On a Matrix Inequality Related to the Distillability Problem

We investigate the distillability problem in quantum information in ℂd⊗ℂd. One case of the problem has been reduced to proving a matrix inequality when d=4. We investigate the inequality for three families of non-normal matrices. We prove the inequality for the first two families with d=4 and for the third family with d≥5. We also present a sufficient condition for the fulfillment of the inequality with d=4.


I. INTRODUCTION
We refer to C n×n as the set of n × n matrices with entries in the complex field, and H n×n as the set of n × n Hermitian matrices.Let I n be the identity matrix in C n×n .We shall omit the subscript of the identity matrix when it is clear in the paper.Let A ∈ C n×n and B ∈ C m×m .The Kronecker sum of A and B is defined as A ⊗ I m + I n ⊗ B, see more facts in [1, section 7.2].Ref. [2] has presented the following conjecture on the Kronecker sum when A and B have the same size.
Conjecture 1.Let A, B, I ∈ C d×d , d ≥ 4, and the matrix where Let σ 1 , • • • , σ d 2 be the singular values of X in the descending order.Then The condition d ≥ 4 is essential, because we will show in Lemma 11 that Conjecture 1 fails for d = 3.It has been shown [2] that Conjecture 1 for d = 4 is a special case of the distillability problem.We will mathematically explain the special case in Appendix A, due to the heavy terminologies from quantum physics.The distillability problem has been a main open problem in quantum information [3] for a long time.It lies at the heart of entanglement theory [4][5][6] and is related to the separability problem extensively studied by the mathematics community recently [7][8][9].We briefly introduce the physical motivation of distillability problem, and will give more details in Appendix A. In quantum physics, a quantum state is mathematically described by a positive semidefinite matrix.The state is pure when it has rank one, otherwise the state is mixed.Pure entangled states play an essential role in most quantum-information tasks such as quantum computation.Nevertheless, there is no pure state in nature due to the unavoidable decoherence between the state and environment.So asymptotically converting initially bipartite entangled mixed states into bipartite pure entangled states under local operations and classical communications (LOCC) is a key step in quantum information processing.The distillability problem [3,10] asks whether the above-mentioned conversion succeeds for any mixed states.There have been some attempts to the problem in the past years [2,3,[10][11][12][13][14][15][16][17].
We return to Conjecture 1.Although it is only a special case of the distillability problem, it has been an open problem for years.Evidently, the matrix X in ( The rest of this paper is organized as follows.We introduce some notations and preliminary results in linear algebra in Sec.II.We investigate Conjecture 1 for two families of non-normal matrices in Sec.III and IV, respectively.We further prove Conjecture 1 for normal matrices in Sec.V.

II. PRELIMINARIES
We shall denote A † as the conjugate transpose of matrix A. Let σ(A) be the spectrum of matrix A, λ i (A) be an eigenvalue of A, and A (i,j) be the (i, j) entry of A. We post some lemmas used in the following sections.The following lemma is clear.
Lemma 2. The following four statements are equivalent.
(ii) Conjecture 1 holds when X is replaced by X T , X * or X † .
(iii) Conjecture 1 holds when X is replaced by (U ⊗ V )X(U † ⊗ V † ) with any unitary matrices U and V .
(iv) Conjecture 1 holds when X is replaced by I ⊗ A + B ⊗ I.
Remark: Using statement (iii) we can assume that A and B in Conjecture 1 are both upper-triangular.In particular, we can assume that they are diagonal if and only if they are normal.
Lemma 3. [1, Fact 4.10.16.](Gershgorin circle theorem) Let A ∈ C n×n .Then, and a corollary is A (i,j) and D(A (i,i) , R i ) be the closed disc centered at A (i,i) with radius R i .Every eigenvalue of A lies within at least one of the Gershgorin discs D(A (i,i) , R i ).
III. CONJECTURE 1 WITH NON-NORMAL MATRICES X: FAMILY 1 In this section we prove Conjecture 1 with a family of non-normal matrices X in Definition 1.We will construct our main result in Theorem 1, followed by two preliminary facts i.e.
Definition 1.Let P be the subset of matrices X with d = 4, such that A, B are normal or have the expressions Since the complex numbers a i , b j satisfy (2), X ∈ P may be normal or non-normal.We present the main result of this section as follows.
then the assertion follows from Proposition 1.If one of A and B is normal then we may assume that it is diagonal by Lemma 2 (iii).So the assertion follows from Proposition 2 and the switch of A, B (if any).If A and B are both normal then the assertion follows from Lemma 1.This completes the proof.
Proposition 1. Eq. (3) holds when We claim that the larger root of each quadratic polynomial in each line of (8) isn't greater We can make the same conclusion in the same way to substitute λ = 4 i=1 ) into other lines of (8).Hence, the larger root of each quadratic polynomial in each line of (8) isn't greater than So any eigenvalue of Y 1 isn't greater than 1  4 .One can show that Y 2 can be evolved from Y 1 by replacing a 1 with a 3 and replacing a 2 with a 4 in Y 1 .Hence, we can get a similar formulation of |λI − Y 2 | like ( 8) by replacing a 1 with a 3 and replacing a 2 with a 4 in ( 8).In the same way, we conclude that any eigenvalue of Y 2 isn't greater than 1  4 .Since X † X = Y 1 ⊕ Y 2 , the sum of the largest two eigenvalues of X † X is at most 1  2 .This completes the proof.
We proceed with the proof of Proposition 2. For this purpose we need the following two preliminary results.The first result is known as one of the basic inequalities.
Proof.Using the basic inequality x + y ≤ 2(x 2 + y 2 ) for any real x, y, we obtain that the lhs of ( 9) is upper bounded by The equality follows from the equation a 2 1 + a 2 2 + b 2 1 + b 2 2 = 1/4.This completes the proof.
Proposition 2. Eq. (3) holds when Proof.Let X = A ⊗ I + I ⊗ B in (1).Since A and B satisfy (2), we have By computation one can show that X † X has the same eigenvalues with that of ⊕ 4 j=1 (Y j ⊕Z j ) where Y j and Z j are order-2 submatrices such that and Let λ and µ be two arbitrary eigenvalues of X † X.Then proving Conjecture 1 is equivalent to proving λ + µ ≤ 1/2.We investigate five cases for λ and µ.
Case 2. λ and µ are the eigenvalues of different Y j 's.Without loss of generality we can assume that λ is the maximum eigenvalue of Y 1 , and µ is the maximum eigenvalue of Y 2 .

By computation one can obtain
So λ + µ is upper bounded by the sum of the rhs of ( 15) and ( 16 Case 3. λ and µ are the eigenvalues of different Z j 's.We can prove Conjecture 1 by following the proof in Case 2, except that we switch a 1 and a 3 , and switch a 2 and a 4 at the same time.
Case 4. λ is the eigenvalue of some Y j , µ is the eigenvalue of some Z k , and j = k.Without loss of generality we may assume that j = 1 and k = 2.By computation one can show that where ).The second inequality in (17) follows from Lemma 6 in which we have set The last inequality in (17) follows from (12).So Conjecture 1 holds.
Case 5. λ is the eigenvalue of some Y j , and µ is the eigenvalue of some Z j .Without loss of generality we may assume that j = 1.Eqs. ( 11) and ( 12) imply that Hence On the other hand, by computation one can show that It  18) and ( 19) imply that where and One can verify that To prove the assertion, one need to obtain the maximum of (20).For this purpose we need to obtain the maximum of the function f 1 in terms of g, and the maximum of the function f 2 in terms of h.The two functions f 1 and f 2 are respectively parabolas of cartesian coordinates (sin 2g, f 1 ) and (sin 2h, f 2 ).The axises of symmetry of f 1 and f 2 are respectively sin 2g = To conclude we have proved Conjecture 1 for the matrices A, B in all five cases for λ, µ.
This completes the proof.
Using the statement of Lemma 2 (iii), we may assume in Conjecture 1 that A or B is diagonal if and only if it is normal.Hence, Proposition 2 implies that Conjecture 1 holds when X ∈ P where one of A and B is normal.

IV. CONJECTURE 1 WITH NON-NORMAL MATRICES X: FAMILY 2
In this section we investigate Conjecture 1 with X defined as follows.

Definition 2. Let A = D
which follows from the second equation of (2). where . By calculation we obtain the eigenpolynomial of X † X as follow One can show that λ 2 − (|a| implies the largest two eigenvalues of X † X must be the roots of the second product of (24).
Hence, the sum of the largest two eigenvalues in (24) can be expressed as follow where We have δ j ≤ 1 d and √ δ k ≤ 1 d , ∀j, k which follow from Eq. ( 23).Hence, Eq. ( 25) implies the sum of the largest two eigenvalues of X † X is at most 2 d which isn't greater than 1 2 for d ≥ 4.This completes the proof.
Using the statement of Lemma 2 (iv), we can make the same conclusion if rank B = 1.
Hence, Eq. (3) holds when A, B satisfy Definition 2 and one of them has rank one.
We have seen that it is not easy to characterize the eigenpolynomial of X † X.In the following lemma, we use Gershgorin circle theorem and Brauer theorem to study Conjecture 1.They are two important theorems in the field of localization of eigenvalues to localize the largest two eigenvalues.The following fact will be used in the proof of Lemma 9.
It follows from (1) that X † X = H 1 + H 2 where Furthermore, the first equation of (2) implies that TrH 2 = 0, and the second equation of (2) implies that TrX † X = TrH 1 = d 2 j=1 σ 2 j = 1.So X † X can be regarded as a normalized quantum state in terms of quantum physics.Proof.H 1 in Eq. ( 26) is diagonal and H 2 in Eq. ( 26) is a Hermitian matrix with zerodiagonal when A and B satisfy Definition 2. There exist two permutations σ and τ with d} which are respectively equivalent to P A and P B .We find They imply that exactly two entries in each row of H 2 can be expressed with a i , b j and their conjugates, for i, j which is the d(i − 1) + j, d(σ(i) − 1) + τ −1 (j) entry of X † X are both in the (d(i − 1) + j)'th row of X † X and also are non-diagonal entries of X † X.Further the diagonal entry of X † X in this row is ( implies that the largest eigenvalue λ 1 of X † X satisfies and the fact σ(k) = k, τ (k) = k, ∀k ∈ {1, • • • , d}.Applying the basic inequality, we obtain Let's recall (26).Eq. ( 7) implies the second largest eigenvalue of X † X satisfy λ 2 (X † X) ≤ λ 2 (H 1 ) + λ 1 (H 2 ), where λ 2 (H 1 ) means the second largest eigenvalue of H 1 and λ 1 (H 2 ) means the largest eigenvlaue of H 2 .We find λ 2 (H 1 ) is the second largest diagonal element of H 1 .Applying Lemma 4 to H 2 , one can show that λ 1 (H 2 ) should not greater than the largest root of these quadratic polynomials Suppose k 1 = d(i − 1) + j and k 2 = d(p − 1) + q with (i, j) = (p, q).Hence, we can bound λ 1 (X † X) + λ 2 (X † X) as follow. where Suppose |a i | and |b j | be in the decreasing order.In order to obtain the upper bound of Eq. 28, it is safe to let  x ).Let d = 5 and the numberical result shows that even though there is no constraint f is also upper bounded by 1  2 .To conclude all dimensions d ≥ 5 have been studied.This completes the proof.

V. CONJECTURE 1 WITH NORMAL X AND d = 4
Reference [2] investigated Conjecture 1 for normal matrices with d = 4, as we have introduced in Lemma 1.In this section we extend Lemma 1 to higher dimensions, so that Conjecture 1 is of more mathematical interest apart from its physical connection to the distillability problem.Proof.The part in the proof of Lemma 1 from the beginning to (B9) applies here.This part applies to any d > 4, and Eq.(B9) is the first place in the proof of Lemma 1 in which d = 4 appears.Based on these facts, we begin our proof with d > 4. According to (B9), we have Proposition 3 implies that (B11) is satisfied.So we have The two inequalities in (B6) and (B7) are saturated.So Lemma 1 holds for d ≥ 4.
Next we show that Lemma 1 no longer holds when d = 4 is replaced by d = 3.
Lemma 11.Let χ d be a subset of normal operators X in (1) satisfying constraints (2).
Then for d = 3, we have where σ 1 and σ 2 are the two largest singular values of operator X.
communications if there exists a Schmidt-rank-two bipartite state |ψ ∈ H ⊗n such that ψ|(ρ ⊗n ) Γ |ψ < 0. Otherwise we say that ρ is n-undistillable.We say that ρ is distillable if it is n-distillable for some n ≥ 1.
Due to the alternative property, we can only consider the left term or the right term of ∨ in the second case.Then we can go further with the (B2) as follows. sup Thus, to prove the Lemma we have to show that the following inequalities hold: (B11)
) and P A , P B are permutation matrices with zero-diagonals which make A, B satisfy the first equation of (2) naturally.Meanwhile, A and B are under the following constraint d i=1

SuchLemma 8 .
X may be normal or non-normal, and may have dimension d ≥ 4. It is different from the set P in Definition 1.We shall prove Conjecture 1 when A, B satisfy Definition 2, and two additional conditions respectively in Lemma 8, 9.In particular, Lemma 9 proves Conjecture 1 for d ≥ 5. Eq. (3) holds when A, B satisfy Definition 2 and rank A = 1.Proof.For rank A = 1, it is safe to fix the only nonzero element a of A in the second entry of the first line of A, since there exists proper permutation matrix P such that the only nonzero entry of matrix P AP † is the second entry of the first line for any A with rank = 1.Let B = D B P B , where D B = diag(b 1 , • • • , b d ) and P B is a permutation matrix with zero-diagonal.By calculation we have

|b 1 |
and x 4 = |b 2 | and other |a i | and |b j | all equal zero.Then our problem can be transformed into an optimization task as follow.

Lemma 10 .
Lemma 1 holds when N 4 is replaced by N d with d > 4.
that all NPT bipartite states can be locally converted into NPT Werner states.Using Definition 3, one can show that the distillability of NPT Werner states is equivalent to that with α = −1/2.So the distillability problem indeed asks whether Werner states with α = −1/2 are distillable.Now we can explain the special case proposed in [2].It means that Conjecture 1 with d = 4 is equivalent to the 2-undistillability of Werner states in C 4 ⊗ C 4 with α = −1/2.