Characterizations of the Reflection Operators

Abstract

An operator P is said to be reflective if $P^*=P$ and $P^2 = I$. In this paper, we study the spectral properties of reflection operators and obtain a matrix representation of the reflection operator pair (P, Q). Some related properties of reflection operator pair (P, Q) are given.

1 Introduction and Preliminaries

Let ${\mathcal {H}}$ and ${\mathcal {K}}$ be separable, infinite dimensional, complex Hilbert spaces. We denote the set of all bounded linear operators from ${\mathcal {H}}$ into ${\mathcal {K}}$ by ${\mathcal {B(H, K)}}$ and by ${\mathcal {B(H)}}$ when ${\mathcal {H}}={\mathcal {K}}$. The set of all the unitary operators on ${\mathcal {H}}$ is denoted by ${\mathcal {U}}({\mathcal {H}}).$ For $A\in {\mathcal {B(H, K)}}$, let $A^*$, $\sigma _P(A)$, $\sigma (A)$, ${{\mathcal {R}}}(A)$ and ${{\mathcal {N}}}(A)$ be the adjoint, the point spectrum, the spectrum, the range and the null space of A, respectively. An operator $A\in {\mathcal {B(H, K)}}$ is densely defined if the domain of A is a dense subset of ${\mathcal {H}}$ and the range of A is contained within ${\mathcal {K}}$. A is said to be positive if $(Ax,x) \ge 0$ for all $x\in {\mathcal {H}}.$$I_{{\mathcal {M}}}$ denotes the identity onto ${\mathcal {M}}$ or I if there is no confusion. An operator $P\in \mathcal {B(H)}$ is said to be a reflection operator if $P^*=P$ and $P^2 = I$. Let P and Q be two reflection operators. Throughout this paper, we assume that neither of P, Q is I or $-I$. The term “subspace” always means a closed linear manifold.

We say $P\in {\mathcal {B}}({\mathcal {H}})$ is an orthogonal projection if $P^2=P=P^*$ and there are a plenty of researches about two orthogonal projections [1,2,3,4,5, 12]. The most primitive representation of two orthogonal projections is usually referred to as Halmos’ two projections theorem and sometimes also as the CS decomposition [9, 10].

The reflection operators have much similarities with the orthogonal projections [6]. Furthermore, the reflection operators have some special properties different from the orthogonal projections, which attracts our attention. In [13], the authors establish an explicit characterization of the spectrum and spectral radius estimates for the reflection operator acting on $L_p$ spaces on an infinite angle in two dimensions. The aim of this paper is to study the spectral properties and the matrix representations of the reflection operators on Hilbert spaces. Some related properties of reflection operator pair (P, Q) are also given.

As we know, if there exists a $U\in {\mathcal {U}}({\mathcal {H}})$ such that $PU=UQ$$(UP=QU),$ we say U is an inner (outer) intertwining operator of P respect to Q and all the inner (outer) intertwining operators of P respect to Q are denoted by $inn_{Q}(P)$$(out_{Q}(P)),$ respectively. If there exists a $U\in {\mathcal {U}}({\mathcal {H}})$ such that $PU=UQ$ and $UP=QU$, we say U is an intertwining operator of P respect to Q and P, Q are unitary equivalence. The set of all the intertwining operators of P respect to Q is denoted by $int_{Q}(P)$ [4, 8]. The related researches on unitary equivalence of two orthogonal projections can also be found in [3, 5, 11,12,13,14,15,16,17]. Based on the matrix representation of reflection operators, we study the unitary equivalence of two reflection operators. The general explicit descriptions for intertwining operators of two reflection operators are established. The paper mainly contains two parts. In Sect. 2, we investigate some spectral properties of reflection operators. In Sect. 3, we present the block operator matrix representations of the reflective operators and obtain some properties of the combinations of reflection operators by using their matrix representations.

2 The Spectral Properties of Reflection Operators

First, we characterize the spectrum of reflection operators. If $P^{2}=I$, $\sigma (P)\subseteq \{e^{\frac{2k\pi }{2}i}: k=0, 1\}=\{1, -1\}$ by the spectral mapping theorem [7]. Observing that if $\lambda \in \sigma (P)$, then $\lambda ^{2}=1$. This shows that each $\lambda \in \sigma (P)$ is a simple root of the equation $\lambda ^{2}=1$.

Theorem 2.1

Let $P\in \mathcal {B(H)}$. Then P is a reflection operator ($P^*=P$ and $P^2=I$) if and only if $P=I_{\mathcal {M}}\oplus -I_{{\mathcal {M}}^\perp }$, where ${\mathcal {M}}={\mathcal {R}}(I+P).$

Theorem 2.2

Let P and Q be the reflection operators.

• (i) If $\lambda \in {\mathbb {C}}\backslash \{0, 2, -2\}$, then $\lambda \in \sigma (P-Q)\Longleftrightarrow 3-\lambda ^2 \in \sigma (P+Q+PQ).$

• (ii) $\sigma [2(P-Q)^2]\cup \{0\}=\sigma [(I-P)(I+Q)(I-P)]\cup \sigma [(I+P)(I-Q)(I+P)]\cup \{0\}.$

Proof

(i) If $\lambda \in {\mathbb {C}}\backslash \{0, 2, -2\}$, then $\lambda \pm 1\ne 1$ or $-1$. By Theorem 2.1, $(\lambda -1)I+P$ and $(\lambda +1)I-Q$ are invertible. Note that

$$\begin{aligned}&[(\lambda -1)I+P][\lambda I-(P-Q)][(\lambda +1)I-Q]\\&\quad =[(\lambda ^2-\lambda -1)I+P(I+Q)+(\lambda -1)Q][(\lambda +1)I-Q]\\&\quad =\lambda [(\lambda ^2-3)I+P+Q+PQ]. \end{aligned}$$

We get that $\lambda I-(P-Q)$ is invertible if and only if $(\lambda ^2-3)I+P+Q+PQ$ is invertible. Hence, the result holds.

(ii) Since $2(P-Q)^2=(I-P)(I+Q)(I-P)+(I+P)(I-Q)(I+P)$, we get that

$$\begin{aligned}&\lambda [\lambda I-2(P-Q)^2]\\&\quad = \lambda ^2 I-2\lambda (P-Q)^2\\&\quad = \lambda ^2 I-\lambda (I-P)(I+Q)(I-P) -\lambda (I+P)(I-Q)(I+P)\\&\quad =[\lambda I-(I-P)(I+Q)(I-P)][\lambda I-(I+P)(I-Q)(I+P)]. \end{aligned}$$

Hence, the result holds. $\square$

The following results are concerned with the commutator of two reflection operators.

Theorem 2.3

Let P and Q be the reflection operators.

• (i) There exists $\lambda \in {\mathbb {C}}$ such that $\lambda \in \sigma (PQ+QP)$ if and only if there exists $\mu \in \sigma (P-Q)$ such that $\lambda =2-\mu ^2.$

• (ii) There exists $\lambda \in {\mathbb {C}}$ such that $\lambda \in \sigma (PQ-QP)$ if and only if there exists $\mu \in \sigma (P-Q)$ such that $\lambda ^2 =\mu ^4-4\mu ^2.$

Proof

Note that

$$\begin{aligned} PQ+QP=2I-(P-Q)^2,\quad (PQ-QP)^2=(P-Q)^4-4(P-Q)^2. \end{aligned}$$

Then there exists $\lambda \in {\mathbb {C}}$ such that $\lambda \in \sigma (PQ+QP)$ if and only if $\lambda \in \sigma [2I-(P-Q)^2]$ if and only if there exists $\mu \in \sigma (P-Q)$ such that $\lambda =2-\mu ^2.$

There exists $\lambda \in {\mathbb {C}}$ such that $\lambda \in \sigma (PQ-QP)$ if and only if $\lambda ^2\in \sigma [(P-Q)^4-4(P-Q)^2]$ if and only if there exists $\mu \in \sigma (P-Q)$ such that $\lambda ^2 =\mu ^4-4\mu ^2.$$\square$

Theorem 2.4

Let P and Q be the reflection operators and $c_1$, $c_2\in {\mathbb {C}}\backslash \{0\}.$ Then $\lambda \in \sigma [(P-Q)^2]$ if and only if there exists $\mu \in {\mathbb {C}}$ such that $\mu \in \sigma (c_1P+c_2Q)$ and $\mu ^2=(c_1+c_2)^2-c_1c_2\lambda$. If there exist $\alpha$, $\beta \in {\mathbb {R}}$ such that $\sigma [(P-Q)^2]\subseteq [\alpha , \beta ]$, then

$$\begin{aligned} |\mu |\le \max \left\{ \sqrt{|(c_1+c_2)^2-c_1c_2\alpha |}, \sqrt{|(c_1+c_2)^2-c_1c_2\beta |}\right\} \end{aligned}$$

for all $\mu \in \sigma (c_1P+c_2Q).$

Proof

Note that

$$\begin{aligned} \begin{array}{rcl}\lambda I-(P-Q)^2&{}=&{}\lambda I +PQ+QP-2I\\ &{}=&{}\frac{c_1c_2(PQ+QP)+2c_1c_2 I}{c_1c_2}-4I+\lambda I\\ &{}=&{}\frac{(c_1P+c_2Q)^2-[(c_1-c_2)^2-(\lambda -4)c_1c_2]I}{c_1c_2}\\ &{}=&{}\frac{(c_1P+c_2Q)^2-[(c_1+c_2)^2-c_1c_2\lambda ]I}{c_1c_2}.\end{array} \end{aligned}$$

Hence, $\lambda \in \sigma [(P-Q)^2]$ if and only if there exists $\mu \in {\mathbb {C}}$ such that $\mu \in \sigma (c_1P+c_2Q)$ and $\mu ^2=(c_1+c_2)^2-c_1c_2\lambda$.

If $\mu \in \sigma (c_1P+c_2Q)$ and $\lambda \in \sigma [(P-Q)^2]\subseteq [\alpha , \beta ]$, then $\mu ^2=(c_1+c_2)^2-c_1c_2\lambda$. We get $\lambda =\frac{(c_1+c_2)^2-\mu ^2}{c_1c_2}\in [\alpha , \beta ].$ Note that $f(x)=z_1x+z_2$, $\forall x\in {\mathbb {R}}$ is a line in complex plane for complex numbers $z_1$, $z_2$. We have $\mu ^2$ is a segment in ${\mathbb {C}}$ with boundary points $(c_1+c_2)^2-c_1c_2\alpha$ and $(c_1+c_2)^2-c_1c_2\beta$ if $\lambda \in [\alpha , \beta ]$. Hence, for all $\mu \in \sigma (c_1P+c_2Q),$

$$\begin{aligned} |\mu |\le \max \left\{ \sqrt{|(c_1+c_2)^2-c_1c_2\alpha |}, \sqrt{|(c_1+c_2)^2-c_1c_2\beta |}\right\} . \end{aligned}$$

$\square$

Observing that $\sigma [(P-Q)^{2}]\subseteq [0,\Vert P-Q\Vert ^{2}]\subseteq [0,4]$ for reflection operators P and Q, we derive the following results.

Corollary 2.1

Let P and Q be the reflection operators and $c_1$, $c_2\in {\mathbb {C}}\backslash \{0\}$. Then

$$\begin{aligned} |\mu |\le max\left\{ |c_{1}+c_{2}|,\sqrt{|(c_{1}+c_{2})^{2}- c_{1}c_{2}\Vert P-Q\Vert ^{2}|}\right\} \end{aligned}$$

for every $\mu \in \sigma (c_{1}P+c_{2}Q).$

Corollary 2.2

Let P and Q be the reflection operators and $c_1$, $c_2\in {\mathbb {C}}\backslash \{0\}$. Then

$$\begin{aligned} |\mu |\le max\{|c_{1}+c_{2}|,|c_{1}-c_{2}|\} \end{aligned}$$

for every $\mu \in \sigma (c_{1}P+c_{2}Q).$

For convenience, we define a subset $\Lambda$ of ${\mathbb {C}}^2$ by

$$\begin{aligned} \Lambda =\left\{ (c_1, c_2)\in {\mathbb {C}}^{2}: c_1\ne 0, c_2\ne 0 \, \ \hbox {and}\ \,c_1+ c_2\ne 0\right\} . \end{aligned}$$

Theorem 2.5

Let P and Q be the reflection operators and let $(c_1, c_2)\in \Lambda$. Then

$$\begin{aligned} \dim {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]=\dim \left\{ {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)\right\} . \end{aligned}$$

Proof

If $x\in {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]$, then

$$\begin{aligned}&(I-P)(I+Q)(I-P)x\\&\quad =c^{-1}_1(I-P)(I+Q)[(c_1+c_2)I+c_1I+c_2Q-(c_1+c_2)I-c_1P-c_2Q]x\\&\quad =c^{-1}_1(I-P)(I+Q)[(c_1+c_2)I+c_1I+c_2Q]x\\&\quad =2c^{-1}_1(I-P)[(c_1+c_2)I+(c_1+c_2)Q]x\\&\quad =2(c_1+c_2)(c_1c_2)^{-1}(I-P)(c_2I+c_2Q)x\\&\quad =2(c_1+c_2)(c_1c_2)^{-1}(I-P)[(c_1+c_2)I+c_1P+c_2Q]x\\&\quad =0. \end{aligned}$$

Since $(I-P)x\in {\mathcal {N}}(I+P)$, we conclude that

$$\begin{aligned} (I-P){\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]\subseteq {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P). \end{aligned}$$

If $x\in {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]$ and $(I-P)x=0$, then $x=Px$ and

$$\begin{aligned} (c_1+c_2)(I+Q)x=\frac{1}{2}(I+Q)[(c_1+c_2)I+c_1P+c_2Q]x=0. \end{aligned}$$

So $(I+Q)x=0$. It follows that $[(c_1+c_2)I+c_1P+c_2Q]x=c_1(I+P)x=0$. Hence, $x=0$ since $x=-Px=Px$. We get that $I-P$ embeds ${\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]$ injectively into ${\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P).$ Thus,

$$\begin{aligned} \dim {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]\le \dim \left\{ {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)\right\} .\nonumber \\ \end{aligned}$$

(1)

On the other hand, if $x\in {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)$, then $x+Px=0$ and $x+Qx=Px+PQx.$ Note that

$$\begin{aligned}&[(c_1+c_2)I+c_1P+c_2Q]\left( 2c_1c^{-1}_2I+I-Q\right) x\\&\quad =[c_1(I+P)+c_2(I+Q)]\left( 2c_1c^{-1}_2I+I-Q\right) x\\&\quad =2c^{2}_1c^{-1}_2(I+P)x+2c_1(I+Q)x+c_1(I+P)(I-Q)x\\&\quad =2c_1(I+Q)x+2c_1(P-Q)x\\&\quad =2c_1(I+P)x\\&\quad =0. \end{aligned}$$

We get

$$\begin{aligned}&\left( 2c_1c^{-1}_2I+I-Q\right) \{{\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)\}\\&\quad \subseteq {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]. \end{aligned}$$

If $x\in {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)$ and $(2c_1c^{-1}_2I+I-Q)x=0$, then $(I+P)x=0$, $(I-P)(I+Q)x=0$ and $(I+Q)x=2(c_1c^{-1}_2I+I)x$. From $(I-P)(I+Q)x=0$, we get $(I+P)(I+Q)x=2(I+Q)x$. Hence,

$$\begin{aligned} 2\left( c_1c^{-1}_2I+I\right) x=(I+Q)x=\frac{1}{2}(I+P)(I+Q)x=\left( c_1c^{-1}_2I+I\right) (I+P)x=0. \end{aligned}$$

We get $x=0$ and thus $2c_1c^{-1}_2I+I-Q$ embeds ${\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)$ injectively into ${\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]$. Hence,

$$\begin{aligned} \dim {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]\ge \dim \left\{ {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)\right\} .\nonumber \\ \end{aligned}$$

(2)

By (1) and (2), we get that

$$\begin{aligned} \dim {\mathcal {N}}[(c_1+c_2)I+c_1P+c_2Q]=\dim \left\{ {\mathcal {N}}[(I-P)(I+Q)]\cap {\mathcal {N}}(I+P)\right\} . \end{aligned}$$

$\square$

3 Representations of reflection operators

As we know, if $A\in {{\mathcal {B}}({\mathcal {H}}\oplus {\mathcal {K}})}$ has the operator matrix form $A=\left( \begin{array}{cc} A_{11}&{}A_{12}\\ A_{21}&{}A_{22}\end{array}\right) ,$ then $A\ge 0$ if and only if $A_{ii}\ge 0$, $i=1,2,$$A_{21}=A_{12}^*$ and there exists a contraction operator D from ${{\mathcal {K}}}$ into ${{\mathcal {H}}}$ such that $A_{12}=A_{11}^{1/2}DA_{22}^{1/2},$ where $A_{ii}^{1/2}$ is the positive square root of $A_{ii}$, $i=1,2$ [9]. Recently, Moslehian, Kian and Xu used the Douglas theorem on equivalence of factorization, range inclusion and majorization of operators to characterize the positivity of $2\times 2$ block operator matrices [14]. In this section, we will give detailed block operator matrices representations of reflection operators.

Let P and Q be two reflection operators. Since we assume that neither of P, Q is I or $-I$, by Theorem 2.1, ${\mathcal {M}}={{\mathcal {R}}}(I+P)$, ${\mathcal {M}}^\perp ={{\mathcal {R}}}(I-P)$, ${\mathcal {N}}={{\mathcal {R}}}(I+Q)$ and ${\mathcal {N}}^\perp ={{\mathcal {R}}}(I-Q)$ are non-degenerate subspaces. Therefore, the reflection operator P as an operator on ${\mathcal {M}}\oplus {\mathcal {M}}^\perp$, and the reflection operator Q as an operator on ${\mathcal {N}}\oplus {\mathcal {N}}^\perp$, have the diagonal matrix forms

$$\begin{aligned} P=I_{{\mathcal {M}}}\oplus -I_{{\mathcal {M}}^\perp }\,\, \hbox {and}\,\, Q=I_{{\mathcal {N}}}\oplus -I_{{\mathcal {N}}^\perp }, \end{aligned}$$

(3)

respectively. Denote

$$\begin{aligned} {{\mathcal {H}}}_1= & {} {\mathcal {M}}\cap {\mathcal {N}},\quad {{\mathcal {H}}}_3={\mathcal {M}}^\perp \cap {\mathcal {N}},\quad {{\mathcal {H}}}_5={\mathcal {M}}\cap [{{\mathcal {H}}}\ominus (\oplus _{i=1}^4{ {\mathcal {H}}}_i)],\nonumber \\ {{\mathcal {H}}}_2= & {} {\mathcal {M}}\cap {\mathcal {N}}^\perp ,\quad {{\mathcal {H}}}_4={\mathcal {M}}^ \perp \cap {\mathcal {N}}^\perp ,\quad {{\mathcal {H}}}_6= {{\mathcal {H}}}\ominus (\oplus _{i=1}^5{{\mathcal {H}}}_i). \end{aligned}$$

(4)

It is clear that ${{\mathcal {H}}}_i\perp { {\mathcal {H}}}_j$, $j\ne i$ and $1\le i,j\le 6.$ The pair $({\mathcal {M}}, {\mathcal {N}})$ of subspaces ${\mathcal {M}}$ and ${\mathcal {N}}$ is said to be regular if ${\mathcal {H}}_i=\{0\}$, $i=1, 2, 3, 4$. Clearly, $({\mathcal {M}}, {\mathcal {N}})$ is regular if and only if $({\mathcal {M}}^\perp , {\mathcal {N}}^\perp )$ is. We say (P, Q) is a reflective regular pair whenever $({\mathcal {M}}, {\mathcal {N}})$ is a non-trivial regular pair. First, we study the matrix structures of a reflective regular pair (P, Q).

Theorem 3.1

Let (P, Q) be a reflective regular pair with ${\mathcal {M}}={{\mathcal {R}}}(I+P)$ and ${\mathcal {N}}={{\mathcal {R}}}(I+Q)$. Then there exist a selfadjoint contraction $A\in \mathcal {B(M)}$ with $1, -1\notin \sigma _P(A)$ and a unitary operator $D\in {\mathcal {B}}({\mathcal {M}}^\perp , {\mathcal {M}})$ such that P and Q have the operator matrix forms

$$\begin{aligned} P=\left( \begin{array}{cc} I&{}\quad 0\\ 0&{}\quad -I\end{array}\right) \quad \text{ and } \quad Q=\left( \begin{array}{cc} A&{}\quad (I-A^2)^{1/2}D\\ D^*(I-A^2)^{1/2}&{}\quad -D^*AD\end{array}\right) \end{aligned}$$

(5)

with respect to the space decomposition ${\mathcal {H}}={\mathcal {M}} \oplus {\mathcal {M}}^\perp ,$ respectively.

Proof

By Theorem 2.1, the reflection operator P, as an operator on ${\mathcal {H}}={\mathcal {M}} \oplus {\mathcal {M}}^\perp$, has the operator matrix form $P=I\oplus -I$. Let Q have the corresponding matrix form

$$\begin{aligned} Q=\left( \begin{array}{cc} A&{}\quad B\\ B^*&{}\quad C\end{array}\right) \end{aligned}$$

with respect to the same space decomposition, where A and C are selfadjoint contractions on ${\mathcal {M}}$ and ${\mathcal {M}}^\perp$, respectively. Denote ${\widetilde{P}}=\frac{1}{2}(I+P)$ and ${\widetilde{Q}}=\frac{1}{2}(I+Q)$. It is easy to see that ${\mathcal {R}}({\widetilde{P}})={\mathcal {M}}$, ${\mathcal {N}}({\widetilde{P}})=\mathcal {M^\perp }$, ${\mathcal {R}}({\widetilde{Q}})={\mathcal {N}}$ and ${\mathcal {N}}({\widetilde{Q}})=\mathcal {N^\perp }$ by (3). From ${\widetilde{Q}}^2=\frac{1}{2}(I+Q)={\widetilde{Q}}$, one gets that

$$\begin{aligned} \frac{1}{4}\left( \begin{array}{cc} (I+A)^2+BB^*&{} \quad (I+A)B+B(I+C)\\ B^*(I+A)+(I+C)B^*&{}\quad (I+C)^2+B^*B\end{array}\right) =\frac{1}{2}\left( \begin{array}{cc} I+A&{}\quad B\\ B^*&{}\quad I+C\end{array}\right) . \end{aligned}$$

Comparing the two sides of the above equation, we have

$$\begin{aligned} \left\{ \begin{array}{l}(I+A)^2+BB^*=2(I+A),\\ (I+A)B+B(I+C)=2B,\\ (I+C)^2+B^*B=2(I+C).\end{array}\right. \end{aligned}$$

(6)

The regularity of $({\mathcal {M}}, {\mathcal {N}})$ implies that $I\pm A$ and $I\pm C$ are injective. ${\widetilde{Q}}\ge 0$ implies that there exists a contraction D from ${\mathcal {M}}^\perp$ into ${\mathcal {M}}$ such that $B=(I+A)^{1/2}D(I+C)^{1/2}$. From the system of equations (6), we get

$$\begin{aligned} D(I+C)D^*=I-A,\quad AD=-DC,\quad D^*(I+A)D=I-C. \end{aligned}$$

It follows that

$$\begin{aligned} C=-D^*AD,\quad B=(I-A^2)^{1/2}D,\quad D^*D=I_{{\mathcal {M}}^\perp }\, \text{ and } \,DD^*=I_{{\mathcal {M}}}, \end{aligned}$$

where $I\pm A$ are injective, i.e., 1 and $-1\notin \sigma _P(A)$. $\square$

Remark 3.1

If (P, Q) is a reflective regular pair, then

$$\begin{aligned} \dim {\mathcal {M}}=\dim {\mathcal {M}}^\perp =\frac{1}{2}\dim { {\mathcal {H}}}\ \,\hbox {and}\ \,\dim {\mathcal {N}}=\dim {\mathcal {N}}^\perp =\frac{1}{2}\dim { {\mathcal {H}}} \end{aligned}$$

since the operator D in Theorem 3.1 is a unitary operator from ${\mathcal {M}}^\perp$ onto ${\mathcal {M}}$.

It is well known that if $T=\left( \begin{array}{cc}A \quad &{}B\\ C \quad &{}D\end{array}\right)$ and A is invertible, then the inverse of T is

$$\begin{aligned} T^{-1}=\left( \begin{array}{cc}A^{-1}+A^{-1}BS^{-1}CA^{-1} &{}\quad -A^{-1}BS^{-1}\\ -S^{-1}CA^{-1} &{}\quad S^{-1}\end{array}\right) \end{aligned}$$

(7)

whenever the Schur complement $S=D-CA^{-1}B$ of A in T is invertible. The expression (7) is called the Banachiewicz–Schur form of operator T and can be found in standard textbooks on linear algebra.

Corollary 3.1

If (P, Q) is a reflective regular pair having the matrix representations (5), then the following statements hold:

• (i) ${\mathcal {R}}(P+Q)$ is dense in ${\mathcal {H}};$

• (ii) $\Vert P+Q\Vert =\sqrt{2}\Vert I+A\Vert ^{1/2}$;

• (iii) ${\mathcal {R}}(P+Q)$ is closed if and only if $P+Q$ is invertible if and only if $-1\notin \sigma (A)$. In this case,

  $$\begin{aligned} (P+Q)^{-1}=\frac{1}{2}\left( \begin{array}{cc} I&{}\quad (I+A)^{-1/2}(I-A)^{1/2}D\\ D^*(I+A)^{-1/2}(I-A)^{1/2}&{}\quad I\end{array}\right) . \end{aligned}$$

Proof

(i) ${{\mathcal {R}}}(P+Q)$ is not dense if and only if $0\in \sigma _P(P+Q)$ since $P+Q$ is selfadjoint. Suppose that there exists a unit vector $x=(x_1, x_2)\in {\mathcal {H}}$ such that $(P+Q)x=0.$ By (5), one has

$$\begin{aligned}\begin{array}{rcl}(P+Q)x&{}=&{}\left( \begin{array}{cc} I+A&{}\quad (I-A^2)^{1/2}D\\ D^*(I-A^2)^{1/2}&{}\quad -D^*(I+A)D\end{array}\right) \left( \begin{array}{c}x_1\\ x_2\end{array}\right) =0.\end{array}\end{aligned}$$

Then,

$$\begin{aligned} \left\{ \begin{array}{l}(I+A)x_1+(I-A^2)^{1/2}Dx_2=0,\\ D^*(I-A^2)^{1/2}x_1-D^*(I+A)Dx_2=0.\end{array}\right. \end{aligned}$$

Observing that $I\pm A$ are injective and D is unitary, we get

$$\begin{aligned} \left\{ \begin{array}{l}(I+A)x_1+(I-A^2)^{1/2}Dx_2=0,\\ (I-A)x_1-(I-A^2)^{1/2}Dx_2=0.\end{array}\right. \end{aligned}$$

Solving the above equations, we obtain that $x_1=0$ and $x_2=0.$ Hence, $0\notin \sigma _P(P+Q)$ and thus ${\mathcal {R}}(P+Q)$ is dense.

(ii) By (5), $(P+Q)^2=2(I+A)\oplus 2D^*(I+A)D$. So, $\Vert P+Q\Vert =\Vert (P+Q)^2\Vert ^{1/2}=\sqrt{2}\Vert I+A\Vert ^{1/2}$.

(iii) By items (i) and (ii), we know that $P+Q$ is invertible if and only if ${\mathcal {R}}(P+Q)$ is closed if and only if $-1\notin \sigma (A)$. The inverse $(P+Q)^{-1}$ can be obtained by applying the representation (7). $\square$

If P and Q are two general reflection operators, by Theorem 3.1, we obtain the following useful representations.

Corollary 3.2

Let P and Q be two reflection operators and ${\mathcal {H}}_i$, $i=1,\ldots ,6$ be defined by (4). Then P and Q have the following operator matrix forms

$$\begin{aligned} \begin{array}{l} P=I_1\oplus I_2\oplus -I_3\oplus -I_4\oplus \left( {\begin{matrix} I_5&{}0\\ 0&{}-I_6\end{matrix}}\right) , \\ Q=I_1\oplus -I_2\oplus I_3\oplus -I_4\oplus \left( {\begin{matrix} A&{}\quad (I_5-A^2)^{1/2}D\\ D^*(I_5-A^2)^{1/2}&{}\quad -D^*AD\end{matrix}}\right) \end{array} \end{aligned}$$

(8)

with respect to the space decomposition ${ {\mathcal {H}}}=\oplus _{i=1}^6{ {\mathcal {H}}}_i$, respectively, where A is a selfadjoint contraction on ${\mathcal {H}}_5$ with that $1,-1\notin \sigma _{P}(A)$, D is a unitary operator from ${\mathcal {H}}_6$ onto ${\mathcal {H}}_5$, and $I_i$ is the identity onto ${\mathcal {H}}_i, i=1,\ldots ,6.$

Note that $\dim {\mathcal {H}}_{5}=\dim {\mathcal {H}}_{6}$ since D is a unitary operator from ${\mathcal {H}}_{6}$ onto ${\mathcal {H}}_{5}.$

Corollary 3.3

Let P and Q be two reflection operators and ${\mathcal {H}}_i$, $i=1,\ldots ,6$ be defined by (4). Then $PQ=QP$ if and only if $\dim {\mathcal {H}}_{5}=\dim {\mathcal {H}}_{6}=0$.

Proof

By Corollary 3.2,

$$\begin{aligned} PQ-QP=0\oplus 0\oplus 0\oplus 0\oplus \left( {\begin{matrix} 0 &{}\quad 2(I_5-A^{2})^{1/2}D \\ -2D^{*}(I_5-A^{2})^{1/2} &{}\quad 0 \\ \end{matrix}} \right) . \end{aligned}$$

(9)

Since $I_5\pm A$ are injective and D is a unitary operator, $PQ=QP$ if and only if $\dim {\mathcal {H}}_{5}=\dim {\mathcal {H}}_{6}=0$. $\square$

Also by Corollary 3.2, we have $PQ+QP=2I_1\oplus -2I_2\oplus -2I_3\oplus 2I_4\oplus 2A\oplus 2D^*AD.$

Theorem 3.2

Let P and Q be two reflection operators on ${\mathcal {H}}$ having the matrix representations (8) and ${\mathcal {H}}_i$, $i=1,\ldots ,6$ be defined by (4). Then the following statements hold.

• (i) $\Gamma =c_1P+c_2Q$, where $c_1, c_2\in {\mathbb {C}}{\setminus }\{0\}$, is invertible if and only if $c_1+c_2\ne 0$ whenever ${\mathcal {H}}_1$ or ${\mathcal {H}}_4\ne \{0\}$, $c_1-c_2\ne 0$ whenever ${\mathcal {H}}_2$ or ${\mathcal {H}}_3\ne \{0\}$ and $-\frac{c^2_1+c^2_2}{2c_1c_2}\notin \sigma (A);$

• (ii) If $|c_1|\ne |c_2|$, then $\Gamma =c_1P+c_2Q$ is invertible for every $c_1, c_2\in {\mathbb {C}}.$

Proof

(i) By Corollary 3.2, one has

$$\begin{aligned} \begin{array}{rcl} \Gamma ^2&{}=&{}(c_1P+c_2Q)^2\\ &{}=&{}\left( c^2_1+c^2_2\right) I+c_1c_2(PQ+QP)\\ &{}=&{}(c_1+c_2)^2I_1\oplus (c_1-c_2)^2I_2\oplus (c_1-c_2)^2I_3\oplus (c_1+c_2)^2I_4\\ &{}&{} \oplus \left[ \left( c^2_1+c^2_2\right) I_5+2c_1c_2A\right] \oplus D^*\left[ (c^2_1+c^2_2)I_5+2c_1c_2A\right] D. \end{array} \end{aligned}$$

(10)

$\Gamma$ is invertible if and only if $\Gamma ^2$ is invertible if and only if $c_1+c_2\ne 0$ whenever ${\mathcal {H}}_1$ or ${\mathcal {H}}_4\ne \{0\}$, $c_1-c_2\ne 0$ whenever ${\mathcal {H}}_2$ or ${\mathcal {H}}_3\ne \{0\}$ and $-\frac{c^2_1+c^2_2}{2c_1c_2}\notin \sigma (A)\subseteq [-1,1].$

(ii) If $|c_1|\ne |c_2|$ and $c_1c_2=0$, then $\Gamma =c_1P+c_2Q$ is obviously invertible by (10).

If $c_1c_2\ne 0$, let $c_1=r_1e^{i\theta _1}\ne 0$ and $c_2=r_2e^{i\theta _2}\ne 0,$ where $r_i$ and $\theta _i$ denote the module and argument of complex numbers $c_i,$$i=1,2,$ respectively. By direct computation,

$$\begin{aligned} \frac{c^2_1+c^2_2}{2c_1c_2}=\frac{\left( r_1^2+r_2^2\right) \cos (\theta _1-\theta _2)+i\left( r_1^2-r_2^2\right) \sin (\theta _1-\theta _2)}{2r_1r_2}. \end{aligned}$$

Since $|c_1|\ne |c_2|$, then $c_1\pm c_2\ne 0$ and $r_1\ne r_2$, obviously.

• (a) If $r_1\ne r_2$, $\theta _1=\theta _2+k\pi , k\in {\mathbb {Z}},$ then $1<|\frac{c^2_1+c^2_2}{2c_1c_2}|\notin \sigma (A)\subseteq [-1,1].$

• (b) If $r_1\ne r_2$, $\theta _1\ne \theta _2+k\pi , k\in {\mathbb {Z}},$ then $-\frac{c^2_1+c^2_2}{2c_1c_2}\notin {\mathbb {R}}.$

As a result, we get $c_1\pm c_2\ne 0$ and $-\frac{c^2_1+c^2_2}{2c_1c_2}\notin \sigma (A).$ By (10), we obtain that $\Gamma =c_1P+c_2Q$ is invertible for every $c_1, c_2\in {\mathbb {C}}{\setminus }\{0\}$ with $|c_1|\ne |c_2|$. $\square$

We have discussed the invertibility of operator $P+Q$ if (P, Q) is a reflective regular pair having matrix representations (5) in Corollary 3.1. In the following, we will talk about the invertibility of $P-Q$ and $P+Q-I$ for the general reflection operators P and Q.

Theorem 3.3

Let P and Q be two reflection operators and ${\mathcal {H}}_i$, $i=1,\ldots ,6$ be defined by (4).

• (i) $P-Q$ is invertible if and only if $\dim {\mathcal {H}}_{1}=\dim {\mathcal {H}}_{4}=0$ and $1\notin \sigma (A)$.

• (ii) $P+Q-I$ is invertible if and only if $3I-(P-Q)^2$ is invertible if and only if $-\frac{1}{2}\notin \sigma (A)$. In this case,

  $$\begin{aligned}&(P+Q-I)^{-1}\\&\quad =\left[ 3I-(P-Q)^2\right] ^{-1}(I+P+Q)\\&\quad =(I+P+Q)\left[ 3I-(P-Q)^2\right] ^{-1}\\&\quad =I_1\oplus -I_2\oplus -I_3\oplus -\frac{1}{3}I_4\oplus \left( {\begin{matrix} (2I_5+A)(I_5+2A)^{-1}&{}\quad (I_5+2A)^{-1}(I_5-A^{2})^{{1/2}}D \\ D^{*}(I_5+2A)^{-1}(I_5-A^{2})^{{1/2}} &{} \quad -D^*A(I_5+2A)^{-1}D \\ \end{matrix}}\right) . \end{aligned}$$

Proof

(i) By Corollary 3.2,

$$\begin{aligned} (P-Q)^2=0\oplus 4I_2\oplus 4I_3\oplus 0\oplus 2(I_5-A)\oplus 2D^*(I_5-A)D. \end{aligned}$$

We get that $P-Q$ is invertible if and only if $\dim {\mathcal {H}}_{1}=\dim {\mathcal {H}}_{4}=0$ and $1\notin \sigma (A)$.

(ii) By Corollary 3.2,

$$\begin{aligned} P+Q-I=I_1\oplus -I_2\oplus -I_3\oplus -3I_4\oplus \left( {\begin{matrix} A&{}\quad (I_5-A^{2})^{{1/2}}D \\ D^{*}(I_5-A^{2})^{{1/2}} &{}\quad -D^*(2I_5+A)D \\ \end{matrix}} \right) . \end{aligned}$$

By Theorem 3.1, $2I_5+A$ is invertible since A is a contraction operator. By (7), $P+Q-I$ is invertible if and only if the Schur complement $S=A+(I_5-A^{2})^{{1/2}}DD^*(2I_5+A)^{-1}DD^{*}(I_5-A^{2})^{{1/2}}=(2I_5+A)^{-1}(I_5+2A)$ is invertible if and only if $I_5+2A$ is invertible if and only if $-\frac{1}{2}\notin \sigma (A)$. Applying the formula in (7), we get that

$$\begin{aligned}&(P+Q-I)^{-1}\\&\quad =I_1\oplus -I_2\oplus -I_3\oplus -\frac{1}{3}I_4\oplus \left( {\begin{matrix} (2I_5+A)(I_5+2A)^{-1}&{}\quad (I_5+2A)^{-1}(I_5-A^{2})^{{1/2}}D \\ D^{*}(I_5+2A)^{-1}(I_5-A^{2})^{{1/2}} &{}\quad -D^*A(I_5+2A)^{-1}D \\ \end{matrix}}\right) . \end{aligned}$$

Note that $(P-Q)^2=0\oplus 4I_2\oplus 4I_3\oplus 0\oplus 2(I_5-A)\oplus 2D^*(I_5-A)D,$

$$\begin{aligned} 3I-(P-Q)^2=3I_1\oplus -I_2\oplus -I_3\oplus 3I_4\oplus (I_5+2A)\oplus D^*(I_5+2A)D \end{aligned}$$

and

$$\begin{aligned} 3I-(P-Q)^2=(P+Q-I)(I+P+Q)=(I+P+Q)(P+Q-I). \end{aligned}$$

We also get $3I-(P-Q)^2$ is invertible if and only if $-\frac{1}{2}\notin \sigma (A)$ if and only if $P+Q-I$ is invertible. In this case,

$$\begin{aligned} (P+Q-I)^{-1}= & {} \left[ 3I-(P-Q)^2\right] ^{-1}(I+P+Q)\\= & {} (I+P+Q)\left[ 3I-(P-Q)^2\right] ^{-1}. \end{aligned}$$

$\square$

Let P and Q be the reflection operators on ${\mathcal {H}}$. According to (9), we observe that ${\mathcal {N}}(PQ-QP)$ and ${\mathcal {N}}(PQ-QP)^\perp$ are invariant subspaces of P and Q. Therefore, P and Q as operators on ${\mathcal {H}}={\mathcal {N}}(PQ-QP)\oplus {\mathcal {N}}(PQ-QP)^\perp$ have the operator matrix representations

$$\begin{aligned} P=P_1\oplus P_2\ \hbox { and }\ Q=Q_1\oplus Q_2, \end{aligned}$$

respectively, where the restrictions $P_i$ and $Q_i$, $i=1, 2$ have the following properties:

$$\begin{aligned} (1).\quad P_1Q_1=Q_1P_1; \quad (2).\quad {\mathcal {N}}(P_2Q_2-Q_2P_2)=\{0\}. \end{aligned}$$

Besides, ${\mathcal {N}}(PQ-QP)$ admits the orthogonal decomposition ${\mathcal {N}}(PQ-QP)=\bigoplus _{i=1}^4{\mathcal {H}}_i.$ Therefore, (P, Q) is a regular pair if and only if ${\mathcal {N}}(PQ-QP)=\{0\}$. For orthogonal projections cases, Halmos presented a nice and tractable characterization of the regular pair in [10, Theorem 2].

Theorem 3.4

Let $\Gamma =c_1P+c_2Q$, where $c_1, c_2\in {\mathbb {R}}{\setminus }\{0\}$ and P, Q are two reflection operators on ${\mathcal {H}}$ with $P\ne \pm Q$. Then $\Gamma$ is a reflection operator if and only if the following conditions hold:

• (i) (P, Q) is the reflective regular pair;

• (ii) $|\frac{1-c_1^2-c_2^{2}}{2c_{1}c_2}|<1$;

• (iii) P and Q, as operators onto ${\mathcal {H}}={\mathcal {R}}(I+P)\oplus {\mathcal {R}}(I-P)$, have the operator matrix forms

  $$\begin{aligned} P=\left( \begin{array}{cc} I&{}0\\ 0&{}-I\end{array}\right) \ \hbox {and}\ Q=\left( \begin{array}{cc} A&{}\quad (I-A^2)^{1/2}D\\ D^*(I-A^2)^{1/2}&{}\quad -D^*AD\end{array}\right) , \end{aligned}$$

  where $A=\frac{1-c_1^2-c_2^2}{2c_1c_2}I$ and D is a unitary operator from ${\mathcal {R}}(I-P)$ onto ${\mathcal {R}}(I+P)$.

Proof

The sufficiency is clear, so it is enough to prove the necessity. Let P and Q have the matrix representations in (8). Consequently, $PQ+QP=2I_1\oplus -2I_2\oplus -2I_3\oplus 2I_4\oplus 2A\oplus 2D^*AD$. If $\Gamma$ is a reflection operator, then

$$\begin{aligned} \Gamma ^2=(c_1P+c_2Q)^2=\left( c^2_1+c^2_2\right) I+c_1c_2(PQ+QP)=I. \end{aligned}$$

Hence,

$$\begin{aligned} PQ+QP=2I_1\oplus -2I_2\oplus -2I_3\oplus 2I_4\oplus 2A\oplus 2D^*AD=\frac{1-c_1^2-c_2^2}{c_1c_2}I. \end{aligned}$$

Now, in view of the equations above, we consider the cases as follows:

If $\frac{1-c_1^2-c_2^2}{c_1c_2}=2,$ then $\dim {\mathcal {H}}_i =0,i=2,3,5,6$ since A is a contraction with $1, -1\notin \sigma _P(A)$. In this case,

$$\begin{aligned} P=I_1\oplus -I_4\ \hbox { and }\ Q=I_1\oplus -I_4, \end{aligned}$$

which contradicts to $P\ne Q.$ Similarly, if $\frac{1-c_1^2-c_2^2}{c_1c_2}=-2,$ then $\dim {\mathcal {H}}_i=0, i=1,4,5,6$. In this case,

$$\begin{aligned} P=I_2\oplus -I_3\ \hbox { and }\ Q=-I_2\oplus I_3, \end{aligned}$$

which is a contradiction to $P\ne -Q.$ Also, if $\frac{1-c_1^2-c_2^2}{c_1c_2}\ne \pm 2,$ then $\dim {\mathcal {H}}_i =0, i=1,2,3,4$. In this case,

$$\begin{aligned} P=I_5\oplus -I_6\ \hbox { and }\ Q=\left( \begin{array}{cc} A&{}\quad (I_5-A^2)^{1/2}D\\ D^*(I_5-A^2)^{1/2}&{}\quad -D^*AD\end{array}\right) , \end{aligned}$$

where $A=\frac{1-c_1^2-c_2^2}{2c_1c_2}I_5$ and D is a unitary operator from ${\mathcal {R}}(I-P)$ onto ${\mathcal {R}}(I+P)$. Hence, (P, Q) is the reflective regular pair and $|\frac{1-c_1^2-c_2^2}{2c_1c_2}|<1$. $\square$

In the following, we will investigate the unitary equivalence of the reflection operators P and Q, which is similar to that of two orthogonal projections (see [4, Theorem 3.1]).

Let the reflective regular pair (P, Q) have the matrix representations (5). It is easy to check that the unitary operator

$$\begin{aligned} {}W_{0}=\frac{\sqrt{2}}{2}\left( \begin{array}{cc} (I+A)^{{1/2}} &{}\quad (I-A)^{{1/2}}D \\ D^{*}(I-A)^{{1/2}} &{}\quad -D^{*}(I+A)^{{1/2}}D \\ \end{array} \right) \end{aligned}$$

(11)

satisfies $W_{0}P=QW_{0}$ and $PW_{0}=W_{0}Q.$ Moreover, we observe that the unitary $U\in inn_{Q}(P)$ if and only if $U^*\in out_{Q}(P).$ In the following, we obtain the general expressions for all the intertwining operators of P respect to Q when (P, Q) is a reflective regular pair.

Theorem 3.5

Let (P, Q) be a reflective regular pair and $W_0$ be defined by (11). Then

$$\begin{aligned} \begin{array}{rcl} out_{Q}(P)&{}=&{}\{W_{0}U: UP=PU, U\in {\mathcal {U}}({\mathcal {H}})\}, \\ int_{Q}(P)&{}=&{}\{W_{0}U: UP=PU, UQ=QU, U\in {\mathcal {U}}({\mathcal {H}})\}. \end{array} \end{aligned}$$

Moreover, if $U\in {\mathcal {U}}({\mathcal {H}})$ such that $UP=PU$, then $W_0U=UW_0$ if and only if $QU=UQ$.

Proof

Clearly, $\{W_{0}U: UP=PU,U\in {\mathcal {U}}({\mathcal {H}})\}\subseteq out_{Q}(P)$. On the other hand, if there exists a unitary operator W such that $WP=QW$, then $U=W_{0}^{*}W$ is a unitary operator and $UP=W_{0}^{*}WP=W_{0}^{*}QW=PW_{0}^{*}W=PU.$ Thus, $W=W_{0}U$, where $U\in {\mathcal {U}}({\mathcal {H}})$ satisfies $UP=PU$, i.e., $out_{Q}(P)\subseteq \{W_{0}U: UP=PU, U\in {\mathcal {U}}({\mathcal {H}})\}.$ Therefore, we obtain that

$$\begin{aligned} out_{Q}(P)=\{W_{0}U: UP=PU, U\in {\mathcal {U}}({\mathcal {H}})\}. \end{aligned}$$

Clearly, $int_{Q}(P)\supseteq \{W_{0}U: UP=PU, UQ=QU, U\in {\mathcal {U}}({\mathcal {H}})\}.$ On the other hand, if $T\in int_{Q}(P)=out_{Q}(P)\cap inn_{Q}(P)$, then $T\in out_{Q}(P)$. Hence, there exists a $U\in {\mathcal {U}}({\mathcal {H}})$ such that $UP=PU$ and $T=W_{0}U$. Since $T=W_{0}U\in inn_{Q}(P)$, we have $PW_{0}U=W_{0}UQ$. Since $PW_{0}=W_{0}Q$ and $W_{0}$ is invertible, we get $UQ=QU$ and

$$\begin{aligned} int_{Q}(P)\subseteq \{W_{0}U: UP=PU, UQ=QU, U\in {\mathcal {U}}({\mathcal {H}})\}. \end{aligned}$$

Therefore, we obtain that

$$\begin{aligned} int_{Q}(P)=\{W_{0}U: UP=PU, UQ=QU, U\in {\mathcal {U}}({\mathcal {H}})\}. \end{aligned}$$

By (5), $P=I\oplus -I$. The relation $UP=PU$ implies that $U=U_{11}\oplus U_{22}$, where $U_{11}$ and $U_{22}$ are unitary operators. If $W_0U=UW_0$, then

$$\begin{aligned}&\left( \begin{array}{cc} (I+A)^{{1/2}}U_{11} &{}\quad (I-A)^{{1/2}}DU_{22}\\ D^{*}(I-A)^{{1/2}}U_{11} &{} \quad -D^{*}(I+A)^{{1/2}}DU_{22} \\ \end{array}\right) \\&\quad =\left( \begin{array}{cc}U_{11}(I+A)^{{1/2}} &{}\quad U_{11}(I-A)^{{1/2}}D \\ U_{22}D^{*}(I-A)^{{1/2}} &{}\quad -U_{22}D^{*}(I+A)^{{1/2}}D\\ \end{array}\right) . \end{aligned}$$

It follows that

$$\begin{aligned}&\left\{ \begin{array}{c} U_{11}(I+A)^{{1/2}} =(I+A)^{{1/2}}U_{11},\\ U_{11}(I-A)^{{1/2}}D =(I-A)^{{1/2}}DU_{22}. \end{array}\right. \\&\quad {\mathop {\Longrightarrow }\limits ^{\mathrm{Theorem}~{\mathrm{3.1}}}}\left\{ \begin{array}{c} U_{11}A=AU_{11}, \\ U_{22}=D^*U_{11}D. \end{array}\right. \Longrightarrow \begin{array}{c} QU=UQ. \end{array} \end{aligned}$$

Similarly, if $QU=UQ$, we can derive that $W_0U=UW_0$. $\square$

Remark 3.2

Let the reflective regular pair (P, Q) have the matrix representations (5) and $W_0$ be defined by (11). Let $B_0=P-Q$ and $M_0=I+PQ+QP$.

• (i) Theorem 3.5 shows that if there exists a unitary operator $W\in {\mathcal {U}}({\mathcal {H}})$ such that $PW=WQ$ and $WP=QW$, then

  $$\begin{aligned} {}W=W_0U=\frac{\sqrt{2}}{2}\left( \begin{array}{cc} (I+A)^{{1/2}} &{} \quad (I-A)^{{1/2}}D \\ D^{*}(I-A)^{{1/2}} &{}\quad -D^{*}(I+A)^{{1/2}}D \\ \end{array} \right) \left( \begin{array}{cc}U_{11} &{}\quad 0 \\ 0 &{}\quad D^{*}U_{11}D\\ \end{array} \right) ,\nonumber \\ \end{aligned}$$

  (12)

  where $U_{11}\in {\mathcal {U}}({\mathcal {M}})$ satisfies $U_{11}A=AU_{11}$.

• (ii) Note that A is a contraction on ${\mathcal {M}}$ with that neither 1 nor $-1$ belongs to the point spectrum of A and D is a unitary operator from ${\mathcal {M}}^\perp$ onto ${\mathcal {M}}$. It is easy to get that

  $$\begin{aligned} B_0= & {} \left( \begin{array}{cc}I-A&{}\quad -(I-A^2)^{1/2}D\\ -D^*(I-A^2)^{1/2}&{}\quad -D^*(I- A)D\end{array}\right) ,\\ 2I-B_0= & {} \left( \begin{array}{cc}I+A&{}\quad (I-A^2)^{1/2}D\\ D^*(I-A^2)^{1/2}&{}\quad D^*(3I- A)D\end{array}\right) \end{aligned}$$

  and

  $$\begin{aligned} 2I+B_0=\left( \begin{array}{cc}3I-A&{}\quad -(I-A^2)^{1/2}D\\ -D^*(I-A^2)^{1/2}&{}\quad D^*(I+A)D\end{array}\right) \end{aligned}$$

  are three injective selfadjoint operators.

• (iii) If $-\frac{1}{2}, \frac{1}{2}\notin \sigma _{P}(A),$ then $M_0=(I+2 A)\oplus D^*(I+2A)D$,

  $$\begin{aligned} M_{0}B_0= & {} \left( \begin{array}{cc}(I-A)(I+2A)&{}\quad -(I-A^2)^{1/2}(I+2A)D\\ -D^*(I-A^2)^{1/2}(I+2A)&{}\quad -D^*(I- A)(I+2A)D\end{array}\right) , \\ 2I-M_{0}B_0= & {} \left( \begin{array}{cc}I-A+2A^{2}&{}\quad (I+2A)(I-A^2)^{1/2}D\\ D^*(I+2A)(I-A^2)^{1/2}&{}\quad D^*(3I+A-2A^{2})D\end{array}\right) \end{aligned}$$

  and

  $$\begin{aligned} 2I+M_{0}B_0=\left( \begin{array}{cc}3I+A-2A^{2}&{}\quad -(I-A^2)^{1/2}(I+2A)D\\ -D^*(I-A^2)^{1/2}(I+2A)&{}\quad D^*(I-A+2A^{2})D\end{array}\right) \end{aligned}$$

  are injective selfadjoint operators.

• (iv) As one example, we only prove that $M_{0}B_0$ is injective if $-\frac{1}{2}\notin \sigma _{P}(A)$. In fact, if there exists $(x,y)\in {\mathcal {H}}$ such that $M_{0}B_0\left( \begin{array}{c} x \\ y \\ \end{array} \right) =0,$ then

  $$\begin{aligned} \left( \begin{array}{cc}(I-A)(I+2A)&{}\quad -(I-A^2)^{1/2}(I+2A)D\\ -D^*(I-A^2)^{1/2}(I+2A)&{}\quad -D^*(I- A)(I+2A)D\end{array}\right) \left( \begin{array}{c} x \\ y \\ \end{array} \right) =0. \end{aligned}$$

  That is,

  $$\begin{aligned} \left\{ \begin{array}{ll} (I-A)(I+2A)x-(I-A^{2})^{{1/2}}(I+2A)Dy=0,\\ D^{*}(I-A^{2})^{{1/2}}(I+2A)x+D^{*}(I-A)(I+2A)Dy=0. \end{array} \right. \end{aligned}$$

  Then,

  $$\begin{aligned} \left\{ \begin{array}{ll} (I-A)(I+2A)x-(I-A^{2})^{{1/2}}(I+2A)Dy=0,\\ (I+A)(I+2A)x+(I-A^{2})^{{1/2}}(I+2A)Dy=0. \end{array} \right. \end{aligned}$$

  Adding the two equations, we can get $2(I+2A)x=0.$ Then $x=0$ since $-\frac{1}{2}\notin \sigma _P(A)$ and $y=0$ since D is a unitary operator. Hence, $M_0B_0$ is injective.

Similarly, we can obtain that $B_0$, $2I-B_0$ and $2I+B_0$ are injective. Moreover, $2I-M_0B_0$ and $2I+M_0B_0$ are injective if $-\frac{1}{2}, \frac{1}{2}\notin \sigma _{P}(A)$.

Let (P, Q) be a pair of reflection operators and ${\mathcal {H}}_i$, $i=1,\ldots , 6$ be defined by (4). If there exists a $U\in {\mathcal {U}}({\mathcal {H}})$ such that $PU=UQ$ and $UP=QU,$ then

$$\begin{aligned} U(P-Q)=-(P-Q)U. \end{aligned}$$

Denote $B:=P-Q=0\oplus 2I_2\oplus -2I_3\oplus 0\oplus B_0.$ Clearly, ${\mathcal {N}}(B)={\mathcal {H}}_1\oplus {\mathcal {H}}_4$, ${\mathcal {N}}(B-2I)={\mathcal {H}}_2$ and ${\mathcal {N}}(B+2I)={\mathcal {H}}_3$ are reduced subspaces of B. Let ${\mathcal {H}}_{0}={\mathcal {H}}_5\oplus {\mathcal {H}}_6,$ then B, as an operator on ${\mathcal {H}}=({\mathcal {H}}_1\oplus {\mathcal {H}}_4)\oplus {\mathcal {H}}_2\oplus {\mathcal {H}}_3\oplus {\mathcal {H}}_{0}$, has the operator matrix form $B=0\oplus 2I_2\oplus -2I_3\oplus B_{0}$. Let $U=(U^0_{ij})_{1\le i,j\le 4}.$ By $UB=-BU,$ we have

$$\begin{aligned} \left( \begin{array}{cc} U^0_{11} &{} \quad E \\ F &{} \quad U_2 \\ \end{array} \right) \left( \begin{array}{cc} 0 &{}\quad 0 \\ 0 &{}\quad B^0 \\ \end{array} \right) =-\left( \begin{array}{cc} 0 &{}\quad 0 \\ 0 &{}\quad B^0 \\ \end{array} \right) \left( \begin{array}{cc} U^0_{11} &{}\quad E \\ F &{}\quad U_2 \\ \end{array} \right) , \end{aligned}$$

(13)

where $E=(U^0_{12}$, $U^0_{13}$, $U^0_{14})$, $F=(U^0_{21}, U^0_{31}, U^0_{41})^\top$, $B^0=2I_2\oplus -2I_3\oplus B_0$ is injective and $U_{2}=(U^0_{ij})_{2\le i,j \le 4}.$ The equations in (13) imply that $E=0$, $F=0$ and $U_{2}B^0=-B^0U_{2}.$ Thus, $U^0_{12},$$U^0_{13},$$U^0_{14},$$U^0_{21},$$U^0_{31}$ and $U^0_{41}$ are zero operators. By $U_{2}B^0=-B^0U_{2},$ we have

$$\begin{aligned} \left( \begin{array}{cccc} 2U^0_{22} &{}\quad -2U^0_{23} &{} \quad U^0_{24}B_{0} \\ 2U^0_{32} &{} \quad -2U^0_{33} &{}\quad U^0_{34}B_{0} \\ 2U^0_{42} &{} \quad -2U^0_{43} &{}\quad U^0_{44}B_{0} \\ \end{array} \right) =\left( \begin{array}{cccc} -2U^0_{22} &{}\quad -2U^0_{23} &{} \quad -2U^0_{24}\\ 2U^0_{32} &{}\quad 2U^0_{33} &{}\quad 2U^0_{34} \\ -B_{0}U^0_{42} &{}\quad -B_{0}U^0_{43} &{} \quad -B_{0}U^0_{44}\\ \end{array} \right) . \end{aligned}$$

Since $2I-B_{0}$ and $2I+B_{0}$ are injective and dense, then $U^0_{22}$, $U^0_{24}$, $U^0_{33}$, $U^0_{34}$, $U^0_{42}$ and $U^0_{43}$ are zero operators. Hence,

$$\begin{aligned} U=U^0_{11}\oplus \left( \begin{array}{cc} 0 &{}\quad U^0_{23} \\ U^0_{32} &{}\quad 0\end{array}\right) \oplus U^0_{44}. \end{aligned}$$

Since $P|_{{\mathcal {N}}(B)}=Q|_{{\mathcal {N}}(B)}=I_1\oplus -I_4$, the multi-commutativity of $P|_{{\mathcal {N}}(B)}$, $Q|_{{\mathcal {N}}(B)}$ and $U^0_{11}$ implies that $U^0_{11}=U_{11}\oplus U_{44}\in {\mathcal {B}}({\mathcal {H}}_{1}\oplus {\mathcal {H}}_{4}).$ Moreover, $\left( \begin{array}{cc} 0&{}\quad U^0_{23} \\ U^0_{32} &{}\quad 0 \\ \end{array} \right)$ is a unitary operator if and only if $U^0_{23}$ and $U^0_{32}$ are unitaries if and only if $\dim {\mathcal {H}}_{2}=\dim {\mathcal {H}}_{3}.$ By Remark 3.2 (i), we get that $U^0_{44}$ has the representation (12). As a consequence, we present the following corollary.

Corollary 3.4

Let (P, Q) be a pair of reflection operators and ${\mathcal {H}}_i$, $i=1,\ldots ,6$ be defined by (4). Then P and Q are unitary equivalence if and only if $\dim {\mathcal {H}}_{2}=\dim {\mathcal {H}}_{3}.$ In addition, if P and Q are denoted by (8) and $W_0$ is denoted by (11), then $U\in \mathcal {U(H)}$ with $UP=QU$ and $PU=UQ$ has the representation

$$\begin{aligned}U=U_{11}\oplus \left( \begin{array}{cc}0 &{}\quad U_{23}\\ U_{32} &{}\quad 0\\ \end{array} \right) \oplus U_{44}\oplus W_0 \left( \begin{array}{cc} U_{55} &{}\quad 0 \\ 0 &{}\quad D^{*}U_{55}D \\ \end{array} \right) , \end{aligned}$$

where $U_{ij}$ are unitary operators from ${\mathcal {H}}_j$ onto ${\mathcal {H}}_i$, $U_{55}A=AU_{55}.$

At last, we talk about the unitary equivalence of products PQP and QPQ.

Theorem 3.6

Let (P, Q) be a pair of reflection operators and ${\mathcal {H}}_i$, $i=1,\ldots ,6$ be defined by (4). If $\pm \frac{1}{2}\notin \sigma _{P}(A),$ then PQP and QPQ are unitary equivalence if and only if $\dim {\mathcal {H}}_{2}=\dim {\mathcal {H}}_{3}.$

Proof

Sufficiency. If $\dim {\mathcal {H}}_{2}=\dim {\mathcal {H}}_{3}$, then there is a $U\in \mathcal {U(H)}$ such that $PU=UQ$ and $UP=QU$ by Corollary 3.4. It follows that $PQPU=UQPQ$ and $UPQP=QPQU$.

Necessity. If there is a $U\in {\mathcal {U}}({\mathcal {H}})$ such that $UPQP=QPQU$ and $PQPU=UQPQ$, then we have

$$\begin{aligned} U(PQP-QPQ)=-(PQP-QPQ)U. \end{aligned}$$

Let ${\widetilde{B}}:=PQP-QPQ,B:=P-Q\ \hbox {and}\ M:=I+PQ+QP.$ Clearly, ${\widetilde{B}}=MB.$ Let ${\mathcal {H}}_{0}={\mathcal {H}}_5\oplus {\mathcal {H}}_6.$ Moreover, B and M as operators on ${\mathcal {H}}={\mathcal {H}}_1\oplus {\mathcal {H}}_2\oplus {\mathcal {H}}_3\oplus {\mathcal {H}}_4\oplus {\mathcal {H}}_{0}$ have the matrix forms

$$\begin{aligned} B=0\oplus 2I_2\oplus -2I_3\oplus 0\oplus B_{0}\ \hbox { and }\ M=3I_1\oplus -I_2\oplus -I_3\oplus 3I_4\oplus M_{0}, \end{aligned}$$

respectively, where $B_0=B|_{{\mathcal {H}}_0}$ and $M_0=M|_{{\mathcal {H}}_0}$. Clearly, ${\mathcal {N}}(B)={\mathcal {H}}_1\oplus {\mathcal {H}}_4$. Furthermore, B and M as operators on ${\mathcal {H}}=({\mathcal {H}}_1\oplus {\mathcal {H}}_4)\oplus {\mathcal {H}}_2\oplus {\mathcal {H}}_3\oplus {\mathcal {H}}_{0}$ have matrix forms

$$\begin{aligned} B=0\oplus 2I_2\oplus -2I_3\oplus B_{0}\ \hbox { and }\ M=3I_{{\mathcal {N}}(B)}\oplus -I_2\oplus -I_3\oplus M_{0}. \end{aligned}$$

Thus,

$$\begin{aligned} {\widetilde{B}}=MB=0\oplus -2I_2\oplus 2I_3\oplus M_{0}B_{0}. \end{aligned}$$

By Remark 3.2 (iii) and (iv), $M_{0}B_{0}$ and $2I\pm M_{0}B_{0}$ are injective. Let $U=(U_{ij})_{1\le i,j\le 4}.$ From $U{\widetilde{B}}=-{\widetilde{B}}U,$ one has

$$\begin{aligned}&\left( \begin{array}{cccc} 0 &{} \quad -2U_{12} &{}\quad 2U_{13} &{} \quad U_{14}M_{0}B_{0} \\ 0 &{} \quad -2U_{22} &{}\quad 2U_{23} &{}\quad U_{24}M_{0}B_{0} \\ 0 &{} \quad -2U_{32} &{}\quad 2U_{33} &{}\quad U_{34}M_{0}B_{0} \\ 0 &{}\quad -2U_{42} &{}\quad 2U_{43} &{}\quad U_{44}M_{0}B_{0} \\ \end{array} \right) \\&\quad =\left( \begin{array}{cccc} 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 2U_{21} &{}\quad 2U_{22} &{}\quad 2U_{23} &{}\quad 2U_{24}\\ -2U_{31} &{} \quad -2U_{32} &{} \quad -2U_{33} &{} \quad -2U_{34} \\ -M_{0}B_{0}U_{41} &{}\quad -M_{0}B_{0}U_{42} &{}\quad -M_{0}B_{0}U_{43} &{} \quad -M_{0}B_{0}U_{44}\\ \end{array}\right) . \end{aligned}$$

It is easy to see that $U_{12}$, $U_{13}$, $U_{21}$, $U_{31}$, $U_{22}$ and $U_{33}$ are zero operators. Besides, $U_{14}$ and $U_{41}$ are zero operators since $M_{0}B_{0}$ is injective and dense. $U_{24}$ and $U_{42}$ are zero operators since $2I-M_{0}B_{0}$ is injective and dense. $U_{34}$ and $U_{43}$ are zero operators since $2I+M_{0}B_{0}$ is injective and dense. Hence, $U=U_{11}\oplus \left( \begin{array}{cc} 0 &{}\quad U_{23} \\ U_{32} &{} \quad 0 \\ \end{array} \right) \oplus U_{44},$ where $U_{11}\in {\mathcal {B}}({\mathcal {N}}(B))$, $U_{44}\in {\mathcal {B}}({\mathcal {H}}_{0})$, $U_{23}\in {\mathcal {B}}({\mathcal {H}}_{3},{\mathcal {H}}_{2})$ and $U_{32}\in {\mathcal {B}}({\mathcal {H}}_{2},{\mathcal {H}}_{3})$ are unitary operators. Therefore, $\dim {\mathcal {H}}_{2}=\dim {\mathcal {H}}_{3}.$$\square$