Trace inequalities

In mathematics, there are many kinds of inequalities involving matrices and linear operators on Hilbert spaces. This article covers some important operator inequalities connected with traces of matrices.^[1][2][3][4]

Basic definitions

Let H_n denote the space of Hermitian n×n matrices, H_n⁺ denote the set consisting of positive semi-definite n×n Hermitian matrices and H_n⁺⁺ denote the set of positive definite Hermitian matrices. For operators on an infinite dimensional Hilbert space we require that they be trace class and self-adjoint, in which case similar definitions apply, but we discuss only matrices, for simplicity.

For any real-valued function f on an interval I ⊂ ℝ, one may define a matrix function f(A) for any operator A ∈ H_n with eigenvalues λ in I by defining it on the eigenvalues and corresponding projectors P as

$f(A)\equiv \sum_j f(\lambda_j)P_j ~,$ given the spectral decomposition $A=\sum_j\lambda_j P_j.$

Operator monotone

A function f: I → ℝ defined on an interval I ⊂ ℝ is said to be operator monotone if ∀n, and all A,B ∈ H_n with eigenvalues in I, the following holds,

$A \geq B \Rightarrow f(A) \geq f(B),$

where the inequality A ≥ B means that the operator A − B ≥ 0 is positive semi-definite. One may check that f(A)=A² is, in fact, not monotone!

Operator convex

A function $f: I \rightarrow \mathbb{R}$ is said to be operator convex if for all $n$ and all A,B ∈ H_n with eigenvalues in I, and $0 < \lambda < 1$, the following holds

$f(\lambda A + (1-\lambda)B) \leq \lambda f(A) + (1 -\lambda)f(B) .$

Note that the operator $\lambda A + (1-\lambda)B$ has eigenvalues in $I$, since $A$ and $B$ have eigenvalues in I.

A function $f$ is operator concave if $-f$ is operator convex, i.e. the inequality above for $f$ is reversed.

Joint convexity

A function $g: I\times J \rightarrow \mathbb{R}$, defined on intervals $I,J\subset \mathbb{R}$ is said to be jointly convex if for all $n$ and all $A_1, A_2\in \mathbf{H}_n$ with eigenvalues in $I$ and all $B_1,B_2\in \mathbf{H}_n$ with eigenvalues in $J$, and any $0\leq \lambda\leq 1$ the following holds

$g(\lambda A_1 + (1-\lambda)A_2,\lambda B_1 + (1-\lambda)B_2 ) \leq \lambda g(A_1, B_1) + (1 -\lambda)g(A_2, B_2).$

A function g is jointly concave if −g is jointly convex, i.e. the inequality above for g is reversed.

Trace function

Given a function f: ℝ → ℝ, the associated trace function on H_n is given by

$A\mapsto {\rm Tr} f(A)=\sum_j f(\lambda_j),$

where A has eigenvalues λ and Tr stands for a trace of the operator.

Convexity and monotonicity of the trace function

Let f: ℝ → ℝ be continuous, and let n be any integer. Then, if $t\mapsto f(t)$ is monotone increasing, so is $A \mapsto {\rm Tr} f(A)$ on H_n.

Likewise, if $t \mapsto f(t)$ is convex, so is $A \mapsto {\rm Tr} f(A)$ 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 $-1\leq p \leq 0$, the function $f(t) = -t^p$ is operator monotone and operator concave.

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

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

$f(t) = \log(t)$ is operator concave and operator monotone, while

$f(t) = t \log(t)$ is operator convex.

The original proof of this theorem is due to K. Löwner who gave a necessary and sufficient condition for f to be operator monotone.^[5] An 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 f: ℝ → ℝ with derivative f ' , or for all positive-definite Hermitian n×n matrices A and B, and all differentiable convex functions f:(0,∞) → ℝ, the following inequality holds,

${\rm Tr}[f(A)- f(B)- (A - B)f'(B)] \geq 0~.$

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 logt, see below.

Proof

Let C = A − B so that, for 0 < t < 1,

$B + tC = (1 -t)B + tA$.

Define

$\phi(t) = {\rm Tr}[f(B + tC)]~.$

By convexity and monotonicity of trace functions, φ is convex, and so for all 0 < t < 1,

$\phi(1) - \phi(0) \geq \frac{\phi(t) - \phi(0)}{t},$

and, in fact, the right hand side is monotone decreasing in t. Taking the limit t→0 yields Klein's inequality.

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

Golden–Thompson inequality

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

For any matrices $A, B\in\mathbf{H}_n$,

${\rm Tr}\, e^{A+B}\leq {\rm Tr}\, e^A e^B.$

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

${\rm Tr} \, e^{\ln A -\ln B+\ln C}\leq \int_0^\infty dt\, {\rm Tr}\, A(B+t)^{-1}C(B+t)^{-1}.$

Peierls–Bogoliubov inequality

Let $R, F\in \mathbf{H}_n$ be such that Tr e^R = 1. Defining g = Tr Fe^R, we have

${\rm Tr}\, e^{F}e^R \geq {\rm Tr}\, e^{F+R}\geq e^g.$

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

Gibbs variational principle

Let $H$ be a self-adjoint operator such that $e^{-H}$ is trace class. Then for any $\gamma\geq 0$ with ${\rm Tr}\,\gamma=1,$

${\rm Tr}\, \gamma H+{\rm Tr}\, \gamma\ln\gamma\geq -\ln {\rm Tr}\, e^{-H},$

with equality if and only if $\gamma={\rm exp}(-H)/{\rm Tr}\, {\rm exp}(-H)$.

Lieb's concavity theorem

The following theorem was proved by E. H. Lieb in.^[9] It proves and generalizes a conjecture of E. P. Wigner, M. M. Yanase and F. J. 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 $m\times n$ matrices $K$, and all $q$ and $r$ such that $0 \leq q\leq 1$ and $0\leq r \leq 1$, with $q + r \leq 1$ the real valued map on $\mathbf{H}^+_m \times \mathbf{H}^+_n$ given by

$F(A,B,K) = {\rm Tr}(K^*A^qKB^r)$

• is jointly concave in $(A,B)$
• is convex in $K$.

Here $K^*$ stands for the adjoint operator of $K.$

Lieb's theorem

For a fixed Hermitian matrix $L\in\mathbf{H}_n$, the function

$f(A)={\rm Tr} \,\exp\{L+\ln A\}$

is concave on $\mathbf{H}_n^{++}$.

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 papers,^[14][15] for a review of this argument.

Ando's convexity theorem

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

For all $m \times n$ matrices $K$, and all $1 \leq q \leq 2$ and $0 \leq r \leq 1$ with $q-r \geq 1$, the real valued map on $\mathbf{H}^{++}_m \times \mathbf{H}^{++}_n$ given by

$(A,B) \mapsto {\rm Tr}(K^*A^qKB^{-r})$

is convex.

Joint convexity of relative entropy

For two operators $A, B\in\mathbf{H}^{++}_n$ define the following map

$R(A\|B):= {\rm Tr}(A\log A) - {\rm Tr}(A\log B).$

For density matrices $\rho$ and $\sigma$, the map $R(\rho\|\sigma)=S(\rho\|\sigma)$ is the Umegaki's quantum relative entropy.

Note that the non-negativity of $R(A\|B)$ follows from Klein's inequality with $f(x)=x\log x$.

Statement

The map $R(A\|B): \mathbf{H}^{++}_n \times \mathbf{H}^{++}_n \rightarrow \mathbf{R}$ is jointly convex.

Proof

For all $0 < p < 1$, $(A,B) \mapsto Tr(B^{1-p}A^p)$ is jointly concave, by Lieb's concavity theorem, and thus

$(A,B)\mapsto \frac{1}{p-1}({\rm Tr}(B^{1-p}A^p)-{\rm Tr}\, A)$

is convex. But

$\lim_{p\rightarrow 1}\frac{1}{p-1}({\rm Tr}(B^{1-p}A^p)-{\rm Tr}\, A)=R(A\|B),$

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 is due to C. Davis.^[17]

A continuous, real function $f$ on an interval $I$ satisfies Jensen's Operator Inequality if the following holds

$f\left(\sum_kA_k^*X_kA_k\right)\leq\sum_k A_k^*f(X_k)A_k,$

for operators $\{A_k\}_k$ with $\sum_k A^*_kA_k=1$ and for self-adjoint operators $\{X_k\}_k$ with spectrum on $I$.

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

Jensen's trace inequality

Let f be a continuous function defined on an interval I and let m and n be natural numbers. If f is convex, we then have the inequality

${\rm Tr}\Bigl(f\Bigl(\sum_{k=1}^nA_k^*X_kA_k\Bigr)\Bigr)\leq {\rm Tr}\Bigl(\sum_{k=1}^n A_k^*f(X_k)A_k\Bigr),$

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

$\sum_{k=1}^nA_k^*A_k=1$.

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 $f$ defined on an interval $I$ the following conditions are equivalent:

• $f$ is operator convex.
• For each natural number $n$ we have the inequality

$f\Bigl(\sum_{k=1}^nA_k^*X_kA_k\Bigr)\leq\sum_{k=1}^n A_k^*f(X_k)A_k,$

for all $(X_1, \ldots , X_n)$ bounded, self-adjoint operators on an arbitrary Hilbert space $\mathcal{H}$ with spectra contained in $I$ and all $(A_1, \ldots , A_n)$ on $\mathcal{H}$ with $\sum_{k=1}^n A^*_kA_k=1$.

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

every self-adjoint operator $X$ with spectrum in $I$.

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

Araki-Lieb-Thirring inequality

E. H. Lieb and W. E. Thirring proved the following inequality in ^[19] in 1976: For any $A\geq 0$, $B\geq 0$ and $r\geq 1,$

${\rm Tr} (B^{1/2}A^{1/2}B^{1/2})^r\leq {\rm Tr} B^{r/2}A^{r/2}B^{r/2}.$

In 1990 ^[20] H. Araki generalized the above inequality to the following one: For any $A\geq 0$, $B\geq 0$ and $q\geq 0,$

${\rm Tr}(B^{1/2}AB^{1/2})^{rq}\leq {\rm Tr}(B^{r/2}A^rB^{r/2})^q,$ for $r\geq 1,$

and

${\rm Tr}(B^{r/2}A^rB^{r/2})^q\leq {\rm Tr}(B^{1/2}AB^{1/2})^{rq},$ for $0\leq r\leq 1.$

Lieb-Thirring inequality also enjoys the following generalization:^[21] for any $A\geq 0$, $B\geq 0$ and $\alpha \in [0,1],$

${\rm Tr} (B^{1/2} A^{\alpha}B^{1/2} B^{1/2} A^{1-\alpha} B^{1/2}) \leq {\rm Tr} (BAB).$

Effros's theorem and its extension

E. Effros in ^[22] proved the following theorem.

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

$g(L, R):=f(LR^{-1})R$

is jointly convex, i.e. if $L=\lambda L_1+(1-\lambda)L_2$ and $R=\lambda R_1+(1-\lambda)R_2$ with $[L_i, R_i]=0$ (i=1,2), $0\leq\lambda\leq 1$,

$g(L,R)\leq \lambda g(L_1,R_1)+(1-\lambda)g(L_2,R_2).$

Ebadian et. al. later removed the restriction $[L,R]=LR-RL=0$ . ^[23]

Von Neumann's trace inequality

Von Neumann's trace inequality, named after its originator John von Neumann, states that for any n × n complex matrices A, B with singular values $\alpha_1 \ge \alpha_2 \ge \cdots \ge \alpha_n$ and $\beta_1 \ge \beta_2 \ge \cdots \ge \beta_n$ respectively,^[24]

$\left| {\rm Tr} (AB) \right| \le \sum_{i=1}^n \alpha_i \beta_i.$

The equality is achieved when $A$ and $B$ are simultaneously unitarily diagonalizable (see trace).

See also

• von Neumann entropy
• Lieb–Thirring inequality

References

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