# Trace inequality

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Concept in Hlibert spaces mathematics

In [mathematics](/source/Mathematics), there are many kinds of [inequalities](/source/Inequality_(mathematics)) involving [matrices](/source/Matrix_(mathematics)) and [linear operators](/source/Linear_operator) on [Hilbert spaces](/source/Hilbert_space). This article covers some important operator inequalities connected with [traces](/source/Trace_(linear_algebra)) of matrices.[1][2][3][4]

## Basic definitions

Let H n {\displaystyle \mathbf {H} _{n}} denote the space of [Hermitian](/source/Hermitian_matrix) n × n {\displaystyle n\times n} matrices, H n + {\displaystyle \mathbf {H} _{n}^{+}} denote the set consisting of [positive semi-definite](/source/Positive-definite_matrix#Negative-definite,_semidefinite_and_indefinite_matrices) n × n {\displaystyle n\times n} Hermitian matrices and H n + + {\displaystyle \mathbf {H} _{n}^{++}} denote the set of [positive definite](/source/Positive-definite_matrix#Negative-definite,_semidefinite_and_indefinite_matrices) Hermitian matrices. For operators on an infinite dimensional Hilbert space we require that they be [trace class](/source/Trace_class) and [self-adjoint](/source/Self-adjoint_operator), in which case similar definitions apply, but we discuss only matrices, for simplicity.

For any real-valued function f {\displaystyle f} on an interval I ⊆ R , {\displaystyle I\subseteq \mathbb {R} ,} one may define a [matrix function](/source/Matrix_function) f ( A ) {\displaystyle f(A)} for any operator A ∈ H n {\displaystyle A\in \mathbf {H} _{n}} with [eigenvalues](/source/Eigenvalues_and_eigenvectors) λ {\displaystyle \lambda } in I {\displaystyle I} by defining it on the eigenvalues and corresponding [projectors](/source/Projection_(linear_algebra)) P {\displaystyle P} as f ( A ) ≡ ∑ j f ( λ j ) P j , {\displaystyle f(A)\equiv \sum _{j}f(\lambda _{j})P_{j}~,} given the [spectral decomposition](/source/Spectral_theorem) A = ∑ j λ j P j . {\displaystyle A=\sum _{j}\lambda _{j}P_{j}.}

### Operator monotone

Main article: [Operator monotone function](/source/Operator_monotone_function)

A function f : I → R {\displaystyle f:I\to \mathbb {R} } defined on an interval I ⊆ R {\displaystyle I\subseteq \mathbb {R} } is said to be **operator monotone** if for all n , {\displaystyle n,} and all A , B ∈ H n {\displaystyle A,B\in \mathbf {H} _{n}} with eigenvalues in I , {\displaystyle I,} the following holds, A ≥ B ⟹ f ( A ) ≥ f ( B ) , {\displaystyle A\geq B\implies f(A)\geq f(B),} where the inequality A ≥ B {\displaystyle A\geq B} means that the operator A − B ≥ 0 {\displaystyle A-B\geq 0} is positive semi-definite. One may check that f ( A ) = A 2 {\displaystyle f(A)=A^{2}} is, in fact, *not* operator monotone!

### Operator convex

A function f : I → R {\displaystyle f:I\to \mathbb {R} } is said to be **operator convex** if for all n {\displaystyle n} and all A , B ∈ H n {\displaystyle A,B\in \mathbf {H} _{n}} with eigenvalues in I , {\displaystyle I,} and 0 < λ < 1 {\displaystyle 0<\lambda <1} , the following holds f ( λ A + ( 1 − λ ) B ) ≤ λ f ( A ) + ( 1 − λ ) f ( B ) . {\displaystyle f(\lambda A+(1-\lambda )B)\leq \lambda f(A)+(1-\lambda )f(B).} Note that the operator λ A + ( 1 − λ ) B {\displaystyle \lambda A+(1-\lambda )B} has eigenvalues in I , {\displaystyle I,} since A {\displaystyle A} and B {\displaystyle B} have eigenvalues in I . {\displaystyle I.}

A function f {\displaystyle f} is **operator concave** if − f {\displaystyle -f} is operator convex;=, that is, the inequality above for f {\displaystyle f} is reversed.

### Joint convexity

A function g : I × J → R , {\displaystyle g:I\times J\to \mathbb {R} ,} defined on intervals I , J ⊆ R {\displaystyle I,J\subseteq \mathbb {R} } is said to be **jointly convex** if for all n {\displaystyle n} and all A 1 , A 2 ∈ H n {\displaystyle A_{1},A_{2}\in \mathbf {H} _{n}} with eigenvalues in I {\displaystyle I} and all B 1 , B 2 ∈ H n {\displaystyle B_{1},B_{2}\in \mathbf {H} _{n}} with eigenvalues in J , {\displaystyle J,} and any 0 ≤ λ ≤ 1 {\displaystyle 0\leq \lambda \leq 1} the following holds g ( λ A 1 + ( 1 − λ ) A 2 , λ B 1 + ( 1 − λ ) B 2 ) ≤ λ g ( A 1 , B 1 ) + ( 1 − λ ) g ( A 2 , B 2 ) . {\displaystyle g(\lambda A_{1}+(1-\lambda )A_{2},\lambda B_{1}+(1-\lambda )B_{2})~\leq ~\lambda g(A_{1},B_{1})+(1-\lambda )g(A_{2},B_{2}).}

A function g {\displaystyle g} is **jointly concave** if − g {\displaystyle g} is jointly convex, i.e. the inequality above for g {\displaystyle g} is reversed.

### Trace function

Given a function f : R → R , {\displaystyle f:\mathbb {R} \to \mathbb {R} ,} the associated **trace function** on H n {\displaystyle \mathbf {H} _{n}} is given by A ↦ Tr ⁡ f ( A ) = ∑ j f ( λ j ) , {\displaystyle A\mapsto \operatorname {Tr} f(A)=\sum _{j}f(\lambda _{j}),} where A {\displaystyle A} has eigenvalues λ {\displaystyle \lambda } and Tr {\displaystyle \operatorname {Tr} } stands for a [trace](/source/Trace_(linear_algebra)) of the operator.

## Convexity and monotonicity of the trace function

Let f : R → R {\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} } be continuous, and let n be any [integer](/source/Integer). Then, if t ↦ f ( t ) {\displaystyle t\mapsto f(t)} is monotone increasing, so is A ↦ Tr ⁡ f ( A ) {\displaystyle A\mapsto \operatorname {Tr} f(A)} on **H***n*.

Likewise, if t ↦ f ( t ) {\displaystyle t\mapsto f(t)} is [convex](/source/Convex_function), so is A ↦ Tr ⁡ f ( A ) {\displaystyle A\mapsto \operatorname {Tr} f(A)} on **H***n*, and it is strictly convex if f is strictly convex.

See proof and discussion in,[1] for example.

## Löwner–Heinz theorem

For − 1 ≤ p ≤ 0 {\displaystyle -1\leq p\leq 0} , the function f ( t ) = − t p {\displaystyle f(t)=-t^{p}} is operator monotone and operator concave.

For 0 ≤ p ≤ 1 {\displaystyle 0\leq p\leq 1} , the function f ( t ) = t p {\displaystyle f(t)=t^{p}} is operator monotone and operator concave.

For 1 ≤ p ≤ 2 {\displaystyle 1\leq p\leq 2} , the function f ( t ) = t p {\displaystyle f(t)=t^{p}} is operator convex. Furthermore,

- f ( t ) = log ⁡ ( t ) {\displaystyle f(t)=\log(t)} is operator concave and operator monotone, while

- f ( t ) = t log ⁡ ( t ) {\displaystyle f(t)=t\log(t)} is operator convex.

The original proof of this theorem is due to [K. Löwner](/source/Karl_L%C3%B6wner) who gave a necessary and sufficient condition for f to be operator monotone.[5] An [elementary proof](/source/Elementary_proof) of the theorem is discussed in [1] and a more general version of it in.[6]

## Klein's inequality

For all Hermitian n×n matrices A and B and all differentiable [convex functions](/source/Convex_function) f : R → R {\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} } with [derivative](/source/Derivative) *f '*, or for all positive-definite Hermitian n×n matrices A and B, and all differentiable convex functions f:(0,∞) → R {\displaystyle \mathbb {R} } , the following inequality holds,

Tr ⁡ [ f ( A ) − f ( B ) − ( A − B ) f ′ ( B ) ] ≥ 0 . {\displaystyle \operatorname {Tr} [f(A)-f(B)-(A-B)f'(B)]\geq 0~.}

In either case, if f is strictly convex, equality holds if and only if A = B. A popular choice in applications is *f*(*t*) = *t* log *t*, see below.

### Proof

Let C = A − B {\displaystyle C=A-B} so that, for t ∈ ( 0 , 1 ) {\displaystyle t\in (0,1)} ,

- B + t C = ( 1 − t ) B + t A {\displaystyle B+tC=(1-t)B+tA} ,

varies from B {\displaystyle B} to A {\displaystyle A} .

Define

- F ( t ) = Tr ⁡ [ f ( B + t C ) ] {\displaystyle F(t)=\operatorname {Tr} [f(B+tC)]} .

By convexity and monotonicity of trace functions, F ( t ) {\displaystyle F(t)} is convex, and so for all t ∈ ( 0 , 1 ) {\displaystyle t\in (0,1)} ,

- F ( 0 ) + t ( F ( 1 ) − F ( 0 ) ) ≥ F ( t ) {\displaystyle F(0)+t(F(1)-F(0))\geq F(t)} ,

which is,

- F ( 1 ) − F ( 0 ) ≥ F ( t ) − F ( 0 ) t {\displaystyle F(1)-F(0)\geq {\frac {F(t)-F(0)}{t}}} ,

and, in fact, the right hand side is monotone decreasing in t {\displaystyle t} .

Taking the limit t → 0 {\displaystyle t\to 0} yields,

- F ( 1 ) − F ( 0 ) ≥ F ′ ( 0 ) {\displaystyle F(1)-F(0)\geq F'(0)} ,

which with rearrangement and substitution is Klein's inequality:

- t r [ f ( A ) − f ( B ) − ( A − B ) f ′ ( B ) ] ≥ 0 {\displaystyle \mathrm {tr} [f(A)-f(B)-(A-B)f'(B)]\geq 0}

Note that if f ( t ) {\displaystyle f(t)} is strictly convex and C ≠ 0 {\displaystyle C\neq 0} , then F ( t ) {\displaystyle F(t)} is strictly convex. The final assertion follows from this and the fact that F ( t ) − F ( 0 ) t {\displaystyle {\tfrac {F(t)-F(0)}{t}}} is monotone decreasing in t {\displaystyle t} .

## Golden–Thompson inequality

Main article: [Golden–Thompson inequality](/source/Golden%E2%80%93Thompson_inequality)

In 1965, S. Golden [7] and C.J. Thompson [8] independently discovered that

For any matrices A , B ∈ H n {\displaystyle A,B\in \mathbf {H} _{n}} ,

- Tr ⁡ e A + B ≤ Tr ⁡ e A e B . {\displaystyle \operatorname {Tr} e^{A+B}\leq \operatorname {Tr} e^{A}e^{B}.}

This inequality can be generalized for three operators:[9] for non-negative operators A , B , C ∈ H n + {\displaystyle A,B,C\in \mathbf {H} _{n}^{+}} ,

- Tr ⁡ e ln ⁡ A − ln ⁡ B + ln ⁡ C ≤ ∫ 0 ∞ Tr ⁡ A ( B + t ) − 1 C ( B + t ) − 1 d ⁡ t . {\displaystyle \operatorname {Tr} e^{\ln A-\ln B+\ln C}\leq \int _{0}^{\infty }\operatorname {Tr} A(B+t)^{-1}C(B+t)^{-1}\,\operatorname {d} t.}

## Peierls–Bogoliubov inequality

Let R , F ∈ H n {\displaystyle R,F\in \mathbf {H} _{n}} be such that Tr e*R* = 1. Defining *g* = Tr *FeR*, we have

- Tr ⁡ e F e R ≥ Tr ⁡ e F + R ≥ e g . {\displaystyle \operatorname {Tr} e^{F}e^{R}\geq \operatorname {Tr} e^{F+R}\geq e^{g}.}

The proof of this inequality follows from the above combined with [Klein's inequality](#Klein's_inequality). Take *f*(*x*) = exp(*x*), *A*=*R* + *F*, and *B* = *R* + *gI*.[10]

## Gibbs variational principle

Let H {\displaystyle H} be a self-adjoint operator such that e − H {\displaystyle e^{-H}} is [trace class](/source/Trace_class). Then for any γ ≥ 0 {\displaystyle \gamma \geq 0} with Tr ⁡ γ = 1 , {\displaystyle \operatorname {Tr} \gamma =1,}

- Tr ⁡ γ H + Tr ⁡ γ ln ⁡ γ ≥ − ln ⁡ Tr ⁡ e − H , {\displaystyle \operatorname {Tr} \gamma H+\operatorname {Tr} \gamma \ln \gamma \geq -\ln \operatorname {Tr} e^{-H},}

with equality if and only if γ = exp ⁡ ( − H ) / Tr ⁡ exp ⁡ ( − H ) . {\displaystyle \gamma =\exp(-H)/\operatorname {Tr} \exp(-H).}

## Lieb's concavity theorem

The following theorem was proved by [E. H. Lieb](/source/Elliott_Lieb) in.[9] It proves and generalizes a conjecture of [E. P. Wigner](/source/Eugene_Wigner), [M. M. Yanase](https://en.wikipedia.org/w/index.php?title=Mutsuo_Yanase&action=edit&redlink=1), and [Freeman Dyson](/source/Freeman_Dyson).[11] Six years later other proofs were given by T. Ando [12] and B. Simon,[3] and several more have been given since then.

For all m × n {\displaystyle m\times n} matrices K {\displaystyle K} , and all q {\displaystyle q} and r {\displaystyle r} such that 0 ≤ q ≤ 1 {\displaystyle 0\leq q\leq 1} and 0 ≤ r ≤ 1 {\displaystyle 0\leq r\leq 1} , with q + r ≤ 1 {\displaystyle q+r\leq 1} the real valued map on H m + × H n + {\displaystyle \mathbf {H} _{m}^{+}\times \mathbf {H} _{n}^{+}} given by

- F ( A , B , K ) = Tr ⁡ ( K ∗ A q K B r ) {\displaystyle F(A,B,K)=\operatorname {Tr} (K^{*}A^{q}KB^{r})}

- is jointly concave in ( A , B ) {\displaystyle (A,B)}

- is convex in K {\displaystyle K} .

Here K ∗ {\displaystyle K^{*}} stands for the [adjoint operator](/source/Hermitian_adjoint) of K . {\displaystyle K.}

## Lieb's theorem

For a fixed Hermitian matrix L ∈ H n {\displaystyle L\in \mathbf {H} _{n}} , the function

- f ( A ) = Tr ⁡ exp ⁡ { L + ln ⁡ A } {\displaystyle f(A)=\operatorname {Tr} \exp\{L+\ln A\}}

is concave on H n + + {\displaystyle \mathbf {H} _{n}^{++}} .

The theorem and proof are due to E. H. Lieb,[9] Thm 6, where he obtains this theorem as a corollary of Lieb's concavity Theorem. The most direct proof is due to H. Epstein;[13] see [M.B. Ruskai](/source/Mary_Beth_Ruskai) papers,[14][15] for a review of this argument.

## Ando's convexity theorem

T. Ando's proof [12] of [Lieb's concavity theorem](#Lieb's_concavity_theorem) led to the following significant complement to it:

For all m × n {\displaystyle m\times n} matrices K {\displaystyle K} , and all 1 ≤ q ≤ 2 {\displaystyle 1\leq q\leq 2} and 0 ≤ r ≤ 1 {\displaystyle 0\leq r\leq 1} with q − r ≥ 1 {\displaystyle q-r\geq 1} , the real valued map on H m + + × H n + + {\displaystyle \mathbf {H} _{m}^{++}\times \mathbf {H} _{n}^{++}} given by

- ( A , B ) ↦ Tr ⁡ ( K ∗ A q K B − r ) {\displaystyle (A,B)\mapsto \operatorname {Tr} (K^{*}A^{q}KB^{-r})}

is convex.

## Joint convexity of relative entropy

For two operators A , B ∈ H n + + {\displaystyle A,B\in \mathbf {H} _{n}^{++}} define the following map

- R ( A ∥ B ) := Tr ⁡ ( A log ⁡ A ) − Tr ⁡ ( A log ⁡ B ) . {\displaystyle R(A\parallel B):=\operatorname {Tr} (A\log A)-\operatorname {Tr} (A\log B).}

For [density matrices](/source/Density_matrix) ρ {\displaystyle \rho } and σ {\displaystyle \sigma } , the map R ( ρ ∥ σ ) = S ( ρ ∥ σ ) {\displaystyle R(\rho \parallel \sigma )=S(\rho \parallel \sigma )} is the Umegaki's [quantum relative entropy](/source/Quantum_relative_entropy).

Note that the non-negativity of R ( A ∥ B ) {\displaystyle R(A\parallel B)} follows from Klein's inequality with f ( t ) = t log ⁡ t {\displaystyle f(t)=t\log t} .

### Statement

The map R ( A ∥ B ) : H n + + × H n + + → R {\displaystyle R(A\parallel B):\mathbf {H} _{n}^{++}\times \mathbf {H} _{n}^{++}\rightarrow \mathbf {R} } is jointly convex.

### Proof

For all 0 < p < 1 {\displaystyle 0<p<1} , ( A , B ) ↦ Tr ⁡ ( B 1 − p A p ) {\displaystyle (A,B)\mapsto \operatorname {Tr} (B^{1-p}A^{p})} is jointly concave, by [Lieb's concavity theorem](#Lieb's_concavity_theorem), and thus

- ( A , B ) ↦ 1 p − 1 ( Tr ⁡ ( B 1 − p A p ) − Tr ⁡ A ) {\displaystyle (A,B)\mapsto {\frac {1}{p-1}}(\operatorname {Tr} (B^{1-p}A^{p})-\operatorname {Tr} A)}

is convex. But

- lim p → 1 1 p − 1 ( Tr ⁡ ( B 1 − p A p ) − Tr ⁡ A ) = R ( A ∥ B ) , {\displaystyle \lim _{p\rightarrow 1}{\frac {1}{p-1}}(\operatorname {Tr} (B^{1-p}A^{p})-\operatorname {Tr} A)=R(A\parallel B),}

and convexity is preserved in the limit.

The proof is due to G. Lindblad.[16]

## Jensen's operator and trace inequalities

The operator version of [Jensen's inequality](/source/Jensen's_inequality) is due to C. Davis.[17]

A continuous, real function f {\displaystyle f} on an interval I {\displaystyle I} satisfies **Jensen's Operator Inequality** if the following holds

- f ( ∑ k A k ∗ X k A k ) ≤ ∑ k A k ∗ f ( X k ) A k , {\displaystyle f\left(\sum _{k}A_{k}^{*}X_{k}A_{k}\right)\leq \sum _{k}A_{k}^{*}f(X_{k})A_{k},}

for operators { A k } k {\displaystyle \{A_{k}\}_{k}} with ∑ k A k ∗ A k = 1 {\displaystyle \sum _{k}A_{k}^{*}A_{k}=1} and for [self-adjoint operators](/source/Self-adjoint_operator) { X k } k {\displaystyle \{X_{k}\}_{k}} with [spectrum](/source/Spectrum_(functional_analysis)) on I {\displaystyle I} .

See,[17][18] for the proof of the following two theorems.

### Jensen's trace inequality

Let f be a [continuous function](/source/Continuous_function) defined on an interval I and let m and n be natural numbers. If f is convex, we then have the inequality

- Tr ⁡ ( f ( ∑ k = 1 n A k ∗ X k A k ) ) ≤ Tr ⁡ ( ∑ k = 1 n A k ∗ f ( X k ) A k ) , {\displaystyle \operatorname {Tr} {\Bigl (}f{\Bigl (}\sum _{k=1}^{n}A_{k}^{*}X_{k}A_{k}{\Bigr )}{\Bigr )}\leq \operatorname {Tr} {\Bigl (}\sum _{k=1}^{n}A_{k}^{*}f(X_{k})A_{k}{\Bigr )},}

for all (X1, ... , X*n*) self-adjoint m × m matrices with spectra contained in I and all (A1, ... , A*n*) of m × m matrices with

- ∑ k = 1 n A k ∗ A k = 1. {\displaystyle \sum _{k=1}^{n}A_{k}^{*}A_{k}=1.}

Conversely, if the above inequality is satisfied for some n and m, where n > 1, then f is convex.

### Jensen's operator inequality

For a continuous function f {\displaystyle f} defined on an interval I {\displaystyle I} the following conditions are equivalent:

- f {\displaystyle f} is operator convex.

- For each natural number n {\displaystyle n} we have the inequality

- f ( ∑ k = 1 n A k ∗ X k A k ) ≤ ∑ k = 1 n A k ∗ f ( X k ) A k , {\displaystyle f{\Bigl (}\sum _{k=1}^{n}A_{k}^{*}X_{k}A_{k}{\Bigr )}\leq \sum _{k=1}^{n}A_{k}^{*}f(X_{k})A_{k},}

for all ( X 1 , … , X n ) {\displaystyle (X_{1},\ldots ,X_{n})} bounded, self-adjoint operators on an arbitrary [Hilbert space](/source/Hilbert_space) H {\displaystyle {\mathcal {H}}} with spectra contained in I {\displaystyle I} and all ( A 1 , … , A n ) {\displaystyle (A_{1},\ldots ,A_{n})} on H {\displaystyle {\mathcal {H}}} with ∑ k = 1 n A k ∗ A k = 1. {\displaystyle \sum _{k=1}^{n}A_{k}^{*}A_{k}=1.}

- f ( V ∗ X V ) ≤ V ∗ f ( X ) V {\displaystyle f(V^{*}XV)\leq V^{*}f(X)V} for each isometry V {\displaystyle V} on an infinite-dimensional Hilbert space H {\displaystyle {\mathcal {H}}} and

every self-adjoint operator X {\displaystyle X} with spectrum in I {\displaystyle I} .

- P f ( P X P + λ ( 1 − P ) ) P ≤ P f ( X ) P {\displaystyle Pf(PXP+\lambda (1-P))P\leq Pf(X)P} for each projection P {\displaystyle P} on an infinite-dimensional Hilbert space H {\displaystyle {\mathcal {H}}} , every self-adjoint operator X {\displaystyle X} with spectrum in I {\displaystyle I} and every λ {\displaystyle \lambda } in I {\displaystyle I} .

## Araki–Lieb–Thirring inequality

Not to be confused with the [Lieb–Thirring inequality](/source/Lieb%E2%80%93Thirring_inequality).

E. H. Lieb and W. E. Thirring proved the following inequality in [19] 1976: For any A ≥ 0 , {\displaystyle A\geq 0,} B ≥ 0 {\displaystyle B\geq 0} and r ≥ 1 , {\displaystyle r\geq 1,} Tr ⁡ ( ( B A B ) r ) ≤ Tr ⁡ ( B r A r B r ) . {\displaystyle \operatorname {Tr} ((BAB)^{r})~\leq ~\operatorname {Tr} (B^{r}A^{r}B^{r}).}

In 1990 [20] H. Araki generalized the above inequality to the following one: For any A ≥ 0 , {\displaystyle A\geq 0,} B ≥ 0 {\displaystyle B\geq 0} and q ≥ 0 , {\displaystyle q\geq 0,} Tr ⁡ ( ( B A B ) r q ) ≤ Tr ⁡ ( ( B r A r B r ) q ) , {\displaystyle \operatorname {Tr} ((BAB)^{rq})~\leq ~\operatorname {Tr} ((B^{r}A^{r}B^{r})^{q}),} for r ≥ 1 , {\displaystyle r\geq 1,} and Tr ⁡ ( ( B r A r B r ) q ) ≤ Tr ⁡ ( ( B A B ) r q ) , {\displaystyle \operatorname {Tr} ((B^{r}A^{r}B^{r})^{q})~\leq ~\operatorname {Tr} ((BAB)^{rq}),} for 0 ≤ r ≤ 1. {\displaystyle 0\leq r\leq 1.}

There are several other inequalities close to the Lieb–Thirring inequality, such as the following:[21] for any A ≥ 0 , {\displaystyle A\geq 0,} B ≥ 0 {\displaystyle B\geq 0} and α ∈ [ 0 , 1 ] , {\displaystyle \alpha \in [0,1],} Tr ⁡ ( B A α B B A 1 − α B ) ≤ Tr ⁡ ( B 2 A B 2 ) , {\displaystyle \operatorname {Tr} (BA^{\alpha }BBA^{1-\alpha }B)~\leq ~\operatorname {Tr} (B^{2}AB^{2}),} and even more generally:[22] for any A ≥ 0 , {\displaystyle A\geq 0,} B ≥ 0 , {\displaystyle B\geq 0,} r ≥ 1 / 2 {\displaystyle r\geq 1/2} and c ≥ 0 , {\displaystyle c\geq 0,} Tr ⁡ ( ( B A B 2 c A B ) r ) ≤ Tr ⁡ ( ( B c + 1 A 2 B c + 1 ) r ) . {\displaystyle \operatorname {Tr} ((BAB^{2c}AB)^{r})~\leq ~\operatorname {Tr} ((B^{c+1}A^{2}B^{c+1})^{r}).} The above inequality generalizes the previous one, as can be seen by exchanging A {\displaystyle A} by B 2 {\displaystyle B^{2}} and B {\displaystyle B} by A ( 1 − α ) / 2 {\displaystyle A^{(1-\alpha )/2}} with α = 2 c / ( 2 c + 2 ) {\displaystyle \alpha =2c/(2c+2)} and using the cyclicity of the trace, leading to Tr ⁡ ( ( B A α B B A 1 − α B ) r ) ≤ Tr ⁡ ( ( B 2 A B 2 ) r ) . {\displaystyle \operatorname {Tr} ((BA^{\alpha }BBA^{1-\alpha }B)^{r})~\leq ~\operatorname {Tr} ((B^{2}AB^{2})^{r}).}

Additionally, building upon the Lieb–Thirring inequality the following inequality was derived: [23] For any A , B ∈ H n , T ∈ C n × n {\displaystyle A,B\in \mathbf {H} _{n},T\in \mathbb {C} ^{n\times n}} and all 1 ≤ p , q ≤ ∞ {\displaystyle 1\leq p,q\leq \infty } with 1 / p + 1 / q = 1 {\displaystyle 1/p+1/q=1} , it holds that | Tr ⁡ ( T A T ∗ B ) | ≤ Tr ⁡ ( T ∗ T | A | p ) 1 p Tr ⁡ ( T T ∗ | B | q ) 1 q . {\displaystyle |\operatorname {Tr} (TAT^{*}B)|~\leq ~\operatorname {Tr} (T^{*}T|A|^{p})^{\frac {1}{p}}\operatorname {Tr} (TT^{*}|B|^{q})^{\frac {1}{q}}.}

## Effros's theorem and its extension

E. Effros in [24] proved the following theorem.

If f ( x ) {\displaystyle f(x)} is an operator convex function, and L {\displaystyle L} and R {\displaystyle R} are commuting bounded linear operators, i.e. the commutator [ L , R ] = L R − R L = 0 {\displaystyle [L,R]=LR-RL=0} , the *perspective*

- g ( L , R ) := f ( L R − 1 ) R {\displaystyle g(L,R):=f(LR^{-1})R}

is jointly convex, i.e. if L = λ L 1 + ( 1 − λ ) L 2 {\displaystyle L=\lambda L_{1}+(1-\lambda )L_{2}} and R = λ R 1 + ( 1 − λ ) R 2 {\displaystyle R=\lambda R_{1}+(1-\lambda )R_{2}} with [ L i , R i ] = 0 {\displaystyle [L_{i},R_{i}]=0} (i=1,2), 0 ≤ λ ≤ 1 {\displaystyle 0\leq \lambda \leq 1} ,

- g ( L , R ) ≤ λ g ( L 1 , R 1 ) + ( 1 − λ ) g ( L 2 , R 2 ) . {\displaystyle g(L,R)\leq \lambda g(L_{1},R_{1})+(1-\lambda )g(L_{2},R_{2}).}

Ebadian et al. later extended the inequality to the case where L {\displaystyle L} and R {\displaystyle R} do not commute .[25]

## Von Neumann's trace inequality and related results

Von Neumann's trace inequality, named after its originator [John von Neumann](/source/John_von_Neumann), states that for any n × n {\displaystyle n\times n} complex matrices A {\displaystyle A} and B {\displaystyle B} with [singular values](/source/Singular_value) α 1 ≥ α 2 ≥ ⋯ ≥ α n {\displaystyle \alpha _{1}\geq \alpha _{2}\geq \cdots \geq \alpha _{n}} and β 1 ≥ β 2 ≥ ⋯ ≥ β n {\displaystyle \beta _{1}\geq \beta _{2}\geq \cdots \geq \beta _{n}} respectively,[26] | Tr ⁡ ( A B ) | ≤ ∑ i = 1 n α i β i , {\displaystyle |\operatorname {Tr} (AB)|~\leq ~\sum _{i=1}^{n}\alpha _{i}\beta _{i}\,,} with equality if and only if A {\displaystyle A} and B † {\displaystyle B^{\dagger }} share singular vectors.[27]

A simple corollary to this is the following result:[28] For [Hermitian](/source/Hermitian_matrix) n × n {\displaystyle n\times n} positive semi-definite complex matrices A {\displaystyle A} and B {\displaystyle B} where now the [eigenvalues](/source/Eigenvalue) are sorted decreasingly ( a 1 ≥ a 2 ≥ ⋯ ≥ a n {\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}} and b 1 ≥ b 2 ≥ ⋯ ≥ b n , {\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n},} respectively), ∑ i = 1 n a i b n − i + 1 ≤ Tr ⁡ ( A B ) ≤ ∑ i = 1 n a i b i . {\displaystyle \sum _{i=1}^{n}a_{i}b_{n-i+1}~\leq ~\operatorname {Tr} (AB)~\leq ~\sum _{i=1}^{n}a_{i}b_{i}\,.}

## See also

- [Lieb–Thirring inequality](/source/Lieb%E2%80%93Thirring_inequality) – Inequality in mathematical physics

- [Schur–Horn theorem](/source/Schur%E2%80%93Horn_theorem) – Characterizes the diagonal of a Hermitian matrix with given eigenvalues

- [Trace identity](/source/Trace_identity) – Equations involving the trace of a matrix

- [von Neumann entropy](/source/Von_Neumann_entropy) – Type of entropy in quantum theory

## References

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- [Scholarpedia](http://www.scholarpedia.org/article/Matrix_and_Operator_Trace_Inequalities#Gibbs_variational_principle) primary source.

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