A NOTE ON A TRACE INEQUALITY FOR POSITIVE
BLOCK MATRICES
ÁDÁM BESENYEI
Abstract. We give a short proof of a trace inequality for 2 × 2 positive
block matrices which is a special case of the subadditivity inequality for
q-entropies.
1. Introduction
In this short note we prove the following remarkable inequality for positivesemidefinite block matrices.
Theorem. Let A, B, C ∈ Cn×n be such that the block matrix
A B
X=
∈ C2n×2n
B∗ C
is positive-semidefinite. Then
Tr AC − Tr B ∗ B ≤ Tr A Tr C − Tr B ∗ Tr B.
There holds equality if and only if X = Y ⊗ Z for some positive-semidefinite
Y ∈ C2×2 , Z ∈ Cn×n such that min(rank Y, rank Z) ≤ 1.
Remark. Clearly, if the equality conditions hold, then both sides are 0.
The inequality of the Theorem comes from the subadditivity of q-entropies
proved in [1]. It states that for any bipartite state ρ in a finite-dimensional
Hilbert space H1 ⊗ H2 it holds
Sq (ρ) ≤ Sq (Tr1 ρ) + Sq (Tr2 ρ)
where
1 − Tr ρq
(q > 1)
q−1
is the so-called q-entropy and the partial traces Tr1 and Tr2 are linear operators defined by Tr1 : X ⊗ Y 7→ Tr(X)Y and Tr2 : X ⊗ Y 7→ Tr(Y )X (see
also [2]). If H1 = C2 and H2 = Cn and Mn denotes the space of n × n
complex matrices, then for a density matrix
A B
ρ=
∈ M2 ⊗ Mn
B∗ C
Sq (ρ) =
the partial traces can be expressed as
Tr1 ρ = A + C ∈ Mn
and
Tr2 ρ =
Tr A Tr B
∈ M2 .
Tr B ∗ Tr C
Date: January 14, 2013.
2010 Mathematics Subject Classification. 15A45.
Key words and phrases. positive block matrix, trace, entropy, inequality.
1
2
ÁDÁM BESENYEI
So for q = 2 the subadditivity inequality has the form
− Tr(A2 +C 2 )−2 Tr B ∗ B ≤ 1−Tr(A+C)2 −(Tr A)2 −(Tr C)2 −2 Tr B ∗ Tr B
which reduces to the inequality of the Theorem by using Tr ρ = Tr(A+C) =
1. The proof of the subadditivity relation in [1] is a bit delicate and therefore
the inequality of the Theorem deserves to have an elementary proof. This
will be provided in the following.
2. Proof of the inequality
We may assume that A is diagonal. Indeed, if U ∗ AU = Λ with unitary
U and diagonal Λ, then
∗
∗
U
0
A B U 0
U AU U ∗ BU
=
0 U ∗ B∗ C 0 U
U ∗ B ∗ U U ∗ CU
and
Tr(U ∗ AU )(U ∗ CU ) − Tr(U ∗ B ∗ U )(U ∗ BU ) = Tr AC,
Tr(U ∗ AU ) Tr(U ∗ CU ) − Tr(U ∗ B ∗ U ) Tr(U ∗ BU ) = Tr A Tr C − Tr B ∗ Tr B.
For diagonal A, the inequality has the form
n
X
aii cii −
i=1
n
X
|bij |2 ≤
i,j=1
n
X
i=1
aii ·
n
X
i=1
2
n
X
cii − bii i=1
which can be simplified to
X
X
X
|bij |2 ≤
(aii cjj + ajj cii ).
2
<(bii bjj ) −
i>j
i>j
i6=j
A B
is positive-semidefinite, A and C
B∗ C
are also positive-semidefinite. Further, the principal subdeterminants of X,
especially the 2 × 2 subdeterminants are non-negative (see [3]) thus aii cii ≥
|bii |2 for i = 1, . . . , n. Therefore, by the inequality of the arithmetic and
geometric means we obtain
q
√
aii cjj + ajj cii ≥ 2 aii cii ajj cjj ≥ 2 |bii |2 |bjj |2 ≥ 2<(bii bjj )
Since the block matrix X =
and the desired inequality follows.
3. The case of equality
If n = 1, there holds equality. Suppose that n ≥ 2 and A is diagonal.
From the above arguments it follows that in the case of equality the following
conditions must hold for i, j = 1, . . . , n, i 6= j:
(i)
(ii)
(iii)
(iv)
aii cii ≥ |bii |2 and aii cii ajj cjj = |bii |2 |bjj |2 ;
aii cjj = ajj cii ;
bij = 0;
|bii bjj | = <(bii bjj ).
TRACE INEQUALITY FOR POSITIVE BLOCK MATRICES
3
Clearly, (iii) implies that B is diagonal. Further, (iv) means that the numbers bii have the same argument.
If aii cii > |bii |2 for some i, then by (i), ajj cjj = |bjj |2 = 0 for all j 6= i.
So by (ii), ajj = cjj = 0 and also bjj = 0 for j 6= i. But C must be positivesemidefinite thus cii cjj ≥ |cij |2 (j 6= i) so C should also be diagonal with at
most one non-zero diagonal
entry,
similarly to A and B. This means that
aii bii
is positive-definite and Z is diagonal with
X = Y ⊗ Z where Y =
bii cii
one non-zero entry, zii = 1.
Now assume that aii cii = |bii |2 for all i Since X is positive-semidefinite,
the following 3 × 3 subdeterminant should be non-negative:
aii 0 bii 0 cjj cji = aii cii − |bii |2 cij − aii |cij |2 = −aij |cij |2 .
bii cji cii We have two cases. If A = 0, then B = 0 and C is anarbitrary positive0 0
semidefinite matrix thus X = Y ⊗ Z where Y =
. Otherwise cij =
0 1
0 (j 6= i) when aii 6= 0. Moreover, by (ii), cii = 0 when aii = 0 and
thus by the positive-semidefiniteness of C, cij = 0 (j 6= i). So C is also
diagonal, proportional to A and therefore by (iv), B = λA, C = |λ|2 A for
1 λ
some constant λ ∈ C. Thus X = Y ⊗ A where Y =
. Then X
λ |λ|2
is positive-semidefinite since it is the tensor product of positive-semidefinite
matrices.
In the general case, if A is not diagonal, then we might consider
∗
U AU U ∗ BU
X̃ =
.
U ∗ B ∗ U U ∗ CU
In the case of equality X̃ = Ỹ ⊗ Z̃ thus
X = (I ⊗ U )(Ỹ ⊗ Z̃)(I ⊗ U ∗ ) = Ỹ ⊗ (U Z̃U ∗ )
where rank U Z̃U ∗ = rank Z̃.
References
[1] K. Audenaert, Subadditivity of q-entropies for q > 1, J. Math. Phys., 48, 083507
(2007).
[2] R. Bhatia, Partial traces and entropy inequalities, Linear Algebra Appl., 375 (2003),
211–220.
[3] R. A. Horn, Ch. R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1990.
Ádám Besenyei
Department of Applied Analysis, Eötvös Loránd University, H-1117 Budapest,
Pázmány P. sétány 1/C, Hungary
E-mail address: [email protected]