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Inequalities related to partial transpose and partial trace
Linear Algebra and its Applications 516 (2017) 1–7
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Inequalities related to partial transpose and partial trace Daeshik Choi Southern Illinois University, Edwardsville, Dept. of Mathematics and Statistics, Box 1653, Edwardsville, IL 62026, United States
a r t i c l e
i n f o
Article history: Received 1 November 2016 Accepted 17 November 2016 Available online 23 November 2016 Submitted by R. Brualdi
a b s t r a c t In this paper, we present inequalities related to partial transpose and partial trace for positive semidefinite matrices. Some interesting results involving traces and eigenvalues are also included. © 2016 Elsevier Inc. All rights reserved.
MSC: 47B65 15A42 15A45 Keywords: Partial transpose Partial trace Positive semidefinite Block matrix
1. Introduction Throughout the paper, we use the following standard notation: • Mn×k is the set of n × k complex matrices; if n = k, we use Mn for Mn×n and if k = 1, we use Cn for Mn×1 . • Mn (Mk ) is the set of n × n block matrices with each block in Mk . E-mail address: [email protected]. http://dx.doi.org/10.1016/j.laa.2016.11.027 0024-3795/© 2016 Elsevier Inc. All rights reserved.
2
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
• In is the n × n identity matrix. • A ◦ B is the Hadamard product of A, B. • A ⊗ B is the Kronecker product of A, B; that is, if A = [ aij ] ∈ Mn and B ∈ Mk , then A ⊗ B ∈ Mn (Mk ) whose (i, j) block is aij B. √ • ||x|| denotes the 2-norm of x ∈ Cn ; that is, ||x|| = x∗ x. n
Given A = [ Ai,j ]i,j=1 ∈ Mn (Mk ), we define the partial transpose of A by n
Aτ = [ Aj,i ]i,j=1 . Note that A ≥ 0 does not necessarily imply Aτ ≥ 0. If both A and Aτ are positive semidefinite, then A is said to be positive partial transpose (PPT for short). We also n define two partial traces tr1 A and tr2 A of A = [ Ai,j ]i,j=1 ∈ Mn (Mk ) by tr1 A =
n
Ai,i ,
i=1 n
tr2 A = [ trAi,j ]i,j=1 , where trX denotes the trace of X; see [5] for more details related to partial traces. It is shown in [1,4] that for any positive semidefinite H ∈ Mn (Mk ), we have In ⊗ tr1 H + tr2 H ⊗ Ik − H ≤ (trH)Ink .
(1.1)
Generally, In ⊗ tr1 H + tr2 H ⊗ Ik ≥ H does not hold. In this paper, we show In ⊗ tr1 H + tr2 (H τ ) ⊗ Ik ≥ H τ for H ∈ Mn (Mk ) with H ≥ 0 (in Theorem 2) and I2 ⊗ tr1 H + tr2 (H) ⊗ Ik ≥ H for H ∈ M2 (Mk ) with H ≥ 0 (in Theorem 4). Moreover, some interesting inequalities involving the trace and the minimum and maximum eigenvalues of a positive semidefinite matrix will be presented. 2. Results and proofs Lemma 1. Let A, B ∈ Mn . If A, B ≥ 0, then sum of all entries of X.
(A ◦ B) ≥ 0, where
(X) denotes the
Proof. By the Schur product theorem, we have A ◦ B ≥ 0. Since (X) = u∗ Xu, where u ∈ Cn is the vector with all entries equal to one, (A◦B) ≥ 0 follows from A◦B ≥ 0. 2
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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Theorem 2. If H ∈ Mn (Mk ) is positive semidefinite, then In ⊗ tr1 H ≥ H τ
tr2 (H τ ) ≥ 0.
and ⎡
⎤ X1 Proof. We may write H = XX ∗ for X = ⎣ ... ⎦ with X1 , . . . , Xn ∈ Mk×m [2, Theorem Xn 7.2.7]. Then ⎡ ⎢ ⎢ ⎢ In ⊗ tr1 H − H τ = ⎢ ⎢ ⎣
Xi Xi∗ −X1 X2∗ .. . ∗ −X1 Xn−1 −X1 Xn∗ i=1
−X X ∗ 2 1 ∗ i=2 Xi Xi .. . ∗ −X2 Xn−1 −X2 Xn∗
−Xn−1 X1∗ −Xn−1 X2∗ .. . ∗ i=n−1 Xi Xi ∗ −Xn−1 Xn
··· ··· .. . ··· ···
−Xn X1∗ ⎤ −Xn X2∗ ⎥ ⎥ ⎥ .. ⎥. . ⎥ ∗ ⎦ −Xn Xn−1 ∗ i=n Xi Xi
We will prove In ⊗ tr1 H − H τ ≥ 0 by induction on n. It is obvious for n = 1. Note In+1 ⊗ tr1 H − H τ = A + B, where ⎡ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎣ ⎡
Xi Xi∗ −X1 X2∗ .. . −X1 Xn∗ 0
i=1,n+1
∗ Xn+1 Xn+1 ⎢ 0 ⎢ ⎢ .. B=⎢ . ⎢ ⎣ 0 ∗ −X1 Xn+1
⎤ −Xn X1∗ 0 −Xn X2∗ 0⎥ ⎥ .. .. ⎥ , . .⎥ ⎥ ∗ 0⎦ i=n,n+1 Xi Xi 0 0 ∗ ⎤ 0 −Xn+1 X1 0 −Xn+1 X2∗ ⎥ ⎥ ⎥ .. .. ⎥. . . ⎥ ∗ ∗ ⎦ Xn+1 Xn+1 −Xn+1 Xn n ∗ ∗ −Xn Xn+1 i=1 Xi Xi
−X2 X1∗ ∗ i=2,n+1 Xi Xi .. . −X2 Xn∗ 0
0 ∗ Xn+1 Xn+1 .. . 0 ∗ −X2 Xn+1
··· ··· .. . ··· ···
··· ··· .. . ··· ···
The inductive assumption provides A ≥ 0. Meanwhile, B ≥ 0 follows from B = CC ∗ , where ⎡
Xn+1 ⎢ .. ⎢ C=⎢ . ⎣ 0 −X1
··· .. . ··· ···
0 .. . Xn+1 −Xn
Next, we will prove tr2 (H τ ) ≥ 0. Let Xi = [ xi,1 tr(xy ∗ ) = y ∗ x for x, y ∈ Cn , we have
⎤ −X1 .. ⎥ . ⎥ ⎥. −Xn ⎦ 0 ···
xi,m ], where xi,j ∈ Ck . Since
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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tr2 (H τ ) =
m
x∗i,l xj,l
i,j
.
l=1
⎡
⎤ x1 Thus we may assume m = 1. Let H = xx∗ , where x = ⎣ ... ⎦ and xi ∈ Ck . For any xn ⎡ ⎤ y1 y = ⎣ ... ⎦ ∈ Cn , we have yn y ∗ tr2 (H τ )y =
x∗i xj yi yj
i,j
= =
([ x∗i xj ] ◦ [ yi yj ]) ((E ∗ E) ◦ (w∗ w)) ,
where E = [ x1 · · · xn ] ∈ Mk×n and w = [ y1 follows by Lemma 1. 2
···
yn ] ∈ M1×n . Thus tr2 (H τ ) ≥ 0
The following results from Theorem 2 by replacing H by H − λmin (H)Ink and λmax (H)Ink − H. We omit the detailed proof. Corollary 3. If H ∈ Mn (Mk ) is Hermitian, then (n − 1)λmin (H)Ink ≤ In ⊗ tr1 H − H τ ≤ (n − 1)λmax (H)Ink and kλmin (H)In ≤ tr2 (H τ ) ≤ kλmax (H)In . By Theorem 2, we have In ⊗ tr1 H + tr2 (H τ ) ⊗ Ik ≥ H τ for any positive semidefinite H ∈ Mn (Mk ). The inequality does not hold for n > 2 ⎡ ⎤ ⎡ ⎤∗ e1 e1 τ when H is replaced by H. For example, if H = ⎣ ... ⎦ ⎣ ... ⎦ ∈ Mn (Mn ), where ei is en en the standard unit (column) vector of Rn whose only non-zero entry is a 1 in the ith position, then it is easy to show that tr1 H = tr2 H = In and therefore In ⊗ tr1 H + ⎡ ⎤ ⎡ ⎤ e1 e1 tr2 (H) ⊗ In − H = 2In − H. Since H ⎣ ... ⎦ = n ⎣ ... ⎦, n is an eigenvalue of H. Therefore en en
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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In ⊗ tr1 H + tr2 (H) ⊗ In − H has the negative eigenvalue 2 − n for n > 2. However, the following shows that it holds for n = 2.
A Theorem 4. If H = B∗
B ∈ M2 (Mk ) is positive semidefinite, then C I2 ⊗ tr1 H + tr2 (H) ⊗ Ik ≥ H,
(2.1)
that is,
X Proof. Since W
Y Z
(trB)Ik − B ≥ 0. A + (trC)Ik
C + (trA)Ik (trB ∗ )Ik − B ∗
Z is *-congruent to Y M=
A + (trC)Ik (trB)Ik − B
(2.2)
W , we will show that X (trB ∗ )Ik − B ∗ C + (trA)Ik
is positive semidefinite. By the same argument used in Theorem 2, we may assume x H= [ x∗ y ∗ ] for x, y ∈ Mk×1 , in which case we have y M=
xx∗ + (y ∗ y)Ik (y ∗ x)Ik − xy ∗
(x∗ y)Ik − yx∗ . yy ∗ + (x∗ x)Ik
Without loss of generality, we assume y ∗ y = 1. Let α = complement of M as follows:
1 |x|2 +1 .
−1
S = yy ∗ + (x∗ x)Ik − ((y ∗ x)Ik − xy ∗ ) (xx∗ + Ik )
We compute the Schur
((x∗ y)Ik − yx∗ )
= yy ∗ + (x∗ x)Ik − ((y ∗ x)Ik − xy ∗ ) (Ik − αxx∗ ) ((x∗ y)Ik − yx∗ ) = yy ∗ + (x∗ x − |y ∗ x|2 )Ik + (y ∗ x)yx∗ + (x∗ y)xy ∗ − (y ∗ y)xx∗ . Note that M ≥ 0 is equivalent to S ≥ 0 [2, Theorem 7.7.9]. For z ∈ Ck with ||z|| = 1, let u = y ∗ x, v = z ∗ x, and w = y ∗ z. Then z ∗ Sz = |y ∗ z|2 + ||x||2 − |y ∗ x|2 + (y ∗ x)(z ∗ y)(x∗ z) + (x∗ y)(z ∗ x)(y ∗ z) − (z ∗ x)(x∗ z) = |w|2 + ||x||2 − |u|2 + uwv + uvw − |v|2 and we have z ∗ Sz ≥ 0 ⇐⇒ |u − vw|2 ≤ |w|2 + ||x||2 − |v|2 + |v|2 |w|2 . Let P = I − zz ∗ . Since P ≥ 0, we have |y ∗ P x|2 ≤ (x∗ P x)(y ∗ P y). Thus
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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|u − vw|2 = |y ∗ P x|2 ≤ (x∗ P x)(y ∗ P y) = (||x||2 − |v|2 )(1 − |w|2 ) ≤ |w|2 + ||x||2 − |v|2 + |v|2 |w|2 .
2
The following can be obtained from Theorem 4.
A B Corollary 5. Let H = ∈ M2 (Mk ) be Hermitian. Then we have the following. B∗ C a) (k + 1)λmin (H)I2k ≤ I2 ⊗ tr1 H + tr2 (H) ⊗ Ik − H ≤ (k + 1)λmax (H)I2k . b) If H ≥ 0, then tr(A + C) ≥ 2(k−1) k+1 |Re(trB)|. 1 k−1 |Re(trB)| λmin (H) ≤ 2k tr(A + C) − k(k+1) c) . 1 k−1 λmax (H) ≥ 2k tr(A + C) + k(k+1) |Re(trB)| where Re(·) denotes the real part of a complex number. Proof. Replacing H by H − λmin (H)I2k and λmax (H)I2k − H in Theorem 4, we get a). From (2.2), we have A + C + tr(A + C)Ik ≥ ± (tr(B + B ∗ )Ik − (B + B ∗ )) .
(2.3)
Taking traces on both sides, we have b). Finally, from the first inequality in a) we have
C + (trA − (k + 1)λmin (H)) Ik (trB ∗ )Ik − B ∗
(trB)Ik − B ≥ 0. A + (trC − (k + 1)λmin (H)) Ik
Using the same argument as in b), we get the first inequality in c). Similarly we can obtain the second inequality in c) from the second inequality in a). 2 Remark 6. Replacing H by H − λmin (H)Ink and λmax (H)Ink − H in (1.1), we obtain for any Hermitian H ∈ Mn (Mk ) that In ⊗ tr1 H + tr2 H ⊗ Ik − H ≥ (trH − (n − 1)(k − 1)λmax (H)) Ink , In ⊗ tr1 H + tr2 H ⊗ Ik − H ≤ (trH − (n − 1)(k − 1)λmin (H)) Ink . In particular, when n = 2, we have In ⊗ tr1 H + tr2 H ⊗ I2 − H ≥ (trH − (k − 1)λmax (H)) I2k
(2.4)
from which we cannot deduce (2.1), since trH ≥ (k − 1)λmax (H) does not generally hold; for example, if H ∈ M2 (Mk ) whose only non-zero entry is a 1 in the (1, 1) position, then trH = 1, λmax (H) = 1, and therefore trH < (k − 1)λmax (H) for k > 2.
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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Using a similar approach as in [3, Proposition 2.5], we can also derive the following result.
A B Corollary 7. If H = ∈ M2 (Mk ) is positive semidefinite, then B∗ C Proof. Let
A B∗
B C
(trA)A + BB ∗ (trB)B ∗ − CA
=
X [ X∗ Y
(trB ∗ )B − AC ≥ 0. (trC)C + B ∗ B
Y ∗] =
XX ∗ Y X∗
XY ∗ YY∗
(2.5)
for X, Y ∈ Mk×m . Then we
have
(trA)A + BB ∗ (trB ∗ )B − AC (trB)B ∗ − CA (trC)C + B ∗ B
tr(X ∗ X)XX ∗ + XY ∗ Y X ∗ tr(X ∗ Y )XY ∗ − XX ∗ Y Y ∗ tr(Y ∗ X)Y X ∗ − Y Y ∗ XX ∗ tr(Y ∗ Y )Y Y ∗ + Y X ∗ XY ∗
∗ X 0 tr(X ∗ X)Im + Y ∗ Y tr(X ∗ Y )Im − X ∗ Y 0 X = . 0 Y∗ 0 Y tr(Y ∗ X)Im − Y ∗ X tr(Y ∗ Y )I + X ∗ X
=
Y∗ [Y Since X∗
Y ∗Y X] = X ∗Y
Y ∗X ≥ 0, we have X ∗X
tr(X ∗ X)Im + Y ∗ Y tr(Y ∗ X)Im − Y ∗ X
tr(X ∗ Y )Im − X ∗ Y tr(Y ∗ Y )I + X ∗ X
≥0
by Theorem 4. Thus we get (2.5). 2 Acknowledgements The author is grateful to Professor M. Lin for his assistance to prepare this manuscript. The author also expresses sincere thanks to Professor T. Ando for [1] before its publication. Finally, the author thanks anonymous reviewer for valuable comments and suggestions on the manuscript. References [1] [2] [3] [4] [5]
T. Ando, Matrix inequalities involving partial traces, preprint. R.A. Horn, C.R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2012. M. Lin, A completely PPT map, Linear Algebra Appl. 459 (2014) 404–410. M. Lin, A determinantal inequality involving partial traces, Canad. Math. Bull. 59 (2016) 585–591. D. Petz, Quantum information theory and quantum statistics, in: Theoretical and Mathematical Physics, Springer-Verlag, Berlin, 2008.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
Inequalities related to partial transpose and partial trace Daeshik Choi Southern Illinois University, Edwardsville, Dept. of Mathematics and Statistics, Box 1653, Edwardsville, IL 62026, United States
a r t i c l e
i n f o
Article history: Received 1 November 2016 Accepted 17 November 2016 Available online 23 November 2016 Submitted by R. Brualdi
a b s t r a c t In this paper, we present inequalities related to partial transpose and partial trace for positive semidefinite matrices. Some interesting results involving traces and eigenvalues are also included. © 2016 Elsevier Inc. All rights reserved.
MSC: 47B65 15A42 15A45 Keywords: Partial transpose Partial trace Positive semidefinite Block matrix
1. Introduction Throughout the paper, we use the following standard notation: • Mn×k is the set of n × k complex matrices; if n = k, we use Mn for Mn×n and if k = 1, we use Cn for Mn×1 . • Mn (Mk ) is the set of n × n block matrices with each block in Mk . E-mail address: [email protected]. http://dx.doi.org/10.1016/j.laa.2016.11.027 0024-3795/© 2016 Elsevier Inc. All rights reserved.
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D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
• In is the n × n identity matrix. • A ◦ B is the Hadamard product of A, B. • A ⊗ B is the Kronecker product of A, B; that is, if A = [ aij ] ∈ Mn and B ∈ Mk , then A ⊗ B ∈ Mn (Mk ) whose (i, j) block is aij B. √ • ||x|| denotes the 2-norm of x ∈ Cn ; that is, ||x|| = x∗ x. n
Given A = [ Ai,j ]i,j=1 ∈ Mn (Mk ), we define the partial transpose of A by n
Aτ = [ Aj,i ]i,j=1 . Note that A ≥ 0 does not necessarily imply Aτ ≥ 0. If both A and Aτ are positive semidefinite, then A is said to be positive partial transpose (PPT for short). We also n define two partial traces tr1 A and tr2 A of A = [ Ai,j ]i,j=1 ∈ Mn (Mk ) by tr1 A =
n
Ai,i ,
i=1 n
tr2 A = [ trAi,j ]i,j=1 , where trX denotes the trace of X; see [5] for more details related to partial traces. It is shown in [1,4] that for any positive semidefinite H ∈ Mn (Mk ), we have In ⊗ tr1 H + tr2 H ⊗ Ik − H ≤ (trH)Ink .
(1.1)
Generally, In ⊗ tr1 H + tr2 H ⊗ Ik ≥ H does not hold. In this paper, we show In ⊗ tr1 H + tr2 (H τ ) ⊗ Ik ≥ H τ for H ∈ Mn (Mk ) with H ≥ 0 (in Theorem 2) and I2 ⊗ tr1 H + tr2 (H) ⊗ Ik ≥ H for H ∈ M2 (Mk ) with H ≥ 0 (in Theorem 4). Moreover, some interesting inequalities involving the trace and the minimum and maximum eigenvalues of a positive semidefinite matrix will be presented. 2. Results and proofs Lemma 1. Let A, B ∈ Mn . If A, B ≥ 0, then sum of all entries of X.
(A ◦ B) ≥ 0, where
(X) denotes the
Proof. By the Schur product theorem, we have A ◦ B ≥ 0. Since (X) = u∗ Xu, where u ∈ Cn is the vector with all entries equal to one, (A◦B) ≥ 0 follows from A◦B ≥ 0. 2
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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Theorem 2. If H ∈ Mn (Mk ) is positive semidefinite, then In ⊗ tr1 H ≥ H τ
tr2 (H τ ) ≥ 0.
and ⎡
⎤ X1 Proof. We may write H = XX ∗ for X = ⎣ ... ⎦ with X1 , . . . , Xn ∈ Mk×m [2, Theorem Xn 7.2.7]. Then ⎡ ⎢ ⎢ ⎢ In ⊗ tr1 H − H τ = ⎢ ⎢ ⎣
Xi Xi∗ −X1 X2∗ .. . ∗ −X1 Xn−1 −X1 Xn∗ i=1
−X X ∗ 2 1 ∗ i=2 Xi Xi .. . ∗ −X2 Xn−1 −X2 Xn∗
−Xn−1 X1∗ −Xn−1 X2∗ .. . ∗ i=n−1 Xi Xi ∗ −Xn−1 Xn
··· ··· .. . ··· ···
−Xn X1∗ ⎤ −Xn X2∗ ⎥ ⎥ ⎥ .. ⎥. . ⎥ ∗ ⎦ −Xn Xn−1 ∗ i=n Xi Xi
We will prove In ⊗ tr1 H − H τ ≥ 0 by induction on n. It is obvious for n = 1. Note In+1 ⊗ tr1 H − H τ = A + B, where ⎡ ⎢ ⎢ ⎢ A=⎢ ⎢ ⎣ ⎡
Xi Xi∗ −X1 X2∗ .. . −X1 Xn∗ 0
i=1,n+1
∗ Xn+1 Xn+1 ⎢ 0 ⎢ ⎢ .. B=⎢ . ⎢ ⎣ 0 ∗ −X1 Xn+1
⎤ −Xn X1∗ 0 −Xn X2∗ 0⎥ ⎥ .. .. ⎥ , . .⎥ ⎥ ∗ 0⎦ i=n,n+1 Xi Xi 0 0 ∗ ⎤ 0 −Xn+1 X1 0 −Xn+1 X2∗ ⎥ ⎥ ⎥ .. .. ⎥. . . ⎥ ∗ ∗ ⎦ Xn+1 Xn+1 −Xn+1 Xn n ∗ ∗ −Xn Xn+1 i=1 Xi Xi
−X2 X1∗ ∗ i=2,n+1 Xi Xi .. . −X2 Xn∗ 0
0 ∗ Xn+1 Xn+1 .. . 0 ∗ −X2 Xn+1
··· ··· .. . ··· ···
··· ··· .. . ··· ···
The inductive assumption provides A ≥ 0. Meanwhile, B ≥ 0 follows from B = CC ∗ , where ⎡
Xn+1 ⎢ .. ⎢ C=⎢ . ⎣ 0 −X1
··· .. . ··· ···
0 .. . Xn+1 −Xn
Next, we will prove tr2 (H τ ) ≥ 0. Let Xi = [ xi,1 tr(xy ∗ ) = y ∗ x for x, y ∈ Cn , we have
⎤ −X1 .. ⎥ . ⎥ ⎥. −Xn ⎦ 0 ···
xi,m ], where xi,j ∈ Ck . Since
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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tr2 (H τ ) =
m
x∗i,l xj,l
i,j
.
l=1
⎡
⎤ x1 Thus we may assume m = 1. Let H = xx∗ , where x = ⎣ ... ⎦ and xi ∈ Ck . For any xn ⎡ ⎤ y1 y = ⎣ ... ⎦ ∈ Cn , we have yn y ∗ tr2 (H τ )y =
x∗i xj yi yj
i,j
= =
([ x∗i xj ] ◦ [ yi yj ]) ((E ∗ E) ◦ (w∗ w)) ,
where E = [ x1 · · · xn ] ∈ Mk×n and w = [ y1 follows by Lemma 1. 2
···
yn ] ∈ M1×n . Thus tr2 (H τ ) ≥ 0
The following results from Theorem 2 by replacing H by H − λmin (H)Ink and λmax (H)Ink − H. We omit the detailed proof. Corollary 3. If H ∈ Mn (Mk ) is Hermitian, then (n − 1)λmin (H)Ink ≤ In ⊗ tr1 H − H τ ≤ (n − 1)λmax (H)Ink and kλmin (H)In ≤ tr2 (H τ ) ≤ kλmax (H)In . By Theorem 2, we have In ⊗ tr1 H + tr2 (H τ ) ⊗ Ik ≥ H τ for any positive semidefinite H ∈ Mn (Mk ). The inequality does not hold for n > 2 ⎡ ⎤ ⎡ ⎤∗ e1 e1 τ when H is replaced by H. For example, if H = ⎣ ... ⎦ ⎣ ... ⎦ ∈ Mn (Mn ), where ei is en en the standard unit (column) vector of Rn whose only non-zero entry is a 1 in the ith position, then it is easy to show that tr1 H = tr2 H = In and therefore In ⊗ tr1 H + ⎡ ⎤ ⎡ ⎤ e1 e1 tr2 (H) ⊗ In − H = 2In − H. Since H ⎣ ... ⎦ = n ⎣ ... ⎦, n is an eigenvalue of H. Therefore en en
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
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In ⊗ tr1 H + tr2 (H) ⊗ In − H has the negative eigenvalue 2 − n for n > 2. However, the following shows that it holds for n = 2.
A Theorem 4. If H = B∗
B ∈ M2 (Mk ) is positive semidefinite, then C I2 ⊗ tr1 H + tr2 (H) ⊗ Ik ≥ H,
(2.1)
that is,
X Proof. Since W
Y Z
(trB)Ik − B ≥ 0. A + (trC)Ik
C + (trA)Ik (trB ∗ )Ik − B ∗
Z is *-congruent to Y M=
A + (trC)Ik (trB)Ik − B
(2.2)
W , we will show that X (trB ∗ )Ik − B ∗ C + (trA)Ik
is positive semidefinite. By the same argument used in Theorem 2, we may assume x H= [ x∗ y ∗ ] for x, y ∈ Mk×1 , in which case we have y M=
xx∗ + (y ∗ y)Ik (y ∗ x)Ik − xy ∗
(x∗ y)Ik − yx∗ . yy ∗ + (x∗ x)Ik
Without loss of generality, we assume y ∗ y = 1. Let α = complement of M as follows:
1 |x|2 +1 .
−1
S = yy ∗ + (x∗ x)Ik − ((y ∗ x)Ik − xy ∗ ) (xx∗ + Ik )
We compute the Schur
((x∗ y)Ik − yx∗ )
= yy ∗ + (x∗ x)Ik − ((y ∗ x)Ik − xy ∗ ) (Ik − αxx∗ ) ((x∗ y)Ik − yx∗ ) = yy ∗ + (x∗ x − |y ∗ x|2 )Ik + (y ∗ x)yx∗ + (x∗ y)xy ∗ − (y ∗ y)xx∗ . Note that M ≥ 0 is equivalent to S ≥ 0 [2, Theorem 7.7.9]. For z ∈ Ck with ||z|| = 1, let u = y ∗ x, v = z ∗ x, and w = y ∗ z. Then z ∗ Sz = |y ∗ z|2 + ||x||2 − |y ∗ x|2 + (y ∗ x)(z ∗ y)(x∗ z) + (x∗ y)(z ∗ x)(y ∗ z) − (z ∗ x)(x∗ z) = |w|2 + ||x||2 − |u|2 + uwv + uvw − |v|2 and we have z ∗ Sz ≥ 0 ⇐⇒ |u − vw|2 ≤ |w|2 + ||x||2 − |v|2 + |v|2 |w|2 . Let P = I − zz ∗ . Since P ≥ 0, we have |y ∗ P x|2 ≤ (x∗ P x)(y ∗ P y). Thus
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|u − vw|2 = |y ∗ P x|2 ≤ (x∗ P x)(y ∗ P y) = (||x||2 − |v|2 )(1 − |w|2 ) ≤ |w|2 + ||x||2 − |v|2 + |v|2 |w|2 .
2
The following can be obtained from Theorem 4.
A B Corollary 5. Let H = ∈ M2 (Mk ) be Hermitian. Then we have the following. B∗ C a) (k + 1)λmin (H)I2k ≤ I2 ⊗ tr1 H + tr2 (H) ⊗ Ik − H ≤ (k + 1)λmax (H)I2k . b) If H ≥ 0, then tr(A + C) ≥ 2(k−1) k+1 |Re(trB)|. 1 k−1 |Re(trB)| λmin (H) ≤ 2k tr(A + C) − k(k+1) c) . 1 k−1 λmax (H) ≥ 2k tr(A + C) + k(k+1) |Re(trB)| where Re(·) denotes the real part of a complex number. Proof. Replacing H by H − λmin (H)I2k and λmax (H)I2k − H in Theorem 4, we get a). From (2.2), we have A + C + tr(A + C)Ik ≥ ± (tr(B + B ∗ )Ik − (B + B ∗ )) .
(2.3)
Taking traces on both sides, we have b). Finally, from the first inequality in a) we have
C + (trA − (k + 1)λmin (H)) Ik (trB ∗ )Ik − B ∗
(trB)Ik − B ≥ 0. A + (trC − (k + 1)λmin (H)) Ik
Using the same argument as in b), we get the first inequality in c). Similarly we can obtain the second inequality in c) from the second inequality in a). 2 Remark 6. Replacing H by H − λmin (H)Ink and λmax (H)Ink − H in (1.1), we obtain for any Hermitian H ∈ Mn (Mk ) that In ⊗ tr1 H + tr2 H ⊗ Ik − H ≥ (trH − (n − 1)(k − 1)λmax (H)) Ink , In ⊗ tr1 H + tr2 H ⊗ Ik − H ≤ (trH − (n − 1)(k − 1)λmin (H)) Ink . In particular, when n = 2, we have In ⊗ tr1 H + tr2 H ⊗ I2 − H ≥ (trH − (k − 1)λmax (H)) I2k
(2.4)
from which we cannot deduce (2.1), since trH ≥ (k − 1)λmax (H) does not generally hold; for example, if H ∈ M2 (Mk ) whose only non-zero entry is a 1 in the (1, 1) position, then trH = 1, λmax (H) = 1, and therefore trH < (k − 1)λmax (H) for k > 2.
D. Choi / Linear Algebra and its Applications 516 (2017) 1–7
7
Using a similar approach as in [3, Proposition 2.5], we can also derive the following result.
A B Corollary 7. If H = ∈ M2 (Mk ) is positive semidefinite, then B∗ C Proof. Let
A B∗
B C
(trA)A + BB ∗ (trB)B ∗ − CA
=
X [ X∗ Y
(trB ∗ )B − AC ≥ 0. (trC)C + B ∗ B
Y ∗] =
XX ∗ Y X∗
XY ∗ YY∗
(2.5)
for X, Y ∈ Mk×m . Then we
have
(trA)A + BB ∗ (trB ∗ )B − AC (trB)B ∗ − CA (trC)C + B ∗ B
tr(X ∗ X)XX ∗ + XY ∗ Y X ∗ tr(X ∗ Y )XY ∗ − XX ∗ Y Y ∗ tr(Y ∗ X)Y X ∗ − Y Y ∗ XX ∗ tr(Y ∗ Y )Y Y ∗ + Y X ∗ XY ∗
∗ X 0 tr(X ∗ X)Im + Y ∗ Y tr(X ∗ Y )Im − X ∗ Y 0 X = . 0 Y∗ 0 Y tr(Y ∗ X)Im − Y ∗ X tr(Y ∗ Y )I + X ∗ X
=
Y∗ [Y Since X∗
Y ∗Y X] = X ∗Y
Y ∗X ≥ 0, we have X ∗X
tr(X ∗ X)Im + Y ∗ Y tr(Y ∗ X)Im − Y ∗ X
tr(X ∗ Y )Im − X ∗ Y tr(Y ∗ Y )I + X ∗ X
≥0
by Theorem 4. Thus we get (2.5). 2 Acknowledgements The author is grateful to Professor M. Lin for his assistance to prepare this manuscript. The author also expresses sincere thanks to Professor T. Ando for [1] before its publication. Finally, the author thanks anonymous reviewer for valuable comments and suggestions on the manuscript. References [1] [2] [3] [4] [5]
T. Ando, Matrix inequalities involving partial traces, preprint. R.A. Horn, C.R. Johnson, Matrix Analysis, 2nd ed., Cambridge University Press, 2012. M. Lin, A completely PPT map, Linear Algebra Appl. 459 (2014) 404–410. M. Lin, A determinantal inequality involving partial traces, Canad. Math. Bull. 59 (2016) 585–591. D. Petz, Quantum information theory and quantum statistics, in: Theoretical and Mathematical Physics, Springer-Verlag, Berlin, 2008.