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

Main article: Operator monotone function

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 f: ℝ → 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 H_n.

Likewise, if ${\displaystyle t\mapsto f(t)}$  is convex, so is ${\displaystyle A\mapsto \operatorname {Tr} f(A)}$  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 ${\displaystyle -1\leq p\leq 0}$ , 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 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,

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

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

Main article: 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 e^R = 1. Defining g = Tr Fe^R, 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 ${\displaystyle 0<p<1}$ , ${\displaystyle (A,B)\mapsto \operatorname {Tr} (B^{1-p}A^{p})}$  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, ... , X_n) self-adjoint m × m matrices with spectra contained in I and all (A₁, ... , A_n) 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

Not to be confused with the 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),

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

 

See also

• Lieb–Thirring inequality
• Schur–Horn theorem – Characterizes the diagonal of a Hermitian matrix with given eigenvalues
• Trace identity – Equations involving the trace of a matrix
• von Neumann entropy – Type of entropy in quantum theory

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