# Trace inequality

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 ${\displaystyle \mathbf {H} _{n}}$ denote the space of Hermitian ${\displaystyle n\times n}$ matrices, ${\displaystyle \mathbf {H} _{n}^{+}}$ denote the set consisting of positive semi-definite ${\displaystyle n\times n}$ Hermitian matrices and ${\displaystyle \mathbf {H} _{n}^{++}}$ 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 ${\displaystyle f}$ on an interval ${\displaystyle I\subseteq \mathbb {R} ,}$ one may define a matrix function ${\displaystyle f(A)}$ for any operator ${\displaystyle A\in \mathbf {H} _{n}}$ with eigenvalues ${\displaystyle \lambda }$ in ${\displaystyle I}$ by defining it on the eigenvalues and corresponding projectors ${\displaystyle P}$ as

${\displaystyle f(A)\equiv \sum _{j}f(\lambda _{j})P_{j}~,}$
given the spectral decomposition ${\displaystyle A=\sum _{j}\lambda _{j}P_{j}.}$

### Operator monotone

A function ${\displaystyle f:I\to \mathbb {R} }$ defined on an interval ${\displaystyle I\subseteq \mathbb {R} }$ is said to be operator monotone if for all ${\displaystyle n,}$ and all ${\displaystyle A,B\in \mathbf {H} _{n}}$ with eigenvalues in ${\displaystyle I,}$ the following holds,

${\displaystyle A\geq B\implies f(A)\geq f(B),}$
where the inequality ${\displaystyle A\geq B}$ means that the operator ${\displaystyle A-B\geq 0}$ is positive semi-definite. One may check that ${\displaystyle f(A)=A^{2}}$ is, in fact, not operator monotone!

### Operator convex

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

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

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

### Joint convexity

A function ${\displaystyle g:I\times J\to \mathbb {R} ,}$ defined on intervals ${\displaystyle I,J\subseteq \mathbb {R} }$ is said to be jointly convex if for all ${\displaystyle n}$ and all ${\displaystyle A_{1},A_{2}\in \mathbf {H} _{n}}$ with eigenvalues in ${\displaystyle I}$ and all ${\displaystyle B_{1},B_{2}\in \mathbf {H} _{n}}$ with eigenvalues in ${\displaystyle J,}$ and any ${\displaystyle 0\leq \lambda \leq 1}$ the following holds

${\displaystyle g(\lambda A_{1}+(1-\lambda )A_{2},\lambda B_{1}+(1-\lambda )B_{2})~\leq ~\lambda g(A_{1},B_{1})+(1-\lambda )g(A_{2},B_{2}).}$

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

### Trace function

Given a function ${\displaystyle f:\mathbb {R} \to \mathbb {R} ,}$ the associated trace function on ${\displaystyle \mathbf {H} _{n}}$ is given by

${\displaystyle A\mapsto \operatorname {Tr} f(A)=\sum _{j}f(\lambda _{j}),}$
where ${\displaystyle A}$ has eigenvalues ${\displaystyle \lambda }$ and ${\displaystyle \operatorname {Tr} }$ stands for a trace of the operator.

## Convexity and monotonicity of the trace function

Let ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }$ be continuous, and let n be any integer. Then, if ${\displaystyle t\mapsto f(t)}$ is monotone increasing, so is ${\displaystyle A\mapsto \operatorname {Tr} f(A)}$ on Hn.

Likewise, if ${\displaystyle t\mapsto f(t)}$ is convex, so is ${\displaystyle A\mapsto \operatorname {Tr} f(A)}$ on Hn, and it is strictly convex if f is strictly convex.

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

## Löwner–Heinz theorem

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

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

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

${\displaystyle f(t)=\log(t)}$ is operator concave and operator monotone, while
${\displaystyle f(t)=t\log(t)}$ 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 ${\displaystyle f:\mathbb {R} \rightarrow \mathbb {R} }$ with derivative f ' , or for all positive-definite Hermitian n×n matrices A and B, and all differentiable convex functions f:(0,∞) → ${\displaystyle \mathbb {R} }$, the following inequality holds,

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

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

${\displaystyle B+tC=(1-t)B+tA}$,

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

Define

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

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

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

which is,

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

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

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

${\displaystyle F(1)-F(0)\geq F'(0)}$,

which with rearrangement and substitution is Klein's inequality:

${\displaystyle \mathrm {tr} [f(A)-f(B)-(A-B)f'(B)]\geq 0}$

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

## Golden–Thompson inequality

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

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

${\displaystyle \operatorname {Tr} e^{A+B}\leq \operatorname {Tr} e^{A}e^{B}.}$

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

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

## Peierls–Bogoliubov inequality

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

${\displaystyle \operatorname {Tr} e^{F}e^{R}\geq \operatorname {Tr} e^{F+R}\geq e^{g}.}$

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

## Gibbs variational principle

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

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

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

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

${\displaystyle F(A,B,K)=\operatorname {Tr} (K^{*}A^{q}KB^{r})}$
• is jointly concave in ${\displaystyle (A,B)}$
• is convex in ${\displaystyle K}$.

Here ${\displaystyle K^{*}}$ stands for the adjoint operator of ${\displaystyle K.}$

## Lieb's theorem

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

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

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

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

${\displaystyle (A,B)\mapsto \operatorname {Tr} (K^{*}A^{q}KB^{-r})}$

is convex.

## Joint convexity of relative entropy

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

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

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

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

### Statement

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

### Proof

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

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

is convex. But

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

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 is due to C. Davis.[17]

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

${\displaystyle f\left(\sum _{k}A_{k}^{*}X_{k}A_{k}\right)\leq \sum _{k}A_{k}^{*}f(X_{k})A_{k},}$

for operators ${\displaystyle \{A_{k}\}_{k}}$ with ${\displaystyle \sum _{k}A_{k}^{*}A_{k}=1}$ and for self-adjoint operators ${\displaystyle \{X_{k}\}_{k}}$ with spectrum on ${\displaystyle I}$.

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

${\displaystyle \operatorname {Tr} {\Bigl (}f{\Bigl (}\sum _{k=1}^{n}A_{k}^{*}X_{k}A_{k}{\Bigr )}{\Bigr )}\leq \operatorname {Tr} {\Bigl (}\sum _{k=1}^{n}A_{k}^{*}f(X_{k})A_{k}{\Bigr )},}$

for all (X1, ... , Xn) self-adjoint m × m matrices with spectra contained in I and all (A1, ... , An) of m × m matrices with

${\displaystyle \sum _{k=1}^{n}A_{k}^{*}A_{k}=1.}$

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 ${\displaystyle f}$ defined on an interval ${\displaystyle I}$ the following conditions are equivalent:

• ${\displaystyle f}$ is operator convex.
• For each natural number ${\displaystyle n}$ we have the inequality
${\displaystyle f{\Bigl (}\sum _{k=1}^{n}A_{k}^{*}X_{k}A_{k}{\Bigr )}\leq \sum _{k=1}^{n}A_{k}^{*}f(X_{k})A_{k},}$

for all ${\displaystyle (X_{1},\ldots ,X_{n})}$ bounded, self-adjoint operators on an arbitrary Hilbert space ${\displaystyle {\mathcal {H}}}$ with spectra contained in ${\displaystyle I}$ and all ${\displaystyle (A_{1},\ldots ,A_{n})}$ on ${\displaystyle {\mathcal {H}}}$ with ${\displaystyle \sum _{k=1}^{n}A_{k}^{*}A_{k}=1.}$

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

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

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

## Araki–Lieb–Thirring inequality

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

${\displaystyle \operatorname {Tr} ((BAB)^{r})~\leq ~\operatorname {Tr} (B^{r}A^{r}B^{r}).}$

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

${\displaystyle \operatorname {Tr} ((BAB)^{rq})~\leq ~\operatorname {Tr} ((B^{r}A^{r}B^{r})^{q}),}$
for ${\displaystyle r\geq 1,}$ and
${\displaystyle \operatorname {Tr} ((B^{r}A^{r}B^{r})^{q})~\leq ~\operatorname {Tr} ((BAB)^{rq}),}$
for ${\displaystyle 0\leq r\leq 1.}$

There are several other inequalities close to the Lieb–Thirring inequality, such as the following:[21] for any ${\displaystyle A\geq 0,}$ ${\displaystyle B\geq 0}$ and ${\displaystyle \alpha \in [0,1],}$

${\displaystyle \operatorname {Tr} (BA^{\alpha }BBA^{1-\alpha }B)~\leq ~\operatorname {Tr} (B^{2}AB^{2}),}$
and even more generally:[22] for any ${\displaystyle A\geq 0,}$ ${\displaystyle B\geq 0,}$ ${\displaystyle r\geq 1/2}$ and ${\displaystyle c\geq 0,}$
${\displaystyle \operatorname {Tr} ((BAB^{2c}AB)^{r})~\leq ~\operatorname {Tr} ((B^{c+1}A^{2}B^{c+1})^{r}).}$
The above inequality generalizes the previous one, as can be seen by exchanging ${\displaystyle A}$ by ${\displaystyle B^{2}}$ and ${\displaystyle B}$ by ${\displaystyle A^{(1-\alpha )/2}}$ with ${\displaystyle \alpha =2c/(2c+2)}$ and using the cyclicity of the trace, leading to
${\displaystyle \operatorname {Tr} ((BA^{\alpha }BBA^{1-\alpha }B)^{r})~\leq ~\operatorname {Tr} ((B^{2}AB^{2})^{r}).}$

Additionally, building upon the Lieb-Thirring inequality the following inequality was derived: [23] For any ${\displaystyle A,B\in \mathbf {H} _{n},T\in \mathbb {C} ^{n\times n}}$ and all ${\displaystyle 1\leq p,q\leq \infty }$ with ${\displaystyle 1/p+1/q=1}$, it holds that

${\displaystyle |\operatorname {Tr} (TAT^{*}B)|~\leq ~\operatorname {Tr} (T^{*}T|A|^{p})^{\frac {1}{p}}\operatorname {Tr} (TT^{*}|B|^{q})^{\frac {1}{q}}.}$

## Effros's theorem and its extension

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

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

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

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

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

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

## Von Neumann's trace inequality and related results

Von Neumann's trace inequality, named after its originator John von Neumann, states that for any ${\displaystyle n\times n}$ complex matrices ${\displaystyle A}$ and ${\displaystyle B}$ with singular values ${\displaystyle \alpha _{1}\geq \alpha _{2}\geq \cdots \geq \alpha _{n}}$ and ${\displaystyle \beta _{1}\geq \beta _{2}\geq \cdots \geq \beta _{n}}$ respectively,[26]

${\displaystyle |\operatorname {Tr} (AB)|~\leq ~\sum _{i=1}^{n}\alpha _{i}\beta _{i}\,,}$
with equality if and only if ${\displaystyle A}$ and ${\displaystyle B}$ share singular vectors.[27]

A simple corollary to this is the following result:[28] For Hermitian ${\displaystyle n\times n}$ positive semi-definite complex matrices ${\displaystyle A}$ and ${\displaystyle B}$ where now the eigenvalues are sorted decreasingly (${\displaystyle a_{1}\geq a_{2}\geq \cdots \geq a_{n}}$ and ${\displaystyle b_{1}\geq b_{2}\geq \cdots \geq b_{n},}$ respectively),

${\displaystyle \sum _{i=1}^{n}a_{i}b_{n-i+1}~\leq ~\operatorname {Tr} (AB)~\leq ~\sum _{i=1}^{n}a_{i}b_{i}\,.}$

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