Trace inequality explained

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 x n matrices, H ^{+} _{n} denote the set consisting of positive semi-definite n x 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 \subseteq \Reals, one may define a matrix function f(A) for any operator A \in H _{n} with eigenvalues λ in I by defining it on the eigenvalues and corresponding projectors P as $$f(A)\; \backslash equiv\; \backslash sum\_j\; f(\backslash lambda\_j)P\_j\; ~,$$ given the spectral decomposition A = \sum _{j} λ _{j} P _{j.}

Operator monotone

See main article: Operator monotone function.

A function

f : I \to \Reals defined on an interval I \subseteq \Reals is said to be operator monotone if for all n, and all A, B \in H _{n} with eigenvalues in I, the following holds,$$A\; \backslash geq\; B\; \backslash implies\; f(A)\; \backslash geq\; f(B),$$where the inequality A \geq B means that the operator A - B \geq 0 is positive semi-definite. One may check that f(A) = A ^{2} is, in fact, not operator monotone!

Operator convex

A function

f : I \to \Reals is said to be operator convex if for all n and all A, B \in H _{n} with eigenvalues in I, and 0 < λ < 1 , the following holds$$f(\backslash lambda\; A\; +\; (1-\backslash lambda)B)\; \backslash leq\; \backslash lambda\; f(A)\; +\; (1\; -\backslash lambda)f(B).$$Note that the operator λ A + ( 1-λ )B has eigenvalues in I, since A and B have eigenvalues in I.

A function

f is if - f is operator convex;=, that is, the inequality above for f is reversed.

Joint convexity

A function

g : I x J \to \Reals, defined on intervals I, J \subseteq \Reals is said to be if for all n and allA _{1,} A _{2} \in H _{n} with eigenvalues in I and all B _{1,} B _{2} \in H _{n} with eigenvalues in J, and any 0 \leq λ \leq 1 the following holds$$g(\backslash lambda\; A\_1\; +\; (1-\backslash lambda)\; A\_2,\; \backslash lambda\; B\_1\; +\; (1-\backslash lambda)\; B\_2)\; ~\backslash leq~\; \backslash lambda\; g(A\_1,\; B\_1)\; +\; (1\; -\backslash lambda)\; g(A\_2,\; B\_2).$$

A function

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

Trace function

Given a function

f : \Reals \to \Reals, the associated trace function on H _{n} is given by$$A\; \backslash mapsto\; \backslash operatorname\; f(A)\; =\; \backslash sum\_j\; f(\backslash lambda\_j),$$where A has eigenvalues λ and \operatorname{Tr} stands for a trace of the operator.

Convexity and monotonicity of the trace function

Let

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

Likewise, if

t \mapsto f(t) is convex, so is A \mapsto \operatorname{Tr} f(A) on H_n, andit is strictly convex if is strictly convex.

See proof and discussion in, 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 to be operator monotone.^[5] An elementary proof of the theorem is discussed in and a more general version of it in.^[6]

Klein's inequality

For all Hermitian × matrices and and all differentiable convex functions

f: R \rarr R with derivative, or for all positive-definite Hermitian × matrices and, and all differentiableconvex functions :(0,∞) → R , the following inequality holds, In either case, if is strictly convex, equality holds if and only if = .A popular choice in applications is, see below.

Proof

Let

C = A - B so that, for t\in ( 0 , 1 ) , B + tC = ( 1 - t)B + tA ,varies from B to A .

Define

F(t) = \operatorname{Tr}[f(B + tC)] .By convexity and monotonicity of trace functions, F(t) is convex, and so for all t\in ( 0 , 1 ) ,F( 0 ) + t(F( 1 ) - F( 0 ))\geq F(t) ,which is,F( 1 ) - F( 0 ) \geq (F(t) - F( 0 ))/(t) ,and, in fact, the right hand side is monotone decreasing in t .

Taking the limit

t\to 0 yields,F( 1 ) - F( 0 ) \geq F'( 0 ) ,which with rearrangement and substitution is Klein's inequality:tr [f(A) - f(B) - (A - B)f'(B)] \geq 0

Note that if

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

Golden–Thompson inequality

See main article: Golden–Thompson inequality.

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

For any matrices

A, B\in H _{n} ,\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\in H ^{+} _{n} ,\operatorname{Tr} e ^{ln} \leq \int ^{inf ty} _{0} \operatorname{Tr} A(B + t) ^{-1} C(B + t) ^{-1} \operatorname{d}t.

Peierls–Bogoliubov inequality

Let

R, F\in H _{n} be such that Tr e^R = 1.Defining, we have \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 .^[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 \operatorname{Tr}\gamma =1 , \operatorname{Tr}\gamma H + \operatorname{Tr}\gamma ln \gamma\geq -ln \operatorname{Tr} e ^{-H} , with equality if and only if \gamma = \exp( - H)/\operatorname{Tr} \exp( - H).

Lieb's concavity theorem

The following theorem was proved by E. H. Lieb in. 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, and several more have been given since then.

For all

m x 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 H ^{+} _{m} x H ^{+} _{n} given byF(A,B,K) = \operatorname{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 H _{n} , the functionf(A) = \operatorname{Tr} \exp\{L +ln A\} is concave on H ^{++} _{n} .

The theorem and proof are due to E. H. Lieb, 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 of Lieb's concavity theorem led to the following significant complement to it:

For all

m x 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 H ^{++} _{m} x H ^{++} _{n} given by(A,B) \mapsto \operatorname{Tr}(K ^{*A} ^{qKB} ^{-r} ) is convex.

Joint convexity of relative entropy

For two operators

A, B\in H ^{++} _{n} define the following mapR(A\parallel B): = \operatorname{Tr}(A log A) - \operatorname{Tr}(A log B). \rho and \sigma , the map R(\rho\parallel\sigma) = S(\rho\parallel\sigma) is the Umegaki's quantum relative entropy.

Note that the non-negativity of

R(A\parallel B) follows from Klein's inequality with f(t) = t log t .

Statement

The map

R(A\parallel B): H ^{++} _{n} x H ^{++} _{n} → R is jointly convex.

Proof

For all

0 < p < 1 , (A,B) \mapsto \operatorname{Tr}(B ^{1-p} A ^{p)} is jointly concave, by Lieb's concavity theorem, and thus(A,B)\mapsto (1)/(p -1) (\operatorname{Tr}(B ^{1-p} A ^{p)-\operatorname{Tr}A)} is convex. But\lim _{p →} (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

f on an interval I satisfies Jensen's Operator Inequality if the following holdsf\left(\sum _{kA} ^{*X} ^{kA} _{k\right)\leq\sum} _{k} A ^{*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,^[18] for the proof of the following two theorems.

Jensen's trace inequality

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

\operatorname{Tr} l (f l (\sum ^{*X} _{kA} _{kr)r)\leq} \operatorname{Tr} l (\sum ^{n} _{k =1} A ^{*f(X} _{k)A} _{kr),} for all (₁, ..., _n) self-adjoint × matrices with spectra contained in andall (₁, ..., _n) of × matrices with \sum ^{*A} _{k =1 .}

Conversely, if the above inequality is satisfied for some and, where > 1, then 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 inequalityf l (\sum ^{*X} _{kA} _{kr)\leq\sum} ^{n} ^{k =1} A ^{*f(X} _{k)A} _{k,} for all (X _{1,} \ldots , X _{n)} bounded, self-adjoint operators on an arbitrary Hilbert space l{H} withspectra contained in I and all (A _{1,} \ldots , A _{n)} on l{H} with \sum ^{n} _{k =1} A ^{*} _{kA} _{k=1.} f(V ^{*XV)} \leq V ^{*f(X)V} for each isometry V on an infinite-dimensional Hilbert space l{H} andevery self-adjoint operator X with spectrum in I .Pf(PXP + λ ( 1 - P))P \leq Pf(X)P for each projection P on an infinite-dimensional Hilbert space l{H} , every self-adjoint operator X with spectrum in I and every λ in I .

Araki–Lieb–Thirring inequality

E. H. Lieb and W. E. Thirring proved the following inequality in ^[19] 1976: For any

A \geq 0 , B \geq 0 and r \geq 1 , $$\backslash operatorname\; ((BAB)^r)\; ~\backslash leq~\; \backslash operatorname\; (B^r\; A^r\; B^r).$$

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 , $$\backslash operatorname((BAB)^)\; ~\backslash leq~\; \backslash operatorname((B^r\; A^r\; B^r)^q),$$ for r \geq 1 , and$$\backslash operatorname((B^r\; A^r\; B^r)^q)\; ~\backslash leq~\; \backslash operatorname((BAB)^),$$ for 0 \leq r \leq 1 .

There are several other inequalities close to the Lieb–Thirring inequality, such as the following:^[21] for any

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

Additionally, building upon the Lieb-Thirring inequality the following inequality was derived: ^[23] For any

A,B\in H _{n,} T\in C ^{n x} and all 1 \leq p,q\leq inf ty with 1 /p +1 /q = 1 , it holds that$$|\backslash operatorname(TAT^*B)|\; ~\backslash leq~\; \backslash operatorname(T^*T|A|^p)^\backslash frac\backslash operatorname(TT^*|B|^q)^\backslash frac.$$

Effros's theorem and its extension

E. Effros in ^[24] 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 perspectiveg(L, R): = f(LR ^{-1} )R is jointly convex, i.e. if L =λ L _{1+(1-λ)L} _{2} and R =λ R _{1+(1-λ)R} _{2} with [L _{i,} R _{i]=0} (i=1,2), 0 \leq λ \leq 1 ,g(L,R)\leq λ g(L _{1,R} _{1)+(1-λ)g(L} _{2,R} _{2).}

Ebadian et al. later extended the inequality to the case where

L and R do not commute .^[25]

Von Neumann's trace inequality and related results

, named after its originator John von Neumann, states that for any

n x n complex matrices A and B with singular values \alpha _{1} \geq \alpha _{2} \geq … \geq \alpha _{n} and \beta _{1} \geq \beta _{2} \geq … \geq \beta _{n} respectively,^[26] $$|\backslash operatorname(A\; B)|\; ~\backslash leq~\; \backslash sum\_^n\; \backslash alpha\_i\; \backslash beta\_i\backslash ,,$$with equality if and only if A and B ^{\dagger} share singular vectors.^[27] n x n positive semi-definite complex matrices A and B where now the eigenvalues are sorted decreasingly (a _{1} \geq a _{2} \geq … \geq a _{n} and b _{1} \geq b _{2} \geq … \geq b _{n,} respectively),$$\backslash sum\_^n\; a\_i\; b\_\; ~\backslash leq~\; \backslash operatorname(A\; B)\; ~\backslash leq~\; \backslash sum\_^n\; a\_i\; b\_i\backslash ,.$$

References

• Scholarpedia primary source.

Notes and References

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