Submultiplicativity of the numerical radius of commuting matrices of order two

TL;DR

This paper provides an elementary proof that the numerical radius of the product of two commuting 2x2 matrices is at most the product of their numerical radii, including a characterization of cases where equality holds.

Contribution

It offers a simple proof of submultiplicativity of the numerical radius for commuting 2x2 matrices and characterizes the cases of equality.

Findings

Proof of $w(AB) \,\leq\, w(A)w(B)$ for commuting 2x2 matrices

Characterization of matrix pairs attaining equality

Elementary approach to a known inequality

Abstract

Denote by $w(T)$ the numerical radius of a matrix $T$. An elementary proof is given to the fact that $w(AB) \leq w(A)w(B)$ for a pair of commuting matrices of order two, and characterization is given for the matrix pairs that attain the quality.

Equations

$$W(A)=\{x^{\ast }Ax:x\in {\mathbb{C}}^n,x^{\ast }x=1\}andw(A)=\max \{|\mu |:\mu \in W(A)\},$$

$$w(A)\le \Vert A\Vert \le 2w(A).$$

$$w(AB)\le \xi w(A)w(B)for\; allA,B\in M_n$$

$$w(A^m)\le w(A{)}^{m}for\; all\; positive\; integersm.$$

$$w(AB)\le \Vert AB\Vert \le \Vert A\Vert \Vert B\Vert =w(A)\Vert B\Vert \le 2w(A)w(B),$$

$$w(BA)\le \Vert BA\Vert \le \Vert B\Vert \Vert A\Vert =\Vert B\Vert w(A)\le 2w(B)w(A).$$

$$w(AB)\le \Vert AB\Vert \le \Vert A\Vert \Vert B\Vert =w(A)w(B).$$

$$w(AB)\le 2w(A)w(B).$$

$$4w(AB)$$

$$w(AB)\le \xi w(A)w(B)$$

$$\{\cos \theta +ir\sin \theta :\theta \in [0,2\pi ]\};$$

$$\{{\alpha }_1+|s_1|\cos \theta +i({\alpha }_2+|s_1|r\sin \theta ):\theta \in [0,2\pi ]\}.$$

$$\cos {\varphi }_1+i\sin {\varphi }_1=({\alpha }_1+|s_1|\cos {\theta }_1)+i({\alpha }_2+|s_1|r\sin {\theta }_1).$$

$$x+iy=({\alpha }_1+|s_1|\cos \theta )+i({\alpha }_2+|s_1|r\sin \theta ),$$

$$(\cos {\theta }_1,\sin {\theta }_1)=\frac{(\cos {\varphi }_1,r\sin {\varphi }_1)}{\sqrt{{\cos }^2{\varphi }_1+r^2{\sin }^2{\varphi }_1}}.$$

$${\alpha }_1=\cos {\varphi }_1-\frac{|s_1|\cos {\varphi }_1}{\sqrt{{\cos }^2{\varphi }_1+r^2{\sin }^2{\varphi }_1}}\ge 0and{\alpha }_2=\sin {\varphi }_1-\frac{|s_1|r^2\sin {\varphi }_1}{\sqrt{{\cos }^2{\varphi }_1+r^2{\sin }^2{\varphi }_1}}.$$

$$x+iy=|s_1|\cos \theta +i(\sin {\varphi }_1(1-|s_1|r)+|s_1|r\sin \theta ).$$

$$0$$

$$A_0=e^{i{\varphi }_1}Iand{A}_1=i(1-r^2)\sin {\varphi }_{1}I+{\nu }_1{\hat{s}}_{1}C,$$

$$({\hat{s}}_1\cos \theta {)}^2+((1-r^2)\sin {\varphi }_1+{\hat{s}}_{1}r\sin \theta {)}^2$$

$$B_0=e^{i{\varphi }_2}Iand{B}_1=i(1-r^2)\sin {\varphi }_{2}I+{\nu }_2{\hat{s}}_{2}C$$

$$w(A_1{B}_1)<1.$$

$${\hat{s}}_i=\sqrt{1-(1-r^2)w_i^2}\text{mbox}fori=1,2.$$

$$A_1{B}_1=(1-r^2)(u{I}_2+ivC),$$

$$u={\nu }_1{\hat{s}}_1{\hat{s}}_2-w_1{w}_2(1-r^2)andv=w_1{\hat{s}}_2+{\nu }_1{w}_2{\hat{s}}_1.$$

$$\frac{1}{1-r^2}w(A_1{B}_1)=w(uI+ivC)<\frac{1}{(1-r^2)}.$$

$$f(\theta )=|u+iv(\cos \theta +ir\sin \theta ){|}^2<\frac{1}{(1-r^2{)}^2}for\; all\theta \in [0,2\pi ].$$

$$f(\theta )$$

Full text

Dedicated to Professor Pei Yuan Wu.

AMS Classification. 47A12, 15A60.

Keywords. Numerical radius, submultiplicative.

1 Introduction

Let $M_{n}$ be the set of $n\times n$ matrices. The numerical range and numerical radius of $A\in M_{n}$ are defined by

[TABLE]

respectively. The numerical range and numerical radius are useful tools in studying matrices and operators. There are strong connection between the algebraic properties of a matrix $A$ and the geometric properties of $W(A)$. For example, $W(A)=\{\mu I\}$ if and only if $A=\mu I$; $W(A)\subseteq{\mathbb{R}}$ if and only if $A=A^{*}$; $W(A)\subseteq[0,\infty)$ if and only if $A$ is positive semi-definite.

The numerical radius is a norm on $M_{n}$, and has been used in the analysis of basic iterative solution methods [2]. Researchers have obtained interesting inequalities related to the numerical radius; for example, see [4, 5, 6, 7, 8]. We mention a few of them in the following. Let $\|A\|$ be the operator norm of $A$. It is known that

[TABLE]

While the spectral norm is submultiplicative, i.e., $\|AB\|\leq\|A\|\|B\|$ for all $A,B\in M_{n}$, the numerical radius is not. In general,

[TABLE]

if and only if $\xi\geq 4$; e.g., see [3]. Despite the fact that the numerical radius is not submultiplicative,

[TABLE]

For a normal matrix $A\in M_{n}$, we have $w(A)=\|A\|$. Thus, for a normal matrix $A$ and any $B\in M_{n}$,

[TABLE]

and also

[TABLE]

In case $A,B\in M_{n}$ are normal matrices,

[TABLE]

Also, for any pairs of commuting matrices $A,B\in M_{n}$,

[TABLE]

To see this, we may assume $w(A)=w(B)=1$, and observe that

[TABLE]

The constant 2 is best (smallest) possible for matrices of order at least 4 because $w(AB)=2w(A)w(B)$ if $A=E_{12}+E_{34}$ and $B=E_{13}+E_{24}$, where $E_{ij}\in M_{n}$ has $1$ at the $(i,j)$ position and [math] elsewhere; see [3, Theorem 3.1].

In connection to the above discussion, there has been interest in studying the best (smallest) constant $\xi>0$ such that

[TABLE]

for all commuting matrices $A,B\in M_{n}$ with $n\leq 3$. For $n=2$, the best constant $\xi$ is one; the existing proofs of the $2\times 2$ case depend on deep theory on analytic functions, von Neumann inequality, and functional calculus on operators with numerical radius equal to one, etc.; for example, see [6, 7].

Researchers have been trying to find an elementary proof for this result in view of the fact that the numerical range of $A\in M_{2}$ is well understood, namely, $W(A)$is an elliptical disk with the eigenvalues $\lambda_{1},\lambda_{2}$ as foci and the length of minor axis $\sqrt{({\mathrm{t}r}\,A^{*}A)-|\lambda_{1}|^{2}-|\lambda_{2}|^{2}}$; for example, see [10, 11] and [8, Theorem 1.3.6].

The purpose of this note is to provide such a proof. Our analysis is based on elementary theory in convex analysis, co-ordinate geometry, and inequalities. Using our approach, we readily give a characterization of commuting pairs of matrices $A,B\in M_{2}$ satisfying $w(AB)=w(A)w(B)$, which was done in [3, Theorem 4.1] using yet another deep result of Ando [1] that a matrix $A$ has numerical radius bounded by one if and only if $A=(I-Z)^{1/2}C(A+Z)^{1/2}$ for some contractions $C$ and $Z$, where $Z=Z^{*}$. Here is our main result.

Theorem 1

Let $A,B\in M_{2}$ be nonzero matrices such that $AB=BA$. Then $w(AB)\leq w(A)w(B)$. The equality holds if and only if one of the following holds.

• (a)

$A$* or $B$ is a scalar matrix, i.e. of the form $\mu I_{2}$ for some $\mu\in{\mathbb{C}}$.*

• (b)

There is a unitary $U$ such that $U^{*}AU={\rm diag}\,(a_{1},a_{2})$ and $U^{*}BU={\rm diag}\,(b_{1},b_{2})$ with $|a_{1}|\geq|a_{2}|$ and $|b_{1}|\geq|b_{2}|$.

One can associate the conditions (a) and (b) in the theorem with the geometry of the numerical range of $A$ and $B$ as follows. Condition (a) means that $W(A)$ or $W(B)$ is a single point; condition (b) means that $W(A)$, $W(B)$, $W(AB)$ are line segments with three sets of end points, $\{a_{1},a_{2}\},\{b_{1},b_{2}\},\{a_{1}b_{1},a_{2}b_{2}\}$, respectively, such that $|a_{1}|\geq|a_{2}|$ and $|b_{1}|\geq|b_{2}|$.

2 Proof of Theorem 1

Let $A,B\in M_{2}$ be commuting matrices. We may replace $(A,B)$ by $(A/w(A),B/w(B))$ and assume that $w(A)=w(B)=1$. We need to show that $w(AB)\leq 1$.

Since $AB=BA$, there is a unitary matrix $U\in M_{2}$ such that both $U^{*}AU$ and$U^{*}BU$ are in triangular form; for example, see [9, Theorem 2.3.3]. We may replace $(A,B)$ by $(U^{*}AU,U^{*}BU)$ and assume that $A=\begin{pmatrix}a_{1}&a_{3}\cr 0&a_{2}\cr\end{pmatrix}$, $B=\begin{pmatrix}b_{1}&b_{3}\cr 0&b_{2}\cr\end{pmatrix}$ and $w(A)=w(B)=1$. The result is clear if $A$ or $B$ is normal. So, we assume that $a_{3},b_{3}\neq 0$. Furthermore, comparing the $(1,2)$ entries on both sides of $AB=BA$, we see that $\displaystyle\frac{a_{1}-a_{2}}{a_{3}}=\displaystyle\frac{b_{1}-b_{2}}{b_{3}}$. Applying a diagonal unitary similarity to both $A$ and $B$, we may further assume that $\gamma=\displaystyle\frac{a_{1}-a_{2}}{a_{3}}\geq 0$. Let $r=\displaystyle\frac{1}{\sqrt{\gamma^{2}+1}}$. We have $0<r\leq 1$. Then $A=z_{1}I+s_{1}C$ and $B=z_{2}I+s_{2}C$ with

$z_{1}=\displaystyle\frac{a_{1}+a_{2}}{2},\quad z_{2}=\displaystyle\frac{b_{1}+b_{2}}{2},\quad s_{1}=\displaystyle\frac{a_{3}}{2r},\quad s_{2}=\displaystyle\frac{b_{3}}{2r}$,  and  $C=\begin{pmatrix}\sqrt{1-r^{2}}&2r\cr 0&-\sqrt{1-r^{2}}\end{pmatrix}$.

Note that $W(C)$ is the elliptical disk with boundary

[TABLE]

see [10] and [8, Theorem 1.3.6]. Replacing $(A,B)$ with $(e^{it_{1}}A,e^{it_{2}}B)$ for suitable $t_{1},t_{2}\in[0,2\pi]$, if necessary, we may assume that ${\rm Re}\,z_{1},\ {\rm Re}\,z_{2}\geq 0$ and $s_{1},s_{2}$ are real.

Suppose $z_{1}=\alpha_{1}+i\alpha_{2}$ with $\alpha_{1}\geq 0$ and the boundary of $W(A)$ touches the unit circle at the point $\cos\phi_{1}+i\sin\phi_{1}$ with $\phi_{1}\in[-\pi/2,\pi/2]$. Then $W(A)$ has boundary

[TABLE]

We claim that the matrix $A$ is a convex combination of $A_{0}=e^{i\phi_{1}}I$ and another matrix $A_{1}$ of the form $A_{1}=i(1-r^{2})\sin\phi_{1}I+\xi C$ for some $\xi\in{\mathbb{R}}$ such that $w(A_{1})\leq 1$.

To prove our claim, we first determine $\theta_{1}\in[-\pi/2,\pi/2]$ satisfying

[TABLE]

Since the boundary of $W(A)$ touches the unit circle at the point $\cos\phi_{1}+i\sin\phi_{1}$, using the parametric equation

[TABLE]

of the boundary of $W(A)$, we see that the direction of the tangent at the intersection point $\cos\phi_{1}+i\sin\phi_{1}$ is $-\sin\theta_{1}+ir\cos\theta_{1}$, which agrees with $-\sin\phi_{1}+i\cos\phi_{1}$, the direction of the tangent line of the unit circle at the same point. As a result, we have

[TABLE]

Furthermore, since $\cos\phi_{1}+i\sin\phi_{1}=(\alpha_{1}+|s_{1}|\cos\theta_{1})+i(\alpha_{2}+|s_{1}|r\sin\theta_{1})$, we have

[TABLE]

**Assertion. **If $\hat{s}_{1}=\sqrt{\cos^{2}\phi_{1}+r^{2}\sin^{2}\phi_{1}}$, then $|s_{1}|\leq\hat{s}_{1}$.

If $\cos\phi_{1}>0$, then $\alpha_{1}=\left(1-\displaystyle\frac{|s_{1}|}{\hat{s}_{1}}\right)\cos\phi_{1}\geq 0$, and hence $|s_{1}|\leq\hat{s}_{1}$.

If $\cos\phi_{1}=0$, then $\sin\phi_{1}=\pm 1$, $\hat{s}_{1}=r$ and $(\alpha_{1},\alpha_{2})=(0,\sin\phi_{1}(1-|s_{1}|r))$ so that the parametric equation of the boundary of $W(A)$ in (1) becomes

[TABLE]

Since $w(A)=1$ and $\sin\phi_{1}=\pm 1$, for all $\theta\in[0,2\pi)$ , we have

[TABLE]

Therefore, $(r-|s_{1}|)\geq 0$, which gives $|s_{1}|\leq r=\hat{s}_{1}$.

Now, we show that our claim holds with

[TABLE]

where $\nu_{1}=1$ if $s_{1}\geq 0$ and $\nu_{1}=-1$ if $s_{1}<0$.

Note that $W(A_{1})$ is the elliptical disk with boundary $\{\hat{s}_{1}\cos\theta+i[(1-r^{2})\sin\phi_{1}+\hat{s}_{1}r\sin\theta):\theta\in[0,2\pi)\}$, and for every $\theta\in[0,2\pi]$, we have

[TABLE]

Therefore, $w(A_{1})\leq 1$. By the Assertion, $|s_{1}|\leq\hat{s}_{1}$. Hence $A=\left(1-\displaystyle\frac{|s_{1}|}{\hat{s}_{1}}\right)A_{0}+\displaystyle\frac{|s_{1}|}{\hat{s}_{1}}A_{1}$ is a convex combination of $A_{0}$ and $A_{1}$.

Similarly, if $W(B)$ touches the unit circle at $e^{i\phi_{2}}$ with $\phi_{2}\in[-\pi/2,\pi/2]$, then $B$ is a convex combination of

[TABLE]

with $\hat{s}_{2}=\sqrt{\cos^{2}\phi_{2}+r^{2}\sin^{2}\phi_{2}}$ and $\nu_{2}\in\{1,-1\}$. Let $U=\begin{pmatrix}-r&\sqrt{1-r^{2}}\cr\sqrt{1-r^{2}}&r\end{pmatrix}$. Then $U^{*}CU=-C$. If $\nu_{2}=-1$, we may replace $(A,B)$ by $(U^{*}AU,U^{*}BU)$ so that $(\nu_{1},\nu_{2})$ will change to $(-\nu_{1},-\nu_{2})$. So, we may further assume that $\nu_{2}=1$.

By the above analysis, $AB$ is a convex combination of $A_{0}B_{0},A_{0}B_{1},A_{1}B_{0}$ and $A_{1}B_{1}$. Since $w(e^{it}T)=w(T)$ for all $t\in{\mathbb{R}}$ and $T\in M_{n}$, the first three matrices have numerical radius 1. We will prove that

[TABLE]

It will then follow that $w(AB)\leq 1$, where the equality holds only when $A=A_{0}$ or $B=B_{0}$.

For simplicity of notation, let $w_{1}=\sin\phi_{1}$ and $w_{2}=\sin\phi_{2}$. Then

[TABLE]

Recall from (2) and (3) that $A_{1}=i(1-r^{2})w_{1}I+\nu_{1}\hat{s}_{1}C$ and $B_{1}=i(1-r^{2})w_{2}I+\hat{s}_{2}C$ because $\nu_{2}=1$. Since $C^{2}=(1-r^{2})I_{2}$, we have

[TABLE]

where

[TABLE]

If $r=1$, then $A_{1}B_{1}=0$. Assume that $0<r<1$. We need to show that

[TABLE]

Because $W(uI+ivC)$ is an elliptical disk with boundary $\{u+iv(\cos\theta+ir\sin\theta):\theta\in[0,2\pi]\}$, it suffices to show that

[TABLE]

Note that

[TABLE]

Consequently, we have $w(A_{1}B_{1})<1$ as asserted in (4). Moreover, by the comment after (4), if $w(AB)=w(A)w(B)$, then $A=A_{0}$ or $B=B_{0}$. Conversely, if $A=A_{0}$ or $B_{0}$, then we clearly have $W(AB)=w(A)w(B)$. The proof of the theorem is complete. $\Box$

Acknowledgment

We would like to thank Professor Pei Yuan Wu, Professor Hwa-Long Gau, and the referee for some helpful comments. Li is an affiliate member of the Institute for Quantum Computing, University of Waterloo, and is an honorary professor of the Shanghai University. His research was supported by USA NSF grant DMS 1331021, Simons Foundation Grant 351047, and NNSF of China Grant 11571220.