An extension of the Golden-Thompson theorem

Abstract

In this paper, we shall prove $|tr{e}^{A+B}|\le tr(|e^A||e^B|)$ for normal matrices A, B. In particular, $tr{e}^{A+B}\le tr(e^A{e}^B)$ if A, B are Hermitian matrices, yielding the Golden-Thompson inequality.

MSC:15A16, 47A63, 15A45.

1 Introduction and preliminaries

The famous Golden-Thompson inequality [1–4] for Hermitian matrices A, B states that $tr{e}^{A+B}\le tr(e^A{e}^B)$. This inequality is a basic tool in quantum statistical mechanics and extensions to infinite dimension have an extensive literature [5, 6]. In this paper, we extend the classical Golden-Thompson theorem to normal matrices.

Throughout this paper, we adopt the following notation. Let $M_n$ be the set of all $n\times n$ complex matrices. For a matrix $A\in M_n$, as usual, its conjugate transpose is denoted by $A^{\ast }$. A matrix A is called Hermitian if $A=A^{\ast }$, normal if $A^{\ast }A=A{A}^{\ast }$, and unitary if $A^{\ast }A=A{A}^{\ast }=I_n$ ($I_n$ is the identity matrix of order n). Given a matrix $A\in M_n$, the eigenvalues and singular values of A are denoted by ${\lambda }_1(A),\dots ,{\lambda }_n(A)$, and $s_1(A),\dots ,s_n(A)$, respectively, where $|{\lambda }_1(A)|\ge \cdots \ge |{\lambda }_n(A)|$ and $s_1(A)\ge \cdots \ge s_n(A)$. In particular, when A is positive semidefinite ($A\ge 0$), then ${\lambda }_1(A)\ge \cdots \ge {\lambda }_n(A)\ge 0$. For simplicity, we denote $\lambda (A)\equiv ({\lambda }_1(A),\dots ,{\lambda }_n(A))$ and $s(A)\equiv (s_1(A),\dots ,s_n(A))$. Recall that the singular values of a matrix $A\in M_n$ are defined to be the eigenvalues of $|A|\equiv {(A^{\ast }A)}^{1/2}$, i.e., $s(A)=\lambda (|A|)$. Here $s_1(A)=\parallel A\parallel$ is the spectral norm of A. It is known that the spectral norm $\parallel \cdot \parallel$ over $M_n$ is unitarily invariant, i.e., $\parallel UAV\parallel =\parallel A\parallel$ for all unitary matrices U, V.

We now recall the concept of majorization (details can be found in [7–9]). We have the following basic majorant relations. For real vectors $x=(x_1,\dots ,x_n)$, $y=(y_1,\dots ,y_n)$ in coordinates in decreasing order, we say that x is weakly majorized by y, denoted by $x{\prec }_{w}y$, if

$$\sum _{j=1}^k{x}_j\le \sum _{j=1}^k{y}_j,k=1,\dots ,n,$$

and the weak log-majorant relation $x{\prec }_{\mathrm{wlog}}y$ means

$$\prod _{j=1}^k{x}_j\le \prod _{j=1}^k{y}_j,k=1,\dots ,n.$$

If in addition to $x{\prec }_{\mathrm{wlog}}y$, ${\prod }_1^n{x}_j={\prod }_1^n{y}_j$ holds, we say that x is log-majorized by y, denoted briefly by symbols $x{\prec }_{\log }y$. The following statement (see [8, 10]) is well known: $x{\prec }_{\mathrm{wlog}}y$ yields $x{\prec }_{w}y$ for vectors $x,y\in R_{+}^n$.

Remark 1.1 For $x=(x_1,\dots ,x_n)$, we denote $|x|=(|x_1|,\dots ,|x_n|)$. Weyl’s majorant theorem [10] says that $|\lambda (A)|{\prec }_{\log }s(A)$ for $A\in M_n$, that is,

$$(|{\lambda }_1(A)|,\dots ,|{\lambda }_n(A)|){\prec }_{\log }(|s_1(A)|,\dots ,|s_n(A)|).$$

The formula above implies that $|\lambda (A)|{\prec }_{w}s(A)$.

2 Lemmas

In this section, we shall propose some lemmas, laying the foundations of our main results in the next section.

Lemma 2.1 [11]

If A, B are positive semidefinite matrices, then

$${\parallel AB\parallel }^t\le \parallel A^t{B}^t\parallel ,\mathit{\text{and}}{\lambda }_1^t(AB)\le {\lambda }_1\left(A^t{B}^t\right),\mathit{\text{for }}t\ge 1.$$

Here note that $\parallel X\parallel =s_1(X)$ is the spectral norm of X.

Lemma 2.2 If $A,B\in M_n$ are normal matrices, then for any integer $m\ge 1$

$${\parallel AB\parallel }^m\le \parallel \left|A^m\right|\cdot {|B|}^m\parallel =\parallel A^m{B}^m\parallel .$$

Proof Take the polar decompositions $A=U|A|$ and $B=V|B|$. Here U, V are unitary matrices. Since A, B are normal, we can derive that $U|A|=|A|U$ and $V|B|=|B|V$ (see [12, 13]). Thus

$$AB=U|A||B|V,A^m{B}^m=U^m{|A|}^m{|B|}^m{V}^m.$$

Since the norm $\parallel \cdot \parallel$ is unitary invariant, we obtain the following:

$${\parallel AB\parallel }^m={\parallel U(|A|\cdot |B|)V\parallel }^m={\parallel (|A|\cdot |B|)\parallel }^m,$$

and

$$\parallel A^m{B}^m\parallel =\parallel U^m\cdot {|A|}^m\cdot {|B|}^m\cdot V^m\parallel =\parallel {|A|}^m\cdot {|B|}^m\parallel .$$

From Lemma 2.1, ${\parallel |A|\cdot |B|\parallel }^m\le \parallel ({|A|}^m\cdot {|B|}^m)\parallel$, we therefore conclude that

$${\parallel AB\parallel }^m\le \parallel A^m{B}^m\parallel .$$

 □

Lemma 2.3 If $A,B\in M_n$ are normal matrices, then

$${\parallel AB\parallel }^2=s_1^2(|A|\cdot |B|)={\lambda }_1({|A|}^2\cdot {|B|}^2).$$

Proof Since A and B are normal, it follows from Lemma 2.2 that $\parallel AB\parallel =\parallel |A|\cdot |B|\parallel$. So we get

$${\parallel AB\parallel }^2={\parallel (|A|\cdot |B|)\parallel }^2=s_1^2(|A|\cdot |B|)={\lambda }_1({(|A|\cdot |B|)}^{\ast }\cdot (|A|\cdot |B|)),$$

as desired. □

Lemma 2.4 If $A,B\in M_n$ are normal matrices, then for integers $m\ge 2$

$$s_1^m(AB)={\parallel AB\parallel }^m\le {\lambda }_1({|A|}^m\cdot {|B|}^m).$$

Proof By Lemma 2.3, we have ${\parallel AB\parallel }^2={\lambda }_1({|A|}^2\cdot {|B|}^2)$, and

$${\parallel AB\parallel }^m={\left({\parallel AB\parallel }^2\right)}^{m/2}={\lambda }_1^{m/2}\left({|A|}^2{|B|}^2\right).$$

Applying Lemma 2.1 to the right side above, we have the following:

$${\lambda }_1^{m/2}({|A|}^2\cdot {|B|}^2)\le {\lambda }_1({|A|}^m\cdot {|B|}^m).$$

Thus we get ${\parallel AB\parallel }^m\le {\lambda }_1({|A|}^m\cdot {|B|}^m)$ for integers $m\ge 2$, as desired. □

Here we note that $|A^m|={|A|}^m$ holds for any normal matrix.

The following lemma needs the notion of the Grassmann power ${\mathrm{\Lambda }}^{k}A$ (or antisymmetric tensor product), which can be found in [[8], p.18].

Lemma 2.5 If $A\in M_n$, $1\le k\le n$, then for any natural number m, the following holds:

$$\prod _{j=1}^k{s}_j\left(A^m\right)\le \prod _{j=1}^k{s}_j^m(A),\mathit{\text{and}}\prod _{j=1}^n{s}_j\left(A^m\right)=\prod _{j=1}^n{s}_j^m(A).$$

i.e.,

$$s\left(A^m\right){\prec }_{\log }{s}^m(A).$$

Proof For $1\le k\le n$, consider the k th antisymmetric tensor product ${\mathrm{\Lambda }}^{k}A$ of $A\in M_n$. It is known [[8], p.18] that ${\mathrm{\Lambda }}^k(A^m)={({\mathrm{\Lambda }}^{k}A)}^m$ and

$$s_1\left({\mathrm{\Lambda }}^k\left(A^m\right)\right)=s_1\left({\left({\mathrm{\Lambda }}^{k}A\right)}^m\right)=\parallel {\left({\mathrm{\Lambda }}^{k}A\right)}^m\parallel \le {\parallel {\mathrm{\Lambda }}^{k}A\parallel }^m={\left(s_1\left({\mathrm{\Lambda }}^{k}A\right)\right)}^m.$$

Thus

$$\prod _{j=1}^k{s}_j\left(A^m\right)\le \prod _{j=1}^k{s}_j^m(A),k=1,\dots ,n.$$

In particular,

$$\prod _{j=1}^n{s}_j\left(A^m\right)=\prod _{j=1}^n{s}_j^m(A)={|det(A)|}^m,$$

which, equivalently, says that $s(A^m){\prec }_{\log }{s}^m(A)$. This completes the proof. □

3 Main results

In this section, we shall present the main results of this paper.

Theorem 3.1 If $A,B\in M_n$ are normal matrices, then

$$s\left(e^{A+B}\right){\prec }_{\log }\lambda (\left|e^A\right|\cdot \left|e^B\right|).$$

Proof Let $A,B\in M_n$ be normal matrices. It is clear that ${\mathrm{\Lambda }}^k{e}^{A/m}$, ${\mathrm{\Lambda }}^k{e}^{B/m}$ are normal for $1\le k\le n$. By replacing A, B by ${\mathrm{\Lambda }}^k{e}^{A/m}$, ${\mathrm{\Lambda }}^k{e}^{B/m}$ in Lemma 2.4, respectively, we can obtain the following for integers $m\ge 2$:

$$\begin{array}{rl}{s}_1^m\left({\mathrm{\Lambda }}^k\left(e^{A/m}{e}^{B/m}\right)\right)& =s_1^m\left({\mathrm{\Lambda }}^k{e}^{A/m}{\mathrm{\Lambda }}^k{e}^{B/m}\right)\\ \le {\lambda }_1\left(\left|{\mathrm{\Lambda }}^k{e}^A\right|\left|{\mathrm{\Lambda }}^k{e}^B\right|\right)={\lambda }_1\left({\mathrm{\Lambda }}^k\left(\left|e^A\right|\left|e^B\right|\right)\right).\end{array}$$

Here we note that $|{\mathrm{\Lambda }}^{k}A|={\mathrm{\Lambda }}^k|A|$ because ${|{\mathrm{\Lambda }}^{k}A|}^2={\mathrm{\Lambda }}^k{|A|}^2$. So we obtain

$$\prod _{j=1}^k{s}_j^m\left(e^{A/m}{e}^{B/m}\right)\le \prod _{j=1}^k{\lambda }_j(\left|e^A\right|\cdot \left|e^B\right|).$$

From Lemma 2.5, we have

$$\prod _{j=1}^k{s}_j\left[{\left(e^{A/m}{e}^{B/m}\right)}^m\right]\le \prod _{j=1}^k{s}_j^m\left(e^{A/m}{e}^{B/m}\right).$$

Thus,

$$\prod _{j=1}^k{s}_j\left[{\left(e^{A/m}{e}^{B/m}\right)}^m\right]\le \prod _{j=1}^k{\lambda }_j\left(\left|e^A\right|\left|e^B\right|\right).$$

The Lie product formula [[8], p.254] says that for any matrices A, B

$$\underset{m\to \mathrm{\infty }}{lim}{\left(e^{A/m}{e}^{B/m}\right)}^m=e^{A+B}.$$

Thus taking $m\to \mathrm{\infty }$ in the inequality above yields

$$\prod _{j=1}^k{s}_j\left(e^{A+B}\right)\le \prod _{j=1}^k{\lambda }_j\left(\left|e^A\right|\left|e^B\right|\right).$$

Finally we note that

$$\prod _{j=1}^n{s}_j\left(e^{A+B}\right)=|det\left(e^{A+B}\right)|=|det\left(e^A{e}^B\right)|=\prod _{j=1}^n{\lambda }_j\left(\left|e^A\right|\left|e^B\right|\right).$$

Thus we get

$$s\left(e^{A+B}\right){\prec }_{\log }\lambda (\left|e^A\right|\cdot \left|e^B\right|).$$

This completes the proof. □

From Theorem 3.1, we know that

$$s\left(e^{A+B}\right){\prec }_{\log }\lambda (\left|e^A\right|\cdot \left|e^B\right|).$$

On the other hand, the following equation holds:

$$\lambda (\left|e^A\right|\cdot \left|e^B\right|)=\lambda ({\left|e^A\right|}^{1/2}\cdot \left|e^B\right|\cdot {\left|e^A\right|}^{1/2})=s({\left|e^A\right|}^{1/2}\cdot \left|e^B\right|\cdot {|e^A|}^{1/2}).$$

The above two inequalities yield the following:

$$s\left(e^{A+B}\right){\prec }_{\log }s({\left|e^A\right|}^{1/2}\cdot \left|e^B\right|\cdot {\left|e^A\right|}^{1/2}).$$

Thus, we can get the following corollary by using the Fan Dominance Principle [[10], p.56].

Corollary 3.2 If $A,B\in M_n$ are normal matrices, then

$$\parallel \left|e^{A+B}\right|\parallel \le \parallel |{\left|e^A\right|}^{1/2}\cdot \left|e^B\right|\cdot {\left|e^A\right|}^{1/2}|\parallel ,$$

for all unitarily invariant norms $\parallel |\cdot |\parallel$.

From Theorem 3.1, we can also have the following result.

Theorem 3.3 If $A,B\in M_n$ are normal matrices, then

$$\left|\lambda \left(e^{A+B}\right)\right|{\prec }_{\log }\lambda (\left|e^A\right|\cdot \left|e^B\right|).$$

Proof By Weyl’s majorant theorem we have $|\lambda (A)|{\prec }_{\log }s(A)$. Hence Theorem 3.1 implies the desired inequality in Theorem 3.3. □

Note that Theorem 3.3 strengthens the Golden-Thompson inequality:

$$|tr\left(e^{A+B}\right)|\le tr\left(e^A{e}^B\right)$$

for Hermitian matrices A, B.

Theorem 3.4 If A, B are normal matrices, then

$$|tr\left(e^{A+B}\right)|\le tr(\left|e^A\right|\cdot \left|e^B\right|).$$

Proof Because $x{\prec }_{\log }y$ implies $x{\prec }_{w}y$, it follows from Theorem 3.3 that

$$\left|\lambda \left(e^{A+B}\right)\right|{\prec }_w\lambda (\left|e^A\right|\cdot \left|e^B\right|).$$

Taking the traces above, we have

$$|tr\left(e^{A+B}\right)|\le tr(\left|e^A\right|\cdot \left|e^B\right|).$$

So we get the desired inequality. This completes the proof. □

Of course, Theorem 3.4 is an extension of Golden-Thompson inequality:

$$tr\left(e^{A+B}\right)\le tr\left(e^A{e}^B\right)$$

for Hermitian matrices A, B.