; c1, c2, c3](A) − A, where A∈ M3, a> 0, ci > 0 for i = 1, 2, 3, and
1[a; c1, c2, c3]([xij]) = ax11+ c0 1x33 ax22+ c0 2x11 00 0 0 ax33+ c3x2,2 .
Kye has shown that when a > 2 and c1c2c3> (a − 3)3this map is positive semi-definite preserving. However, it is not decomposable when 26 a < 3 (Kye actually proves the stronger result that it is not the sum of a 2-positive map and a 2-copositive map when 26 a < 3. See his paper for details and definitions).
As an application of Corollary 3.1, we shall show directly that the positive map
H[a; c1, c2, c3] : M3→ M3is not decomposable when 26 a < 3. To show this, it suffices to find A= (Aij)16i,j63∈ M3(M3) with the property that (Aij)∈ PSD9 and (Aj i)∈ PSD9, but
D H[a, c1, c2, c3](Eij) 16i,j63, A E < 0. Observe that H[a, c1, c2, c3](Eij) 16i,j63 = a− 1 0 0 0 −1 0 0 0 −1 0 c2 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 a− 1 0 0 0 −1 0 0 0 0 0 c3 0 0 0 0 0 0 0 0 0 c1 0 0 0 0 0 0 0 0 0 0 0 −1 0 0 0 −1 0 0 0 a− 1 . Let A= (Aij)16i,j63= 1 0 0 0 1 0 0 0 1 0 a22 0 1 0 0 0 0 0 0 0 a771 0 0 0 1 0 0 0 1 0 a221 0 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 a66 0 1 0 0 0 1 0 0 0 a77 0 0 0 0 0 0 0 1 0 a1 66 0 1 0 0 0 1 0 0 0 1 where a22= α 3c 1/3 1 c 1/3 3 c−2/32 , a66= α 3c 1/3 1 c 1/3 2 c−2/33 , a77= α 3c 1/3 2 c 1/3 3 c1−2/3 with α any positive real number. We observe that both (Aij) and (Aj i) are
positive-semidefinite and that
D (H[a, c1, c2, c3](Eij) 16i,j63 , A E = 3(a − 3) + αc1/3 1 c 1/3 2 c 1/3 3 .
SinceH[a, c1, c2, c3] is a positive map, [7, Theorem 2.1] gives us that c1/31 c1/32 c1/33
> (3 − a). If a < 3, then 3 − a > 0. Thus, there exists β > 0 such that c1/3 1 c
1/3 2 c
1/3 3
= β(3 − a). We have that
h (H[a, c1, c2, c3](Eij))16i,j63, Ai = 3(a − 3) + αβ(3 − a) = (3 − αβ)(a − 3). Since α can be any positive real number, choose α > 3/β, and the above inner prod-uct is negative. By Corollary 3.1, we have thatH[a; c1, c2, c3] is not decomposable.
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