Halmos' two projections theorem for Hilbert $C^*$-module operators and the Friedrichs

TL;DR

This paper extends Halmos' two projections theorem to Hilbert $C^*$-modules using the concept of harmonious projections, providing new characterizations of submodules and a norm equation related to the Friedrichs angle.

Contribution

It introduces harmonious projections in Hilbert $C^*$-modules and generalizes Halmos' theorem to this setting, with applications to Friedrichs angle characterization.

Findings

Extended Halmos' theorem to harmonious projections on Hilbert $C^*$-modules.

Provided new characterizations of closed submodules and projections.

Proved a norm equation related to Friedrichs angle in this framework.

Abstract

Halmos' two projections theorem for Hilbert space operators is one of the fundamental results in operator theory. In this paper, we introduce the term of two harmonious projections in the context of adjointable operators on Hilbert $C^*$-modules, extend Halmos' two projections theorem to the case of two harmonious projections. We also give some new characterizations of the closed submodules and their associated projections. As an application, a norm equation associated to a characterization of the Friedrichs angle is proved to be true in the framework of Hilbert $C^*$-modules.

Equations

$$\displaystyle c(M,N)=\sup\big{\{}|\langle x,y\rangle|:x\in\widetilde{M},y\in\widetilde{N},\|x\|\leq 1,\|y\|\leq 1\big{\}},$$

$$c(M,N)=\big{\|}P_{M}P_{N}\big{(}I-P_{M\cap N}\big{)}\|,$$

$$\big{\|}PQ\big{(}I-P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}\big{)}\big{\|}=\big{\|}(I-P)(I-Q)\big{(}I-P_{\mathcal{N}(P)\cap\mathcal{N}(Q)}\big{)}\big{\|},$$

$$c(M,N)=c(M^{\perp },N^{\perp }).$$

$$\Vert x{\Vert }^2=\Vert P(x){\Vert }^2+\Vert (I-P)x{\Vert }^2,$$

$$\mathcal{R}(P)\cap \mathcal{R}(Q)=\mathcal{R}(P)\cap \mathcal{N}(Q)=\mathcal{N}(P)\cap \mathcal{R}(Q)=\mathcal{N}(P)\cap \mathcal{N}(Q)=\{0\}.$$

$$\overline{\mathcal{R}(P+Q)},\overline{\mathcal{R}(P+I-Q)},\overline{\mathcal{R}(I-P+Q)}\text{mbox}nor\overline{\mathcal{R}(I-P+I-Q)}$$

$$P=\left(\begin{array}[]{cc}I&0\\ 0&0\end{array}\right)\ \mbox{and}\ Q=\left(\begin{array}[]{cc}C^{2}&CDW^{-1}\\ WCD&WD^{2}W^{-1}\end{array}\right)\,.$$

$$H_1\oplus H_2:=\left\{(h_1,h_2{)}^T:h_i\in H_i,i=1,2\right\},$$

$$\left<(x_{1},y_{1})^{T},(x_{2},y_{2})^{T}\right>=\big{<}x_{1},x_{2}\big{>}+\big{<}y_{1},y_{2}\big{>}$$

$$H_1\dotplus H_2:=\{h_1+h_2:h_i\in H_i,i=1,2\}.$$

$$M^{\perp}=\big{\{}x\in H:\langle x,y\rangle=0\ \mbox{for every }\ y\in M\big{\}}.$$

$$\big{(}M^{\bot}\big{)}^{\bot}=\mathcal{R}\big{(}I-(I-P)\big{)}=\mathcal{R}(P)=M.$$

$$\overline{\mathcal{R}(T^{\ast }T)}=\overline{\mathcal{R}(T^{\ast })}\text{mbox}and\overline{\mathcal{R}(T{T}^{\ast })}=\overline{\mathcal{R}(T)}.$$

$$\overline{\mathcal{R}(P)+\mathcal{R}(Q)}=\overline{\mathcal{R}(P+Q)}.$$

$$\mathcal{R}(P)\subseteq \overline{\mathcal{R}(P+Q)}\text{mbox}and\mathcal{R}(Q)\subseteq \overline{\mathcal{R}(P+Q)}.$$

$$\displaystyle T=\left(\begin{array}[]{cc}0&0\\ P&Q\\ \end{array}\right)\in\mathcal{L}(H\oplus H).$$

$$\displaystyle\mathcal{R}(T)=\{0\}\oplus\big{(}\mathcal{R}(P)+\mathcal{R}(Q)\big{)},\quad\mathcal{R}(TT^{*})=\{0\}\oplus\big{(}\mathcal{R}(P+Q)\big{)}.$$

$$\displaystyle\lim_{n\to\infty}\|T_{n}x-P_{\overline{\mathcal{R}(T)}}x\|=0\ \mbox{for all $x$ in $H$}.$$

$$A_n=P-P{T}_{n}P,B_n=P{T}_{n}Q,C_n=Q-Q{T}_{n}Q.$$

$$\big{(}\mathcal{R}(P)\cap\mathcal{R}(Q)\big{)}^{\perp}=\overline{\mathcal{R}(2I-P-Q)};$$

$$\underset{n\to \infty }{\lim }{A}_n=\underset{n\to \infty }{\lim }{B}_n=\underset{n\to \infty }{\lim }{C}_n=P_{\mathcal{R}(P)\cap \mathcal{R}(Q)}\text{mbox}inthestricttopology.$$

$$P_{\overline{\mathcal{R}(P+Q)}}P=P{P}_{\overline{\mathcal{R}(P+Q)}}=P,P_{\overline{\mathcal{R}(P+Q)}}Q=Q{P}_{\overline{\mathcal{R}(P+Q)}}=Q.$$

$$\underset{n\to \infty }{\lim }(P+Q)T_n=P_{\overline{\mathcal{R}(P+Q)}}\text{mbox}inthestricttopology,$$

$$\displaystyle\lim\limits_{n\to\infty}\big{[}(P+Q)T_{n}P-P\big{]}=0,\quad\lim\limits_{n\to\infty}\big{[}(P+Q)T_{n}Q-Q\big{]}=0.$$

$$\displaystyle A_{n}=\big{[}P-(P+Q)T_{n}P\big{]}+B_{n}^{*},\quad C_{n}=\big{[}Q-(P+Q)T_{n}Q\big{]}+B_{n},$$

$$\displaystyle B_{n}=\big{[}(P+Q)T_{n}Q-Q\big{]}+\big{[}Q-QT_{n}(P+Q)\big{]}+B_{n}^{*}.$$

$$\underset{n\to \infty }{\lim }{A}_n=\underset{n\to \infty }{\lim }{B}_n=\underset{n\to \infty }{\lim }{C}_n=E\text{mbox}inthestricttopology.$$

$$\Big{(}P+Q+\frac{1}{n}I\Big{)}x=\Big{(}2+\frac{1}{n}\Big{)}x\ \mbox{for every $n\in\mathbb{N}$},$$

$$2Ex=2\underset{n\to \infty }{\lim }{B}_{n}x=x.$$

Full text

1. Introduction

Let $M$ and $N$ be two closed subspaces of a Hilbert space $H$. The Friedrichs angle [9], denoted by $\alpha(M,N)$, is the unique angle in $[0,\frac{\pi}{2}]$ whose cosine is equal to $c(M,N)$, where

[TABLE]

in which $\widetilde{M}=M\cap(M\cap N)^{\bot}$ and $\widetilde{N}=N\cap(M\cap N)^{\bot}$. It is known [6] that

[TABLE]

where the notation $P_{E}$ stands for the projection from $H$ onto its closed subspace $E$. Given any projection $P$ on $H$, let $\mathcal{R}(P)$ and $\mathcal{N}(P)$ denote the range and the null space of $P$, respectively. It is proved in [6] that for every two projections $P$ and $Q$ on $H$,

[TABLE]

which gives a characterization of the Friedrichs angle as

[TABLE]

The proof of equation (1.2) given in [6] relies on the Pythagorean theorem, that is,

[TABLE]

where $P$ is any projection on $H$ and $x\in H$ is arbitrary.

The Hilbert $C^{*}$-module is the generalization of the Hilbert space by allowing the inner product to take values in certain $C^{*}$-algebra $A$ instead of the complex field $\mathbb{C}$. The purpose of this paper is to investigate the validity of (1.2) in the Hilbert $C^{*}$-module case. Let $H$ be a Hilbert $C^{*}$-module and $M$ be a closed submodule of $H$. It can be verified directly that the notation $P_{M}$ is meaningful if and only if $M$ is orthogonally complemented in $H$, and in this case $(M^{\bot})^{\bot}=M$. So, for projections $P$ and $Q$ on $H$, one can associate the determination of the orthogonal complementarity of $\mathcal{R}(P)\cap\mathcal{R}(Q)$ and $\mathcal{N}(P)\cap\mathcal{N}(Q)$ to (1.2).

Let $P$ and $Q$ be two projections on a Hilbert $C^{*}$-module. It is clear that $\overline{\mathcal{R}(P+Q)}^{\bot}=\mathcal{N}(P)\cap\mathcal{N}(Q)$, so $P_{\mathcal{N}(P)\cap\mathcal{N}(Q)}$ is meaningful whenever $\overline{\mathcal{R}(P+Q)}$ is orthogonally complemented. This observation together with (1.2) lead us to study such a topic: Assume that $\overline{\mathcal{R}(P+Q)}$ is orthogonally complemented, under what conditions $\overline{\mathcal{R}(I-P+I-Q)}$ is also orthogonally complemented. In Section 2, we will give some necessary and sufficient conditions on this topic. Another observation is that the orthogonal complementarity of $\mathcal{R}(P)\cap\mathcal{R}(Q)$ is obviously guaranteed if $\mathcal{R}(P)\cap\mathcal{R}(Q)=\{0\}$. So we turn to consider the projections $P$ and $Q$ such that $\mathcal{R}(P)$ and $\mathcal{R}(Q)$ are in generic position [10], that is, $P$ and $Q$ satisfy

[TABLE]

In Section 3, we have managed to construct such two projections $P$ and $Q$ on some Hilbert $C^{*}$-module $H$ such that neither

[TABLE]

is equal to $H$ ensuring that none of them is orthogonally complemented in $H$. We call such $P$ and $Q$ extremely discomplementable projections.

It is notable that the Pythagorean theorem is no longer true for a general Hilbert $A$-module $H$, since for a projection $P$ on $H$ and an element $x\in H$, the associated two positive elements $a=\langle Px,x\rangle$ and $b=\langle(I-P)x,x\rangle$ in the given $C^{*}$-algebra $A$ will only satisfy the inequality $\|a+b\|\leq\|a\|+\|b\|$ rather than the equality $\|a+b\|=\|a\|+\|b\|$. This leads us to study the validity of (1.2) by constructing unitary operators based on the generalized Halmos’ two projections theorem.

Halmos’ two projections theorem [10] for Hilbert space operators is one of the fundamental results in operator theory, see [1], [7] and [20]. It says that if $P$ and $Q$ are two projections on a Hilbert space such that $\mathcal{R}(P)$ and $\mathcal{R}(Q)$ are in generic position, then there exist a unitary map $W$ from $M_{1}=\mathcal{R}(P)$ onto $M_{2}=\mathcal{N}(P)$, and two positive operators $C$ and $D$ on $M_{1}$ with $C^{2}+D^{2}=I,CD=DC$ so that under the orthogonal decomposition $H=M_{1}\dotplus M_{2}$, we have

[TABLE]

It has applications in many areas such as the $cs$-decomposition, characterizations of the closedness of the sum of two subspaces, derivations of von Neumann’s formula and the Feldman–Krupnik–Markus formulas, as well as computations of various angles and gaps between two subspaces of a Hilbert space. For the details, the reader is referred to the excellent survey [3] and the references therein; see also [2, 4, 19]. As far as we know, little has been done in the generalization of Halmos’ two projections theorem for Hilbert $C^{*}$-module operators, which is the concern of this paper. In Section 4, we have made some new characterizations of the closed submodules and the associated projections; see (4.1), (4.5), (4.6), (4.16) and (4.17), respectively. Here the key point is the introduction of the harmonious projections described in Definition 4.1. It is proved in Theorem 4.6 that Halmos’ two projections theorem remains to be true for every two harmonious projections. As an application, in Section 5 we show that equation (1.2) is true firstly in Lemma 5.1 for every two harmonious projections, secondly in Lemma 5.4 for every two projections $P$ and $Q$ such that $\overline{\mathcal{R}(P+Q)}$ and $\overline{\mathcal{R}(2I-P-Q)}$ are both orthogonally complemented, and finally in Theorem 5.12 for every two projections $P$ and $Q$ whenever $\mathcal{R}(P)\cap\mathcal{R}(Q)$ and $\mathcal{N}(P)\cap\mathcal{N}(Q)$ are both orthogonally complemented111Note that the notations $P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}$ and $P_{\mathcal{N}(P)\cap\mathcal{N}(Q)}$ in equation (1.2) are meaningful if and only if such an orthogonal complementarity condition is satisfied.. Overall, it helps us to extend our viewpoint of the geometry of Hilbert $C^{*}$-modules; see, e.g., [8, 14, 17]

Let us briefly recall some basic knowledge about Hilbert $C^{*}$-modules and adjointable operators. An inner-product module over a $C^{*}$-algebra $A$ is a right $A$-module $H$ equipped with an $A$-valued inner product $\langle\cdot,\cdot\rangle:H\times H\to A$ that is $\mathbb{C}$-linear and $A$-linear in the second variable and satisfies $\langle x,y\rangle^{*}=\langle y,x\rangle$ as well as $\langle x,x\rangle\geq 0$ with equality if and only if $x=0$. An inner-product $A$-module $H$ which is complete with respect to the induced norm $\|x\|=\sqrt{\|\langle x,x\rangle\|}\,\,(x\in H)$ is called a (right) Hilbert $A$-module.

Suppose that $H$ and $K$ are two Hilbert $A$-modules, let $\mathcal{L}(H,K)$ be the set of operators $T:H\to K$ for which there is an operator $T^{*}:K\to H$ such that $\langle Tx,y\rangle=\langle x,T^{*}y\rangle$ for all $x\in H$ and $y\in K$. Each member in $\mathcal{L}(H,K)$ is called an adjointable operator. When $H=K$, $\mathcal{L}(H,H)$, abbreviated to $\mathcal{L}(H)$, is a $C^{*}$-algebra. By a positive operator we mean an operator $T\in\mathcal{L}(H)$ such that $\langle Tx,x\rangle\geq 0$ for all $x\in H$ [11, Lemma 4.1]. The strict topology (strong∗ topology) on $\mathcal{L}(H,K)$ is defined to be the topology determined by the seminorms $T\mapsto\|Tx\|\,\,(x\in H)$ and $T\mapsto\|T^{*}y\|\,\,(y\in K)$. For every $T\in{\mathcal{L}}(H,K)$, the range and the null space of $T$ are designated by $\mathcal{R}(T)$ and $\mathcal{N}(T)$, respectively. By $I_{H}$ (or simply $I$) we denote the identity operator on $H$. The reader is referred to [11, 16] for some other basic notions related to Hilbert $C^{*}$-modules.

In this paper, the notations of “$\oplus$” and “$\dotplus$” are used with different meanings for the sake of reader’s convenience. For Hilbert $A$-modules $H_{1}$ and $H_{2}$, let

[TABLE]

which is also a Hilbert $A$-module whose $A$-valued inner product is given by

[TABLE]

for $x_{i}\in H_{1},y_{i}\in H_{2},i=1,2$. If both $H_{1}$ and $H_{2}$ are submodules of a Hilbert $A$-module $H$ such that $H_{1}\cap H_{2}=\{0\}$, then we set

[TABLE]

2. Orthogonal complementarity of closed submodules associated to two projections

Throughout the rest of this paper, $A$ is a $C^{*}$-algebra, $H$ and $K$ are Hilbert $A$-modules. By a projection, we mean an operator $P\in\mathcal{L}(H)$ such that $P=P^{2}=P^{*}$. Recall that a closed submodule $M$ of $H$ is said to be orthogonally complemented in $H$ if $H=M\dotplus M^{\perp}$, where

[TABLE]

In this case, the projection from $H$ onto $M$ is denoted by $P_{M}$.

Remark 2.1**.**

Let $M$ be a closed submodule of $H$. Then $M$ is orthogonally complemented in $H$ if and only if there exists a projection $P\in\mathcal{L}(H)$ such that $\mathcal{R}(P)=M$. In this case,

[TABLE]

Lemma 2.2**.**

[11, Proposition 3.7]* Let $T\in\mathcal{L}(H,K)$. Then*

[TABLE]

Lemma 2.3**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections. Then

[TABLE]

In particular,

[TABLE]

Proof.

Put

[TABLE]

Clearly,

[TABLE]

The equations above together with Lemma 2.2 yield (2.2), which gives (2.3) immediately. ∎

Lemma 2.4**.**

[12, Proposition 2.5]* Let $T\in\mathcal{L}(H)$ be positive such that $\overline{\mathcal{R}(T)}$ is orthogonally complemented in $H$. For every $n\in\mathbb{N}$, let $T_{n}:=\left(\frac{1}{n}I+T\right)^{-1}T$. Then $\lim\limits_{n\to\infty}T_{n}=P_{\overline{\mathcal{R}(T)}}$ in the strict topology, that is,*

[TABLE]

Suppose that $P,Q\in\mathcal{L}(H)$ are two projections such that $\overline{\mathcal{R}(P+Q)}$ is orthogonally complemented in $H$. It is interesting to determine conditions under which $\overline{\mathcal{R}(I-P+I-Q)}$ is still orthogonally complemented in $H$. We provide such a result as follows.

Theorem 2.5**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections such that $\overline{\mathcal{R}(P+Q)}$ is orthogonally complemented in $H$. For every $n\in\mathbb{N}$, let $T_{n}=(P+Q+\frac{1}{n}I)^{-1}$ and

[TABLE]

Then the following statements are equivalent:

1. (i)

one of $\{A_{n}\},\{B_{n}\}$, and $\{C_{n}\}$ is convergent in the strict topology; 2. (ii)

all of $\{A_{n}\},\{B_{n}\}$, and $\{C_{n}\}$ are convergent to the same limit in the strict topology; 3. (iii)

$\mathcal{R}(P)\cap\mathcal{R}(Q)$* is orthogonally complemented in $H$ such that*

[TABLE] 4. (iv)

$\overline{\mathcal{R}(2I-P-Q)}$* is orthogonally complemented in $H$.*

In each case,

[TABLE]

Proof.

According to (2.3), we have

[TABLE]

Also, by Lemma 2.4 we have

[TABLE]

which is combined with (2.10) to conclude that in the strict topology,

[TABLE]

Note that for each $n\in\mathbb{N}$,

[TABLE]

and

[TABLE]

Hence the equivalence of (i) and (ii) can be derived from the equations above together with (2.11), (2.12), and (2.13). Furthermore, if any of $A_{n},B_{n}$ and $C_{n}$ has the limit in the strict topology, then all of them will have the same limit in the strict topology.

(ii)$\Longrightarrow$(iii): Suppose that

[TABLE]

From the definitions of $B_{n}$ and $C_{n}$ in (2.8), we have $\mathcal{R}(B_{n})\subseteq\mathcal{R}(P)$ and $\mathcal{R}(C_{n})\subseteq\mathcal{R}(Q)$ for each $n$. Employing (2.14) we get $\mathcal{R}(E)\subseteq\mathcal{R}(P)\cap\mathcal{R}(Q)$. Conversely, given each $x\in\mathcal{R}(P)\cap\mathcal{R}(Q)$, we have

[TABLE]

which means that $T_{n}x=\frac{1}{2+\frac{1}{n}}x$ and thus $B_{n}x=\frac{1}{2+\frac{1}{n}}x$. Hence

[TABLE]

This ensures that $\mathcal{R}(E)=\mathcal{R}(P)\cap\mathcal{R}(Q)$.

Now for every $x\in H$, since $Ex\in\mathcal{R}(P)\cap\mathcal{R}(Q)$, from (2.15) we can get

[TABLE]

Hence the operator $2E$ is an idempotent. Moreover, as

[TABLE]

we see that $2E$ is also self-adjoint. Therefore, $2E$ is actually a projection. It follows that $\mathcal{R}(P)\cap\mathcal{R}(Q)$ is orthogonally complemented in $H$ and

[TABLE]

Next, we prove that (2.9) is valid. It is obvious that

[TABLE]

On the other hand, we have $\lim\limits_{n\to\infty}B_{n}^{*}=E^{*}=E$ in the strict topology. Therefore, given any $x\in H$, we have

[TABLE]

Due to (2.16) and (2.2), we have

[TABLE]

The proof of (2.9) is then finished.

(iii)$\Longrightarrow$(iv) It is illustrated by Remark 2.1.

(iv)$\Longrightarrow$(i): Assume that $\overline{\mathcal{R}(2I-P-Q)}$ is orthogonally complemented in $H$. Since $\overline{\mathcal{R}(2I-P-Q)}^{\bot}=\mathcal{N}(I-P)\cap\mathcal{N}(I-Q)=\mathcal{R}(P)\cap\mathcal{R}(Q)$, by (2.1) we know that $\mathcal{R}(P)\cap\mathcal{R}(Q)$ is also orthogonally complemented in $H$ such that (2.9) is satisfied. So the notation $P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}$ is meaningful, and $H$ can be decomposed orthogonally as

[TABLE]

In what follows, we prove that

[TABLE]

First, given any $x\in\mathcal{R}(2I-P-Q)$, there exists some $u\in H$ such that $x=(2I-P-Q)u=(I-P+I-Q)u$. Then

[TABLE]

Note that for every $n\in\mathbb{N}$, we have

[TABLE]

so $\|PT_{n}^{\frac{1}{2}}\|=\|(PT_{n}^{\frac{1}{2}})^{*}\|=\|T_{n}^{\frac{1}{2}}P\|<1$. Similarly, $\|T_{n}^{\frac{1}{2}}Q\|<1$. As a result,

[TABLE]

The boundedness of $\{B_{n}\}$ together with (2.19) indicates

[TABLE]

Next, from the proof of (ii)$\Longrightarrow$(iii) we know that

[TABLE]

The assertion $\lim\limits_{n\to\infty}2B_{n}x=P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}x\,\,(x\in H)$ follows from (2.17), (2.20) and (2.21). Similarly, one can show that $\lim\limits_{n\to\infty}2B_{n}^{*}x=P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}x\,\,(x\in H)$. Thus (2.18) holds true. ∎

Remark 2.6**.**

It is remarkable that there exist a Hilbert $C^{*}$-module $H$ and two projections $P,Q\in\mathcal{L}(H)$ such that $\mathcal{R}(P)\cap\mathcal{R}(Q)$ is orthogonally complemented in $H$, whereas (2.9) is not true. Such an example is constructed in the next section.

3. An example of extremely discomplementable projections

Inspired by [15, Section 3], we construct a Hilbert $C^{*}$-module $H$ and two extremely discomplementable projections on it as follows.

Let $M_{2}(\mathbb{C})$ be the set of all $2\times 2$ complex matrices and $\|\cdot\|$ be the spectral norm on $M_{2}(\mathbb{C})$. Let A=C\big{(}[0,1];M_{2}(\mathbb{C})\big{)} be the set of all continuous matrix-valued functions from $[0,1]$ to $M_{2}(\mathbb{C})$. Set

[TABLE]

With the $*$-operation above together with the usual algebraic operations, $A$ is a unital $C^{*}$-algebra. Therefore, $A$ itself becomes a Hilbert $A$-module with the usual $A$-valued inner product given by

[TABLE]

Let $e$ be the unit of $A$, that is, e(t)=\left(\begin{array}[]{cc}1&0\\ 0&1\\ \end{array}\right) for every $t\in[0,1]$. It is known that $A\cong\mathcal{L}(A)$ via $a\to L_{a}$, where L_{a}(x)=ax\ \mbox{for every a,x\in A}; furthermore $(L_{a})^{*}=L_{a^{*}}$. Indeed, for every $T\in\mathcal{L}(A)$ and $x\in A$, we have

[TABLE]

For simplicity, we put

[TABLE]

Let $\widetilde{P},\widetilde{Q}\in A$ be determined by the matrix-valued functions

[TABLE]

Then both $P=L_{\widetilde{P}}$ and $Q=L_{\widetilde{Q}}$ are projections in $\mathcal{L}(A)$. Let $x\in A$ be determined by x(t)=\big{(}x_{ij}(t)\big{)}_{1\leq i,j\leq 2}, where each $x_{ij}(t)$ is a continuous complex-valued function on $[0,1]$. If $x\in\mathcal{N}(P)$, then obviously $x_{11}(t)=0$ and $x_{12}(t)=0$. So if furthermore $x\in\mathcal{N}(Q)$, then

[TABLE]

Note that $s_{t}\neq 0$ for every $t\in(0,1]$, so the equations above together with the continuity of both $x_{21}(t)$ and $x_{22}(t)$ at $t=0$ yield $x_{21}(t)=0$ and $x_{22}(t)=0$. Therefore, $\mathcal{N}(P)\cap\mathcal{N}(Q)=\{0\}$; or equivalently, $\mathcal{N}(P+Q)=\{0\}$.

Consider an element $x\in A$ having the form x(t)=\big{(}x_{ij}(t)\big{)}_{1\leq i,j\leq 2}. Since $s_{0}=0$, we have

[TABLE]

It follows immediately that $e\notin\overline{\mathcal{R}(P+Q)}$, whence

[TABLE]

Therefore, $\overline{\mathcal{R}(P+Q)}$ is not orthogonally complemented.

Similarly, it can be proved that

[TABLE]

and the unit $e$ is also not contained in any one of the remaining three closures in (1.4).

4. Halmos’ two projections theorem for Hilbert $C^{*}$-module operators

The purpose of this section is to generalize Halmos’ two projections theorem to the case of the Hilbert $C^{*}$-module. We start this section with the following lemma.

Lemma 4.1**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections such that $\overline{\mathcal{R}(I-Q+P)}$ is orthogonally complemented in $H$. Then $\overline{\mathcal{R}(QP)}$ is also orthogonally complemented in $H$ and

[TABLE]

Proof.

First, we prove that

[TABLE]

Indeed, it is clear that

[TABLE]

which leads to the orthogonal decomposition of $H$ as

[TABLE]

since $\overline{\mathcal{R}(I-Q+P)}$ is orthogonally complemented. As a result, the notation of $P_{\mathcal{R}(Q)\cap\mathcal{N}(P)}$ is meaningful.

Now, given any $x\in H$, $x$ can be decomposed as

[TABLE]

Thus

[TABLE]

As $u\in\overline{\mathcal{R}(I-Q+P)}$, there exists a sequence $\{x_{n}\}\subseteq H$ such that $(I-Q+P)x_{n}\rightarrow u.$ Then $QPx_{n}=Q(I-Q+P)x_{n}\rightarrow Qu$, and hence $Qu\in\overline{\mathcal{R}(QP)}$. It follows from (4.3) that

[TABLE]

which gives (4.2) since $x\in H$ is arbitrary and $\mathcal{R}(QP)\subseteq\mathcal{R}(Q)$.

Next, we prove that

[TABLE]

In fact, given any $s\in H$ and $t\in\mathcal{R}(Q)\cap\mathcal{N}(P)$, we have

[TABLE]

The proof of (4.4) is then finished by the continuity of the $A$-valued inner product with respect to each variable.

Finally, it follows from (4.2) and (4.4) that

[TABLE]

Hence (4.1) is satisfied. ∎

Lemma 4.2**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections such that $\overline{\mathcal{R}(2I-P-Q)}$ is orthogonally complemented in $H$. Then $\mathcal{R}(P)\cap\mathcal{R}(Q)$ is also orthogonally complemented in $H$ and an orthogonal decomposition of $\overline{\mathcal{R}(QP)}$ can be given as

[TABLE]

Proof.

Since $\overline{\mathcal{R}(2I-P-Q)}$ is orthogonally complemented in $H$, the notation $P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}$ is meaningful, and $H$ can be decomposed orthogonally as (2.17). Replacing $Q$ and $P$ in Lemma 4.1 with $P$ and $I-Q$ respectively, by (4.1) we obtain

[TABLE]

which clearly gives

[TABLE]

Note that \overline{\mathcal{R}\big{(}QP(I-Q)\big{)}} and $\mathcal{R}(P)\cap\mathcal{R}(Q)$ are orthogonal to each other, so their sum is closed. Hence

[TABLE]

since $QPx=x$ for every $x\in\mathcal{R}(P)\cap\mathcal{R}(Q)$. The proof of (4.5) is then finished. ∎

A direct application of Lemmas 4.1 and 4.2 is as follows.

Corollary 4.3**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections. If both $\overline{\mathcal{R}(I-Q+P)}$ and $\overline{\mathcal{R}(2I-P-Q)}$ are orthogonally complemented in $H$, then \overline{\mathcal{R}\big{(}QP(I-Q)\big{)}} is also orthogonally complemented in $H$ and

[TABLE]

Definition 4.1**.**

Two projections $P,Q\in\mathcal{L}(H)$ are said to be harmonious if the four closures in (1.4) are all orthogonally complemented in $H$.

Suppose that $P,Q\in\mathcal{L}(H)$ are two harmonious projections. Let

[TABLE]

Since $\overline{\mathcal{R}(P+Q)}^{\bot}=H_{4}$ and $\overline{\mathcal{R}(P+Q)}$ is orthogonally complemented in $H$, we conclude from (2.1) that $H_{4}$ is also orthogonally complemented in $H$. Similarly, $H_{1},H_{2}$ and $H_{3}$ are all orthogonally complemented in $H$. Let

[TABLE]

and put

[TABLE]

With the notations given above, a unitary operator $U_{P,Q}:H\to\oplus_{i=1}^{6}H_{i}$ is given by

[TABLE]

with the property that

[TABLE]

It follows that

[TABLE]

where

[TABLE]

in which $P_{5}QP_{5}|_{H_{5}}$ is the restriction of the operator $P_{5}QP_{5}$ on $H_{5}$. The same convention is taken for $P_{5}QP_{6}|_{H_{6}}$ and $P_{6}QP_{6}|_{H_{6}}$.

Lemma 4.4**.**

Suppose that $P,Q\in\mathcal{L}(H)$ are two harmonious projections. Let $H_{i},P_{i}$ $(1\leq i\leq 6$) be defined by (4.7)–(4.11), respectively. Then

[TABLE]

Proof.

Exchanging $P$ with $Q$, we observe from (4.6) and (4.10) that

[TABLE]

Moreover, we have $(I-P_{1})(I-P)=I-P$ and

[TABLE]

which shows that

[TABLE]

The inclusions of the above sets together with (4.18) yield $\overline{\mathcal{R}(P_{5}Q)}=H_{5}$.

Note that $I-Q=(I-P_{1})(I-Q)$ and

[TABLE]

so

[TABLE]

Meanwhile, by (4.6), we have

[TABLE]

so \overline{\mathcal{R}(QP_{5})}\subseteq\overline{\mathcal{R}\big{(}QP(I-Q)\big{)}}. Therefore \overline{\mathcal{R}(QP_{5})}=\overline{\mathcal{R}\big{(}QP(I-Q)\big{)}}. This completes the proof of (4.16).

Replacing the pair $(P,Q)$ with $(I-P,Q)$, we have

[TABLE]

Similarly,

[TABLE]

Note that $Q(I-P)(I-Q)=-QP(I-Q)$, so from (4.16) we obtain

[TABLE]

Definition 4.2**.**

[15] An operator $T\in\mathcal{L}(H,K)$ is said to be semi-regular if $\overline{\mathcal{R}(T)}$ and $\overline{\mathcal{R}(T^{*})}$ are orthogonally complemented in $K$ and $H$, respectively.

Lemma 4.5**.**

([12, Lemma 3.6] and [21, Proposition 15.3.7])* Let $T\in\mathcal{L}(H,K)$ be semi-regular. Then there exists a partial isometry $U\in\mathcal{L}(H,K)$ such that*

[TABLE]

Halmos’ two projections theorem for Hilbert space operators has several equivalent versions [3], one of which turns out to be [5, Theorem 1.4]. It can be generalized for Hilbert $C^{*}$-module operators as follows.

Theorem 4.6**.**

(cf. [5, Theorem 1.4])* Suppose that $P,Q\in\mathcal{L}(H)$ are two harmonious projections. Let $H_{i},P_{i}$ $(1\leq i\leq 6$) be defined by (4.7)–(4.11), respectively. Then the operator $T$ given by (4.15) can be characterized as*

[TABLE]

where $U_{0}\in\mathcal{L}(H_{6},H_{5})$ is a unitary operator, $Q_{0}$ is the restriction of $P_{5}QP_{5}$ on $H_{5}$, and both $Q_{0}$ and $I_{H_{5}}-Q_{0}$ are positive, injective and contractive.

Proof.

By Lemma 4.4 and (4.6), all $\overline{\mathcal{R}(P_{5}Q)}$, $\overline{\mathcal{R}(QP_{5})}$, $\overline{\mathcal{R}(P_{6}Q)}$ and $\overline{\mathcal{R}(QP_{6})}$ are orthogonally complemented in $H$. Therefore, by Lemma 4.5, there exist partial isometries $U,V\in\mathcal{L}(H)$ such that

[TABLE]

Let $Q_{0}\in\mathcal{L}(H_{5})$ and $Q_{1}\in\mathcal{L}(H_{6})$ be defined respectively by

[TABLE]

Clearly, $Q_{0}$ is positive and contractive. Furthermore, we have

[TABLE]

In fact, if $x=P_{5}x\in H_{5}$ is given such that $Q_{0}x=0$, then $Qx=QP_{5}x=0$, which means that $x\in H_{5}\cap\mathcal{N}(Q)=H_{5}\cap\mathcal{R}(P)\cap\mathcal{N}(Q)=H_{5}\cap H_{2}=\{0\}$. Similarly, if $y\in H_{5}$ is such that $(I_{H_{5}}-Q_{0})y=0$, then $P_{5}(I-Q)P_{5}y=0$, and hence $(I-Q)y=(I-Q)P_{5}y=0$, which gives $y\in H_{5}\cap H_{1}=\{0\}$. This completes the proof of (4.25).

Let $U_{0}=UV|_{H_{6}}$ be the restriction of $UV$ on $H_{6}$. Then by (4.21) and (4.22), we see that $U_{0}\in\mathcal{L}(H_{6},H_{5})$ with $U_{0}^{*}=V^{*}U^{*}|_{H_{5}}$. We prove that $U_{0}$ is a unitary. Indeed, for every $x\in H_{6}$, by (4.22) and (4.23) we have

[TABLE]

which means that $U_{0}^{*}U_{0}=I_{H_{6}}$. Similarly, for every $y\in H_{5}$, by (4.21) and (4.23) we have

[TABLE]

whence $U_{0}U_{0}^{*}=I_{H_{5}}$. Thus $U_{0}\in\mathcal{L}(H_{6},H_{5})$ is a unitary.

Next, we prove that

[TABLE]

where $Q_{0}$ and $Q_{1}$ are defined by (4.24). In fact, from (4.20) and (4.21) we have

[TABLE]

therefore the operator $T$ defined by (4.15) can be expressed alternately as

[TABLE]

which, in virtue of $T^{2}=T$, gives

[TABLE]

The above equation together with (4.25) yields

[TABLE]

Taking $*$-operation, we get

[TABLE]

Once again, by the injectivity of $Q_{0}^{\frac{1}{2}}$, we can obtain $Q_{0}+U_{0}Q_{1}U_{0}^{*}=I_{H_{5}}$, which clearly leads to (4.26). Since $U_{0}$ is a unitary, from (4.26) we can obtain

[TABLE]

Formula (4.19) for $T$ then follows from (4.26)–(4.28). ∎

5. The norm equation concerning the characterization of the Friedrichs angle

In this section, we focus on the study of the validity of equation (1.2). First, we give a partial positive answer as follows.

Lemma 5.1**.**

Equation (1.2) is true for every two harmonious projections $P$ and $Q$.

Proof.

Let $P,Q\in\mathcal{L}(H)$ be two harmonious projections. It needs only to prove that

[TABLE]

Let $H_{i},P_{i}$ $(1\leq i\leq 6$) be defined by (4.7)–(4.11), respectively. Denote $I_{H_{i}}$ simply by $I$ for $1\leq i\leq 6$. Let $U_{P,Q}$ be defined by (4.12) and for each $n\in\mathbb{N}$, put

[TABLE]

It can be deduced directly from (4.13), (4.14) and (4.19) that

[TABLE]

where

[TABLE]

Using the above equations and (5.2), we obtain

[TABLE]

Similarly, we have

[TABLE]

where

[TABLE]

Using the above equations and (5.3), we obtain

[TABLE]

Let $\widetilde{U_{0}}\in\mathcal{L}(H_{5}\oplus H_{6})$ be the unitary defined by

[TABLE]

Then $\widetilde{U_{0}}^{*}=-\widetilde{U_{0}}$ and

[TABLE]

Hence $\widetilde{U_{0}}X_{n}^{-1}\widetilde{U_{0}}^{*}=Y_{n}^{-1}$. Therefore, by (5.4)–(5.7), we have

[TABLE]

where $\widetilde{U}$ is the unitary defined by

[TABLE]

In view of (5.2), (5.3), (5.8) and Lemma 2.4, we observe that for every $x\in H$,

[TABLE]

which gives (5.1), since $U^{*}_{P,Q}\widetilde{U}^{*}U_{P,Q}$ is unitary and $x\in H$ is arbitrary. ∎

Corollary 5.2**.**

Equation (1.2) is true for every two projections $P$ and $Q$ on a Hilbert space.

Proof.

Since any two projections on a Hilbert space are harmonious, the conclusion follows immediately from Lemma 5.1. ∎

Lemma 5.1 can in fact be improved. To this end, we need a lemma as follows.

Lemma 5.3**.**

Let $T\in\mathcal{L}(H)$ be positive such that $\overline{\mathcal{R}(T)}$ is orthogonally complemented in $H$. For every $n\in\mathbb{N}$, let $T_{n}$ be defined as in Lemma 2.4. Then

[TABLE]

Proof.

Clearly, $0\leq T_{n}^{2}\leq T_{n+1}^{2}\leq P_{\overline{\mathcal{R}(T)}}$ for each $n\in\mathbb{N}$, so

[TABLE]

which gives by [18, Proposition 1.3.5] that

[TABLE]

Hence $\lim\limits_{n\to\infty}\|ST_{n}\|$ exists and $\lim\limits_{n\to\infty}\|ST_{n}\|\leq\|SP_{\overline{\mathcal{R}(T)}}\|$. On the other hand, given any $x\in H$, by Lemma 2.4 we have

[TABLE]

therefore $\|SP_{\overline{\mathcal{R}(T)}}\|\leq\lim\limits_{n\to\infty}\|ST_{n}\|$. This completes the proof of (5.9). ∎

A generalization of Lemma 5.1 is as follows.

Lemma 5.4**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections such that $\overline{\mathcal{R}(P+Q)}$ and $\overline{\mathcal{R}(2I-P-Q)}$ are both orthogonally complemented in $H$. Then equation (1.2) is valid.

Proof.

It needs only to prove that (5.1) is true. Since $\mathcal{L}(H)$ is a unital $C^{*}$-algebra, there exists a Hilbert space $E$ and a $C^{*}$-morphism $\pi:\mathcal{L}(H)\to\mathcal{L}(E)$ such that $\pi$ is faithful [18, Corollary 3.7.5]. Replacing $E$ with $\pi(I_{H})E$ if necessary, we may assume that $\pi$ is unital. For every $n\in\mathbb{N}$, let

[TABLE]

Then according to Lemma 5.3 and Corollary 5.2, we have

[TABLE]

The proof of (5.1) is then finished. ∎

It is remarkable that Lemma 5.4 above can be generalized furthermore. Indeed, we will prove that equation (1.2) is always true whenever $\mathcal{R}(P)\cap\mathcal{R}(Q)$ and $\mathcal{N}(P)\cap\mathcal{N}(Q)$ are both orthogonally complemented in $H$. To this end, we need a couple of lemmas.

Recall that the Moore-Penrose inverse $T^{\dagger}$ of an operator $T\in\mathcal{L}(H,K)$ is the unique element $X\in\mathcal{L}(K,H)$ which satisfies

[TABLE]

If such an operator $T^{\dagger}$ exists, then $T$ is said to be M-P invertible.

Lemma 5.5**.**

[22, Theorem 2.2]* For every $T\in\mathcal{L}(H,K)$, $T$ is M-P invertible if and only if $\mathcal{R}(T)$ is closed.*

Remark 5.6**.**

Let $T\in\mathcal{L}(H,K)$ be such that $\mathcal{R}(T)$ is closed. Then both $P=TT^{\dagger}$ and $Q=T^{\dagger}T$ are projections such that $\mathcal{R}(P)=\mathcal{R}(T)$ and $\mathcal{R}(Q)=\mathcal{R}(T^{*})$. So in this case, $\mathcal{R}(T)$ and $\mathcal{R}(T^{*})$ are orthogonally complemented in $K$ and $H$, respectively.

Lemma 5.7**.**

(cf. [11, Theorem 3.2] and [22, Remark 1.1])* Let $T\in\mathcal{L}(H,K)$. Then the closedness of any one of the following sets implies the closedness of the remaining three sets:*

[TABLE]

If $\mathcal{R}(T)$ is closed, then $\mathcal{R}(T)=\mathcal{R}(TT^{*})$, $\mathcal{R}(T^{*})=\mathcal{R}(T^{*}T)$ and the following orthogonal decompositions hold:

[TABLE]

Remark 5.8**.**

Suppose that $P,Q\in\mathcal{L}(H)$ are projections. Let $T$ be defined by (2.6). Then it follows from (2.7) and Lemma 5.7 that

[TABLE]

and in this case $\mathcal{R}(P)+\mathcal{R}(Q)=\mathcal{R}(P+Q)$.

Lemma 5.9**.**

[13, Proposition 4.6]* Let $P,Q\in\mathcal{L}(H)$ be projections such that $\mathcal{R}(P)+\mathcal{R}(Q)$ is closed. Then*

[TABLE]

We provide a technical lemma of this section as follows.

Lemma 5.10**.**

Let $P,Q\in\mathcal{L}(H)$ be projections. Then the following statements are equivalent:

1. (i)

$\|PQ\|<1$; 2. (ii)

$\mathcal{R}(P)\cap\mathcal{R}(Q)=\{0\}$* and $\mathcal{R}(P)+\mathcal{R}(Q)$ is closed;* 3. (iii)

$\mathcal{R}(I-P)+\mathcal{R}(I-Q)=H$.

Proof.

“(i)$\Longrightarrow$(ii)”: Assume that $\|PQ\|<1$. If $\mathcal{R}(P)\cap\mathcal{R}(Q)\neq\{0\}$, then there exists $x_{0}\in\mathcal{R}(P)\cap\mathcal{R}(Q)$ such that $\|x_{0}\|=1$. Thus, $1>\|PQ\|\geq\|PQx_{0}\|=1$, which is a contradiction. Therefore, $\mathcal{R}(P)\cap\mathcal{R}(Q)=\{0\}$.

Given any $y\in\overline{\mathcal{R}(P)+\mathcal{R}(Q)}$, there exist sequences $\{x_{n}\}$ and $\{y_{n}\}$ in $H$ such that

[TABLE]

which gives

[TABLE]

Since $\|PQ\|<1$, the operator $I-PQ$ is invertible. Hence

[TABLE]

Taking limits together with (5.12) yield

[TABLE]

from which we get $y=Pv+Qu\in\mathcal{R}(P)+\mathcal{R}(Q)$. This completes the proof of the closedness of $\mathcal{R}(P)+\mathcal{R}(Q)$.

“(ii)$\Longrightarrow$(i)”: From $\|PQ\|^{2}=\|PQQ^{*}P^{*}\|=\|PQP\|$ and the positivity of $PQP$, we conclude that

[TABLE]

Assume that $\mathcal{R}(P)\cap\mathcal{R}(Q)=\{0\}$ and $\mathcal{R}(P)+\mathcal{R}(Q)$ is closed. We show that $I-PQP$ is invertible.

Injectivity: Let $x\in H$ be such that $(I-PQP)x=0$. Then $x=PQPx\in\mathcal{R}(P)$, so $x=Px$ and thus $P(I-Q)Px=0$. Hence $(I-Q)Px=0$. Therefore, $x=Px=QPx\in\mathcal{R}(P)\cap\mathcal{R}(Q)=\{0\}$.

Surjectivity: By (5.10), we know that $\mathcal{R}(P+Q)$ is closed and $\mathcal{R}(P)+\mathcal{R}(Q)=\mathcal{R}(P+Q)$. Given any y\in\overline{\mathcal{R}\big{(}(I-Q)P\big{)}}, there exists a sequence $\{x_{n}\}$ in $H$ such that

[TABLE]

since both $\mathcal{R}(I-Q)$ and $\mathcal{R}(P+Q)$ are closed. Then

[TABLE]

The process above shows that \mathcal{R}\big{(}(I-Q)P\big{)} is closed. In view of Lemma 5.7, we infer that \mathcal{R}\big{(}P(I-Q)P\big{)}=\mathcal{R}\Big{(}\big{(}(I-Q)P\big{)}^{*}(I-Q)P\Big{)} is also closed.

Clearly, \mathcal{N}(P)\subseteq\mathcal{N}\big{(}P(I-Q)P\big{)}. Conversely, given any $x\in H$ such that $P(I-Q)Px=0$, we arrive at $(I-Q)Px=0$. Hence

[TABLE]

Therefore, \mathcal{N}\big{(}P(I-Q)P\big{)}\subseteq\mathcal{N}(P). This completes the proof that \mathcal{N}(P)=\mathcal{N}\big{(}P(I-Q)P\big{)}.

Accordingly, by Lemma 5.7, $H$ can be orthogonally decomposed as

[TABLE]

Now, given any $x\in H$, there exist $u\in H$ and $v\in\mathcal{N}(P)$ such that

[TABLE]

so that $Px=P(I-Q)Pu$. Therefore,

[TABLE]

This completes the proof that $\mathcal{R}(I-PQP)=H$.

“(ii)$\Longrightarrow$(iii)”: The conclusion is directly deduced from (5.11) in Lemma 5.9.

“(iii)$\Longrightarrow$(ii)”: Assume that $\mathcal{R}(I-P)+\mathcal{R}(I-Q)=H$. Replacing $P$ and $Q$ in Lemma 5.9 with $I-P$ and $I-Q$, respectively, we infer that $\mathcal{R}(P)+\mathcal{R}(Q)$ is closed. From (5.11) we get \big{(}\mathcal{R}(P)\cap\mathcal{R}(Q)\big{)}^{\bot}=H, which can happen only if $\mathcal{R}(P)\cap\mathcal{R}(Q)=\{0\}$. ∎

Lemma 5.11**.**

Let $P,Q\in\mathcal{L}(H)$ be projections such that $\mathcal{R}(P)\cap\mathcal{R}(Q)$ is orthogonally complemented in $H$. Then the following statements are equivalent:

1. (i)

$\|PQ-P_{\mathcal{R}(P)\cap\mathcal{R}(Q)}\|<1$; 2. (ii)

\mathcal{R}(P)\cap\big{(}\mathcal{R}(P)\cap\mathcal{R}(Q)\big{)}^{\bot}+\mathcal{R}(Q)\cap\big{(}\mathcal{R}(P)\cap\mathcal{R}(Q)\big{)}^{\bot}* is closed;* 3. (iii)

$\mathcal{R}(P)+\mathcal{R}(Q)$* is closed;* 4. (iv)

$\mathcal{R}(I-P)+\mathcal{R}(I-Q)$* is closed.*

Proof.

For simplicity, we put $\mathcal{R}(P)\cap\mathcal{R}(Q)=\Omega$. It is obvious that both $P$ and $Q$ commute with $I-P_{\Omega}$. Hence, if we put

[TABLE]

then $P_{1}$ and $Q_{1}$ are projections such that

[TABLE]

Furthermore, it is clear that

[TABLE]

“(i)$\Longleftrightarrow$(ii)”: From (5.14), (5.15) and Lemma 5.10, we conclude that

[TABLE]

“(ii)$\Longleftrightarrow$(iii)”: Let $P_{1},Q_{1}$ be defined by (5.13), and put

[TABLE]

Since both $P_{1}$ and $Q_{1}$ are projections, by (5.10) and (5.14), we see that $\mathcal{R}(P)\cap\Omega^{\bot}+\mathcal{R}(Q)\cap\Omega^{\bot}$ is closed if and only $\mathcal{R}(T)$ is closed.

Suppose that $\mathcal{R}(T)$ is closed. Given any $x\in\overline{\mathcal{R}(P)+\mathcal{R}(Q)}=\overline{\mathcal{R}(P+Q)}$, there exist $u_{n}\in H\,\,(n\in\mathbb{N})$ such that $(P+Q)u_{n}\to x$ as $n\to\infty$. Then

[TABLE]

Now (5.16) ensures that

[TABLE]

This completes the proof of the closedness of $\mathcal{R}(P)+\mathcal{R}(Q)$.

Conversely, suppose that $\mathcal{R}(P)+\mathcal{R}(Q)$ is closed. By (5.10), $\mathcal{R}(P+Q)$ is also closed. Given any $x\in\overline{\mathcal{R}(T)}$, there exist $u_{n}\in H(n\in\mathbb{N})$ such that $Tu_{n}\to x$ as $n\to\infty$. Then

[TABLE]

Meanwhile, since $\mathcal{R}(P+Q)$ is closed, from (5.16) we get

[TABLE]

It follows that $x=(I-P_{\Omega})(P+Q)u=Tu$. This completes the proof of the closedness of $\mathcal{R}(T)$.

“(iii)$\Longleftrightarrow$(iv)”: The conclusion is directly deduced from Lemma 5.9. ∎

Now, we are in the position to give to the main result of this section as follows.

Theorem 5.12**.**

Let $P,Q\in\mathcal{L}(H)$ be two projections such that $\mathcal{R}(P)\cap\mathcal{R}(Q)$ and $\mathcal{N}(P)\cap\mathcal{N}(Q)$ are both orthogonally complemented in $H$. Then equation (1.2) is valid.

Proof.

Two cases are to be taken into consideration.

Case 1 $\mathcal{R}(P)+\mathcal{R}(Q)$ is closed. In this case, $\mathcal{R}(I-P)+\mathcal{R}(I-Q)$ is also closed by Lemma 5.11. Therefore, by (5.10) and Remark 5.6 we know that $\mathcal{R}(P+Q)$ and $\mathcal{R}(I-P+I-Q)$ are both orthogonally complemented in $H$, hence equation (1.2) is valid by Lemma 5.4.

Case 2 $\mathcal{R}(P)+\mathcal{R}(Q)$ is not closed. In this case, $\mathcal{R}(I-P)+\mathcal{R}(I-Q)$ is also not closed by Lemma 5.11. Therefore, by Lemma 5.11 we conclude that

[TABLE]

So, in this case, equation (1.2) is valid. ∎

Remark 5.13**.**

Let $M$ and $N$ be two closed submodules of $H$ such that $M\cap N$ is orthogonally complemented in $H$. As in the Hilbert space case, we can define the Friedrichs angle $\alpha(M,N)$ through $c(M,N)$, which is formulated by (1.1). If furthermore $M^{\bot}\cap N^{\bot}$ is also orthogonally complemented in $H$, then from (1.1) and Theorem 5.12 we can conclude that the characterization (1.3) of the Friedrichs angle is also true.

Acknowledgement. The authors would like to sincerely thank the anonymous referee for carefully reading the paper and for useful comments.