The Moore–Penrose Inverse and Product Decomposition of Idempotent Operators on Hilbert C*-Modules

Abstract

We study the Moore–Penrose inverse of idempotent operators on Hilbert $C^{*}$-modules. First, we extend the computation of the Moore–Penrose inverse of an idempotent operator and its difference from the range projection to this setting. This leads to an explicit formula for the Moore–Penrose inverse of the sum of an idempotent and its adjoint. Furthermore, we establish a decomposition of an idempotent operator into a product of two commuting idempotents and clarify the relationship between their Moore–Penrose inverses and that of the original operator. We also analyze spectral properties and operator norms, obtaining sharp norm bounds.

1. Introduction

The concept of a generalized inverse, extending the classical notion of an inverse to singular or rectangular matrices, has evolved over more than a century. While related ideas for integral operators can be traced back to Fredholm (1903) and Hilbert (1904), the definition for arbitrary matrices was first introduced by E.H. Moore in 1920 [1]. This work remained relatively obscure until the 1950s, when it was independently rediscovered by Arne Bjerhammar and later rigorously defined by Roger Penrose (1955) [2]. Penrose’s seminal contribution was to define the unique matrix $X=A^{+}$ satisfying the four famous equations $AXA=A$, $XAX=X$, ${\left(AX\right)}^{*}=AX$, and ${\left(XA\right)}^{*}=XA$, thereby firmly establishing the Moore–Penrose (MP) inverse. The subsequent decades saw an explosion in the theory and applications of the MP inverse across diverse fields, including statistics, optimization, and system theory.

The investigation of the MP inverse for specific matrix classes is a natural and fruitful line of inquiry. Idempotent matrices and particularly orthogonal projectors form one of the most fundamental classes.

Research has extensively explored the precise conditions and properties of the MP inverse within the realm of idempotents. A significant milestone was the systematic characterization of matrices whose MP inverse is also idempotent. Ref. [3] conducted a thorough investigation into this class, deriving its essential properties and establishing new characteristics, connecting it to concepts like EP matrices. Beyond single operators, the behavior of the MP inverse for products, commutators, and anticommutators of idempotent operators has been a rich topic.

The research landscape has expanded beyond the classical MP inverse to study how idempotent structures interact with broader generalized inverses and operator relations. The introduction of the core inverse by Baksalary and Trenkler in 2010 offered an alternative for matrices of index one, with properties closely linked to idempotent projectors like $P_F=F{F}^{†}$ [4]. This line of thinking was further generalized with the introduction of the weak core inverse in [5], extending the notion to a broader class of matrices. Zhu (2019) extended the notions of DMP inverses and m-EP elements from matrix theory to unital *-rings, characterized their existence criteria and properties, examined their interrelations, and addressed the research gap in the systematic investigation of relevant generalized inverses and special ring elements [6]. Recent work continues to examine the interactions between idempotency and generalized inverses, such as similarities between the numerical range of an operator and that of its various generalized inverses under specific conditions [7].

Parallel to these theoretical developments, significant advances have been made in computational methods for the Moore–Penrose inverse. Ref. [8] proposed massively parallel algorithms for efficient computation on modern hardware architectures, while [9] explored FPGA implementations for real-time applications. Machine learning approaches have also emerged, with [10] developing robust neural networks for time-varying matrix equations and [11] proposing efficient second-order neural network models. More broadly, Ref. [12] examined generalized matrix inversion through a machine learning lens. These computational advances are complemented by theoretical work on error bounds and perturbation analysis, including [13] on error-bound-based algorithms, [14] on tensor perturbation theory via the Einstein product, and [15] on perturbations of both Moore–Penrose and dual Moore–Penrose generalized inverses.

In Section 3, using the Halmos decomposition, we compute the Moore–Penrose inverse of an idempotent operator and that of the difference between an idempotent operator and its range projection. These results are then extended to the setting of Hilbert $C^{*}$-modules. Furthermore, by applying previous results from our earlier work, we compute the Moore–Penrose inverse of the sum of an idempotent operator and its adjoint in Hilbert $C^{*}$-modules. We also provide counterexamples to demonstrate that the existence of the Moore–Penrose inverses for both the difference between an idempotent operator and its range projection and the sum of an idempotent operator and its adjoint requires certain conditions, and an example matrix is given to illustrate the correctness of our theorem. Finally, we extend the work of O. M. Baksalary and G. Trenkler to idempotent operators in Hilbert $C^{*}$-modules, investigating the idempotency of the Moore–Penrose inverse of an idempotent operator. Similarly, we provide numerical examples to prove the correctness of the theorem. In Section 4, we study the decomposition of an idempotent operator as a product of two commuting idempotent operators. Based on this decomposition, we derive explicit expressions for the Moore–Penrose inverses of the resulting idempotent operators and establish their relationship with the original idempotent operator. Building on the study of the Moore–Penrose inverse of idempotent operators, in Section 5, we investigate the spectral properties and norms of operators related to the Moore–Penrose inverse of an idempotent operator. Finally, we add a numerical example. This result not only confirms the validity of our theoretical conclusions but also demonstrates good compatibility with classical matrix analysis.

2. Preliminaries

Let $\mathcal{A}$ be a $C^{*}$-algebra. For two Hilbert $\mathcal{A}$-modules $\mathcal{E}$ and $\mathcal{F}$, we write $\mathcal{L}(\mathcal{E},\mathcal{F})$ for the set of all adjointable operators from $\mathcal{E}$ to $\mathcal{F}$; when $\mathcal{E}=\mathcal{F}$, we simply denote it by $\mathcal{L}\left(\mathcal{E}\right)$. The identity operator on a module is denoted by I. For any $T\in \mathcal{L}(\mathcal{E},\mathcal{F})$, its range and kernel are denoted by $\mathcal{R}\left(T\right)$ and $\mathcal{N}\left(T\right)$, respectively. General references for the theory of Hilbert $C^{*}$-modules can be found in [16,17,18].

An operator $T\in \mathcal{L}\left(\mathcal{E}\right)$ is called idempotent if it satisfies $T^2=T$. If, in addition, T is self-adjoint ($T=T^{*}$), then it is a projection. A submodule $\mathcal{M}$ of $\mathcal{E}$ is said to be orthogonally complemented if $\mathcal{E}=\mathcal{M}\oplus {\mathcal{M}}^{\perp }$, where ${\mathcal{M}}^{\perp }=\{x\in \mathcal{E}:⟨x,y⟩=0\mathrm{for}\mathrm{all}y\in \mathcal{M}\}$. In that case, $\mathcal{M}$ is closed, and we write $P_{\mathcal{M}}$ for the corresponding projection onto $\mathcal{M}$. It is important to note that, unlike the Hilbert space setting, a closed submodule of a Hilbert $C^{*}$-module may fail to be orthogonally complemented.

In this paper, the notations “⊕” and “∔” are used with different meanings. For Hilbert $\mathcal{A}$-modules $H_1$ and $H_2$, let

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

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’s two-projections theorem as a mathematical tool. Halmos’ two-projections theorem was originally obtained in [19], and now we give a brief introduction.

Suppose that P and Q are two projections on Hilbert space H. Let

$$H_1=\mathcal{R}\left(P\right)\cap \mathcal{R}\left(Q\right),H_2=\mathcal{R}\left(P\right)\cap \mathcal{N}\left(Q\right),$$

(1)

$$H_3=\mathcal{N}\left(P\right)\cap \mathcal{R}\left(Q\right),H_4=\mathcal{N}\left(P\right)\cap \mathcal{N}\left(Q\right)$$

(2)

Furthermore, let

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

(3)

and put

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

(4)

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

(5)

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

$$U_{P,Q}\left(x\right)={\left(P_1\left(x\right),P_2\left(x\right),\cdots ,P_6\left(x\right)\right)}^T\mathrm{for\; any} x\in H,$$

(6)

with the property that

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

It follows that

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

(7)

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

(8)

where

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

(9)

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 taken for $P_{5}Q{P}_6{|}_{H_6}$ and $P_{6}Q{P}_6{|}_{H_6}$.

Lemma 1

([20], Theorem 1.4). Suppose that P and Q are two projections on Hilbert space H. Let $H_i,P_i$ $(1\le i\le 6$) be defined by (1)–(5), respectively. The operator T formulated by (9) can then be characterized as follows:

$$T=\left(\begin{array}{cc}{Q}_0& Q_0^{{\displaystyle \frac{1}{2}}}{(I_{H_5}-Q_0)}^{{\displaystyle \frac{1}{2}}}{U}_0\\ U_0^{*}{Q}_0^{{\displaystyle \frac{1}{2}}}{(I_{H_5}-Q_0)}^{{\displaystyle \frac{1}{2}}}& U_0^{*}(I_{H_5}-Q_0)U_0\end{array}\right)\in \mathcal{L}(H_5\oplus H_6),$$

(10)

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.

Lemma 2

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

Lemma 3

([22], Remark 1.1). For an operator $T\in \mathcal{L}(\mathcal{E},\mathcal{F})$, the ranges $\mathcal{R}\left(T\right)$, $\mathcal{R}\left(T{T}^{*}\right)$, $\mathcal{R}\left(T^{*}\right)$, and $\mathcal{R}\left(T^{*}T\right)$ are either all closed or all non-closed simultaneously. Moreover, $\mathcal{R}\left(T\right)=\mathcal{R}\left(T{T}^{*}\right)$ and $\mathcal{R}\left(T^{*}\right)=\mathcal{R}\left(T^{*}T\right)$.

Lemma 4

([22], 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. The Moore–Penrose Inverse of T is denoted by $T^{†}$.

Lemma 5

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

Lemma 6

([24], Theorem 2.4). Let $T\in \mathcal{L}\left(\mathcal{E}\right)$ be a positive operator. The following statements are then equivalent:

(i) 

$\mathcal{R}\left(T\right)$ is closed.

(ii) 

$\mathcal{R}\left(T^{\alpha }\right)$ is closed for all $\alpha >0$.

(iii) 

$\mathcal{R}\left(T^{\alpha }\right)$ is closed for some $\alpha >0$.

Moreover, in any of the above cases, $\mathcal{R}\left(T\right)=\mathcal{R}\left(T^{\alpha }\right)$ holds for all $\alpha >0$.

Lemma 7

([25]). Let A be a positive operator on a Hilbert space with closed range. For any real number k, the Moore–Penrose inverse of $A^k$ satisfies ${\left(A^k\right)}^{†}={\left(A^{†}\right)}^k$.

Proof. 

This equality can be rigorously proven via spectral decomposition: Let $A={\int }_0^{\infty }\lambda dE\left(\lambda \right)$. Then,

$$A^k={\int }_0^{\infty }{\lambda }^{k}dE\left(\lambda \right),A^{+}={\int }_{\lambda >0}{\lambda }^{-1}dE\left(\lambda \right),{\left(A^{+}\right)}^k={\int }_{\lambda >0}{\lambda }^{-k}dE\left(\lambda \right),$$

while ${\left(A^k\right)}^{+}={\int }_{\lambda >0}{\left({\lambda }^k\right)}^{-1}dE\left(\lambda \right)={\int }_{\lambda >0}{\lambda }^{-k}dE\left(\lambda \right)$. Therefore, the two are equal. □

3. Moore–Penrose Inverse of Various Adjointable Idempotents

We first employ the Halmos decomposition to obtain the Moore–Penrose inverse of idempotent operators on Hilbert spaces, and then we extend these results to the setting of Hilbert $C^{*}$-modules.

Theorem 1.

For any idempotents Ψ on Hilbert space H, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. Then, ${\mathsf{\Psi }}^{†}={\mathsf{\Psi }}^{*}{\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}=P-QP$.

Proof. 

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

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

which imply that $\mathcal{R}(I-P)\cap \mathcal{R}(I-Q)=0$. Let $H_i$ and $P_i$ ($1\le i\le 6$) be defined as in (1)–(5). 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 (4), 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

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

(11)

Combining $U_{P,Q}$ (defined in (6)) with the decompositions (7), (8), and Lemma 5, we obtain

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

(12)

Using the four defining equations of the Moore–Penrose inverse, we can compute ${\mathsf{\Psi }}^{†}$ via matrix operations as follows:

$$\begin{array}{c}\hfill U_{P,Q}{\mathsf{\Psi }}^{†}{U}_{P,Q}^{*}=I_{H_2}\oplus 0\oplus \left(\begin{array}{cc}{(I_{H_5}+V{V}^{*})}^{-1}& 0\\ V^{*}{(I_{H_5}+V{V}^{*})}^{-1}& 0\end{array}\right).\end{array}$$

(13)

Clearly, we have

$$\begin{array}{c}\hfill P_{5}V{P}_6=\mathsf{\Psi }-P,({\mathsf{\Psi }}^{*}-P)(P_2+I-P)=0.\end{array}$$

(14)

Moreover,

$$I_{H_5}+V{V}^{*}=\left(P_5+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right){|}_{H_5}.$$

Then, it follows that

$$\begin{array}{c}\hfill P_5{(I_{H_5}+V{V}^{*})}^{-1}{P}_5={\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}-P_2-(I-P).\end{array}$$

(15)

This identity holds because

$$\begin{array}{cc}\hfill & \left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)\left(P_2+I-P+P_5{\left(P_5+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}{|}_{H_5}{P}_5\right)\hfill \\ \hfill =& \left(P_2+I-P+P_5+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)\left(P_2+I-P+P_5{\left(P_5+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right|}_{H_5}{)}^{-1}{P}_5\right)\hfill \\ \hfill =& I.\hfill \end{array}$$

Furthermore, from

$$(I-P)\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)=I-P,$$

we deduce that

$$\begin{array}{c}\hfill (I-P){\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}=I-P.\end{array}$$

(16)

Combined with Lemma 5, Equations (13) and (15), and the relation ${\mathsf{\Psi }}^{*}{P}_2=P_2$, we obtain

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}^{†}& =P_2+P_5{(I_{H_5}+V{V}^{*})}^{-1}{P}_5+P_6{V}^{*}{(I_{H_5}+V{V}^{*})}^{-1}{P}_5\hfill \\ \hfill & =P_2+({\mathsf{\Psi }}^{*}+I-P)\left({\left(I+(\mathsf{\Psi }+P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}-P_2-(I-P)\right)\hfill \\ \hfill & =({\mathsf{\Psi }}^{*}+I-P)\left({\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}-(I-P)\right)\hfill \end{array}$$

(17)

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

(18)

□

Theorem 2.

For any idempotents Ψ on Hilbert space H, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. If $\mathcal{R}(\mathsf{\Psi }-P)$ is closed, then

$${(\mathsf{\Psi }-P)}^{†}=({\mathsf{\Psi }}^{*}-P){\left(PQP\right)}^{†}(I-PQP)=(PQP-QP){\left(PQP\right)}^{†}.$$

Proof. 

Based on Lemma 5, we first compute

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

Using the representation $\mathsf{\Psi }={(I-PQ)}^{-1}P(I-PQ)$, we obtain

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

(19)

Consequently,

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

(20)

By Lemma 3, we have

$$\mathcal{R}\left(PQ\right)=\mathcal{R}\left(PQP\right)=\mathcal{R}(\mathsf{\Psi }-P)$$

closed. Let $H_i$ and $P_i$ ($1\le i\le 6$) be defined as in (1)–(5); hence, ${\left(P_{5}Q{P}_5\right)}^{†}={\left(PQP\right)}^{†}$ is well defined.

With $Q_0=P_{5}Q{P}_5{|}_{H_5}$, (11) and (14) give

$$\mathsf{\Psi }-P=-P_5{Q}_0^{1/2}{(I_{H_5}-Q_0)}^{-1/2}{U}_0{P}_6.$$

Combined with Lemma 7, Lemma 6, and (19), we obtain

$$\begin{array}{cc}\hfill {(\mathsf{\Psi }-P)}^{†}& =-P_6{U}_0^{*}{\left(Q_0^{1/2}\right)}^{†}{(I_{H_5}-Q_0)}^{1/2}{P}_5\hfill \\ \hfill & =-P_6{U}_0^{*}{Q}_0^{1/2}{(I_{H_5}-Q_0)}^{-1/2}{\left(Q_0^{1/2}\right)}^{†}{\left(Q_0^{1/2}\right)}^{†}(I_{H_5}-Q_0)P_5\hfill \\ \hfill & =({\mathsf{\Psi }}^{*}-P){\left(Q_0\right)}^{†}(I_{H_5}-Q_0)P_5\hfill \\ \hfill & =({\mathsf{\Psi }}^{*}-P){\left(PQP\right)}^{†}(I-PQP)\hfill \\ \hfill & =(PQP-QP){(I-PQP)}^{-1}{\left(PQP\right)}^{†}(I-PQP)\hfill \\ \hfill & =(PQP-QP){\left(PQP\right)}^{†}.\hfill \end{array}$$

□

In Hilbert $C^{*}$-modules, the Halmos decomposition generally does not hold. However, the results of the Hilbert-space theorems presented above can offer insight for computing the Moore–Penrose inverse of idempotent operators in Hilbert C-modules. We thus obtain the following result:

Theorem 3.

For any idempotents $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. Then, ${\mathsf{\Psi }}^{†}={\mathsf{\Psi }}^{*}{\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}=P-QP$.

Proof. 

Let $T={\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}$. We prove that ${\mathsf{\Psi }}^{*}T$ is the Moore–Penrose inverse of $\mathsf{\Psi }$. Using (16) and the relation $\mathsf{\Psi }P=P$, we compute

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

Because $P\mathsf{\Psi }=\mathsf{\Psi }$, we obtain $\mathsf{\Psi }{\mathsf{\Psi }}^{*}T\mathsf{\Psi }=\mathsf{\Psi }$. Clearly, ${\left({\mathsf{\Psi }}^{*}T\mathsf{\Psi }\right)}^{*}={\mathsf{\Psi }}^{*}T\mathsf{\Psi }$ because T is positive. Moreover,

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}^{*}T\mathsf{\Psi }{\mathsf{\Psi }}^{*}T& ={\mathsf{\Psi }}^{*}{\left(\mathsf{\Psi }{\mathsf{\Psi }}^{*}T\right)}^{*}T\hfill \\ \hfill & ={\mathsf{\Psi }}^{*}PT\hfill \\ \hfill & ={\mathsf{\Psi }}^{*}T,\hfill \end{array}$$

which confirms that ${\mathsf{\Psi }}^{*}T$ satisfies all four Penrose equations; therefore, ${\mathsf{\Psi }}^{*}T={\mathsf{\Psi }}^{†}$.

Next, we show that $P-QP$ also equals ${\mathsf{\Psi }}^{†}$. Given $\mathcal{R}\left(Q\right)=\mathcal{N}\left(\mathsf{\Psi }\right)$, we have $\mathcal{R}(I-Q)=\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)$. Consequently,

$$(I-Q){\mathsf{\Psi }}^{*}={\mathsf{\Psi }}^{*},\mathsf{\Psi }(I-Q)=\mathsf{\Psi }.$$

Using these relations, we obtain

$$\begin{array}{cc}\hfill \mathsf{\Psi }(P-QP)& =\mathsf{\Psi }(I-Q)P=\mathsf{\Psi }P=P,\hfill \\ \hfill (P-QP)\mathsf{\Psi }& =(I-Q)P\mathsf{\Psi }=(I-Q)\mathsf{\Psi }=\mathsf{\Psi }.\hfill \end{array}$$

It is then straightforward to verify that

$$\mathsf{\Psi }(P-QP)\mathsf{\Psi }=\mathsf{\Psi },(P-QP)\mathsf{\Psi }(P-QP)=P-QP.$$

Thus, $P-QP$ likewise satisfies the four Penrose equations and, hence, coincides with ${\mathsf{\Psi }}^{†}$. □

Theorem 4.

For any idempotents $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. If $\mathcal{R}(\mathsf{\Psi }-P)$ is closed, then

$${(\mathsf{\Psi }-P)}^{†}=({\mathsf{\Psi }}^{*}-P){\left(PQP\right)}^{†}(I-PQP)=(PQP-QP){\left(PQP\right)}^{†}.$$

Proof. 

By Lemma 3 and (20), we have

$$\begin{array}{c}\hfill \mathcal{R}\left(PQ\right)=\mathcal{R}\left(PQP\right)=\mathcal{R}(\mathsf{\Psi }-P),\end{array}$$

(21)

which is closed; therefore, ${\left(PQP\right)}^{†}$ is well defined. Let $T={\left(PQP\right)}^{†}(I-PQP)$. Since $PQP$ is self-adjoint, clearly $T^{*}=T$.

We now verify that $({\mathsf{\Psi }}^{*}-P)T$ is the Moore–Penrose inverse of $\mathsf{\Psi }-P$. Using (20),

$$\begin{array}{cc}\hfill (\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)T& ={(I-PQP)}^{-1}PQP{\left(PQP\right)}^{†}(I-PQP)\hfill \\ \hfill & =PQP{\left(PQP\right)}^{†},\hfill \\ \hfill ({\mathsf{\Psi }}^{*}-P)T(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)T& =({\mathsf{\Psi }}^{*}-P){\left(PQP\right)}^{†}PQP{\left(PQP\right)}^{†}(I-PQP)\hfill \\ \hfill & =({\mathsf{\Psi }}^{*}-P){\left(PQP\right)}^{†}(I-PQP)\hfill \\ \hfill & =({\mathsf{\Psi }}^{*}-P)T.\hfill \end{array}$$

Furthermore, because $\mathcal{R}\left(PQP\right)=\mathcal{R}(\mathsf{\Psi }-P)$,

$$(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)T(\mathsf{\Psi }-P)=PQP{\left(PQP\right)}^{†}(\mathsf{\Psi }-P)=\mathsf{\Psi }-P.$$

Finally,

$${\left(({\mathsf{\Psi }}^{*}-P)T(\mathsf{\Psi }-P)\right)}^{*}=({\mathsf{\Psi }}^{*}-P)T(\mathsf{\Psi }-P)$$

follows from the self-adjointness of T and the fact that ${(\mathsf{\Psi }-P)}^{*}={\mathsf{\Psi }}^{*}-P$. Thus, $({\mathsf{\Psi }}^{*}-P)T$ satisfies all four Penrose equations and equals ${(\mathsf{\Psi }-P)}^{†}$.

To obtain the simplified form, note from (19) that

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

Hence, ${(\mathsf{\Psi }-P)}^{†}=(PQP-QP){\left(PQP\right)}^{†}$. □

The requirement that $\mathcal{R}(\mathsf{\Psi }-P)$ be closed in the theorem above is meaningful, because this condition often fails to hold, as demonstrated in the following counterexample:

Example 1.

Let $H={\ell }^2\oplus {\ell }^2$. Consider the closed subspace $M={\ell }^2\oplus \left\{0\right\}$ and let $K={\ell }^2$.

Define the bounded linear operator $T:K\to K$ by

$$T\left(z\right)={\left({\displaystyle \frac{z_n}{n}}\right)}_{n=1}^{\infty },z=\left(z_n\right)\in {\ell }^2.$$

T is compact because it is a diagonal operator with eigenvalues $1/n\to 0$. The range of T is

$$\mathcal{R}\left(T\right)=\left\{{\left({\displaystyle \frac{z_n}{n}}\right)}_{n=1}^{\infty }:z\in {\ell }^2\right\}=\left\{x\in {\ell }^2:\sum _{n=1}^{\infty }{n}^2{\left|x_n\right|}^2<\infty \right\}.$$

Take $x^{\left(N\right)}=\left(1,{\displaystyle \frac{1}{2}},\dots ,{\displaystyle \frac{1}{N}},0,0,\dots \right)$. Then, $x^{\left(N\right)}\in \mathcal{R}\left(T\right)$ and $x^{\left(N\right)}\to x={\left({\displaystyle \frac{1}{n}}\right)}_{n=1}^{\infty }$ in ${\ell }^2$. If x were in $\mathcal{R}\left(T\right)$, there would exist $z\in {\ell }^2$ such that $z_n=n·{\displaystyle \frac{1}{n}}=1$ for all n, so $z=(1,1,1,\dots )\notin {\ell }^2$, a contradiction. Hence, $x\in \overline{\mathcal{R}\left(T\right)}\setminus \mathcal{R}\left(T\right)$, so $\mathcal{R}\left(T\right)$ is not closed.

Define the closed subspace

$$N=\left\{(Tz,z):z\in K\right\}\subset H.$$

Then, $M\cap N=\left\{0\right\}$, and every $(m,k)\in H$ can be written uniquely as

$$(m,k)=(m-Tk,0)+(Tk,k),$$

with $(m-Tk,0)\in M$ and $(Tk,k)\in N$. Hence, $H=M\dotplus N$. The idempotent $A:H\to H$ that projects onto M along N is given by

$$A(m,k)=(m-Tk,0),(m,k)\in H.$$

Clearly, $A^2=A,\mathcal{R}\left(A\right)=M$ and A is bounded. Let P be the orthogonal projection onto M, such that

$$P(m,k)=(m,0).$$

Compute

$$(A-P)(m,k)=(m-Tk,0)-(m,0)=(-Tk,0).$$

Thus,

$$\mathcal{R}(A-P)=\left\{\right(-Tk,0):k\in K\}\cong \mathcal{R}\left(T\right),$$

which is not closed because $\mathcal{R}\left(T\right)$ is not closed.

We have therefore constructed a bounded idempotent A and its projection P such that the range of $A-P$ is not closed.

In my previous work on Hilbert $C^{*}$-modules, for an idempotent operator $\mathsf{\Psi }$ and its range projection P, I established the following lemma concerning the relationship between the Moore–Penrose inverse of $\mathsf{\Psi }+{\mathsf{\Psi }}^{*}$ and that of $\mathsf{\Psi }-P$:

Lemma 8

([26], Theorem 3). Let $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent operator and P be the projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$. Then, $\mathsf{\Psi }+{\mathsf{\Psi }}^{*}$ is MP-invertible if and only if ${\mathsf{\Psi }}^{*}-P$ is MP-invertible. The following identities hold: ${({\mathsf{\Psi }}^{*}-P)}^{†}=P{(\mathsf{\Psi }+{\mathsf{\Psi }}^{*})}^{†}(I-P)$ and

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

Combining the expression for the Moore–Penrose inverse of $\mathsf{\Psi }-P$ obtained in Theorem 4, we can further derive an explicit expression for the Moore–Penrose inverse of $\mathsf{\Psi }+{\mathsf{\Psi }}^{*}$.

Theorem 5.

For any idempotents $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. If $\mathcal{R}(\mathsf{\Psi }-P)$ is closed, then

$${(\mathsf{\Psi }+{\mathsf{\Psi }}^{*})}^{†}={\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}(PQP-2QP){\left({\left(PQP\right)}^{†}\right)}^2(PQP-2PQ).$$

Proof. 

By (21), $\mathcal{R}\left(PQ\right)$ is closed, ${\left(PQP\right)}^{†}$ is well defined, and $P_{\mathcal{R}\left(PQ\right)}{({\mathsf{\Psi }}^{*}-P)}^{†}={({\mathsf{\Psi }}^{*}-P)}^{†}$ since $\mathcal{R}\left({({\mathsf{\Psi }}^{*}-P)}^{†}\right)=\mathcal{R}(\mathsf{\Psi }-P)=\mathcal{R}\left(PQ\right)$. Let $T=\mathsf{\Psi }-P$. By Theorem 4, Lemma 8, and $TP=0$, we obtain

$$\begin{array}{cc}\hfill {(\mathsf{\Psi }+{\mathsf{\Psi }}^{*})}^{†}& =\left({\displaystyle \frac{2I-T}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}\right)\left({\displaystyle \frac{2I-T^{*}}{2}}\right).\hfill \\ \hfill & =\left({\displaystyle \frac{2I-T}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}+{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}{T}^{*}\right)\hfill \\ \hfill & =\left({\displaystyle \frac{2I-T}{2}}\right)\left({\displaystyle \frac{P}{2}}-2{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}+{\left(T\right)}^{†}\right)\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-2{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}+{\left(T\right)}^{†}+T{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}+{\displaystyle \frac{1}{2}}T{\left(T\right)}^{†}\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-2{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}+{\left(T\right)}^{†}+{\left(T^{*}\right)}^{†}+{\displaystyle \frac{1}{2}}T{\left(T\right)}^{†}\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-2{\left(T\right)}^{†}{\left(T^{*}\right)}^{†}+{\left(T\right)}^{†}+{\left(T^{*}\right)}^{†}+{\displaystyle \frac{1}{2}}{P}_{\mathcal{R}\left(PQ\right)}\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}\left(2T^{†}-P_{\mathcal{R}\left(PQ\right)}\right)\left(2{\left(T^{*}\right)}^{†}-P_{\mathcal{R}\left(PQ\right)}\right)\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}\left(2T^{†}-PQP{\left(PQP\right)}^{†}\right)\left(2{\left(T^{*}\right)}^{†}-PQP{\left(PQP\right)}^{†}\right)\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}\left(2(PQP-QP){\left(PQP\right)}^{†}-PQP{\left(PQP\right)}^{†}\right)\hfill \\ \hfill & \times \left(2{\left(PQP\right)}^{†}(PQP-PQ)-PQP{\left(PQP\right)}^{†}\right)\hfill \\ \hfill & ={\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}(PQP-2QP){\left({\left(PQP\right)}^{†}\right)}^2(PQP-2PQ).\hfill \end{array}$$

□

Now, we provide an example of a matrix to illustrate that the above theorem is correct in the context of matrices.

Example 2.

Consider the matrix

$$\mathsf{\Psi }=\left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 0& 0& 0\end{array}\right).$$

One readily verifies that ${\mathsf{\Psi }}^2=\mathsf{\Psi }$, confirming that Ψ is idempotent with $rank\left(\mathsf{\Psi }\right)=2$. Since Ψ is real, ${\mathsf{\Psi }}^{*}={\mathsf{\Psi }}^{\top }$. Hence,

$$\mathsf{\Psi }+{\mathsf{\Psi }}^{*}=\left(\begin{array}{ccc}2& 0& 1\\ 0& 2& 0\\ 1& 0& 0\end{array}\right).$$

This matrix is invertible (its determinant is $-2$), so its Moore–Penrose inverse coincides with its ordinary inverse:

$${(\mathsf{\Psi }+{\mathsf{\Psi }}^{*})}^{†}=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0.5& 0\\ 1& 0& -2\end{array}\right).$$

The range $\mathcal{R}\left(\mathsf{\Psi }\right)$ is spanned by the vectors $\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)$ and $\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)$. Thus,

$$P=\left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right).$$

The null space $\mathcal{N}\left(\mathsf{\Psi }\right)$ consists of vectors $\mathbf{x}={(x_1,x_2,x_3)}^{\top }$ satisfying $x_1+x_3=0$ and $x_2=0$. A unit normal vector is ${\displaystyle \frac{1}{\sqrt{2}}}{(-1,0,1)}^{\top }$, so

$$Q={\displaystyle \frac{1}{2}}\left(\begin{array}{ccc}1& 0& -1\\ 0& 0& 0\\ -1& 0& 1\end{array}\right).$$

We now compute the components appearing in the formula

$${(\mathsf{\Psi }+{\mathsf{\Psi }}^{*})}^{†}={\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}(PQP-2QP){\left({\left(PQP\right)}^{†}\right)}^2(PQP-2PQ).$$

First, compute the products involving P and Q:

$$\begin{array}{cc}\hfill PQP& =\left(\begin{array}{ccc}0.5& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),\hfill \\ \hfill {\left(PQP\right)}^{†}& =\left(\begin{array}{ccc}2& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),{\left({\left(PQP\right)}^{†}\right)}^2=\left(\begin{array}{ccc}4& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),\hfill \\ \hfill PQ& =\left(\begin{array}{ccc}0.5& 0& -0.5\\ 0& 0& 0\\ 0& 0& 0\end{array}\right),QP=\left(\begin{array}{ccc}0.5& 0& 0\\ 0& 0& 0\\ -0.5& 0& 0\end{array}\right).\hfill \end{array}$$

Then,

$$\begin{array}{cc}\hfill PQP-2QP& =\left(\begin{array}{ccc}-0.5& 0& 0\\ 0& 0& 0\\ 1& 0& 0\end{array}\right),\hfill \\ \hfill PQP-2PQ& =\left(\begin{array}{ccc}-0.5& 0& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right).\hfill \end{array}$$

Multiplying these three matrices gives

$$(PQP-2QP){\left({\left(PQP\right)}^{†}\right)}^2(PQP-2PQ)=\left(\begin{array}{ccc}1& 0& -2\\ 0& 0& 0\\ -2& 0& 4\end{array}\right).$$

Hence,

$${\displaystyle \frac{1}{2}}(PQP-2QP){\left({\left(PQP\right)}^{†}\right)}^2(PQP-2PQ)=\left(\begin{array}{ccc}0.5& 0& -1\\ 0& 0& 0\\ -1& 0& 2\end{array}\right).$$

Now, ${\displaystyle \frac{P}{2}}=\left(\begin{array}{ccc}0.5& 0& 0\\ 0& 0.5& 0\\ 0& 0& 0\end{array}\right)$, and subtracting the above matrix yields

$${\displaystyle \frac{P}{2}}-{\displaystyle \frac{1}{2}}(PQP-2QP){\left({\left(PQP\right)}^{†}\right)}^2(PQP-2PQ)=\left(\begin{array}{ccc}0& 0& 1\\ 0& 0.5& 0\\ 1& 0& -2\end{array}\right).$$

This result exactly matches the directly computed ${(\mathsf{\Psi }+{\mathsf{\Psi }}^{*})}^{†}$. Therefore, the theorem’s formula holds for this three-dimensional idempotent matrix, confirming its validity in the matrix case.

In [3], O. M. Baksalary and G. Trenkler studied the class of operators whose Moore–Penrose inverses are idempotent. We now focus specifically on the idempotency of the Moore–Penrose inverse of an idempotent operator.

Theorem 6.

For any idempotents $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. Then ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$ if and only if $PQP=0$.

Proof. 

By Theorem 3, we have ${\mathsf{\Psi }}^{†}=P-QP$. Therefore, ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$ if and only if ${(P-QP)}^2=P-QP$, which is equivalent to $QPQP=PQP$. This implies that $\mathcal{R}\left(PQP\right)\subseteq \mathcal{R}\left(Q\right)=\mathcal{R}(I-\mathsf{\Psi })$ and, consequently, $PQP=0$.

Conversely, if $PQP=0$, then it is straightforward to verify that ${(P-QP)}^2=P-QP$, which means that ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$. □

Theorem 7.

For any idempotents $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$, let P and Q be two projections from H to $\mathcal{R}\left(\mathsf{\Psi }\right)$ and $\mathcal{N}\left(\mathsf{\Psi }\right)$, respectively. Then ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$ if and only if ${\left(PQP\right)}^2=PQP$.

Proof. 

If ${\left(PQP\right)}^2=PQP$, then by the properties of the norm, either $\parallel PQP\parallel =0$ or $\parallel PQP\parallel \ge 1$. Combining this with Lemma 5, we can conclude that $\parallel PQP\parallel =0$, which implies that $PQP=0$. Hence, by Theorem 6, we have ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$.

Conversely, if ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$, then Theorem 6 yields $PQP=0$ and, consequently, ${\left(PQP\right)}^2=0$. □

We now provide matrix counterexamples to illustrate both directions of Theorem 6.

Example 3.

Counterexample for the equivalence: Consider the idempotent matrix

$$\mathsf{\Psi }=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right),$$

which satisfies ${\mathsf{\Psi }}^2=\mathsf{\Psi }$. Its Moore–Penrose inverse is

$${\mathsf{\Psi }}^{†}=\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& 0\\ {\displaystyle \frac{1}{2}}& 0\end{array}\right).$$

Computing ${\left({\mathsf{\Psi }}^{†}\right)}^2$ gives

$${\left({\mathsf{\Psi }}^{†}\right)}^2=\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& 0\\ {\displaystyle \frac{1}{2}}& 0\end{array}\right)\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& 0\\ {\displaystyle \frac{1}{2}}& 0\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{1}{4}}& 0\\ {\displaystyle \frac{1}{4}}& 0\end{array}\right)\ne {\mathsf{\Psi }}^{†},$$

so ${\left({\mathsf{\Psi }}^{†}\right)}^2\ne {\mathsf{\Psi }}^{†}$.

Now, the range $\mathcal{R}\left(\mathsf{\Psi }\right)$ is spanned by $\left(\begin{array}{c}1\\ 0\end{array}\right)$, and the null space $\mathcal{N}\left(\mathsf{\Psi }\right)$ is spanned by $\left(\begin{array}{c}1\\ -1\end{array}\right)$. The orthogonal projections onto these subspaces are

$$P=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),Q={\displaystyle \frac{1}{2}}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right).$$

Computing $PQP$ yields

$$PQP=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right){\displaystyle \frac{1}{2}}\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)={\displaystyle \frac{1}{2}}\left(\begin{array}{cc}1& 0\\ -1& 0\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& 0\\ 0& 0\end{array}\right)\ne 0.$$

Thus, we have $PQP\ne 0$ and ${\left({\mathsf{\Psi }}^{†}\right)}^2\ne {\mathsf{\Psi }}^{†}$, confirming the necessity part of the theorem.

Satisfying the equivalence, consider the orthogonal projection matrix

$$\mathsf{\Psi }=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),$$

which is idempotent. Its Moore–Penrose inverse is ${\mathsf{\Psi }}^{†}=\mathsf{\Psi }$, so ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^2=\mathsf{\Psi }={\mathsf{\Psi }}^{†}$.

The range $\mathcal{R}\left(\mathsf{\Psi }\right)$ is spanned by $\left(\begin{array}{c}1\\ 0\end{array}\right)$, and the null space $\mathcal{N}\left(\mathsf{\Psi }\right)$ is spanned by $\left(\begin{array}{c}0\\ 1\end{array}\right)$. The corresponding orthogonal projections are

$$P=\mathsf{\Psi }=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),Q=I-\mathsf{\Psi }=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right).$$

Computing $PQP$ gives

$$PQP=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)=0.$$

Thus, we have $PQP=0$ and ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$, confirming the sufficiency part of the theorem.

These two examples together validate Theorem 6: for an idempotent operator Ψ, ${\left({\mathsf{\Psi }}^{†}\right)}^2={\mathsf{\Psi }}^{†}$ if and only if $PQP=0$.

4. Decomposition of Idempotent Operators into Products of Commuting Idempotents and the Moore–Penrose Inverse of the Resulting Idempotents

If an idempotent operator $\mathsf{\Psi }$ can be decomposed into a product of two commuting idempotent operators, it is a natural question to ask how the Moore–Penrose inverses of the resulting idempotents relate to $\mathsf{\Psi }$. To address this question, we first need to derive explicit expressions for the decomposed idempotent operators. For this purpose, we introduce the following lemma from our earlier work:

Lemma 9

([26], Theorem 7). Let Ψ be a linear idempotent operator on $\mathcal{E}$, and let P be the projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$. Then, $\mathsf{\Psi }={\mathsf{\Psi }}_1{\mathsf{\Psi }}_2$ for idempotents ${\mathsf{\Psi }}_1,{\mathsf{\Psi }}_2$ 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:

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

(22)

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

(23)

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

(24)

$$\mathcal{R}\left(Y_2\right)\subseteq \mathcal{N}\left(P\right),\mathcal{N}\left(Y_2\right)\supseteq \mathcal{R}\left(P\right).$$

(25)

(2) 

Zero-product relation:

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

(3) 

Factorization expressions:

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

For idempotent operators (adjointable or not) on Hilbert $C^{*}$-modules that can be decomposed into a product of two commuting idempotent operators, we present the following decomposition representation:

Theorem 8.

Let Ψ be a linear idempotent operator on $\mathcal{E}$, and let P be the projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$. Then, $\mathsf{\Psi }={\mathsf{\Psi }}_1{\mathsf{\Psi }}_2={\mathsf{\Psi }}_2{\mathsf{\Psi }}_1$ for idempotents ${\mathsf{\Psi }}_1$ and ${\mathsf{\Psi }}_2$ on $\mathcal{E}$ if and only if there exist idempotents X and Y on $\mathcal{E}$ satisfying the following: ${\mathsf{\Psi }}_1=\mathsf{\Psi }+(I-\mathsf{\Psi })X$, ${\mathsf{\Psi }}_2=\mathsf{\Psi }+(I-\mathsf{\Psi })Y$, $XY=0$ and $YX=0$.

(1) 

Range and null space conditions:

$$\mathcal{R}\left(X\right)\subseteq \mathcal{R}(I-P),\mathcal{N}\left(X\right)⫆\mathcal{R}\left(P\right),$$

(26)

$$\mathcal{R}\left(Y\right)\subseteq \mathcal{R}(I-P),\mathcal{N}\left(Y\right)⫆\mathcal{R}\left(P\right),$$

(27)

(2) 

Zero-product relation:

$$XY=0,YX=0,$$

(3) 

Factorization expressions:

$$\begin{array}{c}{\mathsf{\Psi }}_1=\mathsf{\Psi }+(I-\mathsf{\Psi })X,\\ {\mathsf{\Psi }}_2=\mathsf{\Psi }+(I-\mathsf{\Psi })Y.\end{array}$$

Proof. 

(⟹) By Theorem 9, there exist operators $X_1,Y_1$ and idempotents $X_2,Y_2$ on $\mathcal{E}$ satisfying the range and null space conditions (22)–(25) such that

$${\mathsf{\Psi }}_1=P+X_1+X_2,{\mathsf{\Psi }}_2=\mathsf{\Psi }+(I-\mathsf{\Psi })Y_1\mathsf{\Psi }+(I-\mathsf{\Psi })Y_2.$$

Using these conditions and the identity $\mathsf{\Psi }={\mathsf{\Psi }}_2{\mathsf{\Psi }}_1$, we can deduce that

$$\begin{array}{c}\hfill \mathsf{\Psi }-P-X_1-\mathsf{\Psi }{X}_2=(I-\mathsf{\Psi })(Y_{1}P+Y_1{X}_1+Y_1\mathsf{\Psi }{X}_2+Y_2{X}_2).\end{array}$$

Since $\mathcal{R}\left(X_1\right)\subseteq \mathcal{R}\left(P\right)$, $\mathcal{R}\left(Y_1\right)\subseteq \mathcal{R}(I-P)$, $\mathcal{R}\left(Y_2\right)\subseteq \mathcal{R}(I-P)$, and $(I-P)(I-\mathsf{\Psi })=I-P$, we obtain

$$\begin{array}{cc}\hfill \mathsf{\Psi }-P-X_1-\mathsf{\Psi }{X}_2& =0,\hfill \\ \hfill Y_{1}P+Y_1{X}_1+Y_1\mathsf{\Psi }{X}_2+Y_2{X}_2& =0.\hfill \end{array}$$

Combining the first equation with $P+X_1+\mathsf{\Psi }{X}_2=\mathsf{\Psi }{\mathsf{\Psi }}_1$ yields

$$\begin{array}{cc}\hfill \mathsf{\Psi }& =\mathsf{\Psi }{\mathsf{\Psi }}_1,\hfill \\ \hfill Y_1\mathsf{\Psi }& =-Y_2{X}_2.\hfill \end{array}$$

(28)

Furthermore, using $Y_1{Y}_2=0$ and $Y_1^2=Y_1$, we get $Y_1\mathsf{\Psi }=0$. Since $\mathcal{N}\left(Y_1\right)\supseteq \mathcal{N}\left(P\right)$ and $\mathcal{R}\left(\mathsf{\Psi }\right)=\mathcal{R}\left(P\right)$, it follows that $Y_1=0$. Consequently,

$${\mathsf{\Psi }}_2=\mathsf{\Psi }+(I-\mathsf{\Psi })Y_2.$$

On the other hand, from (28), we have

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

Set $X=X_2$ and $Y=Y_2$. The commutativity condition ${\mathsf{\Psi }}_1{\mathsf{\Psi }}_2={\mathsf{\Psi }}_2{\mathsf{\Psi }}_1$ then implies that $XY=0$ and $YX=0$.

(⟸) The converse follows by direct verification. □

Next, we investigate the relationship between the idempotent operator $\mathsf{\Psi }$ and the Moore–Penrose inverses of the idempotent operators arising from its product decomposition.

Theorem 9.

Let $\mathsf{\Psi }\in \mathcal{L}\left(\mathcal{E}\right)$ be an idempotent operator and P be the projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$. Suppose that there exist idempotents ${\mathsf{\Psi }}_1,{\mathsf{\Psi }}_2\in \mathcal{L}\left(\mathcal{E}\right)$ such that $\mathsf{\Psi }={\mathsf{\Psi }}_1{\mathsf{\Psi }}_2={\mathsf{\Psi }}_2{\mathsf{\Psi }}_1$. Define $X=(I-P)({\mathsf{\Psi }}_1-\mathsf{\Psi })$ and $Y=(I-P)({\mathsf{\Psi }}_2-\mathsf{\Psi })$. Then, the Moore–Penrose inverses of ${\mathsf{\Psi }}_1$ and ${\mathsf{\Psi }}_2$ are given by

$${\mathsf{\Psi }}_1^{†}=P_{\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(X^{*}\right)}{P}_{\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(X\right)},{\mathsf{\Psi }}_2^{†}=P_{\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(Y^{*}\right)}{P}_{\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(Y\right)}.$$

Proof. 

By Theorem 8, there exist idempotents X and Y on $\mathcal{E}$ such that

$${\mathsf{\Psi }}_1=\mathsf{\Psi }+(I-\mathsf{\Psi })X=\mathsf{\Psi }(I-X)+X,{\mathsf{\Psi }}_2=\mathsf{\Psi }+(I-\mathsf{\Psi })Y=\mathsf{\Psi }(I-Y)+Y.$$

From (26) and (27), we directly obtain $X=(I-P)({\mathsf{\Psi }}_1-\mathsf{\Psi })$ and $Y=(I-P)({\mathsf{\Psi }}_2-\mathsf{\Psi })$.

Clearly, $\mathcal{R}\left({\mathsf{\Psi }}_1\right)\subseteq \mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(X\right)$ and $\mathcal{R}\left({\mathsf{\Psi }}_2\right)\subseteq \mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(Y\right)$. Using (26) and (27), for any $x,y\in \mathcal{E}$, we have

$${\mathsf{\Psi }}_1\mathsf{\Psi }x=\mathsf{\Psi }x,{\mathsf{\Psi }}_{1}Xy=Xy,{\mathsf{\Psi }}_2\mathsf{\Psi }x=\mathsf{\Psi }x,{\mathsf{\Psi }}_{2}Yy=Yy.$$

This implies that $\mathcal{R}\left({\mathsf{\Psi }}_1\right)=\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(X\right)$ and $\mathcal{R}\left({\mathsf{\Psi }}_2\right)=\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(Y\right)$; hence, the projections $P_{\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(X\right)}$ and $P_{\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(Y\right)}$ are well defined.

Similarly, we have ${\mathsf{\Psi }}_1^{*}={\mathsf{\Psi }}^{*}+X^{*}(I-{\mathsf{\Psi }}^{*})$ and ${\mathsf{\Psi }}_2^{*}={\mathsf{\Psi }}^{*}+Y^{*}(I-{\mathsf{\Psi }}^{*})$. Again, by (26) and (27), for any $x,y\in \mathcal{E}$,

$${\mathsf{\Psi }}_1^{*}{\mathsf{\Psi }}^{*}x={\mathsf{\Psi }}^{*}x,{\mathsf{\Psi }}_1^{*}{X}^{*}y=X^{*}(I-{\mathsf{\Psi }}^{*})X^{*}y=X^{*}y,$$

$${\mathsf{\Psi }}_2^{*}{\mathsf{\Psi }}^{*}x={\mathsf{\Psi }}^{*}x,{\mathsf{\Psi }}_2^{*}{Y}^{*}y=Y^{*}(I-{\mathsf{\Psi }}^{*})Y^{*}y=Y^{*}y.$$

Thus, $\mathcal{R}\left({\mathsf{\Psi }}_1^{*}\right)=\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(X^{*}\right)$ and $\mathcal{R}\left({\mathsf{\Psi }}_2^{*}\right)=\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(Y^{*}\right)$, so the projections $P_{\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(X^{*}\right)}$ and $P_{\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(Y^{*}\right)}$ are also well defined.

Applying Theorem 3, we finally obtain

$$\begin{array}{cc}\hfill {\mathsf{\Psi }}_1^{†}& =P_{\mathcal{R}\left({\mathsf{\Psi }}_1^{*}\right)}{P}_{\mathcal{R}\left({\mathsf{\Psi }}_1\right)}=P_{\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(X^{*}\right)}{P}_{\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(X\right)},\hfill \\ \hfill {\mathsf{\Psi }}_2^{†}& =P_{\mathcal{R}\left({\mathsf{\Psi }}_2^{*}\right)}{P}_{\mathcal{R}\left({\mathsf{\Psi }}_2\right)}=P_{\mathcal{R}\left({\mathsf{\Psi }}^{*}\right)+\mathcal{R}\left(Y^{*}\right)}{P}_{\mathcal{R}\left(\mathsf{\Psi }\right)+\mathcal{R}\left(Y\right)}.\hfill \end{array}$$

□

5. On the Spectrum and Spectral Norm of Moore–Penrose Inverses of Idempotents

The perturbation theory for operator generalized inverses constitutes a crucial research topic, as evidenced by investigations into error-bound-based algorithms [13], tensor perturbation theory via the Einstein product [14], and perturbations of both Moore–Penrose and dual Moore–Penrose generalized inverses [15]. Since perturbation analysis inherently depends on the study of various operator norms, it becomes essential to further examine the associated spectral properties and spectral norm related to the Moore–Penrose inverse of idempotent operators.

Theorem 10.

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

$$\sigma \left({\left({\mathsf{\Psi }}^{†}\right)}^{*}{\mathsf{\Psi }}^{†}\right)=\sigma \left(P(I-Q)P\right),$$

$$\parallel {\mathsf{\Psi }}^{†}\parallel =max\left\{t^{1/2}\mid t\in \sigma \left(P(I-Q)P\right)\right\}={\parallel P(I-Q)P\parallel }^{1/2}.$$

Moreover,

$$\sigma \left({\left({\mathsf{\Psi }}^{†}\right)}^{*}(I-P){\mathsf{\Psi }}^{†}\right)=\left\{t-t^2\mid t\in \sigma \left(PQP\right)\right\},$$

and

$$\parallel (I-P){\mathsf{\Psi }}^{†}\parallel =max\left\{{(t-t^2)}^{1/2}\mid t\in \sigma \left(PQP\right)\right\}\le {\displaystyle \frac{1}{2}}.$$

Proof. 

By Theorem 3 and (20), we have

$${\left({\mathsf{\Psi }}^{†}\right)}^{*}{\mathsf{\Psi }}^{†}={(P-PQ)}^2=P(I-Q)P.$$

Hence,

$$\sigma \left({\left({\mathsf{\Psi }}^{†}\right)}^{*}{\mathsf{\Psi }}^{†}\right)=\sigma \left(P(I-Q)P\right),$$

and by Lemma 5,

$$\parallel {\mathsf{\Psi }}^{†}\parallel =max\left\{t^{1/2}\mid t\in \sigma \left(P(I-Q)P\right)\right\}={\parallel P(I-Q)P\parallel }^{1/2}.$$

Similarly, using Theorem 3 and (20), we compute

$$\begin{array}{cc}\hfill {\left({\mathsf{\Psi }}^{†}\right)}^{*}(I-P){\mathsf{\Psi }}^{†}& ={\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}\mathsf{\Psi }(I-P){\mathsf{\Psi }}^{*}{\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}\hfill \\ \hfill & ={\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}(\mathsf{\Psi }{\mathsf{\Psi }}^{*}-P){\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}\hfill \\ \hfill & ={\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}\left((\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right){\left(I+(\mathsf{\Psi }-P)({\mathsf{\Psi }}^{*}-P)\right)}^{-1}\hfill \\ \hfill & =PQP(I-PQP).\hfill \end{array}$$

Therefore, $\lambda I-{\left({\mathsf{\Psi }}^{†}\right)}^{*}(I-P){\mathsf{\Psi }}^{†}$ is invertible if and only if $\lambda I-PQP(I-PQP)$ is invertible.

Let $\mathcal{A}=C^{*}(I,PQP)$ be the unital commutative $C^{*}$-algebra generated by I and $PQP$. By the Gelfand transform, $\mathcal{A}$ is isometrically *-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, $\lambda I-PQP(I-PQP)$ is invertible if and only if

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

which is equivalent to

$$\lambda \notin \left\{t-t^2\mid t\in \sigma \left(PQP\right)\right\}.$$

Consequently,

$$\sigma \left({\left({\mathsf{\Psi }}^{†}\right)}^{*}(I-P){\mathsf{\Psi }}^{†}\right)=\left\{t-t^2\mid t\in \sigma \left(PQP\right)\right\}.$$

Finally,

$$\begin{array}{cc}\hfill \parallel (I-P){\mathsf{\Psi }}^{†}\parallel & =max\left\{{(t-t^2)}^{1/2}\mid t\in \sigma \left(PQP\right)\right\}\hfill \\ \hfill & =max\left\{{\left(-{(t-\frac{1}{2})}^2+\frac{1}{4}\right)}^{1/2}\mid t\in \sigma \left(PQP\right)\right\}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}.\hfill \end{array}$$

□

The following matrix example demonstrates the correctness of the inequality $\parallel (I-P){\mathsf{\Psi }}^{†}\parallel \le {\displaystyle \frac{1}{2}}$ and shows that this bound is sharp:

Example 4.

Let $\mathsf{\Psi }\in {\mathbb{C}}^{n\times n}$ be an idempotent matrix, and let P be the orthogonal projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$. Since Ψ is idempotent, it can be expressed in the Schur form:

$$\mathsf{\Psi }=Q\left(\begin{array}{cc}{I}_r& S\\ 0& 0\end{array}\right)Q^{*},$$

where Q is unitary, $r=rank\left(\mathsf{\Psi }\right)$, and $S\in {\mathbb{C}}^{r\times (n-r)}$.

The Moore–Penrose inverse of Ψ is given by

$${\mathsf{\Psi }}^{†}=Q\left(\begin{array}{cc}{(I_r+S{S}^{*})}^{-1}& 0\\ S^{*}{(I_r+S{S}^{*})}^{-1}& 0\end{array}\right)Q^{*}.$$

The orthogonal projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$ is

$$P=\mathsf{\Psi }{\mathsf{\Psi }}^{†}=Q\left(\begin{array}{cc}{I}_r& 0\\ 0& 0\end{array}\right)Q^{*}.$$

Thus,

$$(I-P){\mathsf{\Psi }}^{†}=Q\left(\begin{array}{cc}0& 0\\ S^{*}{(I_r+S{S}^{*})}^{-1}& 0\end{array}\right)Q^{*}.$$

Let $A=S^{*}$. Then, the spectral norm of $(I-P){\mathsf{\Psi }}^{†}$ equals the spectral norm of the matrix

$$M=\left(\begin{array}{cc}0& 0\\ A{(I_r+A^{*}A)}^{-1}& 0\end{array}\right).$$

Using the identity $A{(I_r+A^{*}A)}^{-1}={(I_{n-r}+A{A}^{*})}^{-1}A$, we have

$$M=\left(\begin{array}{cc}0& 0\\ {(I_{n-r}+A{A}^{*})}^{-1}A& 0\end{array}\right).$$

Let $A=U\mathsf{\Sigma }{V}^{*}$ be the singular value decomposition of A, where U and V are unitary, and Σ is an $(n-r)\times r$ matrix with non-negative diagonal entries ${\sigma }_1\ge {\sigma }_2\ge \cdots \ge {\sigma }_r\ge 0$. Then,

$${(I_{n-r}+A{A}^{*})}^{-1}A=U{(I_{n-r}+\mathsf{\Sigma }{\mathsf{\Sigma }}^{*})}^{-1}\mathsf{\Sigma }{V}^{*}.$$

The matrix ${(I_{n-r}+\mathsf{\Sigma }{\mathsf{\Sigma }}^{*})}^{-1}\mathsf{\Sigma }$ has nonzero singular values ${\displaystyle \frac{{\sigma }_i}{1+{\sigma }_i^2}}$ for $i=1,\dots ,r$. Therefore,

$$\parallel (I-P){\mathsf{\Psi }}^{†}\parallel =\underset{1\le i\le r}{max}{\displaystyle \frac{{\sigma }_i}{1+{\sigma }_i^2}}.$$

Define the function $f\left(x\right)={\displaystyle \frac{x}{1+x^2}}$ for $x\ge 0$. Its maximum is attained at $x=1$, with $f\left(1\right)={\displaystyle \frac{1}{2}}$. Hence,

$$\parallel (I-P){\mathsf{\Psi }}^{†}\parallel \le {\displaystyle \frac{1}{2}}.$$

To show that inequality can be strict, consider a $2\times 2$ idempotent matrix

$$\mathsf{\Psi }=\left(\begin{array}{cc}1& 2\\ 0& 0\end{array}\right).$$

Its Moore–Penrose inverse is

$${\mathsf{\Psi }}^{†}=\left(\begin{array}{cc}{\displaystyle \frac{1}{5}}& 0\\ {\displaystyle \frac{2}{5}}& 0\end{array}\right).$$

The projection onto $\mathcal{R}\left(\mathsf{\Psi }\right)$ is

$$P=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right).$$

Then,

$$(I-P){\mathsf{\Psi }}^{†}=\left(\begin{array}{cc}0& 0\\ {\displaystyle \frac{2}{5}}& 0\end{array}\right),$$

whose spectral norm is ${\displaystyle \frac{2}{5}}=0.4<{\displaystyle \frac{1}{2}}$. This demonstrates that the inequality can be strict.

To show that the bound is sharp, consider the $2\times 2$ example:

$$\mathsf{\Psi }=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right),{\mathsf{\Psi }}^{†}=\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& 0\\ {\displaystyle \frac{1}{2}}& 0\end{array}\right),P=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right).$$

Then,

$$(I-P){\mathsf{\Psi }}^{†}=\left(\begin{array}{cc}0& 0\\ {\displaystyle \frac{1}{2}}& 0\end{array}\right),$$

whose singular values are 0 and ${\displaystyle \frac{1}{2}}$. Thus, $\parallel (I-P){\mathsf{\Psi }}^{†}\parallel ={\displaystyle \frac{1}{2}}$.

6. Discussion

This paper presents a systematic investigation of the Moore–Penrose inverse of idempotent operators in Hilbert $C^{*}$-modules. Our main contributions are summarized below.

6.1. Summary of Contributions

We began by extending classical results on computing the Moore–Penrose inverse of an idempotent operator and characterizing its difference from the range projection to the setting of Hilbert $C^{*}$-modules. Building on this foundation, we derived an explicit formula for the Moore–Penrose inverse of the sum of an idempotent operator and its adjoint, and we illustrated the necessity of the underlying conditions with counterexamples. Furthermore, we generalized the work of Baksalary and Trenkler on the idempotency of the Moore–Penrose inverse to Hilbert $C^{*}$-modules, supporting our theoretical findings with numerical examples. In the second part, we established a decomposition of an idempotent operator into a product of two commuting idempotents. Using this decomposition, we obtained explicit expressions for the Moore–Penrose inverses of the factors in terms of the original operator. Finally, we analyzed the spectral properties and norms of operators associated with the Moore–Penrose inverse of an idempotent operator, obtaining sharp estimates such as $\parallel (I-P){\mathsf{\Psi }}^{†}\parallel \le 1/2$. The validity of these estimates and the sharpness of the bounds were demonstrated through concrete matrix examples.

6.2. Theoretical and Practical Implications

Our results provide a unified framework for studying idempotent operators and their generalized inverses within Hilbert $C^{*}$-modules. The decomposition theorems and spectral analysis offer new structural insights and facilitate the computation of Moore–Penrose inverses. The norm estimates and spectral characterizations are applicable to perturbation theory, numerical analysis, and optimization problems involving idempotent operators. Moreover, this work bridges classical operator-theoretic results with the more general setting of Hilbert $C^{*}$-modules, underscoring the robustness and adaptability of these concepts.

6.3. Future Research Directions

Several natural extensions of this work merit further investigation:

• Extending the analysis of the Moore–Penrose inverse for idempotent operators to more general operator algebras, such as von Neumann algebras.
• Applying the decomposition theory developed here to other classes of operators, such as tripotents or partial isometries.
• Further exploring the functional calculus and numerical range of operators related to the Moore–Penrose inverse, building on the spectral properties obtained in Section 5.
• Investigating applications of these results to iterative methods, least squares problems, and quantum information theory, which would be of both theoretical and practical interest.