Decomposition of Idempotent Operators on Hilbert C*-Modules

Abstract

This study advances the application of the generalized Halmos’ two projections theorem to idempotent operators on Hilbert $C^{*}$-modules through a comprehensive study of sums involving adjointable idempotents and their adjoints. We establish fundamental properties including the closedness, orthogonal complementability, Moore–Penrose inverses, and spectral norms of such sums. For arbitrary (not necessarily adjointable) idempotent operators that admit a decomposition into linear combinations or products of two idempotents, we derive explicit representations for all such decompositions. A numerical example is given to show how our main theorem allows for the decomposition of idempotent matrices into linear combinations of two idempotent matrices, and two concrete examples on Hilbert $C^{*}$-modules validate the theoretical significance of our framework.

1. Introduction

Recent advances in operator theory have highlighted the fundamental roles of idempotent operators across mathematical physics, where their norm properties and generalized inverses enable critical applications. Studies such as [1] have revealed deep connections between algebraic equivalence, similarity, and norm behavior in nest algebras, while [2] established geometric relationships between idempotent norms and subspace configurations in Hilbert spaces. Further investigations by [3] led to the derivation of norm lower bounds for idempotent functions on locally compact groups, and [4] uncovered concentration phenomena for integral norms of trigonometric polynomials. Constructive approaches in [5] yielded idempotents with controlled diagonals, whereas [6] developed explicit expressions for Drazin inverses of operator sums. Complementary work by [7] allowed for the characterization of matrices with idempotent Moore–Penrose inverses, and [8] quantified Frobenius distances to generalized inverses.

Quantum systems derive profound advantages from these developments, where idempotent norm constraints govern the interchangeability of projections and explain fundamental quantization phenomena such as integer quantum conductance in the Hall effect [9]. Complementary perturbation analyses [10] have demonstrated how the exponential decay properties of Fermi projections induce Anderson orthogonality with positive probability in localized systems. Beyond these phenomena, generalized inverses enable crucial advancements in quantum information processing, facilitating near-optimal reversal operations that are adaptive to noise and initial states while preserving the fidelity of both quantum entanglement and classical information for efficient state recovery in noisy channels [11]. These techniques further extend to non-Hermitian quantum systems, where generalized inverses support the construction of metric-operator frameworks that ensure essential compatibility between probability conservation and real spectra in pseudo-Hermitian systems [12]. Building upon these foundations, we establish the complete spectral characterization of conjugatable idempotent sums on Hilbert $C^{*}$-modules, determining both the Moore–Penrose inverse and spectral norm for the sum formed by a conjugatable idempotent operator and its adjoint.

Research on linear combinations of idempotents has yielded substantial insights, particularly through the complete characterization of idempotency for two-matrix combinations in [13]. This foundation was extended to three-idempotent combinations with pairwise commutativity in [14], while Drazin inverses for operator sums/differences emerged through the Hilbert space constraints presented in [6] and the Banach algebra representations in [15]. Despite these advances, decomposition theory for idempotents remains comparatively underdeveloped. The decomposition of certain idempotent matrices demonstrated by [16], the proof in [17] that every square matrix expresses as a three-idempotent combination, and the establishment that diagonal operators decompose into three-idempotent combinations (requiring four projections for self-adjoint cases) in [18] collectively highlight a persistent gap: the unresolved characterization of two-term linear decompositions. This study addresses this theoretical deficiency within Hilbert $C^{*}$-module theory, providing comprehensive decomposition characterizations for arbitrary idempotents (including non-adjointable cases).

Factorization problems have also attracted considerable attention. Prior work has extensively studied operators that are expressible as products of projections [19,20,21,22,23]. In [24], the factorization of idempotents into products of two idempotents were examined, deriving explicit representations. In this paper, we extend these factorization results to the Hilbert $C^{*}$-module setting, significantly broadening their applicability.

The remainder of this paper is structured as follows: Section 2 presents a generalized version of Halmos’ two-projections theorem for Hilbert $C^{*}$-modules and essential preliminaries. To enable the application of this generalized Halmos framework to idempotent operators, Section 3 establishes necessary and sufficient conditions for the closedness and orthogonal complementability of ranges for sums of adjointable idempotents and their adjoints, supplemented by counterexamples demonstrating the failure of these properties in general cases. Section 4 provides explicit characterizations of the Moore–Penrose inverse, spectral points, and a spectral norm for such sums. Section 5 presents advances in two directions: (1) For adjointable idempotents on Hilbert $C^{*}$-modules, we derive concrete representations of linearly decomposed idempotents; and (2) we extend these results to non-adjointable idempotents, obtaining precise representations for both linear and multiplicative decompositions. Finally, Section 6 presents: (i) Matrix decomposition examples illustrating computational applications of our theorems; and (ii) Hilbert $C^{*}$-module counterexamples validating the necessity of adjointability assumptions for range properties, Moore–Penrose inverses, and spectral norms.

2. Preliminaries

Throughout this paper, $\mathcal{A}$ denotes a $C^{*}$-algebra. We assume that $\mathcal{E}$ and $\mathcal{F}$ are Hilbert $C^{*}$-modules over $\mathcal{A}.$ The set of all adjointable operators from $\mathcal{E}$ to $\mathcal{F}$ is represented by $\mathcal{L}(\mathcal{E},\mathcal{F})$, with the abbreviation $\mathcal{L}\left(\mathcal{E}\right)$ if $\mathcal{E}=\mathcal{F}.$ The identity element of an algebra is denoted by $I.$ The range and nullity of an operator T are denoted by $\mathcal{R}\left(T\right)$ and $\mathcal{N}\left(T\right)$, respectively. For further information on Hilbert $C^{*}$-modules and their geometry, see [25,26,27].

An operator $T\in \mathcal{L}\left(\mathcal{E}\right)$ is said to be idempotent if $T=T^2$. By a projection, we mean an operator $P\in \mathcal{L}\left(\mathcal{E}\right)$ such that $P=P^2=P^{*}$. Recall that a submodule $\mathcal{M}\subseteq \mathcal{E}$ is said to be orthogonally complemented in $\mathcal{E}$ if $\mathcal{M}\mathrm{\oplus }{\mathcal{M}}^{\perp }=\mathcal{E}$, where ${\mathcal{M}}^{\perp }=\{x\in \mathcal{E}:⟨x,y⟩=0,\mathrm{for}\mathrm{all}y\in \mathcal{M}\}$. In this case, $\mathcal{M}$ is closed, and we refer to the projection from $\mathcal{E}$ onto $\mathcal{M}$ as $P_{\mathcal{M}}$. Unlike Hilbert spaces, a closed submodule is not necessarily orthogonally complemented. H In this paper, the notations “⊕" and “∔" are used with different meanings. For Hilbert $\mathcal{A}$-modules $H_1$ and $H_2$, let

$$H_1\mathrm{\oplus }{H}_2=\left\{{(h_1,h_2)}^T:h_i\in H_i,i=1,2\right\},$$

which is also a Hilbert $\mathcal{A}$-module whose $\mathcal{A}$-valued inner product is given by

$$⟨{(x_1,y_1)}^T,{(x_2,y_2)}^T⟩=⟨x_1,x_2⟩+⟨y_1,y_2⟩$$

for any $x_i\in H_1, y_i\in H_2, i=1,2.$

Let $H_1$ and $H_2$ be submodules of a Hilbert $\mathcal{A}$-module $\mathcal{E}$. If $⟨x,y⟩=0$ for all $x\in H_1$ and $y\in H_2$, we define the orthogonal sum as follows:

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

This study utilizes Halmos’ two projections theorem as a mathematical tool. Halmos’ two projections theorem was originally obtained in [28], more applications can be found in [29,30], and we generalized it to harmonious projection pairs on Hilbert $C^{*}$-modules in [31]. A brief introduction is provided in the following.

Two projections $P,Q\in \mathcal{L}\left(\mathcal{E}\right)$ are said to be harmonious ([31], Definition 4.1) if the four closures

$$\begin{array}{c}\hfill \overline{\mathcal{R}(P+Q)},\overline{\mathcal{R}(P+I-Q)},\overline{\mathcal{R}(I-P+Q)},\mathrm{and}\overline{\mathcal{R}(I-P+I-Q)}\end{array}$$

(1)

are all orthogonally complemented in H.

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

$$\begin{array}{c}{H}_1=\mathcal{R}\left(P\right)\cap \mathcal{R}\left(Q\right),H_2=\mathcal{R}\left(P\right)\cap \mathcal{N}\left(Q\right),\hfill \end{array}$$

(2)

$$\begin{array}{c}{H}_3=\mathcal{N}\left(P\right)\cap \mathcal{R}\left(Q\right),H_4=\mathcal{N}\left(P\right)\cap \mathcal{N}\left(Q\right).\hfill \end{array}$$

(3)

Since ${\overline{\mathcal{R}(P+Q)}}^{\perp }=H_4$ and $\overline{\mathcal{R}(P+Q)}$ is orthogonally complemented in $\mathcal{E}$, we conclude that $H_4$ is likewise orthogonally complemented in $\mathcal{E}$. Similarly, $H_1$, $H_2$, and $H_3$ are all orthogonally complemented in $\mathcal{E}$.

Furthermore, let

$$P_{H_i}\mathrm{be} \mathrm{denoted} \mathrm{simply} \mathrm{by} P_i \mathrm{for} i=1,2,3,4,$$

(4)

and define

$$\begin{array}{c}{P}_5=P-P_1-P_2\mathrm{and}{H}_5=\mathcal{R}\left(P_5\right),\hfill \end{array}$$

(5)

$$\begin{array}{c}{P}_6=(I-P)-P_3-P_4\mathrm{and}{H}_6=\mathcal{R}\left(P_6\right).\hfill \end{array}$$

(6)

With the notation above, a unitary operator $U_{P,Q}:H\to {\mathrm{\oplus }}_{i=1}^6{H}_i$ can be induced as follows:

$$U_{P,Q}\left(x\right)={\left(P_1\left(x\right),P_2\left(x\right),\dots ,P_6\left(x\right)\right)}^T\mathrm{for} \mathrm{any} x\in \mathcal{E},$$

(7)

with the property that

$$U_{P,Q}^{*}\left({(x_1,x_2,\dots ,x_6)}^T\right)=\sum _{i=1}^6{x}_i\mathrm{for} \mathrm{any} x_i\in H_i,i=1,2,\dots ,6.$$

It follows that

$$\begin{array}{cc}\hfill U_{P,Q}P{U}_{P,Q}^{*}& =I_{H_1}\mathrm{\oplus }{I}_{H_2}\mathrm{\oplus }0\mathrm{\oplus }0\mathrm{\oplus }{I}_{H_5}\mathrm{\oplus }0,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill U_{P,Q}Q{U}_{P,Q}^{*}& =I_{H_1}\mathrm{\oplus }0\mathrm{\oplus }{I}_{H_3}\mathrm{\oplus }0\mathrm{\oplus }T,\hfill \end{array}$$

(9)

where

$$T=\left(\begin{array}{cc}{P}_{5}Q{P}_5{|}_{H_5}& P_{5}Q{P}_6{|}_{H_6}\\ {\left(P_{5}Q{P}_6{|}_{H_6}\right)}^{*}& P_{6}Q{P}_6{|}_{H_6}\end{array}\right)\in \mathcal{L}(H_5\mathrm{\oplus }{H}_6),$$

(10)

in which $P_{5}Q{P}_5{|}_{H_5}$ is the restriction of the operator $P_{5}Q{P}_5$ on $H_5$. The same convention can be used for $P_{5}Q{P}_6{|}_{H_6}$ and $P_{6}Q{P}_6{|}_{H_6}$.

Lemma 1

([31], Theorem 4.6). Suppose that $P,Q\in \mathcal{L}\left(\mathcal{E}\right)$ are two harmonious projections. Let $H_i,P_i$$(1\le i\le 6$) be defined by (2)–(6), respectively. Then, the operator T formulated by (10) can be characterized as

$$T=\left(\begin{array}{cc}{Q}_0& Q_0^{{\displaystyle \frac{1}{2}}}{(I_{H_5}-Q_0)}^{{\displaystyle \frac{1}{2}}}{U}_0\\ U_0^{*}{Q}_0^{{\displaystyle \frac{1}{2}}}{(I_{H_5}-Q_0)}^{{\displaystyle \frac{1}{2}}}& U_0^{*}(I_{H_5}-Q_0)U_0\end{array}\right)\in \mathcal{L}(H_5\mathrm{\oplus }{H}_6),$$

(11)

where $U_0\in \mathcal{L}(H_6,H_5)$ is a unitary operator, $Q_0$ is the restriction of $P_{5}Q{P}_5$ on $H_5$, and both $Q_0$ and $I_{H_5}-Q_0$ are positive, injective, and contractive.

The following lemmas are crucial for the decomposition of idempotent operators, the closedness of operator ranges, and orthogonal complementability.

Lemma 2

([31], Lemma 2.3). Let $P,Q\in \mathcal{L}\left(\mathcal{E}\right)$ be two projections. Then, $\overline{\mathcal{R}\left(P\right)+\mathcal{R}\left(Q\right)}=\overline{\mathcal{R}(P+Q)}.$

Lemma 3

([32], Theorem 3.2). Let $T\in \mathcal{L}(\mathcal{E},\mathcal{F})$ have a closed range. Then, $\mathcal{R}\left(T\right)$ and $\mathcal{N}\left(T\right)$ are orthogonally complemented.

Lemma 4

([33], Theorem 2.2). An operator $T\in \mathcal{L}(\mathcal{E},\mathcal{F})$ is MP-invertible if and only if $\mathcal{R}\left(T\right)$ is closed.

Lemma 5

([34], Theorem 1.3). For any idempotents $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$ and $\mathcal{N}(\mathrm{\Psi })$, respectively. Then, $\parallel PQ\parallel <1$ and $\mathrm{\Psi }={(I-PQ)}^{-1}P(I-PQ)$.

Lemma 6.

Let $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent and $T\in \mathcal{L}\left(\mathcal{E}\right)$ such that $\mathcal{R}\left(T\right)\subseteq \mathcal{R}(\mathrm{\Psi })$. Then, $\mathrm{\Psi }T=T$.

Proof. 

As $\mathcal{R}\left(T\right)\subseteq \mathcal{R}(\mathrm{\Psi })$, for any $x\in \mathcal{E}$, there exists $y\in \mathcal{E}$ such that $Tx=\mathrm{\Psi }y$. Thus, $\mathrm{\Psi }Tx=\mathrm{\Psi }\mathrm{\Psi }y=\mathrm{\Psi }y=Tx$. □

3. Closedness and Orthogonal Complementability of the Sum of Adjointable Idempotents and Their Adjoints

For any idempotent $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$ and $\mathcal{N}(\mathrm{\Psi })$, respectively. Then, we have that

$$\begin{array}{c}\hfill \mathcal{R}(\mathrm{\Psi })=\mathcal{R}\left(P\right),\mathcal{N}(\mathrm{\Psi })=\mathcal{R}\left(Q\right),\mathcal{N}\left({\mathrm{\Psi }}^{*}\right)=\mathcal{R}(I-P),\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)=\mathcal{R}(I-Q).\end{array}$$

Based on Lemma 2, we deduce that

$$\begin{array}{cc}\hfill \overline{\mathcal{R}(P+Q)}& =\overline{\mathcal{R}\left(P\right)+\mathcal{R}\left(Q\right)}=\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{N}(\mathrm{\Psi })}=\mathcal{E}.\hfill \\ \hfill \overline{\mathcal{R}(I-P+I-Q)}& =\overline{\mathcal{R}(I-P)+\mathcal{R}(I-Q)}=\overline{\mathcal{N}\left({\mathrm{\Psi }}^{*}\right)+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}=\mathcal{E}.\hfill \\ \hfill \overline{\mathcal{R}(I-P+Q)}& =\overline{\mathcal{R}(I-P)+\mathcal{R}\left(Q\right)}=\overline{\mathcal{N}\left({\mathrm{\Psi }}^{*}\right)+\mathcal{N}(\mathrm{\Psi })}.\hfill \\ \hfill \overline{\mathcal{R}(P+I-Q)}& =\overline{\mathcal{R}\left(P\right)+\mathcal{R}(I-Q)}=\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}.\hfill \end{array}$$

Therefore, to apply Halmos’ two projections theorem to the study of adjointable idempotents on Hilbert $C^{*}$-modules, according to (1), it is necessary to investigate the orthogonal complementability of both $\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$ and $\overline{\mathcal{R}(I-\mathrm{\Psi })+\mathcal{R}(I-{\mathrm{\Psi }}^{*})}$. Lemma 3 establishes that the closedness of $\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$ implies its orthogonal complementability. Consequently, we establish necessary and sufficient conditions for the closedness and orthogonal complementability of $\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$ in the following theorem.

Theorem 1.

Let $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent and P be a projection from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$. Then, $\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$ is closed if and only if $\mathcal{R}({\mathrm{\Psi }}^{*}-P)$ is closed. Furthermore, $\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$ is orthogonally complemented if and only if $\overline{\mathcal{R}({\mathrm{\Psi }}^{*}-P)}$ is orthogonally complemented.

Proof. 

$\mathcal{R}({\mathrm{\Psi }}^{*}-P)\cap \mathcal{R}\left(P\right)=\left\{0\right\}$ and $\mathcal{R}({\mathrm{\Psi }}^{*}-P)\subseteq \mathcal{R}(I-P)$ can be derived from $(I-P)({\mathrm{\Psi }}^{*}-P)=({\mathrm{\Psi }}^{*}-P).$ Then, $\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)+\mathcal{R}(\mathrm{\Psi })=\mathcal{R}({\mathrm{\Psi }}^{*}-P+P)+\mathcal{R}(\mathrm{\Psi })\subseteq \mathcal{R}({\mathrm{\Psi }}^{*}-P)\dotplus \mathcal{R}\left(P\right)\subseteq \mathcal{R}\left({\mathrm{\Psi }}^{*}\right)+\mathcal{R}(\mathrm{\Psi }).$ This leads to

$$\begin{array}{c}\hfill \mathcal{R}\left({\mathrm{\Psi }}^{*}\right)+\mathcal{R}(\mathrm{\Psi })=\mathcal{R}({\mathrm{\Psi }}^{*}-P)\dotplus \mathcal{R}\left(P\right)\end{array}$$

(12)

and it is clear that $\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$ is closed if and only if $\mathcal{R}({\mathrm{\Psi }}^{*}-P)$ is closed.

Now, suppose $\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$ is orthogonally complemented. Then, combined with (12), we have the following:

$$\begin{array}{cc}\hfill \mathcal{E}& =\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}\dotplus \mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left({\mathrm{\Psi }}^{*}\right)\hfill \\ \hfill & =\overline{R({\mathrm{\Psi }}^{*}-P)}\dotplus R\left(P\right)\dotplus \mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left({\mathrm{\Psi }}^{*}\right).\hfill \end{array}$$

(13)

Let $P_{\mathrm{\Omega }}=I-P-P_{\mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left({\mathrm{\Psi }}^{*}\right)}=I-P-P_{\mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left(P\right)}$. Observe that $P_{\mathrm{\Omega }}$ satisfies $P_{\mathrm{\Omega }}^2=P_{\mathrm{\Omega }}$, making it a projection. The conditions $P_{\mathrm{\Omega }}P=0$ further imply $\mathcal{N}\left(P_{\mathrm{\Omega }}\right)\supseteq \mathcal{R}\left(P\right)$. For any $x\in \mathcal{R}(I-P)$ such that $P_{\mathrm{\Omega }}x=0$, then $(I-P_{\mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left(P\right)})x=0$, which means that $x\in \mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left(P\right)$, yielding $\mathcal{N}\left(P_{\mathrm{\Omega }}\right)=R\left(P\right)\dotplus \mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left({\mathrm{\Psi }}^{*}\right)$. Consequently, based on (13), we have that $\mathcal{R}\left(P_{\mathrm{\Omega }}\right)=\overline{R({\mathrm{\Psi }}^{*}-P)}$ is orthogonally complemented.

On the other hand, if $\overline{R({\mathrm{\Psi }}^{*}-P)}$ is orthogonally complemented, let $P_{\omega }=I-P-P_{\overline{R({\mathrm{\Psi }}^{*}-P)}}$. Observe that $P_{\omega }$ satisfies $P_{\omega }^2=P_{\omega }$, making it a projection. The conditions $P_{\omega }P=0$ further imply $\mathcal{N}\left(P_{\omega }\right)\supseteq \mathcal{R}\left(P\right)$. For any $x\in \mathcal{R}(I-P)$ such that $P_{\omega }x=0$, then $(I-P_{\overline{R({\mathrm{\Psi }}^{*}-P)}})x=0$. This means that $x\in \overline{R({\mathrm{\Psi }}^{*}-P)}$, yielding $\mathcal{N}\left(P_{\omega }\right)=R\left(P\right)\dotplus \mathcal{R}\left(\overline{R({\mathrm{\Psi }}^{*}-P)}\right)$. Consequently, based on (13), we have that $\mathcal{R}(I-P_{\omega })=\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$ is orthogonally complemented. □

Corollary 1.

Let $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent and Q be projection from $\mathcal{E}$ to $\mathcal{R}(I-\mathrm{\Psi })$. Then, $\mathcal{R}(I-\mathrm{\Psi })+\mathcal{R}(I-{\mathrm{\Psi }}^{*})$ is closed if and only if $\mathcal{R}(I-{\mathrm{\Psi }}^{*}-P)$ is closed. Furthermore, $\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$ is orthogonally complemented if and only if $\overline{\mathcal{R}(I-{\mathrm{\Psi }}^{*}-Q)}$ is orthogonally complemented.

Proof. 

Substituting $\mathrm{\Psi },{\mathrm{\Psi }}^{*}$, and P with $I-\mathrm{\Psi },I-{\mathrm{\Psi }}^{*}$ and Q, respectively, in the proof of Theorem 1 immediately completes the argument. □

Remark 1.

The necessary and sufficient conditions for closedness and orthogonal complementability established in Theorem 1 are significant. The following example demonstrates the existence of an idempotent operator $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ and a projection P from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$ for which $\mathcal{R}({\mathrm{\Psi }}^{*}-P)$ is not closed and $\overline{\mathcal{R}({\mathrm{\Psi }}^{*}-P)}$ is not orthogonally complemented.

Example 1.

Let $\mathcal{A}=C[0,1]$ be the $C^{*}$-algebra of all continuous complex-valued functions on [0, 1]. Let $\mathcal{E}=\mathcal{A}\mathrm{\oplus }\mathcal{A}$ be the canonical Hilbert $\mathcal{A}$-module with $\mathcal{A}$-valued inner product

$$⟨{(x_1,y_1)}^T,{(x_2,y_2)}^T⟩=⟨x_1,x_2⟩+⟨y_1,y_2⟩=x_1^{*}{x}_2+y_1^{*}{y}_2,(x_1,y_1,x_2,y_2\in \mathcal{A}),$$

where $x_1^{*}\left(t\right)=\overline{x_1\left(t\right)}$. Let $\pi \in \mathcal{L}\left(\mathcal{E}\right)$ be defined as $\pi {(x_1,y_1)}^T={(0,f·x_1+y_1)}^T$, where $f\left(t\right)=t\in \mathcal{A}$ and $(f·x_1)\left(t\right)=f\left(t\right)·x_1\left(t\right)=t{x}_1\left(t\right)$. Obviously, ${\pi }^2=\pi$. Now, we prove that π is adjointable.

$$\begin{array}{c}\hfill ⟨\pi {(x_1,y_1)}^T,{(x_2,y_2)}^T⟩=⟨{(0,f{x}_1+y_1)}^T,{(x_2,y_2)}^T⟩=t{x}_1^{*}{y}_2+y_1^{*}{y}_2.\end{array}$$

Assume there exists ${\pi }^{*}\in \mathcal{L}\left(\mathcal{E}\right)$ such that

$$⟨\pi {(x_1,y_1)}^T,{(x_2,y_2)}^T⟩=⟨{(x_1,y_1)}^T,{\pi }^{*}{(x_2,y_2)}^T⟩.$$

Let ${\pi }^{*}{(x_2,y_2)}^T={(u,v)}^T$. Then,

$$\begin{array}{c}\hfill ⟨{(x_1,y_1)}^T,{\pi }^{*}{(x_2,y_2)}^T⟩=⟨{(x_1,y_1)}^T,{(u,v)}^T⟩=x_1^{*}u+y_1^{*}v=t{x}_1^{*}{y}_2+y_1^{*}{y}_2\end{array}$$

for any $x_1,y_1,x_2,y_2\in \mathcal{A}=C[0,1]$. Then, $u=t{y}_2$ can be deduced by choosing $x_1=t,y_1=0$, and $v=y_2$ can be deduced by choosing $x_1=0, y_1=t$. Through direct compoutation, we can verify that ${\pi }^{*}{(x_2,y_2)}^T={(t{y}_2,y_2)}^T={(f{y}_2,y_2)}^T.$

$$\begin{array}{c}\hfill \mathcal{R}\left(\pi \right)=\{\pi {(x,y)}^T\mid {(x,y)}^T\in \mathcal{E}\}=\{{(0,tx+y)}^T\mid x,y\in \mathcal{A}\}=\left\{0\right\}\mathrm{\oplus }\mathcal{A}.\end{array}$$

Let P be a projection on $\mathcal{R}\left(\pi \right)$. Then, it is clear that $P{(x,y)}^T={(0,y)}^T$ for any ${(x,y)}^T\in \mathcal{E}$. Consequently, we have

$$\begin{array}{cc}\hfill \mathcal{R}(\pi -P)& =\{\pi {(x,y)}^T-P{(x,y)}^T\mid {(x,y)}^T\in \mathcal{E}\}\hfill \\ \hfill & =\{{(0,tx+y)}^T-{(0,y)}^T\mid x,y\in \mathcal{A}\}\hfill \\ \hfill & =\{{(0,tx)}^T\mid x\in C[0,1]\}\hfill \\ \hfill & =\left\{0\right\}\mathrm{\oplus }M,\hfill \end{array}$$

where $M=\{tx\mid x\in C[0,1\left]\right\}.$ Now, we prove that M is not closed. Let $G_n\left(t\right)=t·min\{n,t^{-{\displaystyle \frac{1}{2}}}\}$. Clearly,

$$G_n\left(t\right)=\left\{\begin{array}{cc}t·n\hfill & \mathit{if}0\le t\le n^{-2}\hfill \\ \sqrt{t}\hfill & \mathit{if}{n}^{-2}<t\le 1.\hfill \end{array}\right.$$

It follows that $G_n\left(t\right)$ converges uniformly to $t^{{\displaystyle \frac{1}{2}}}\notin M$, which means that neither M nor $\mathcal{R}(\pi -P)$ is closed.

We now prove that $\overline{\mathcal{R}(\pi -P)}$ is not orthogonally complemented.

$$\overline{\mathcal{R}(\pi -P)}=\overline{0\mathrm{\oplus }M}=0\mathrm{\oplus }\overline{M},$$

where $\overline{M}=\{x\mid x\in C[0,1],x\left(0\right)=0\}$. If there exists a projection Q from $\mathcal{E}$ onto $\overline{\mathcal{R}(\pi -P)}$, then, for any ${(x,y)}^T\in \mathcal{E}$, there is $Q{(x,y)}^T={(0,q(x,y))}^T,q(x,y)\in \overline{M}$. Now, for $(0,1)\in \mathcal{E}$, where $1\left(t\right)=1$ for any $t\in [0,1]$, assume that $Q(0,1)=(0,a),a\in \overline{M}$. Then,

$$\begin{array}{c}\hfill 0=⟨{((0,1)-Q(0,1))}^T,{(0,b)}^T⟩={(1-a)}^{*}b\mathit{for}\mathit{any}b\in \overline{M}.\end{array}$$

As a is a continuous complex-valued function on [0,1], setting $b=t$ yields $a\left(t\right)=1$ for any $t\in [0,1]$. This contradicts the condition $a\left(0\right)=0$, demonstrating that no orthogonal projection onto $\overline{\mathcal{R}(\pi -P)}$ exists.

Based on the preceding analysis, we formally define the class of idempotent operators for which Halmos’ two projections theorem is applicable.

Definition 1.

$\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$ denotes an idempotent $\mathrm{\Psi }\in \mathcal{L}N\left(\mathcal{E}\right)$ such that both $\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$ and $\overline{\mathcal{R}(I-\mathrm{\Psi })+\mathcal{R}(I-{\mathrm{\Psi }}^{*})}$ are orthogonally complemented.

4. Moore–Penrose Inverse and Spectral Norm of the Sum of Adjointable Idempotents and Their Adjoints

For any $T\in \mathcal{L}\left(\mathcal{E}\right)$, the Moore–Penrose inverse of T is denoted by $T^{†}$. Lemma 4 establishes that the existence of the Moore–Penrose inverse for an idempotent operator $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ is intimately connected to its closedness and the orthogonal complementability of its range. We now establish a sufficient condition for the orthogonal complementability of $\overline{\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)}$. Our analysis begins with operators $\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$, subsequently extending to the general case of arbitrary idempotents $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right).$

Theorem 2.

For any idempotent $\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$ such that $\mathcal{R}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$ is closed, let P and Q be two projections from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$ and $\mathcal{N}(\mathrm{\Psi })$, respectively. Then, $\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$ is closed, ${({\mathrm{\Psi }}^{*}-P)}^{†}=P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P)$, and ${(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}=\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{(\mathrm{\Psi }-P)}^{†}{({\mathrm{\Psi }}^{*}-P)}^{†}\right)\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right)$.

Proof. 

Clearly, $\mathcal{R}\left(P\right)\cap \mathcal{R}\left(Q\right)=\mathcal{R}(\mathrm{\Psi })\cap \mathcal{N}(\mathrm{\Psi })=0$. As $\mathcal{E}=\mathcal{R}(\mathrm{\Psi })\dotplus \mathcal{N}\left({\mathrm{\Psi }}^{*}\right)=\mathcal{N}(\mathrm{\Psi })\dotplus \mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$, we obtain

$$\begin{array}{c}\hfill \mathcal{R}(I-P)=\mathcal{N}\left({\mathrm{\Psi }}^{*}\right)\mathrm{and}\mathcal{R}(I-Q)=\mathcal{R}\left({\mathrm{\Psi }}^{*}\right),\end{array}$$

which imply $\mathcal{R}(I-P)\cap \mathcal{R}(I-Q)=0$. Let $H_i$ and $P_i$ ($1\le i\le 6$) be defined as in (2)–(6). This gives

$$\begin{array}{c}\hfill H_1=0,H_4=0,\mathrm{and}{H}_2\dotplus H_3\dotplus H_5\dotplus H_6=\mathcal{E}.\end{array}$$

Lemma 5 implies $\parallel PQ\parallel <1$, which yields $\parallel QPQ\parallel <1$. Combined with (5), we have

$$\begin{array}{c}\hfill \parallel Q{P}_{5}Q\parallel <1\mathrm{and},\mathrm{thus},\parallel P_{5}Q{P}_5\parallel <1.\end{array}$$

Define $Q_0$ and the unitary operator $U_0\in \mathcal{L}(H_6,H_5)$ via Lemma 1. From $Q_0=P_{5}Q{P}_5|H_5$, it follows that $\parallel Q_0\parallel <1$. Therefore, $I_{H_5}-Q_0$ is invertible when $H_5\ne 0$. Note that $P_5=0$ if $H_5=0$. We extend this by defining $I_{H_5}=0$ when $H_5=0$. Set

$$\begin{array}{c}\hfill V=-Q_0^{1/2}{(I_{H_5}-Q_0)}^{-1/2}{U}_0.\end{array}$$

(14)

Combining $U_{P,Q}$ (defined in (7)) with the decompositions (8), (9), and Lemma 5, we obtain

$$U_{P,Q}·\mathrm{\Psi }·U_{P,Q}^{*}=I_{H_2}\mathrm{\oplus }0\mathrm{\oplus }\left(\begin{array}{cc}{I}_{H_5}& V\\ 0& 0\end{array}\right).$$

(15)

Additionally,

$$U_{P,Q}·(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})·U_{P,Q}^{*}=2I_{H_2}\mathrm{\oplus }0\mathrm{\oplus }\left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right).$$

(16)

Based on Lemma 4, the Moore–Penrose inverse of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$ exists. This yields the representation

$$U_{P,Q}·{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}·U_{P,Q}^{*}={\displaystyle \frac{1}{2}}{I}_{H_2}\mathrm{\oplus }0\mathrm{\oplus }\left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right),\mathrm{where}a=a^{*},c=c^{*}.$$

(17)

It can be deduced, from $(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})=(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$, that

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right)& =\left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right)\left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right)\left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}4a+2V{b}^{*}+2b{V}^{*}+Vc{V}^{*}& 2aV+V{b}^{*}V\\ 2V^{*}a+V^{*}b{V}^{*}& V^{*}aV\end{array}\right).\hfill \end{array}$$

(18)

Thus, equating the (2,1)- and (2,2)-entries, we obtain that

$$\begin{array}{c}\hfill 2V^{*}a+V^{*}b{V}^{*}=V^{*},V^{*}aV=0.\end{array}$$

(19)

The Moore–Penrose inverse property ${(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})=(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}$ implies the matrix commutation relation

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right)\left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right)& =\left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right)\left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right).\hfill \end{array}$$

Performing matrix multiplication yields the following:

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}2a+b{V}^{*}& aV\\ 2b^{*}+c{V}^{*}& b^{*}V\end{array}\right)& =\left(\begin{array}{cc}2a+V{b}^{*}& 2b+Vc\\ V^{*}a& V^{*}b\end{array}\right).\hfill \end{array}$$

Equating the (1,2)-entries gives the following:

$$\begin{array}{c}\hfill aV=2b+Vc.\end{array}$$

(20)

Equating the (1,1)- and (2,2)-entries gives

$$\begin{array}{c}\hfill b{V}^{*}=V{b}^{*}\mathrm{and}{b}^{*}V=V^{*}b.\end{array}$$

(21)

Furthermore, the Moore–Penrose inverse property ${(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}={(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$${(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}$ implies that

$$\begin{array}{cc}\hfill \left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right)& =\left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right)\left(\begin{array}{cc}2I_{H_5}& V\\ V^{*}& 0\end{array}\right)\left(\begin{array}{cc}a& b\\ b^{*}& c\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}2a^2+aV{b}^{*}+b{V}^{*}a& 2ab+b{V}^{*}b+aVc\\ 2b^{*}a+c{V}^{*}a+b^{*}V{b}^{*}& 2b^{*}b+c{V}^{*}b+b^{*}Vc\end{array}\right).\hfill \end{array}$$

(22)

Left-multiplying by $V^{*}$ and right-multiplying by V at the (1,1)-entries of both matrices, we obtain $V^{*}{a}^{2}V=V^{*}aV=0$ via (19), implying $aV=0$. Substituting $aV=0$ into (20) yields

$$\begin{array}{c}\hfill 2b=-Vc,\end{array}$$

(23)

which gives $ab=0$. Substituting $aV=0$ and $ab=0$ into the (1,2)-entries of the matrix yields the following:

$$\begin{array}{c}\hfill b{V}^{*}b=b.\end{array}$$

(24)

Similarly, substituting $aV=0$ into (19) produces the following:

$$\begin{array}{c}\hfill V^{*}b{V}^{*}=V^{*}.\end{array}$$

(25)

Combining (21), (24), and (25), we deduce that b is the Moore–Penrose inverse of $V^{*}$.

Based on (16) and (17), we have

$$\begin{array}{cc}\hfill P_6{V}^{*}{P}_5& =P_6(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})P_5=(I-P)(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})P={\mathrm{\Psi }}^{*}-P,\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill P_{5}b{P}_6& =P_5{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}{P}_6=P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P),\hfill \end{array}$$

(27)

which establishes

$${({\mathrm{\Psi }}^{*}-P)}^{†}=P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P).$$

(28)

Theorem 1 implies that $\mathcal{R}(\mathrm{\Psi })+\mathcal{R}\left({\mathrm{\Psi }}^{*}\right)$ is closed.

Substituting (23) into the (2,2)-position of (22), yields the following:

$$\begin{array}{c}\hfill c=-2b^{*}b.\end{array}$$

Substituting (21) and (23) into the (1,1)-position of (18) gives the following:

$$\begin{array}{c}\hfill a={\displaystyle \frac{I_{H_5}-b{V}^{*}}{2}}.\end{array}$$

Combining the Moore–Penrose inverse property $b={\left(V^{*}\right)}^{†}$ with $P=P_2+P_5$, (17), (26), (27), and (28), we derive

$$\begin{array}{cc}\hfill {(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}& ={\displaystyle \frac{P_2}{2}}+{\displaystyle \frac{P_5-P_{5}b{P}_6{P}_6{V}^{*}{P}_6}{2}}+P_{5}b{P}_6+P_6{b}^{*}{P}_5-2P_6{b}^{*}{P}_5{P}_{5}b{P}_6\hfill \\ \hfill & =(I-{\displaystyle \frac{P_{5}V{P}_6}{2}})({\displaystyle \frac{P}{2}}-2P_6{b}^{*}{P}_5{P}_{5}b{P}_6)(I-{\displaystyle \frac{P_6{V}^{*}{P}_5}{2}})\hfill \\ \hfill & =\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{(\mathrm{\Psi }-P)}^{†}{({\mathrm{\Psi }}^{*}-P)}^{†}\right)\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right).\hfill \end{array}$$

□

Theorem 3.

Let $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent operator and P be the projection onto $\mathcal{R}(\mathrm{\Psi })$. Then, $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$ is MP-invertible if and only if ${\mathrm{\Psi }}^{*}-P$ is MP-invertible. The following identities hold: ${({\mathrm{\Psi }}^{*}-P)}^{†}=P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P)$ and

$$\begin{array}{c}\hfill {(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}=\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{(\mathrm{\Psi }-P)}^{†}{({\mathrm{\Psi }}^{*}-P)}^{†}\right)\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right).\end{array}$$

Proof. 

We begin by establishing the necessary condition. If $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$ is MP-invertible, inspired by Theorem 2, let $A=P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P)$. We now prove that A is the MP-inverse of ${\mathrm{\Psi }}^{*}-P$. Applying left-multiplication by P and $I-P$ to both sides of the Moore–Penrose inverse identity

$$(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}={(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}),$$

we obtain

$$\begin{array}{cc}\hfill (\mathrm{\Psi }+P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}& =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}),\hfill \\ \hfill ({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}& =(I-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}).\hfill \end{array}$$

(29)

Then, combined with $\mathcal{R}(\mathrm{\Psi }-P)\subseteq \mathcal{R}\left(P\right)$ and $\mathcal{R}(\mathrm{\Psi }+P)\subseteq \mathcal{R}\left(P\right)$, we have

$$\begin{array}{cc}\hfill {\left(A({\mathrm{\Psi }}^{*}-P)\right)}^{*}& =(\mathrm{\Psi }-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}P\hfill \\ \hfill & =\left((\mathrm{\Psi }+P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}-2P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}\right)P\hfill \\ \hfill & =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})P-2P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}P,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\left(({\mathrm{\Psi }}^{*}-P)A\right)}^{*}& =(I-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)\hfill \\ \hfill & =(I-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})(I-P).\hfill \end{array}$$

Consequently,

$$\begin{array}{c}\hfill {\left(A({\mathrm{\Psi }}^{*}-P)\right)}^{*}=A({\mathrm{\Psi }}^{*}-P),{\left(({\mathrm{\Psi }}^{*}-P)A\right)}^{*}=({\mathrm{\Psi }}^{*}-P)A.\end{array}$$

(30)

Applying left-multiplication by $I-P$ and right-multiplication by $I-P$ to the Moore–Penrose identity

$$(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})=\mathrm{\Psi }+{\mathrm{\Psi }}^{*},$$

we obtain

$$\begin{array}{c}\hfill ({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)=0.\end{array}$$

(31)

Similarly, applying left-multiplication by $I-P$ and right-multiplication by P to the same identity

$$(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})=\mathrm{\Psi }+{\mathrm{\Psi }}^{*},$$

yields

$$\begin{array}{c}\hfill ({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}({\mathrm{\Psi }}^{*}+P)={\mathrm{\Psi }}^{*}-P.\end{array}$$

(32)

Combined with (31) and the Moore–Penrose identity

$${(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}={(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†},$$

we have

$$\begin{array}{cc}\hfill 0& =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)\hfill \\ \hfill & =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)\hfill \\ \hfill & =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P+{\mathrm{\Psi }}^{*}-P+2P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)\hfill \\ \hfill & =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}\left(2P\right){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P),\hfill \end{array}$$

which means

$$\begin{array}{c}\hfill P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)=0.\end{array}$$

(33)

Together with (29), this yields

$$\begin{array}{cc}\hfill A({\mathrm{\Psi }}^{*}-P)A& =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P)\hfill \\ \hfill & =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})(I-P)\hfill \\ \hfill & =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)\hfill \\ \hfill & =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }-P)\hfill \\ \hfill & =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})(I-P)\hfill \\ \hfill & =P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P)\hfill \\ \hfill & =A.\hfill \end{array}$$

(34)

Similarly, based on (33) and (32), we have the following:

$$\begin{array}{cc}\hfill ({\mathrm{\Psi }}^{*}-P)A({\mathrm{\Psi }}^{*}-P)& =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}({\mathrm{\Psi }}^{*}-P)\hfill \\ \hfill & =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}({\mathrm{\Psi }}^{*}-P+2P)\hfill \\ \hfill & =({\mathrm{\Psi }}^{*}-P){(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}{\mathrm{\Psi }}^{*}+P)\hfill \\ \hfill & ={\mathrm{\Psi }}^{*}-P.\hfill \end{array}$$

Combining (34) and (30), we conclude that $A=P{(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})}^{†}(I-P)$ is the Moore–Penrose inverse of ${\mathrm{\Psi }}^{*}-P$.

We now prove sufficiency for this theorem. For notational simplicity, we set $\beta ={({\mathrm{\Psi }}^{*}-P)}^{†}$ and, obviously, ${\beta }^{*}={(\mathrm{\Psi }-P)}^{†}$. Based on the definition of the Moore–Penrose inverses, it is well-known that

$$\begin{array}{cc}\hfill \mathcal{R}\left(\beta \right)& =\mathcal{R}(\mathrm{\Psi }-P),\mathcal{N}\left(\beta \right)=\mathcal{N}(\mathrm{\Psi }-P)\supseteq \mathcal{R}\left(P\right),\hfill \end{array}$$

(35)

$$\begin{array}{cc}\hfill \mathcal{R}\left({\beta }^{*}\right)& =\mathcal{R}({\mathrm{\Psi }}^{*}-P),\mathcal{N}\left({\beta }^{*}\right)=\mathcal{N}({\mathrm{\Psi }}^{*}-P)\supseteq \mathcal{R}(I-P),\hfill \end{array}$$

(36)

$$\begin{array}{cc}\hfill \beta ({\mathrm{\Psi }}^{*}-P)& =(\mathrm{\Psi }-P){\beta }^{*},({\mathrm{\Psi }}^{*}-P)\beta ={\beta }^{*}(\mathrm{\Psi }-P),\hfill \end{array}$$

(37)

$$\begin{array}{cc}\hfill {\beta }^{*}(\mathrm{\Psi }-P)& =({\mathrm{\Psi }}^{*}-P)\beta ,(\mathrm{\Psi }-P){\beta }^{*}=\beta ({\mathrm{\Psi }}^{*}-P).\hfill \end{array}$$

(38)

Inspired by Theorem 2, let

$$B=\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta \right)\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right).$$

We now prove that B is the Moore–Penrose inverse of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$. To prevent cumbersome expressions in subsequent derivations, let $T=\left({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta \right)\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right)$. Combined with (35)–(38), we have

$$\begin{array}{c}\hfill T={\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +{\beta }^{*}\end{array}$$

(39)

and

$$\begin{array}{cc}\hfill T\left(2P\right)T^{*}& =({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +{\beta }^{*})\left(2P\right)({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +\beta )\hfill \\ \hfill & =({\displaystyle \frac{P}{2}}+{\beta }^{*})\left(2P\right)({\displaystyle \frac{P}{2}}+\beta )\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}+2{\beta }^{*}\beta +{\beta }^{*}+\beta ,\hfill \end{array}$$

(40)

$$\begin{array}{cc}\hfill T(\mathrm{\Psi }-P)T^{*}& =({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +{\beta }^{*})(\mathrm{\Psi }-P)({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +\beta )\hfill \\ \hfill & =({\displaystyle \frac{\mathrm{\Psi }-P}{2}}+{\beta }^{*}(\mathrm{\Psi }-P))({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +\beta )\hfill \\ \hfill & =-\beta -2{\beta }^{*}\beta .\hfill \end{array}$$

(41)

Right-multiplying both sides of (39) by $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$ yields the following:

$$\begin{array}{cc}\hfill T(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})& =({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +{\beta }^{*})(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\hfill \\ \hfill & =({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta +{\beta }^{*})(\mathrm{\Psi }-P+{\mathrm{\Psi }}^{*}-P+2P)\hfill \\ \hfill & ={\displaystyle \frac{\mathrm{\Psi }-P}{2}}+P+{\beta }^{*}(\mathrm{\Psi }-P).\hfill \end{array}$$

Left-multiplying both sides of the above equation by $\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)$ yields the following:

$$\begin{array}{cc}\hfill B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})& =\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)\left({\displaystyle \frac{\mathrm{\Psi }-P}{2}}+P+{\beta }^{*}(\mathrm{\Psi }-P)\right)\hfill \\ \hfill & =P+{\beta }^{*}(\mathrm{\Psi }-P).\hfill \end{array}$$

(42)

Then, left-multiplying both sides of the above equation by $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$, combined with (35)–(38), yields the following:

$$\begin{array}{cc}\hfill (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})& =(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})(P+{\beta }^{*}(\mathrm{\Psi }-P))\hfill \\ \hfill & =(\mathrm{\Psi }-P+{\mathrm{\Psi }}^{*}-P+2P)(P+{\beta }^{*}(\mathrm{\Psi }-P))\hfill \\ \hfill & =\mathrm{\Psi }-P+{\mathrm{\Psi }}^{*}-P+2P\hfill \\ \hfill & =\mathrm{\Psi }+{\mathrm{\Psi }}^{*}.\hfill \end{array}$$

(43)

Based on (39), (40), and (41), we have the following:

$$\begin{array}{cc}\hfill T(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})T^{*}& =T(\mathrm{\Psi }-P+{\mathrm{\Psi }}^{*}-P+2P)T^{*}\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta .\hfill \end{array}$$

Furthermore, combined with (35)–(38), this yields the following:

$$\begin{array}{cc}\hfill B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})B& =\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)T(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})T^{*}\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right)\hfill \\ \hfill & =\left({\displaystyle \frac{2I-(\mathrm{\Psi }-P)}{2}}\right)({\displaystyle \frac{P}{2}}-2{\beta }^{*}\beta )\left({\displaystyle \frac{2I-({\mathrm{\Psi }}^{*}-P)}{2}}\right)\hfill \\ \hfill & =B.\hfill \end{array}$$

(44)

Based on (37) and (42), we have the following:

$$\begin{array}{c}\hfill B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})=P+{\beta }^{*}(\mathrm{\Psi }-P)=P+({\mathrm{\Psi }}^{*}-P)\beta ={\left(B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\right)}^{*}.\end{array}$$

(45)

As $B=B^{*}$, we obtain that

$$\begin{array}{c}\hfill (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})B={\left(B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\right)}^{*}=B(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})={\left((\mathrm{\Psi }+{\mathrm{\Psi }}^{*})B\right)}^{*}.\end{array}$$

(46)

Combining (43)–(46), we establish that B is the Moore–Penrose inverse of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$. □

Corollary 2.

Let $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent operator and Q be the projection onto $\mathcal{R}(I-\mathrm{\Psi })$. Then, $I-\mathrm{\Psi }+I-{\mathrm{\Psi }}^{*}$ is MP-invertible if and only if $I-{\mathrm{\Psi }}^{*}-Q$ is MP-invertible. The following identities hold:

$$\begin{array}{cc}\hfill {(I-{\mathrm{\Psi }}^{*}-Q)}^{†}& =Q{(I-\mathrm{\Psi }+I-{\mathrm{\Psi }}^{*})}^{†}(I-Q),\hfill \\ \hfill {(I-\mathrm{\Psi }+I-{\mathrm{\Psi }}^{*})}^{†}& =\left({\displaystyle \frac{2I-(I-\mathrm{\Psi }-Q)}{2}}\right)C\left({\displaystyle \frac{2I-(I-{\mathrm{\Psi }}^{*}-Q)}{2}}\right),\hfill \end{array}$$

where $C={\displaystyle \frac{Q}{2}}-2{(I-\mathrm{\Psi }-Q)}^{†}{(I-{\mathrm{\Psi }}^{*}-Q)}^{†}.$

Proof. 

Substituting $\mathrm{\Psi }$, ${\mathrm{\Psi }}^{*}$, and P with $I-\mathrm{\Psi }$, $I-{\mathrm{\Psi }}^{*}$, and Q, respectively, in Theorem 3 yields the proof. □

After investigating the Moore–Penrose inverse of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$, it is natural to consider its spectrum and norm. We denote the spectrum of $T\in \mathcal{E}$ by $\sigma \left(T\right)=\{\lambda \in \mathbb{C}\mid \lambda I-T\mathrm{is}\mathrm{not}\mathrm{invertible}\}$, which is used in the following theorem.

Theorem 4.

For any idempotent operator $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q denote the projections onto $\mathcal{R}(\mathrm{\Psi })$ and $\mathcal{N}(\mathrm{\Psi })$, respectively. Then,

$$\sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\subseteq \left\{0,2,1+{(1-t)}^{-1/2},1-{(1-t)}^{-1/2}\mid t\in \sigma \left(PQP\right)\right\}$$

and

$$\parallel \mathrm{\Psi }+{\mathrm{\Psi }}^{*}{\parallel =1+(1-\parallel PQ\parallel }^2{)}^{-1/2}.$$

Proof. 

Based on Lemma 5, we obtain the following:

$$\begin{array}{cc}\hfill {(I-PQ)}^{-1}P& =(I+PQ+PQPQ+\dots )P\hfill \\ \hfill & =P+PQP+PQPQP+\dots \hfill \\ \hfill & =(P-I)+(I+PQP+PQPQP+\dots )\hfill \\ \hfill & ={(I-PQP)}^{-1}-(I-P).\hfill \end{array}$$

Combined with $\mathrm{\Psi }={(I-PQ)}^{-1}P(I-PQ)$, it follows that

$$\begin{array}{cc}\hfill \mathrm{\Psi }-P& ={(I-PQ)}^{-1}P(I-PQ)-{(I-PQ)}^{-1}(I-PQ)P\hfill \\ \hfill & ={(I-PQ)}^{-1}(PQP-PQ)\hfill \\ \hfill & =\left({(I-PQP)}^{-1}-(I-P)\right)(PQP-PQ)\hfill \\ \hfill & ={(I-PQP)}^{-1}(PQP-PQ).\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}\hfill (\mathrm{\Psi }-P)({\mathrm{\Psi }}^{*}-P)& ={(I-PQP)}^{-1}(PQP-PQ)(PQP-QP){(I-PQP)}^{-1}\hfill \\ \hfill & ={(I-PQP)}^{-1}\left(PQP-{\left(PQP\right)}^2\right){(I-PQP)}^{-1}\hfill \\ \hfill & =PQP{(I-PQP)}^{-1}.\hfill \end{array}$$

(47)

For any $\lambda \in \mathbb{C}\setminus \{0,2\}$, the operators

$$I+\frac{1}{\lambda }(\mathrm{\Psi }-P)\mathrm{and}\frac{P}{\lambda -2}+\frac{I-P}{\lambda }+\frac{{\mathrm{\Psi }}^{*}-P}{\lambda (\lambda -2)}$$

are both invertible, as

$$\begin{array}{cc}\hfill I& =\left(I+\frac{1}{\lambda }(\mathrm{\Psi }-P)\right)\left(I-\frac{1}{\lambda }(\mathrm{\Psi }-P)\right),\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill I& =\left(\frac{P}{\lambda -2}+\frac{I-P}{\lambda }+\frac{{\mathrm{\Psi }}^{*}-P}{\lambda (\lambda -2)}\right)\left((\lambda -2)P+\lambda (I-P)-({\mathrm{\Psi }}^{*}-P)\right).\hfill \end{array}$$

(49)

Define

$$\begin{array}{c}\hfill T=\left(\lambda I-(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\right)\left(\frac{P}{\lambda -2}+\frac{I-P}{\lambda }+\frac{{\mathrm{\Psi }}^{*}-P}{\lambda (\lambda -2)}\right).\end{array}$$

(50)

Using ${(\mathrm{\Psi }-P)}^2=0$, $\mathcal{R}(\mathrm{\Psi }-P)\subseteq \mathcal{R}\left(P\right)$, and $\mathcal{R}({\mathrm{\Psi }}^{*}-P)\subseteq \mathcal{R}(I-P)$, we compute

$$\begin{array}{cc}\hfill T& =\left[(\lambda -2)P+\lambda (I-P)-(\mathrm{\Psi }-P)-({\mathrm{\Psi }}^{*}-P)\right]\left(\frac{P}{\lambda -2}+\frac{I-P}{\lambda }+\frac{{\mathrm{\Psi }}^{*}-P}{\lambda (\lambda -2)}\right)\hfill \\ \hfill & =I-\frac{\mathrm{\Psi }-P}{\lambda }-\frac{(\mathrm{\Psi }-P)({\mathrm{\Psi }}^{*}-P)}{\lambda (\lambda -2)}.\hfill \end{array}$$

Left-multiplying T by $I+{\displaystyle \frac{1}{\lambda }}(\mathrm{\Psi }-P)$ and using (47) yields the following:

$$\begin{array}{cc}\hfill \left(I+\frac{1}{\lambda }(\mathrm{\Psi }-P)\right)T& =\left(I+\frac{1}{\lambda }(\mathrm{\Psi }-P)\right)\left(I-\frac{\mathrm{\Psi }-P}{\lambda }-\frac{(\mathrm{\Psi }-P)({\mathrm{\Psi }}^{*}-P)}{\lambda (\lambda -2)}\right)\hfill \\ \hfill & =I-\frac{(\mathrm{\Psi }-P)({\mathrm{\Psi }}^{*}-P)}{\lambda (\lambda -2)}\hfill \\ \hfill & ={\displaystyle \frac{\lambda (\lambda -2)I-PQP{(I-PQP)}^{-1}}{\lambda (\lambda -2)}}.\hfill \end{array}$$

Combining (48), (49), and (50), we conclude that $\lambda I-(\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$ is invertible if and only if $\lambda (\lambda -2)I-PQP{(I-PQP)}^{-1}$ is invertible.

Let $\mathcal{A}=C^{*}(I,PQP)$ be the unital commutative $C^{*}$-algebra generated by I and $PQP$. Using the Gelfand transform ∧, $\mathcal{A}$ is isomorphic to $C\left(\sigma \right(PQP\left)\right)$, with

$$\widehat{PQP}\left(t\right)=t,\widehat{I}\left(t\right)=1\mathrm{for}\mathrm{all}t\in \sigma \left(PQP\right).$$

Thus, $\lambda (\lambda -2)I-PQP{(I-PQP)}^{-1}$ is invertible if and only if

$$\lambda (\lambda -2)-t{(1-t)}^{-1}\ne 0\mathrm{for}\mathrm{all}t\in \sigma \left(PQP\right),$$

which holds precisely when

$$\lambda \ne 1\pm {(1-t)}^{-1/2}\mathrm{for}\mathrm{all}t\in \sigma \left(PQP\right).$$

Having excluded $\lambda =0$ and $\lambda =2$ in our derivation, we obtain

$$\sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\subseteq \left\{0,2,1+{(1-t)}^{-1/2},1-{(1-t)}^{-1/2}\mid t\in \sigma \left(PQP\right)\right\}.$$

Finally, as $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$ is self-adjoint and $\parallel PQ\parallel <1$, based on Lemma 5,

$$\begin{array}{cc}\hfill \parallel \mathrm{\Psi }+{\mathrm{\Psi }}^{*}\parallel & =sup\left\{\left|z\right|:z\in \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\right\}\hfill \\ \hfill & =max\left\{1+{(1-t)}^{-1/2}:t\in \sigma \left(PQP\right)\right\}\hfill \\ \hfill & =1+{(1-\parallel PQP\parallel )}^{-1/2}\hfill \\ \hfill & ={1+(1-\parallel PQ\parallel }^2{)}^{-1/2}.\hfill \end{array}$$

□

Remark 2.

Note that 0 and 2 may not always be spectral points of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$. The necessary and sufficient conditions for 0 or 2 not being spectral points of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$ are established below. Let P be the projection onto $\mathcal{R}(\mathrm{\Psi })$.

Case 1: $0\notin \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$

When $0\notin \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$, the operator

$$\begin{array}{c}\hfill P-{\displaystyle \frac{({\mathrm{\Psi }}^{*}-P)(\mathrm{\Psi }-P)}{4}}=\left(I-{\displaystyle \frac{{\mathrm{\Psi }}^{*}-P}{2}}\right)(-\mathrm{\Psi }-{\mathrm{\Psi }}^{*})\left(-{\displaystyle \frac{I}{2}}-{\displaystyle \frac{\mathrm{\Psi }-P}{4}}\right)\end{array}$$

is invertible, as

$$\begin{array}{cc}\hfill \left(I-{\displaystyle \frac{{\mathrm{\Psi }}^{*}-P}{2}}\right)\left(I+{\displaystyle \frac{{\mathrm{\Psi }}^{*}-P}{2}}\right)& =I,\hfill \\ \hfill \left(-{\displaystyle \frac{I}{2}}-{\displaystyle \frac{\mathrm{\Psi }-P}{4}}\right)\left(-2I-(\mathrm{\Psi }-P)\right)& =I.\hfill \end{array}$$

Given $\mathcal{R}({\mathrm{\Psi }}^{*}-P)\subseteq \mathcal{R}(I-P)$, the operator $P-{\displaystyle \frac{({\mathrm{\Psi }}^{*}-P)(\mathrm{\Psi }-P)}{4}}$ satisfies the following conditions:

• injectivity holds if and only if $\mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left({\mathrm{\Psi }}^{*}\right)=\left\{0\right\}$;
• bijectivity holds if and only if $\mathcal{R}({\mathrm{\Psi }}^{*}-P)=\mathcal{R}(I-P)$.

Based on (13), $\mathcal{R}({\mathrm{\Psi }}^{*}-P)=\mathcal{R}(I-P)$ implies $\mathcal{N}(\mathrm{\Psi })\cap \mathcal{N}\left({\mathrm{\Psi }}^{*}\right)=\left\{0\right\}$. Thus,

$$0\notin \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\iff \mathcal{R}({\mathrm{\Psi }}^{*}-P)=\mathcal{R}(I-P).$$

Example: For ${\mathrm{\Psi }}_1=I$, is it clear that $0\notin \sigma ({\mathrm{\Psi }}_1+{\mathrm{\Psi }}_1^{*})$.

Case 2: $2\notin \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$

When $2\notin \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})$, the operator

$$\begin{array}{c}\hfill I-P+{\displaystyle \frac{(\mathrm{\Psi }-P)({\mathrm{\Psi }}^{*}-P)}{4}}=\left(I+{\displaystyle \frac{\mathrm{\Psi }-P}{2}}\right)(2I-(\mathrm{\Psi }+{\mathrm{\Psi }}^{*}))\left({\displaystyle \frac{I}{2}}+{\displaystyle \frac{{\mathrm{\Psi }}^{*}-P}{4}}\right)\end{array}$$

is invertible, with inverses

$$\begin{array}{cc}\hfill \left(I+{\displaystyle \frac{\mathrm{\Psi }-P}{2}}\right)\left(I-{\displaystyle \frac{\mathrm{\Psi }-P}{2}}\right)& =I,\hfill \\ \hfill \left({\displaystyle \frac{I}{2}}+{\displaystyle \frac{{\mathrm{\Psi }}^{*}-P}{4}}\right)\left(2I-({\mathrm{\Psi }}^{*}-P)\right)& =I.\hfill \end{array}$$

Given $\mathcal{R}(\mathrm{\Psi }-P)\subseteq \mathcal{R}\left(P\right)$, the operator $I-P+{\displaystyle \frac{(\mathrm{\Psi }-P)({\mathrm{\Psi }}^{*}-P)}{4}}$ satisfies the following conditions:

• injectivity holds if and only if $\mathcal{R}(\mathrm{\Psi })\cap \mathcal{R}\left({\mathrm{\Psi }}^{*}\right)=\left\{0\right\}$;
• bijectivity holds if and only if $\mathcal{R}(\mathrm{\Psi }-P)=\mathcal{R}\left(P\right)$.

Thus,

$$2\notin \sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})\iff \mathcal{R}(\mathrm{\Psi })\cap \mathcal{R}\left({\mathrm{\Psi }}^{*}\right)=\left\{0\right\}\mathrm{and}\mathcal{R}(\mathrm{\Psi }-P)=\mathcal{R}\left(P\right).$$

Example: For ${\mathrm{\Psi }}_2=0$, it is clear that $2\notin \sigma ({\mathrm{\Psi }}_2+{\mathrm{\Psi }}_2^{*})$.

Finally, for any $\mathrm{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ where $\mathcal{R}({\mathrm{\Psi }}^{*}-P)\ne \mathcal{R}(I-P)$ and either $\mathcal{R}(\mathrm{\Psi })\cap \mathcal{R}\left({\mathrm{\Psi }}^{*}\right)\ne \left\{0\right\}$ or $\mathcal{R}(\mathrm{\Psi }-P)\ne \mathcal{R}\left(P\right)$, we can obtain

$$\sigma (\mathrm{\Psi }+{\mathrm{\Psi }}^{*})=\left\{0,2,1+{(1-t)}^{-1/2},1-{(1-t)}^{-1/2}\mid t\in \sigma \left(PQP\right)\right\}.$$

5. Decomposition of Adjointable Idempotent Operators on Hilbert ${\mathit{C}}^{*}$-Modules

We perform a linear decomposition of the idempotent operators defined on $\mathcal{I}\left(\mathcal{E}\right)$ (see Definition 1), then we provide representations for the resulting idempotent operators after decomposition. Finally, we extend these results to arbitrary linear idempotent operators (without requiring adjointability) on Hilbert $C^{*}$-modules. In order to obtain the main results, we require that the set of all idempotents is a poset.

Definition 2

([35]). The idempotents on a Hilbert $C^{*}$-module $\mathcal{E}$ form a poset if the order ≤ defined by ${\mathrm{\Psi }}_1\le {\mathrm{\Psi }}_2$ if and only if ${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2={\mathrm{\Psi }}_2{\mathrm{\Psi }}_1={\mathrm{\Psi }}_1$ holds for any idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$.

Let $\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$. For any ${\mathrm{\Psi }}_x\ge \mathrm{\Psi }$ or ${\mathrm{\Psi }}_y\le \mathrm{\Psi }$, the following lemma provides a characterization of ${\mathrm{\Psi }}_x$ and ${\mathrm{\Psi }}_y$, which yields a constructive method.

Lemma 7.

For any $\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$, let P be a projection from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$. Then,

(i)

For any linear idempotents ${\mathrm{\Psi }}_x$ on $\mathcal{E}$, ${\mathrm{\Psi }}_x\ge \mathrm{\Psi }$ if and only if there exists linear idempotents ${\mathrm{\Psi }}_{x1}$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_{x1}\le I-P$ and ${\mathrm{\Psi }}_x=\mathrm{\Psi }+{\mathrm{\Psi }}_{x1}-\mathrm{\Psi }{\mathrm{\Psi }}_{x1}$.

(ii)

For any linear idempotents ${\mathrm{\Psi }}_y$ on $\mathcal{E}$, ${\mathrm{\Psi }}_y\le \mathrm{\Psi }$ if and only if there exists linear idempotents ${\mathrm{\Psi }}_{y1}$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_{y1}\le P$ and ${\mathrm{\Psi }}_y={\mathrm{\Psi }}_{y1}\mathrm{\Psi }$.

Proof. 

(i) Necessity: Let the subspaces $H_i$ and projections $P_i$ ($1\le i\le 6$) be defined by (2)–(6). Let Q be the projection onto $\mathcal{N}(\mathrm{\Psi })$. Let $Q_0$ and the unitary operator $U_0\in \mathcal{L}(H_6,H_5)$ be defined by Lemma 1, with V defined in (14) and $U_{P,Q}$ in (7). Based on (15), we have

$$U_{P,Q}\mathrm{\Psi }{U}_{P,Q}^{*}=I_{H_2}\mathrm{\oplus }0\mathrm{\oplus }\left(\begin{array}{cc}{I}_{H_5}& V\\ 0& 0\end{array}\right).$$

Represent ${\mathrm{\Psi }}_x$ as

$$U_{P,Q}{\mathrm{\Psi }}_x{U}_{P,Q}^{*}=\left(\begin{array}{cccc}{\alpha }_{11}& {\alpha }_{12}& {\alpha }_{13}& {\alpha }_{14}\\ {\alpha }_{21}& {\alpha }_{22}& {\alpha }_{23}& {\alpha }_{24}\\ {\alpha }_{31}& {\alpha }_{32}& {\alpha }_{33}& {\alpha }_{34}\\ {\alpha }_{41}& {\alpha }_{42}& {\alpha }_{43}& {\alpha }_{44}\end{array}\right)=\left(\begin{array}{c}{\alpha }_{ij}\end{array}\right).$$

The condition ${\mathrm{\Psi }}_x\mathrm{\Psi }=\mathrm{\Psi }{\mathrm{\Psi }}_x=\mathrm{\Psi }$ implies

$$\begin{array}{cc}\hfill \left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& 0\end{array}\right)\left(\begin{array}{c}{\alpha }_{ij}\end{array}\right)& =\left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& 0\end{array}\right),\hfill \\ \hfill \left(\begin{array}{c}{\alpha }_{ij}\end{array}\right)\left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& 0\end{array}\right)& =\left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& 0\end{array}\right).\hfill \end{array}$$

Direct computation yields

$$U_{P,Q}{\mathrm{\Psi }}_x{U}_{P,Q}^{*}=\left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& {\alpha }_{22}& 0& {\alpha }_{24}\\ 0& {\alpha }_{32}& I_{H_5}& {\alpha }_{34}\\ 0& {\alpha }_{42}& 0& {\alpha }_{44}\end{array}\right),$$

(51)

subject to

$$\begin{array}{cc}\hfill {\alpha }_{34}+V{\alpha }_{44}& =V,\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill {\alpha }_{32}+V{\alpha }_{42}& =0.\hfill \end{array}$$

(53)

The idempotence relation ${\mathrm{\Psi }}_x^2={\mathrm{\Psi }}_x$ gives

$$\begin{array}{cc}\hfill {\alpha }_{22}^2+{\alpha }_{24}{\alpha }_{42}& ={\alpha }_{22},\hfill \end{array}$$

(54)

$$\begin{array}{cc}\hfill {\alpha }_{22}{\alpha }_{24}+{\alpha }_{24}{\alpha }_{44}& ={\alpha }_{24},\hfill \end{array}$$

(55)

$$\begin{array}{cc}\hfill {\alpha }_{42}{\alpha }_{22}+{\alpha }_{44}{\alpha }_{42}& ={\alpha }_{42},\hfill \end{array}$$

(56)

$$\begin{array}{cc}\hfill {\alpha }_{42}{\alpha }_{24}+{\alpha }_{44}^2& ={\alpha }_{44}.\hfill \end{array}$$

(57)

From (15) and (51), we can decompose

$${\mathrm{\Psi }}_x=\mathrm{\Psi }+{\mathrm{\Psi }}_{x1}+{\mathrm{\Psi }}_{x2},$$

where

$$\begin{array}{cc}\hfill U_{P,Q}{\mathrm{\Psi }}_{x1}{U}_{P,Q}^{*}& =\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& {\alpha }_{22}& 0& {\alpha }_{24}\\ 0& 0& 0& 0\\ 0& {\alpha }_{42}& 0& {\alpha }_{44}\end{array}\right),\hfill \end{array}$$

(58)

$$\begin{array}{cc}\hfill U_{P,Q}{\mathrm{\Psi }}_{x2}{U}_{P,Q}^{*}& =\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& {\alpha }_{32}& 0& {\alpha }_{34}-V\\ 0& 0& 0& 0\end{array}\right).\hfill \end{array}$$

Based on (54)–(57), ${\mathrm{\Psi }}_{x1}$ is idempotent. The form in (58) implies ${\mathrm{\Psi }}_{x1}\le I-P$, and, so, $P{\mathrm{\Psi }}_{x1}=0$. From (15), we have

$$\begin{array}{c}\hfill \mathrm{\Psi }=P+P_{5}V{P}_6,\end{array}$$

(59)

and, based on (52) and (53),

$${\mathrm{\Psi }}_{x2}=-P_{5}V{P}_6{\mathrm{\Psi }}_{x1}=(P-\mathrm{\Psi }){\mathrm{\Psi }}_{x1}=-\mathrm{\Psi }{\mathrm{\Psi }}_{x1}.$$

Thus, ${\mathrm{\Psi }}_x=\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_{x1}$.

Sufficiency: As ${\mathrm{\Psi }}_{x1}\le I-P$, we have $\mathcal{R}\left(P\right)\subseteq \mathcal{N}\left({\mathrm{\Psi }}_{x1}\right)$. With $\mathcal{R}\left(P\right)=\mathcal{R}(\mathrm{\Psi })$,

$${\mathrm{\Psi }}_x\mathrm{\Psi }=(\mathrm{\Psi }+{\mathrm{\Psi }}_{x1}-\mathrm{\Psi }{\mathrm{\Psi }}_{x1})\mathrm{\Psi }=\mathrm{\Psi }.$$

As $\mathrm{\Psi }{\mathrm{\Psi }}_x=\mathrm{\Psi }$ and ${\mathrm{\Psi }}_x^2={\mathrm{\Psi }}_x$, we conclude ${\mathrm{\Psi }}_x\ge \mathrm{\Psi }$.

(ii) Necessity: Set

$$U_{P,Q}{\mathrm{\Psi }}_y{U}_{P,Q}^{*}=\left(\begin{array}{cccc}{\beta }_{11}& {\beta }_{12}& {\beta }_{13}& {\beta }_{14}\\ {\beta }_{21}& {\beta }_{22}& {\beta }_{23}& {\beta }_{24}\\ {\beta }_{31}& {\beta }_{32}& {\beta }_{33}& {\beta }_{34}\\ {\beta }_{41}& {\beta }_{42}& {\beta }_{43}& {\beta }_{44}\end{array}\right)=\left(\begin{array}{c}{\beta }_{ij}\end{array}\right).$$

The condition ${\mathrm{\Psi }}_y\mathrm{\Psi }=\mathrm{\Psi }{\mathrm{\Psi }}_y={\mathrm{\Psi }}_y$ implies the following:

$$\begin{array}{cc}\hfill \left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& 0\end{array}\right)\left({\beta }_{ij}\right)& =\left({\beta }_{ij}\right),\hfill \\ \hfill \left({\beta }_{ij}\right)\left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& 0\end{array}\right)& =\left({\beta }_{ij}\right).\hfill \end{array}$$

This simplifies to the following:

$$U_{P,Q}{\mathrm{\Psi }}_y{U}_{P,Q}^{*}=\left(\begin{array}{cccc}{\beta }_{11}& 0& {\beta }_{13}& {\beta }_{13}V\\ 0& 0& 0& 0\\ {\beta }_{31}& 0& {\beta }_{33}& {\beta }_{33}V\\ 0& 0& 0& 0\end{array}\right).$$

(60)

The idempotence relation ${\mathrm{\Psi }}_y^2={\mathrm{\Psi }}_y$ yields the following:

$$\begin{array}{cc}\hfill {\beta }_{11}^2+{\beta }_{13}{\beta }_{31}& ={\beta }_{11},\hfill \end{array}$$

(61)

$$\begin{array}{cc}\hfill {\beta }_{11}{\beta }_{13}+{\beta }_{13}{\beta }_{33}& ={\beta }_{13},\hfill \end{array}$$

(62)

$$\begin{array}{cc}\hfill {\beta }_{31}{\beta }_{11}+{\beta }_{33}{\beta }_{31}& ={\beta }_{31},\hfill \end{array}$$

(63)

$$\begin{array}{cc}\hfill {\beta }_{31}{\beta }_{13}+{\beta }_{33}^2& ={\beta }_{33}.\hfill \end{array}$$

(64)

We decompose

$${\mathrm{\Psi }}_y={\mathrm{\Psi }}_{y1}+{\mathrm{\Psi }}_{y2},$$

where

$$\begin{array}{cc}\hfill U_{P,Q}{\mathrm{\Psi }}_{y1}{U}_{P,Q}^{*}& =\left(\begin{array}{cccc}{\beta }_{11}& 0& {\beta }_{13}& 0\\ 0& 0& 0& 0\\ {\beta }_{31}& 0& {\beta }_{33}& 0\\ 0& 0& 0& 0\end{array}\right),\hfill \end{array}$$

(65)

$$\begin{array}{cc}\hfill U_{P,Q}{\mathrm{\Psi }}_{y2}{U}_{P,Q}^{*}& =\left(\begin{array}{cccc}0& 0& 0& {\beta }_{13}V\\ 0& 0& 0& 0\\ 0& 0& 0& {\beta }_{33}V\\ 0& 0& 0& 0\end{array}\right).\hfill \end{array}$$

Based on (61)–(64), ${\mathrm{\Psi }}_{y1}$ is idempotent. The form in (65) implies ${\mathrm{\Psi }}_{y1}\le P$, and, so, ${\mathrm{\Psi }}_{y1}P={\mathrm{\Psi }}_{y1}$. Using (59),

$${\mathrm{\Psi }}_{y2}={\mathrm{\Psi }}_{y1}{P}_{5}V{P}_6={\mathrm{\Psi }}_{y1}(\mathrm{\Psi }-P)={\mathrm{\Psi }}_{y1}\mathrm{\Psi }-{\mathrm{\Psi }}_{y1},$$

and, so, ${\mathrm{\Psi }}_y={\mathrm{\Psi }}_{y1}\mathrm{\Psi }$.

Sufficiency: As $\mathcal{R}\left({\mathrm{\Psi }}_{y1}\right)\subseteq \mathcal{R}\left(P\right)=\mathcal{R}(\mathrm{\Psi })$, Lemma 6 gives the following:

$$\mathrm{\Psi }{\mathrm{\Psi }}_y=\mathrm{\Psi }({\mathrm{\Psi }}_{y1}\mathrm{\Psi })=(\mathrm{\Psi }{\mathrm{\Psi }}_{y1})\mathrm{\Psi }={\mathrm{\Psi }}_{y1}\mathrm{\Psi }={\mathrm{\Psi }}_y.$$

As ${\mathrm{\Psi }}_y\mathrm{\Psi }={\mathrm{\Psi }}_y$ and ${\mathrm{\Psi }}_y^2={\mathrm{\Psi }}_y$, we conclude ${\mathrm{\Psi }}_y\le \mathrm{\Psi }$. □

Motivated by the results obtained by Baksalary [13], we pay attention to the decomposition of idempotents. We also use their results to prove a decomposition theorem. This is a parallel generalization of Baksalary’s results to Hilbert $C^{*}$-modules.

Lemma 8

(cf. [13], Theorem 1). Let ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ be linear idempotents on $\mathcal{E}$ such that ${\mathrm{\Psi }}_1\ne 0,$${\mathrm{\Psi }}_2\ne 0,{\mathrm{\Psi }}_1\ne {\mathrm{\Psi }}_2$. Let Ψ be their linear combination of the form $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+c_2{\mathrm{\Psi }}_2$ with non-zero scalars $c_1$ and $c_2$. Then, there are exactly four conditions:

(i)

${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2={\mathrm{\Psi }}_2{\mathrm{\Psi }}_1,c_1=1,c_2=1,{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2=0$;

(ii)

${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2={\mathrm{\Psi }}_2{\mathrm{\Psi }}_1,c_1=1,c_2=-1,{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2={\mathrm{\Psi }}_2$;

(iii)

${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2={\mathrm{\Psi }}_2{\mathrm{\Psi }}_1,c_1=-1,c_2=1,{\mathrm{\Psi }}_1{\mathrm{\Psi }}_2={\mathrm{\Psi }}_1$;

(iv)

${\mathrm{\Psi }}_1{\mathrm{\Psi }}_2\ne {\mathrm{\Psi }}_2{\mathrm{\Psi }}_1,c_1\ne 0\mathrm{and}{c}_1\ne 1,c_1+c_2=1,{({\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2)}^2=0$.

Proof. 

The proof is the same as Baksalary’s proof in ([13], Theorem 1). □

Now, for any linear idempotents $\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$, we try to characterize all idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$ and non-zero scalars $c_1$ and $c_2$ such that $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+c_2{\mathrm{\Psi }}_2$, where ${\mathrm{\Psi }}_1\ne 0,$${\mathrm{\Psi }}_2\ne 0,{\mathrm{\Psi }}_1\ne {\mathrm{\Psi }}_2$. Obviously, ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2,c_1$, and $c_2$ must satisfy one condition in Lemma 8. The main theorem is as follows.

Theorem 5.

For any $\mathrm{\Psi }\in \mathcal{I}\left(\mathcal{E}\right)$, let P be a projection from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$. Then,

(i)

If $\mathrm{\Psi }\ne 0$, there exist two non-zero idempotents ${\mathrm{\Psi }}_1$ and ${\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$ if and only if there exists non-zero idempotents ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_0\le P$, ${\mathrm{\Psi }}_0\ne P$, ${\mathrm{\Psi }}_1={\mathrm{\Psi }}_0\mathrm{\Psi }$, and ${\mathrm{\Psi }}_2=\mathrm{\Psi }-{\mathrm{\Psi }}_0\mathrm{\Psi }$;

(ii)

If $\mathrm{\Psi }\ne 0$ and $\mathrm{\Psi }\ne I$, there exist two non-zero idempotents ${\mathrm{\Psi }}_1$ and ${\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$ if and only if there exists non-zero idempotents ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_0\le I-P$, ${\mathrm{\Psi }}_1=\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_0$, and ${\mathrm{\Psi }}_2=(I-\mathrm{\Psi }){\mathrm{\Psi }}_0$;

(iii)

For $\mathrm{\Psi }\ne 0$, $\mathrm{\Psi }\ne I$, and a non-zero scalar $c_1\ne 1$, the following are equivalent:

(a) 

There exist distinct non-zero idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+$$(1-c_1){\mathrm{\Psi }}_2$.

(b) 

There exist operators $A_1,A_2$ on $\mathcal{E}$ satisfying

i. 

$\mathcal{R}\left(A_1\right)\subseteq \mathcal{N}\left(P\right)$, $\mathcal{N}\left(A_1\right)\supseteq \mathcal{N}\left(P\right)$;

ii. 

$\mathcal{R}\left(A_2\right)\subseteq \mathcal{R}\left(P\right)$, $\mathcal{N}\left(A_2\right)\supseteq \mathcal{R}\left(P\right)$;

iii. 

$A_1{A}_2=A_2{A}_1=0$;

iv. 

$(I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2\ne 0$

with explicit expressions

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1& =\mathrm{\Psi }-(I-\mathrm{\Psi })A_1\mathrm{\Psi }-A_2,\hfill \\ \hfill {\mathrm{\Psi }}_2& =\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}\left((I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2\right).\hfill \end{array}$$

Proof. 

(i) Assume there exist idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$. Based on Lemma 8(i),

$$\begin{array}{cc}\hfill \mathrm{\Psi }{\mathrm{\Psi }}_1& =({\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2){\mathrm{\Psi }}_1={\mathrm{\Psi }}_1,\hfill \\ \hfill {\mathrm{\Psi }}_1\mathrm{\Psi }& ={\mathrm{\Psi }}_1({\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2)={\mathrm{\Psi }}_1,\hfill \end{array}$$

establishing

$$\begin{array}{c}\hfill {\mathrm{\Psi }}_1\le \mathrm{\Psi },{\mathrm{\Psi }}_1\ne \mathrm{\Psi }\mathrm{and}{\mathrm{\Psi }}_1\ne 0\end{array}$$

(66)

as ${\mathrm{\Psi }}_1\ne 0$ and ${\mathrm{\Psi }}_2\ne 0$.

Conversely, for any idempotent ${\mathrm{\Psi }}_1$ on $\mathcal{E}$ satisfying ${\mathrm{\Psi }}_1\le \mathrm{\Psi }$ and ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$,

$${(\mathrm{\Psi }-{\mathrm{\Psi }}_1)}^2={\mathrm{\Psi }}^2-\mathrm{\Psi }{\mathrm{\Psi }}_1-{\mathrm{\Psi }}_1\mathrm{\Psi }+{\mathrm{\Psi }}_1^2=\mathrm{\Psi }-{\mathrm{\Psi }}_1,$$

which implies that ${\mathrm{\Psi }}_2=\mathrm{\Psi }-{\mathrm{\Psi }}_1$ is idempotent. Thus, $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$ with ${\mathrm{\Psi }}_2\ne 0$.

Therefore, the decomposition $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$ exists if and only if there exists a non-zero idempotent ${\mathrm{\Psi }}_1$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_1\le \mathrm{\Psi }$ and ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$. Using Lemma 7(ii), we determine that $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$ exists if and only if there exists a non-zero idempotent ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_0\le P,{\mathrm{\Psi }}_0\ne P$, and ${\mathrm{\Psi }}_1={\mathrm{\Psi }}_0\mathrm{\Psi }$. Note that 0 cannot be decomposed as $0={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$, considering that only $0\le 0$ holds.

(ii) Suppose there exist idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$. Based on Lemma 8(ii),

$$\begin{array}{cc}\hfill \mathrm{\Psi }{\mathrm{\Psi }}_1& =({\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2){\mathrm{\Psi }}_1={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2=\mathrm{\Psi },\hfill \\ \hfill {\mathrm{\Psi }}_1\mathrm{\Psi }& ={\mathrm{\Psi }}_1({\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2)={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2=\mathrm{\Psi },\hfill \end{array}$$

establishing $\mathrm{\Psi }\le {\mathrm{\Psi }}_1$. As ${\mathrm{\Psi }}_2\ne 0$, we have ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$.

Conversely, for any idempotents ${\mathrm{\Psi }}_1$ on $\mathcal{E}$ satisfying ${\mathrm{\Psi }}_1\ge \mathrm{\Psi }$ and ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$,

$${({\mathrm{\Psi }}_1-\mathrm{\Psi })}^2={\mathrm{\Psi }}_1^2-{\mathrm{\Psi }}_1\mathrm{\Psi }-\mathrm{\Psi }{\mathrm{\Psi }}_1+{\mathrm{\Psi }}^2={\mathrm{\Psi }}_1-\mathrm{\Psi },$$

which implies that ${\mathrm{\Psi }}_2={\mathrm{\Psi }}_1-\mathrm{\Psi }$ is idempotent. Thus, $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$ with ${\mathrm{\Psi }}_2\ne 0$.

Therefore, the decomposition $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$ exists if and only if there exists an idempotent operator ${\mathrm{\Psi }}_1$ on $\mathcal{E}$ such that

$$\begin{array}{c}\hfill {\mathrm{\Psi }}_1\ge \mathrm{\Psi }\mathrm{and}{\mathrm{\Psi }}_1\ne \mathrm{\Psi }.\end{array}$$

(67)

Using Lemma 7(i), we obtain that the decomposition $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$ exists if and only if there exists a non-zero idempotent ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_0\le I-P$ and ${\mathrm{\Psi }}_1=\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_0$. Note that I cannot be decomposed as $I={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$, given that only $I\ge I$ holds, and 0 cannot be decomposed as $0={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$, given that ${\mathrm{\Psi }}_1\ne {\mathrm{\Psi }}_2$.

(iii) Assume ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ are idempotents on $\mathcal{E}$ satisfying $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+(1-c_1){\mathrm{\Psi }}_2$. Set ${\mathrm{\Psi }}_0=\mathrm{\Psi }-{\mathrm{\Psi }}_1$. Based on Lemma 8(iv),

$${\mathrm{\Psi }}_0^2={(c_1{\mathrm{\Psi }}_1+(1-c_1){\mathrm{\Psi }}_2-{\mathrm{\Psi }}_1)}^2={(1-c_1)}^2{({\mathrm{\Psi }}_2-{\mathrm{\Psi }}_1)}^2=0.$$

As ${\mathrm{\Psi }}_1\ne {\mathrm{\Psi }}_2$ and $c_1\notin \{0,1\}$, we have ${\mathrm{\Psi }}_0\ne 0$.

Conversely, for ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ with ${\mathrm{\Psi }}_0\ne 0$, ${\mathrm{\Psi }}_0^2=0$, and ${(\mathrm{\Psi }-{\mathrm{\Psi }}_0)}^2=\mathrm{\Psi }-{\mathrm{\Psi }}_0$, define

$${\mathrm{\Psi }}_1=\mathrm{\Psi }-{\mathrm{\Psi }}_0,{\mathrm{\Psi }}_2=\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}{\mathrm{\Psi }}_0$$

for $c_1\in \mathbb{C}\setminus \{0,1\}$. The condition implies

$$\begin{array}{c}\hfill {\mathrm{\Psi }}_0=\mathrm{\Psi }{\mathrm{\Psi }}_0+{\mathrm{\Psi }}_0\mathrm{\Psi }\end{array}$$

(68)

and, thus,

$${\mathrm{\Psi }}_2^2={\left(\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}{\mathrm{\Psi }}_0\right)}^2={\mathrm{\Psi }}_2.$$

The decomposition is $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+(1-c_1){\mathrm{\Psi }}_2$.

Therefore, the decomposition exists if and only if there exists ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ satisfying

$$\begin{array}{c}\hfill {\mathrm{\Psi }}_0\ne 0,{\mathrm{\Psi }}_0^2=0,{(\mathrm{\Psi }-{\mathrm{\Psi }}_0)}^2=\mathrm{\Psi }-{\mathrm{\Psi }}_0.\end{array}$$

(69)

Note that $\mathrm{\Psi }\notin \{0,I\}$ here; otherwise, it is straightforward to derive a contradiction through computation.

To characterize ${\mathrm{\Psi }}_0$, let Q be the projection onto $\mathcal{N}(\mathrm{\Psi })$, with subspaces $H_i$ and projections $P_i$ ($1\le i\le 6$) defined in (2)–(6). Let $U_{P,Q}$ be as in (7), and $Q_0$, $U_0\in \mathcal{L}(H_6,H_5)$ from Lemma 1, with V defined in (14). Represent ${\mathrm{\Psi }}_0$ as

$$U_{P,Q}{\mathrm{\Psi }}_0{U}_{P,Q}^{*}=\left(\begin{array}{cccc}{\alpha }_{11}& {\alpha }_{12}& {\alpha }_{13}& {\alpha }_{14}\\ {\alpha }_{21}& {\alpha }_{22}& {\alpha }_{23}& {\alpha }_{24}\\ {\alpha }_{31}& {\alpha }_{32}& {\alpha }_{33}& {\alpha }_{34}\\ {\alpha }_{41}& {\alpha }_{42}& {\alpha }_{43}& {\alpha }_{44}\end{array}\right).$$

(70)

Using (15),

$$\begin{array}{cc}\hfill \mathrm{\Psi }{\mathrm{\Psi }}_0& =\left(\begin{array}{cccc}{\alpha }_{11}& {\alpha }_{12}& {\alpha }_{13}& {\alpha }_{14}\\ 0& 0& 0& 0\\ {\alpha }_{31}+V{\alpha }_{41}& {\alpha }_{32}+V{\alpha }_{42}& ·& ·\\ 0& 0& 0& 0\end{array}\right),\hfill \\ \hfill {\mathrm{\Psi }}_0\mathrm{\Psi }& =\left(\begin{array}{cccc}{\alpha }_{11}& 0& {\alpha }_{13}& {\alpha }_{13}V\\ {\alpha }_{21}& 0& {\alpha }_{23}& {\alpha }_{23}V\\ {\alpha }_{31}& 0& {\alpha }_{33}& {\alpha }_{33}V\\ {\alpha }_{41}& 0& {\alpha }_{43}& {\alpha }_{43}V\end{array}\right).\hfill \end{array}$$

From (68), we obtain

$$\begin{array}{cc}\hfill {\alpha }_{11}& =0,{\alpha }_{13}=0,{\alpha }_{22}=0,{\alpha }_{24}={\alpha }_{23}V,\hfill \\ \hfill {\alpha }_{31}& =-V{\alpha }_{41},{\alpha }_{42}=0,{\alpha }_{33}=-V{\alpha }_{43},{\alpha }_{44}={\alpha }_{43}V.\hfill \end{array}$$

Thus,

$$U_{P,Q}{\mathrm{\Psi }}_0{U}_{P,Q}^{*}=\left(\begin{array}{cccc}0& {\alpha }_{12}& 0& {\alpha }_{14}\\ {\alpha }_{21}& 0& {\alpha }_{23}& {\alpha }_{23}V\\ -V{\alpha }_{41}& {\alpha }_{32}& -V{\alpha }_{43}& {\alpha }_{34}\\ {\alpha }_{41}& 0& {\alpha }_{43}& {\alpha }_{43}V\end{array}\right).$$

Define

$$\begin{array}{c}\hfill T:U_{P,Q}T{U}_{P,Q}^{*}=\left(\begin{array}{cccc}{I}_{H_2}& 0& 0& 0\\ 0& I_{H_3}& 0& 0\\ 0& 0& I_{H_5}& V\\ 0& 0& 0& I_{H_6}\end{array}\right),\end{array}$$

and the operators

$$\begin{array}{c}\hfill A_1:U_{P,Q}{A}_1{U}_{P,Q}^{*}=\left(\begin{array}{cccc}0& 0& 0& 0\\ {\alpha }_{21}& 0& {\alpha }_{23}& 0\\ 0& 0& 0& 0\\ {\alpha }_{41}& 0& {\alpha }_{43}& 0\end{array}\right),\end{array}$$

$$\begin{array}{c}\hfill A_2:U_{P,Q}{A}_2{U}_{P,Q}^{*}=\left(\begin{array}{cccc}0& {\alpha }_{12}& 0& {\alpha }_{14}\\ 0& 0& 0& 0\\ 0& {\alpha }_{32}& 0& {\alpha }_{34}+V{\alpha }_{43}V\\ 0& 0& 0& 0\end{array}\right).\end{array}$$

Then, ${\mathrm{\Psi }}_0=T^{-1}(A_1+A_2)T$ with

$$\begin{array}{cc}\hfill \mathcal{R}\left(A_1\right)& \subseteq \mathcal{N}\left(P\right),\mathcal{N}\left(A_1\right)\supseteq \mathcal{N}\left(P\right),\hfill \\ \hfill \mathcal{R}\left(A_2\right)& \subseteq \mathcal{R}\left(P\right),\mathcal{N}\left(A_2\right)\supseteq \mathcal{R}\left(P\right),\hfill \end{array}$$

such that $A_1^2=0$ and $A_2^2=0$. As ${\mathrm{\Psi }}_0^2=0$, we have

$$\begin{array}{cc}\hfill A_1{A}_2+A_2{A}_1& ={(A_1+A_2)}^2-A_1^2-A_1^2=0.\hfill \end{array}$$

Furthermore, $\mathcal{R}\left(A_1\right)\cap \mathcal{R}\left(A_2\right)=\left\{0\right\}$ implies $A_1{A}_2=A_2{A}_1=0$. Using (59),

$$\begin{array}{cc}\hfill T^{-1}{A}_{1}T& =(I-P_{5}V{P}_6)A_1(I+P_{5}V{P}_6)\hfill \\ \hfill & =(I-P-P_{5}V{P}_6)A_1(P+P_{5}V{P}_6)\hfill \\ \hfill & =(I-\mathrm{\Psi })A_1\mathrm{\Psi },\hfill \\ \hfill T^{-1}{A}_{2}T& =(I-P_{5}V{P}_6)A_2(I+P_{5}V{P}_6)\hfill \\ \hfill & =\left(I\right)A_2\left(I\right)\hfill \\ \hfill & =A_2.\hfill \end{array}$$

Thus, ${\mathrm{\Psi }}_0=(I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2$.

Conversely, for $A_1,A_2$ on $\mathcal{E}$ satisfying

• $\mathcal{R}\left(A_1\right)\subseteq \mathcal{N}\left(P\right)$, $\mathcal{N}\left(A_1\right)\supseteq \mathcal{N}\left(P\right)$;
• $\mathcal{R}\left(A_2\right)\subseteq \mathcal{R}\left(P\right)$, $\mathcal{N}\left(A_2\right)\supseteq \mathcal{R}\left(P\right)$;
• $A_1{A}_2=A_2{A}_1=0$;
• $(I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2\ne 0$;

set ${\mathrm{\Psi }}_0=(I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2$. Then, ${\mathrm{\Psi }}_0\ne 0$, ${\mathrm{\Psi }}_0^2=0$ and, based on Lemma 6,

$${(\mathrm{\Psi }-{\mathrm{\Psi }}_0)}^2=\mathrm{\Psi }-{\mathrm{\Psi }}_0.$$

This completes the proof by (69). □

Subsequently, we extend the above theorem to arbitrary idempotent operators (i.e., without requiring adjointability) on $\mathcal{E}$.

Theorem 6.

For any Ψ on $\mathcal{E}$, let P be a projection from $\mathcal{E}$ to $\mathcal{R}(\mathrm{\Psi })$. Then,

(i)

If $\mathrm{\Psi }\ne 0$, there exist two non-zero idempotents ${\mathrm{\Psi }}_1$ and ${\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$ if and only if there exists non-zero idempotents ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_0\le P$, ${\mathrm{\Psi }}_0\ne P$, ${\mathrm{\Psi }}_1={\mathrm{\Psi }}_0\mathrm{\Psi }$, and ${\mathrm{\Psi }}_2=\mathrm{\Psi }-{\mathrm{\Psi }}_0\mathrm{\Psi }$;

(ii)

If $\mathrm{\Psi }\ne 0$ and $\mathrm{\Psi }\ne I$, there exist two non-zero idempotents ${\mathrm{\Psi }}_1$ and ${\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$ if and only if there exists non-zero idempotents ${\mathrm{\Psi }}_0$ on $\mathcal{E}$ such that ${\mathrm{\Psi }}_0\le I-P$, ${\mathrm{\Psi }}_1=\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_0$, and ${\mathrm{\Psi }}_2=(I-\mathrm{\Psi }){\mathrm{\Psi }}_0$;

(iii)

For $\mathrm{\Psi }\ne 0$, $\mathrm{\Psi }\ne I$, and a non-zero scalar $c_1\ne 1$, the following are equivalent:

(a) 

There exist distinct non-zero idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$ such that $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+(1-c_1){\mathrm{\Psi }}_2$.

(b) 

There exist operators $A_1,A_2$ on $\mathcal{E}$ satisfying

i. 

$\mathcal{R}\left(A_1\right)\subseteq \mathcal{N}\left(P\right)$, $\mathcal{N}\left(A_1\right)\supseteq \mathcal{N}\left(P\right)$;

ii. 

$\mathcal{R}\left(A_2\right)\subseteq \mathcal{R}\left(P\right)$, $\mathcal{N}\left(A_2\right)\supseteq \mathcal{R}\left(P\right)$;

iii. 

$A_1{A}_2=A_2{A}_1=0$;

iv. 

$(I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2\ne 0$

with explicit expressions

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1& =\mathrm{\Psi }-(I-\mathrm{\Psi })A_1\mathrm{\Psi }-A_2,\hfill \\ \hfill {\mathrm{\Psi }}_2& =\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}\left((I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2\right).\hfill \end{array}$$

Proof. 

(i) Necessity: If there exists a decomposition $\mathrm{\Psi }={\mathrm{\Psi }}_1+{\mathrm{\Psi }}_2$, then based on (66), we have ${\mathrm{\Psi }}_1\le \mathrm{\Psi }$. Now, define ${\mathrm{\Psi }}_0={\mathrm{\Psi }}_{1}P$. Then, ${\mathrm{\Psi }}_0^2={\mathrm{\Psi }}_0$ and ${\mathrm{\Psi }}_0\le P$. To see that ${\mathrm{\Psi }}_0\ne 0$, if ${\mathrm{\Psi }}_0=0$ holds, then ${\mathrm{\Psi }}_1={\mathrm{\Psi }}_1\mathrm{\Psi }={\mathrm{\Psi }}_{1}P\mathrm{\Psi }={\mathrm{\Psi }}_0\mathrm{\Psi }=0$, contradicting the hypothesis ${\mathrm{\Psi }}_1\ne 0$. Moreover, ${\mathrm{\Psi }}_0\ne P$, as assuming equality would imply ${\mathrm{\Psi }}_2=\mathrm{\Psi }-P$ and, consequently, ${\mathrm{\Psi }}_2={\mathrm{\Psi }}_2^2={(\mathrm{\Psi }-P)}^2=0$, contradicting the hypothesis ${\mathrm{\Psi }}_2\ne 0$. Furthermore, based on calculation, we have

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_0\mathrm{\Psi }& ={\mathrm{\Psi }}_{1}P\mathrm{\Psi }={\mathrm{\Psi }}_1\mathrm{\Psi }={\mathrm{\Psi }}_1,\hfill \\ \hfill \mathrm{\Psi }-{\mathrm{\Psi }}_0\mathrm{\Psi }& =\mathrm{\Psi }-{\mathrm{\Psi }}_1={\mathrm{\Psi }}_2.\hfill \end{array}$$

(i) Sufficiency: As ${\mathrm{\Psi }}_0\le P$, we have $P{\mathrm{\Psi }}_0={\mathrm{\Psi }}_0$, which implies $\mathrm{\Psi }{\mathrm{\Psi }}_0=\mathrm{\Psi }P{\mathrm{\Psi }}_0=P{\mathrm{\Psi }}_0={\mathrm{\Psi }}_0$. Consequently,

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1^2& =({\mathrm{\Psi }}_0\mathrm{\Psi })({\mathrm{\Psi }}_0\mathrm{\Psi })={\mathrm{\Psi }}_0(\mathrm{\Psi }{\mathrm{\Psi }}_0)\mathrm{\Psi }={\mathrm{\Psi }}_0{\mathrm{\Psi }}_0\mathrm{\Psi }={\mathrm{\Psi }}_0\mathrm{\Psi }={\mathrm{\Psi }}_1,\hfill \\ \hfill {\mathrm{\Psi }}_2^2& ={(\mathrm{\Psi }-{\mathrm{\Psi }}_0\mathrm{\Psi })}^2=\mathrm{\Psi }-{\mathrm{\Psi }}_0\mathrm{\Psi }={\mathrm{\Psi }}_2.\hfill \end{array}$$

Moreover, ${\mathrm{\Psi }}_1\ne 0$, if equality holds, then ${\mathrm{\Psi }}_0\mathrm{\Psi }=0$ would imply ${\mathrm{\Psi }}_0={\mathrm{\Psi }}_{0}P={\mathrm{\Psi }}_0\mathrm{\Psi }P=0$, contradicting ${\mathrm{\Psi }}_0\ne 0$. Similarly, to see ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$, if equality holds, then ${\mathrm{\Psi }}_0\mathrm{\Psi }=\mathrm{\Psi }$ would imply ${\mathrm{\Psi }}_0={\mathrm{\Psi }}_{0}P={\mathrm{\Psi }}_0\mathrm{\Psi }P=\mathrm{\Psi }P=P$, contradicting ${\mathrm{\Psi }}_0\ne P$.

(ii) Necessity: Assuming there exists a decomposition $\mathrm{\Psi }={\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2$, based on (67), we have ${\mathrm{\Psi }}_1\ge \mathrm{\Psi }$, which implies ${\mathrm{\Psi }}_1\mathrm{\Psi }=\mathrm{\Psi }$. Consequently,

$${\mathrm{\Psi }}_{1}P={\mathrm{\Psi }}_1\mathrm{\Psi }P=\mathrm{\Psi }P=P.$$

Define

$${\mathrm{\Psi }}_0=(I-P){\mathrm{\Psi }}_1(I-P)={\mathrm{\Psi }}_1-P{\mathrm{\Psi }}_1.$$

Clearly, ${\mathrm{\Psi }}_0\le I-P$. Direct computation yields the following:

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_0^2& =({\mathrm{\Psi }}_1-P{\mathrm{\Psi }}_1)({\mathrm{\Psi }}_1-P{\mathrm{\Psi }}_1)\hfill \\ \hfill & ={\mathrm{\Psi }}_1^2-{\mathrm{\Psi }}_{1}P{\mathrm{\Psi }}_1-P{\mathrm{\Psi }}_1^2+P{\mathrm{\Psi }}_{1}P{\mathrm{\Psi }}_1\hfill \\ \hfill & ={\mathrm{\Psi }}_1-P{\mathrm{\Psi }}_1={\mathrm{\Psi }}_0.\hfill \end{array}$$

Moreover, ${\mathrm{\Psi }}_0\ne 0$, as, if equality held, then $P{\mathrm{\Psi }}_1={\mathrm{\Psi }}_1$ would imply

$${\mathrm{\Psi }}_2={\mathrm{\Psi }}_1-\mathrm{\Psi }=P{\mathrm{\Psi }}_1-\mathrm{\Psi }=\mathrm{\Psi }P{\mathrm{\Psi }}_1-\mathrm{\Psi }=\mathrm{\Psi }{\mathrm{\Psi }}_1-\mathrm{\Psi }=0,$$

contradicting ${\mathrm{\Psi }}_2\ne 0$. Finally, simple calculations give

$$\begin{array}{cc}\hfill \mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_0& =\mathrm{\Psi }+(I-\mathrm{\Psi })({\mathrm{\Psi }}_1-P{\mathrm{\Psi }}_1)={\mathrm{\Psi }}_1,\hfill \\ \hfill (I-\mathrm{\Psi }){\mathrm{\Psi }}_0& ={\mathrm{\Psi }}_1-\mathrm{\Psi }={\mathrm{\Psi }}_2.\hfill \end{array}$$

(ii) Sufficiency: As ${\mathrm{\Psi }}_0\le I-P$, we have ${\mathrm{\Psi }}_0(I-P)={\mathrm{\Psi }}_0$, which leads to ${\mathrm{\Psi }}_{0}P=0$ and

$${\mathrm{\Psi }}_0\mathrm{\Psi }={\mathrm{\Psi }}_{0}P\mathrm{\Psi }=0.$$

Thus,

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1^2& ={(\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_0)}^2=\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_0={\mathrm{\Psi }}_1,\hfill \\ \hfill {\mathrm{\Psi }}_2^2& =(I-\mathrm{\Psi }){\mathrm{\Psi }}_0(I-\mathrm{\Psi }){\mathrm{\Psi }}_0=(I-\mathrm{\Psi }){\mathrm{\Psi }}_0={\mathrm{\Psi }}_2.\hfill \end{array}$$

Note that ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$ as, if equality holds, then $\mathrm{\Psi }{\mathrm{\Psi }}_0={\mathrm{\Psi }}_0$ would imply ${\mathrm{\Psi }}_0={\mathrm{\Psi }}_0^2={\mathrm{\Psi }}_0\mathrm{\Psi }{\mathrm{\Psi }}_0=0$, contradicting ${\mathrm{\Psi }}_0\ne 0$.

(iii) Necessity: Assuming there exists a decomposition $\mathrm{\Psi }=c_1{\mathrm{\Psi }}_1+(1-c_1){\mathrm{\Psi }}_2$, Lemma 8 implies ${({\mathrm{\Psi }}_1-{\mathrm{\Psi }}_2)}^2=0$. Substituting ${\mathrm{\Psi }}_2={\displaystyle \frac{\mathrm{\Psi }-c_1{\mathrm{\Psi }}_1}{1-c_1}}$ yields

$$\begin{array}{c}\hfill 0={({\mathrm{\Psi }}_1-\mathrm{\Psi })}^2={\mathrm{\Psi }}_1+\mathrm{\Psi }-{\mathrm{\Psi }}_1\mathrm{\Psi }-\mathrm{\Psi }{\mathrm{\Psi }}_1.\end{array}$$

(71)

From (71), we directly deduce

$$\begin{array}{c}\hfill (I-\mathrm{\Psi }){\mathrm{\Psi }}_1=-(I-{\mathrm{\Psi }}_1)\mathrm{\Psi }.\end{array}$$

(72)

Left-multiplying (71) by $\mathrm{\Psi }$ and ${\mathrm{\Psi }}_1$, respectively, gives

$$\begin{array}{c}\hfill {\mathrm{\Psi }}_1\mathrm{\Psi }{\mathrm{\Psi }}_1={\mathrm{\Psi }}_1\mathrm{and}\mathrm{\Psi }{\mathrm{\Psi }}_1\mathrm{\Psi }=\mathrm{\Psi }.\end{array}$$

(73)

Define

$$A_1=-(I-P){\mathrm{\Psi }}_{1}P\mathrm{and}{A}_2=-\mathrm{\Psi }{\mathrm{\Psi }}_1(I-\mathrm{\Psi }).$$

These satisfy

$$\begin{array}{c}\hfill \mathcal{R}\left(A_1\right)\subseteq \mathcal{N}\left(P\right),\mathcal{N}\left(A_1\right)\supseteq \mathcal{N}\left(P\right),\\ \hfill \mathcal{R}\left(A_2\right)\subseteq \mathcal{R}\left(P\right),\mathcal{N}\left(A_2\right)\supseteq \mathcal{R}\left(P\right).\end{array}$$

Combining (72) and (73) yields the following

$$\begin{array}{cc}\hfill A_1{A}_2& =(I-P){\mathrm{\Psi }}_{1}P\mathrm{\Psi }{\mathrm{\Psi }}_1(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P){\mathrm{\Psi }}_1(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P)\mathrm{\Psi }(I-{\mathrm{\Psi }}_1)\hfill \\ \hfill & =0,\hfill \\ \hfill A_2{A}_1& =\mathrm{\Psi }{\mathrm{\Psi }}_1(I-\mathrm{\Psi })(I-P){\mathrm{\Psi }}_{1}P\hfill \\ \hfill & =\mathrm{\Psi }{\mathrm{\Psi }}_1(I-\mathrm{\Psi }){\mathrm{\Psi }}_{1}P\hfill \\ \hfill & =0.\hfill \end{array}$$

Note that

$$\begin{array}{cc}\hfill (I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2& =-(I-\mathrm{\Psi }){\mathrm{\Psi }}_1\mathrm{\Psi }-\mathrm{\Psi }{\mathrm{\Psi }}_1(I-\mathrm{\Psi })\hfill \\ \hfill & =-{\mathrm{\Psi }}_1\mathrm{\Psi }+\mathrm{\Psi }-\mathrm{\Psi }{\mathrm{\Psi }}_1+\mathrm{\Psi }\hfill \\ \hfill & =(I-{\mathrm{\Psi }}_1)\mathrm{\Psi }-\mathrm{\Psi }{\mathrm{\Psi }}_1+\mathrm{\Psi }\hfill \\ \hfill & =-(I-\mathrm{\Psi }){\mathrm{\Psi }}_1-\mathrm{\Psi }{\mathrm{\Psi }}_1+\mathrm{\Psi }\hfill \\ \hfill & =\mathrm{\Psi }-{\mathrm{\Psi }}_1\hfill \\ \hfill & \ne 0.\hfill \end{array}$$

From this expression, it follows that

$$\mathrm{\Psi }-(I-\mathrm{\Psi })A_1\mathrm{\Psi }-A_2={\mathrm{\Psi }}_1$$

and, consequently,

$$\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}\left((I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2\right)={\displaystyle \frac{\mathrm{\Psi }-c_1{\mathrm{\Psi }}_1}{1-c_1}}={\mathrm{\Psi }}_2.$$

(iii) Sufficiency: Let ${\mathrm{\Psi }}_0=(I-\mathrm{\Psi })A_1\mathrm{\Psi }+A_2$. Then, based on $\mathcal{R}\left(A_2\right)\subseteq \mathcal{R}\left(P\right)$, $\mathcal{N}\left(A_2\right)\supseteq \mathcal{R}\left(P\right)$, and $A_1{A}_2=A_2{A}_1=0$, we have

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_0^2& =(I-\mathrm{\Psi })A_1\mathrm{\Psi }{A}_2+A_2(I-\mathrm{\Psi })A_1\mathrm{\Psi }\hfill \\ \hfill & =(I-\mathrm{\Psi })A_1{A}_2+A_2{A}_1\mathrm{\Psi }\hfill \\ \hfill & =0,\hfill \end{array}$$

and

$$\begin{array}{c}\hfill \mathrm{\Psi }{\mathrm{\Psi }}_0+{\mathrm{\Psi }}_0\mathrm{\Psi }=\mathrm{\Psi }{A}_2+(I-\mathrm{\Psi })A_1\mathrm{\Psi }={\mathrm{\Psi }}_0.\end{array}$$

Thus,

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1^2& ={(\mathrm{\Psi }-{\mathrm{\Psi }}_0)}^2\hfill \\ \hfill & =\mathrm{\Psi }-\mathrm{\Psi }{\mathrm{\Psi }}_0-{\mathrm{\Psi }}_0\mathrm{\Psi }+{\mathrm{\Psi }}_0^2\hfill \\ \hfill & =\mathrm{\Psi }-{\mathrm{\Psi }}_0\hfill \\ \hfill & ={\mathrm{\Psi }}_1,\hfill \\ \hfill {\mathrm{\Psi }}_2^2& ={(\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}{\mathrm{\Psi }}_0)}^2\hfill \\ \hfill & =\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}(\mathrm{\Psi }{\mathrm{\Psi }}_0+{\mathrm{\Psi }}_0\mathrm{\Psi })\hfill \\ \hfill & =\mathrm{\Psi }+{\displaystyle \frac{c_1}{1-c_1}}{\mathrm{\Psi }}_0\hfill \\ \hfill & ={\mathrm{\Psi }}_2.\hfill \end{array}$$

Note that ${\mathrm{\Psi }}_1\ne \mathrm{\Psi }$ since ${\mathrm{\Psi }}_0\ne 0$. Furthermore, to see ${\mathrm{\Psi }}_1\ne 0$, if equality holds, then ${\mathrm{\Psi }}_0=\mathrm{\Psi }$ would imply $\mathrm{\Psi }={\mathrm{\Psi }}^2={\mathrm{\Psi }}_0^2=0$, contradicting $\mathrm{\Psi }\ne 0$. □

Remark 3.

Let H be a Hilbert space. For any idempotent operator $\mathrm{\Pi }\in \mathcal{L}\left(H\right)$ satisfying $\mathrm{\Pi }\ne 0$ and $\mathrm{\Pi }\ne I$, there always exists a decomposition of the forms described in Theorem 6(ii) and Theorem 6(iii). Consider the specific decomposition where ${\mathrm{\Pi }}_1=I$ and ${\mathrm{\Pi }}_2=I-\mathrm{\Pi }$, giving $\mathrm{\Pi }={\mathrm{\Pi }}_1-{\mathrm{\Pi }}_2$.

Let P be the orthogonal projection from H onto $\mathcal{R}\left(\mathrm{\Pi }\right)$. As $\mathrm{\Pi }\ne 0$ and $\mathrm{\Pi }\ne I$, we have $P\ne 0$ and $I-P\ne 0$. Thus, there exist $a_1\in \mathcal{R}(I-P)$ and $a_2\in \mathcal{R}\left(P\right)$ with $\parallel a_1\parallel =1$ and $\parallel a_2\parallel =1$. Let $\tilde{H}$ be the one-dimensional closed subspace generated by $a_1$, and define $W\in \mathcal{L}(\tilde{H},H)$ by $W{a}_1=a_2$. Denote the composition $W{P}_{\tilde{H}}$ by ${\mathrm{\Pi }}_0$. Then, ${\mathrm{\Pi }}_0\in \mathcal{L}\left(H\right)$ satisfies ${\mathrm{\Pi }}_0\ne 0$,${\mathrm{\Pi }}_0^2=0$, $\mathcal{R}\left({\mathrm{\Pi }}_0\right)\subseteq \mathcal{R}\left(P\right)$, and $\mathcal{N}\left({\mathrm{\Pi }}_0\right)\supseteq \mathcal{R}\left(P\right)$. These properties, combined with Lemma 6, yield

$${(\mathrm{\Pi }-{\mathrm{\Pi }}_0)}^2=\mathrm{\Pi }-\mathrm{\Pi }{\mathrm{\Pi }}_0-{\mathrm{\Pi }}_0\mathrm{\Pi }=\mathrm{\Pi }-{\mathrm{\Pi }}_0.$$

Now, define

$${\mathrm{\Pi }}_1=\mathrm{\Pi }-{\mathrm{\Pi }}_0,{\mathrm{\Pi }}_2=\mathrm{\Pi }+{\displaystyle \frac{c_1}{1-c_1}}{\mathrm{\Pi }}_0,$$

where $c_1$ is any complex number not equal to 0 or 1. This gives the decomposition

$$\mathrm{\Pi }=c_1{\mathrm{\Pi }}_1+(1-c_1){\mathrm{\Pi }}_2.$$

However, as demonstrated by the example in the next section, not every idempotent Π admits a decomposition as the sum of two idempotents.

Having completed our investigation into the linear decomposition of idempotent operators on Hilbert $C^{*}$-modules, we now examine their product decompositions. In [24], Theorem 3.2, we established the factorization of an idempotent operator into a product of two idempotents on Hilbert spaces and derived explicit representations for such factorizations. We now extend these results to the framework of Hilbert $C^{*}$-modules.

Theorem 7.

Let Ψ be an linear idempotent operator on $\mathcal{E}$ and P be the projection onto $\mathcal{R}(\mathrm{\Psi })$. Then, $\mathrm{\Psi }={\mathrm{\Psi }}_1{\mathrm{\Psi }}_2$ for idempotents ${\mathrm{\Psi }}_1,{\mathrm{\Psi }}_2$ on $\mathcal{E}$ if and only if there exist operators $X_1,Y_1$ on $\mathcal{E}$ and idempotents $X_2,Y_2$ on $\mathcal{E}$ satisfying the following:

(1) 

Range and null space conditions

$$\begin{array}{cc}\hfill & \mathcal{R}\left(X_1\right)\subseteq \mathcal{R}\left(P\right),\mathcal{N}\left(X_1\right)\supseteq \mathcal{R}\left(P\right),\hfill \\ \hfill & \mathcal{R}\left(Y_1\right)\subseteq \mathcal{N}\left(P\right),\mathcal{N}\left(Y_1\right)\supseteq \mathcal{N}\left(P\right),\hfill \\ \hfill & \mathcal{R}\left(X_2\right)\subseteq \mathcal{N}\left(P\right),\mathcal{N}\left(X_2\right)\supseteq \mathcal{R}\left(P\right),\hfill \\ \hfill & \mathcal{R}\left(Y_2\right)\subseteq \mathcal{N}\left(P\right),\mathcal{N}\left(Y_2\right)\supseteq \mathcal{R}\left(P\right).\hfill \end{array}$$

(2) 

Zero-product relation

$$\begin{array}{cc}\hfill & X_1{X}_2=0,Y_2{Y}_1=0,(X_1-\mathrm{\Psi })Y_1=0,\hfill \\ \hfill & (X_1-\mathrm{\Psi })Y_2=0,X_2{Y}_1=0,X_2{Y}_2=0.\hfill \end{array}$$

(3) 

Factorization expressions

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1& =P+X_1+X_2,\hfill \\ \hfill {\mathrm{\Psi }}_2& =\mathrm{\Psi }+(I-\mathrm{\Psi })Y_1\mathrm{\Psi }+(I-\mathrm{\Psi })Y_2.\hfill \end{array}$$

Proof. 

Necessity: Assuming there exists a product decomposition $\mathrm{\Psi }={\mathrm{\Psi }}_1{\mathrm{\Psi }}_2$, we have $\mathcal{R}\left({\mathrm{\Psi }}_1\right)\supseteq \mathcal{R}(\mathrm{\Psi })$. Based on Lemma 6, this implies

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1\mathrm{\Psi }& =\mathrm{\Psi },\hfill \end{array}$$

(74)

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_{1}P& =P.\hfill \end{array}$$

(75)

Define the operators

$$\begin{array}{cc}\hfill X_1& =P{\mathrm{\Psi }}_1(I-P),\hfill \\ \hfill X_2& =(I-P){\mathrm{\Psi }}_1(I-P),\hfill \\ \hfill Y_1& =(I-P){\mathrm{\Psi }}_{2}P,\hfill \\ \hfill Y_2& =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi }).\hfill \end{array}$$

Using (74) and Lemma 6, we derive the idempotency properties

$$\begin{array}{cc}\hfill X_2^2& =(I-P){\mathrm{\Psi }}_1(I-P){\mathrm{\Psi }}_1(I-P)\hfill \\ \hfill & =(I-P)({\mathrm{\Psi }}_1-P-P{\mathrm{\Psi }}_1+P)\hfill \\ \hfill & =(I-P){\mathrm{\Psi }}_1\hfill \\ \hfill & =X_2.\hfill \end{array}$$

$$\begin{array}{cc}\hfill Y_2^2& =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi }){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P)({\mathrm{\Psi }}_2-{\mathrm{\Psi }}_2\mathrm{\Psi })(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =Y_2.\hfill \end{array}$$

The zero-product relations are established as follows:

$$\begin{array}{cc}\hfill X_1{X}_2& =P{\mathrm{\Psi }}_1(I-P)(I-P){\mathrm{\Psi }}_1(I-P)\hfill \\ \hfill & =P{({\mathrm{\Psi }}_1-P)}^2\hfill \\ \hfill & =0;\hfill \\ \hfill (X_1-{\mathrm{\Psi }}_1)Y_1& =(P{\mathrm{\Psi }}_1-P-{\mathrm{\Psi }}_1)(I-P){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =P{\mathrm{\Psi }}_1-{\mathrm{\Psi }}_1(I-P){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =(P{\mathrm{\Psi }}_1-P-{\mathrm{\Psi }}_1+P){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =P{\mathrm{\Psi }}_{1}P-{\mathrm{\Psi }}_{1}P\hfill \\ \hfill & =0;\hfill \\ \hfill Y_2{Y}_1& =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })(I-P){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi }){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =(I-P)({\mathrm{\Psi }}_2-{\mathrm{\Psi }}_2\mathrm{\Psi })P\hfill \\ \hfill & =(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })P\hfill \\ \hfill & =0;\hfill \\ \hfill (X_1-{\mathrm{\Psi }}_1)Y_2& =(P{\mathrm{\Psi }}_1-P-{\mathrm{\Psi }}_1)(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =(P{\mathrm{\Psi }}_1-{\mathrm{\Psi }}_1){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =({\mathrm{\Psi }}_1-{\mathrm{\Psi }}_1)(I-\mathrm{\Psi })\hfill \\ \hfill & =0;\hfill \\ \hfill X_2{Y}_1& =(I-P){\mathrm{\Psi }}_1(I-P)(I-\mathrm{\Psi })(I-P){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =(I-P)({\mathrm{\Psi }}_1-P){\mathrm{\Psi }}_{2}P\hfill \\ \hfill & =(I-P)\mathrm{\Psi }P\hfill \\ \hfill & =0;\hfill \\ \hfill X_2{Y}_2& =(I-P){\mathrm{\Psi }}_1(I-P)(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P)({\mathrm{\Psi }}_1-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\hfill \\ \hfill & =(I-P)\mathrm{\Psi }(I-\mathrm{\Psi })\hfill \\ \hfill & =0.\hfill \end{array}$$

The factorization expressions are established through the following derivations:

$$\begin{array}{cc}\hfill P+X_1+X_2& =P+P{\mathrm{\Psi }}_1(I-P)+(I-P){\mathrm{\Psi }}_1(I-P)\hfill \\ \hfill & =P+{\mathrm{\Psi }}_1(I-P)\hfill \\ \hfill & ={\mathrm{\Psi }}_1,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & {\mathrm{\Psi }}_1+(I-{\mathrm{\Psi }}_1)Y_1{\mathrm{\Psi }}_1+(I-{\mathrm{\Psi }}_1)Y_2\hfill \\ \hfill & =\mathrm{\Psi }+(I-\mathrm{\Psi })(Y_1{\mathrm{\Psi }}_1+Y_2)\hfill \\ \hfill & =\mathrm{\Psi }+(I-\mathrm{\Psi })\left[(I-P){\mathrm{\Psi }}_2\mathrm{\Psi }+(I-P){\mathrm{\Psi }}_2(I-\mathrm{\Psi })\right]\hfill \\ \hfill & =\mathrm{\Psi }+(I-\mathrm{\Psi })(I-P){\mathrm{\Psi }}_2\hfill \\ \hfill & =\mathrm{\Psi }+(I-\mathrm{\Psi }){\mathrm{\Psi }}_2\hfill \\ \hfill & ={\mathrm{\Psi }}_2.\hfill \end{array}$$

Sufficiency: The idempotency and product relations are verified through direct computation

$$\begin{array}{cc}\hfill {\mathrm{\Psi }}_1^2& ={(P+X_1+X_2)}^2\hfill \\ \hfill & =P+X_1+X_1{X}_2+X_2^2\hfill \\ \hfill & =P+X_1+X_2\hfill \\ \hfill & ={\mathrm{\Psi }}_1;\hfill \\ \hfill {\mathrm{\Psi }}_2^2& ={[\mathrm{\Psi }+(I-\mathrm{\Psi })Y_1\mathrm{\Psi }+(I-\mathrm{\Psi })Y_2]}^2\hfill \\ \hfill & =\mathrm{\Psi }+(I-\mathrm{\Psi })Y_1\mathrm{\Psi }+(I-\mathrm{\Psi })Y_2(I-\mathrm{\Psi })+{\mathrm{\Psi }}_2(I-\mathrm{\Psi })Y_2\hfill \\ \hfill & ={\mathrm{\Psi }}_2;\hfill \\ \hfill {\mathrm{\Psi }}_1{\mathrm{\Psi }}_2& =(P+X_1+X_2)[\mathrm{\Psi }+(I-\mathrm{\Psi })Y_1\mathrm{\Psi }+(I-\mathrm{\Psi })Y_2]\hfill \\ \hfill & =\mathrm{\Psi }-\mathrm{\Psi }{Y}_1-\mathrm{\Psi }{Y}_2+X_1{Y}_1\mathrm{\Psi }+X_1{Y}_2+X_2{Y}_1\mathrm{\Psi }+X_2{Y}_2\hfill \\ \hfill & =\mathrm{\Psi }+(X_1-\mathrm{\Psi })Y_1\mathrm{\Psi }+(X_1-\mathrm{\Psi })Y_2\hfill \\ \hfill & =\mathrm{\Psi }.\hfill \end{array}$$

□

6. Two Examples

To demonstrate the utility of Theorem 6 in decomposing idempotents, we provide the following example, which relies on a key lemma:

Lemma 9

([36], Theorem 8). For projections $P,Q$ on a finite-dimensional complex vector space M, the projection onto $\mathcal{R}\left(P\right)\cap \mathcal{R}\left(Q\right)$ is $2P{(P+Q)}^{†}Q$, where ${(P+Q)}^{†}$ denotes the Moore–Penrose inverse.

Example 2.

Consider the following idempotent matrix on ${\mathbb{C}}^4$:

$$\tilde{\mathrm{\Pi }}=\left(\begin{array}{cccc}{c}^2-2c^2{s}^2& 0& cs+2s{c}^3& 0\\ 0& 0& 0& 0\\ cs-2c{s}^3& 0& s^2+2c^2{s}^2& 0\\ 0& 0& 0& 0\end{array}\right),$$

where $c=cos(\pi /3)$ and $s=sin(\pi /3)$. Let P and Q be projections onto $\mathcal{R}\left(\tilde{\mathrm{\Pi }}\right)$ and $\mathcal{N}\left(\tilde{\mathrm{\Pi }}\right)$, respectively, and let $H_i,P_i$$(1\le i\le 6$) be defined as in (2)–(6), and $Q_0$ and the unitary operator $U_0\in \mathcal{L}(H_6,H_5)$ as defined in Lemma 1. Furthermore, let $V=-Q_0^{{\displaystyle \frac{1}{2}}}{(I_{H_5}-Q_0)}^{-{\displaystyle \frac{1}{2}}}{U}_0$. Then, $P_i$$(1\le i\le 6)$ can be calculated using Lemma 9 combined with (59). In particular, we compute the following:

$$\begin{array}{cc}\hfill P_1& =P_2=P_4=0,\hfill \\ \hfill P_3& =\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right),\hfill \\ \hfill P_5& =\left(\begin{array}{cccc}{c}^2& 0& cs& 0\\ 0& 0& 0& 0\\ cs& 0& s^2& 0\\ 0& 0& 0& 0\end{array}\right),\hfill \\ \hfill P_6& =\left(\begin{array}{cccc}{s}^2& 0& -cs& 0\\ 0& 0& 0& 0\\ -cs& 0& c^2& 0\\ 0& 0& 0& 0\end{array}\right),\hfill \\ \hfill \tilde{\mathrm{\Pi }}-P& =\left(\begin{array}{cccc}-2c^2{s}^2& 0& 2s{c}^3& 0\\ 0& 0& 0& 0\\ -2c{s}^3& 0& 2c^2{s}^2& 0\\ 0& 0& 0& 0\end{array}\right).\hfill \end{array}$$

Note that $P=P_5$ and $I-P=P_3+P_6$.

Non-existence of additive decomposition: Assume there exists a decomposition $\tilde{\mathrm{\Pi }}={\mathrm{\Pi }}_1+{\mathrm{\Pi }}_2$ as in Theorem 6(i). Then, there exist ${\mathrm{\Pi }}_0\le P$ with ${\mathrm{\Pi }}_0\ne P$. As $\mathcal{R}\left(P_5\right)$ is one-dimensional and spanned by $\mathbf{e}={(c^2,0,cs,0)}^T$, we have ${\mathrm{\Pi }}_0\mathbf{e}=k\mathbf{e}$. Idempotence implies $k=1$ and, so, ${\mathrm{\Pi }}_0=P$—a contradiction.

MATLAB R2022a(9.12) verification confirmed that no solution exists for ${\mathrm{\Pi }}_1^2={\mathrm{\Pi }}_1$, ${\mathrm{\Pi }}_2^2={\mathrm{\Pi }}_2$, ${\mathrm{\Pi }}_1+{\mathrm{\Pi }}_2=\tilde{\mathrm{\Pi }}$.

Existence of alternative decompositions: As $P_3\le I-P$, Theorem 6(ii) gives the following:

$$\tilde{\mathrm{\Pi }}=(\tilde{\mathrm{\Pi }}+P_3)-P_3.$$

Considering Theorem 6(iii), let ${\mathrm{\Pi }}_0={\displaystyle \frac{1}{2}}(\tilde{\mathrm{\Pi }}-P)$ and $c_1\in \mathbb{C}\setminus \{0,1\}$

$$\begin{array}{cc}\hfill {\mathrm{\Pi }}_1& =\tilde{\mathrm{\Pi }}-{\mathrm{\Pi }}_0=\left(\begin{array}{cccc}{c}^4& 0& cs+s{c}^3& 0\\ 0& 0& 0& 0\\ c^{3}s& 0& s^2+c^2{s}^2& 0\\ 0& 0& 0& 0\end{array}\right),\hfill \\ \hfill {\mathrm{\Pi }}_2& =\tilde{\mathrm{\Pi }}+{\displaystyle \frac{c_1}{1-c_1}}{\mathrm{\Pi }}_0=\left(\begin{array}{cccc}{c}^2-{\displaystyle \frac{2-c_1}{1-c_1}}{c}^2{s}^2& 0& cs+{\displaystyle \frac{2-c_1}{1-c_1}}s{c}^3& 0\\ 0& 0& 0& 0\\ cs-{\displaystyle \frac{2-c_1}{1-c_1}}c{s}^3& 0& s^2+{\displaystyle \frac{2-c_1}{1-c_1}}{c}^2{s}^2& 0\\ 0& 0& 0& 0\end{array}\right).\hfill \end{array}$$

This yields $\tilde{\mathrm{\Pi }}=c_1{\mathrm{\Pi }}_1+(1-c_1){\mathrm{\Pi }}_2$.

Throughout Section 3 and Section 4, we focused exclusively on adjointable idempotent operators on Hilbert $C^{*}$-modules. This restriction is necessary due to a fundamental difference between Hilbert $C^{*}$-modules and Hilbert spaces: unlike operators on Hilbert spaces, idempotent operators on Hilbert $C^{*}$-modules may not admit an adjoint. A simple counterexample illustrating this phenomenon is presented below.

Example 3.

Let $B=C\left(\right[0,1\left]\right)$ be the algebra of complex-valued continuous functions on $[0,1]$. Let $I=\{f\in B\mid f(0)=0\}$ be a closed ideal in B. Define the Hilbert B-module $\mathcal{E}=I\mathrm{\oplus }B$ with the following inner product:

$${⟨{(x_1,y_1)}^T,{(x_2,y_2)}^T⟩}_{\mathcal{E}}=x_1^{*}{x}_2+y_1^{*}{y}_2.$$

Define the idempotent operator $A:\mathcal{E}\to \mathcal{E}$ as follows:

$$A{(x,y)}^T={(0,x+y)}^T.$$

Verification of idempotency

$$A^2{(x,y)}^T=A{(0,x+y)}^T={(0,x+y)}^T=A{(x,y)}^T.$$

Verification thatAhas no adjoint

Assume there exists an adjoint $A^{*}:\mathcal{E}\to \mathcal{E}$ satisfying

$${⟨A{(x_1,y_1)}^T,{(x_2,y_2)}^T⟩}_{\mathcal{E}}={⟨{(x_1,y_1)}^T,A^{*}{(x_2,y_2)}^T⟩}_{\mathcal{E}}$$

for all ${(x_1,y_1)}^T,{(x_2,y_2)}^T\in \mathcal{E}$. The left side is

$${⟨A{(x_1,y_1)}^T,{(x_2,y_2)}^T⟩}_{\mathcal{E}}={⟨{(0,x_1+y_1)}^T,{(x_2,y_2)}^T⟩}_{\mathcal{E}}={(x_1+y_1)}^{*}{y}_2.$$

For the right side, let $A^{*}{(x_2,y_2)}^T={(u,v)}^T$. Then,

$${⟨{(x_1,y_1)}^T,{(u,v)}^T⟩}_{\mathcal{E}}=x_1^{*}u+y_1^{*}v.$$

Equating both sides yields

$$x_1^{*}u+y_1^{*}v=(x_1^{*}+y_1^{*})y_2\forall {(x_1,y_1)}^T\in \mathcal{E}.$$

(76)

Fixing $y_1=0$ gives

$$x_1^{*}u=x_1^{*}{y}_2\forall x_1\in I.$$

Taking $x_1\left(t\right)=t\in I$ gives

$$tu\left(t\right)=t{y}_2\left(t\right)\forall t\in [0,1].$$

For $t\ne 0$, we have $u\left(t\right)=y_2\left(t\right)$. By continuity, $u=y_2$. Substituting into (76) yields the following:

$$x_1^{*}{y}_2+y_1^{*}v=x_1^{*}{y}_2+y_1^{*}{y}_2\Rightarrow y_1^{*}v=y_1^{*}{y}_2\forall y_1\in B.$$

Thus, $v=y_2$, implying that $A^{*}{(x,y)}^T={(y,y)}^T$. Note that $A^{*}{(x,y)}^T={(y,y)}^T$ requires $y\in I$, as $\mathcal{E}=I\mathrm{\oplus }B$ requires the first component in I. However, $y\in B$ need not belong to I; for example, the constant function $y\left(t\right)=1$ satisfies $1\notin I$. Hence, $A^{*}$ does not map into $\mathcal{E}$, giving a contradiction.

7. Conclusions

This study developed a complete decomposition theory for idempotent operators on Hilbert $C^{*}$-modules. The key contributions include:

• Characterization of closedness and orthogonal complementability for ranges of idempotent sums;
• Explicit formulas for Moore–Penrose inverses and spectral norms of $\mathrm{\Psi }+{\mathrm{\Psi }}^{*}$;
• Complete linear decomposition theorems for adjointable and non-adjointable idempotents;
• Multiplicative decomposition into products of idempotents;
• Counterexamples validating the necessity of adjointability assumptions;
• Numerical examples demonstrating practical applications.

The results resolve long-standing decomposition problems and create new connections between projection theory, operator algebras, and spectral analysis in the context of Hilbert $C^{*}$-modules.