Maximal Norms of Orthogonal Projections and Closed-Range Operators

Abstract

Using the Dixmier angle between two closed subspaces of a complex Hilbert space $\mathcal{H}$, we establish the necessary and sufficient conditions for the operator norm of the sum of two orthogonal projections, $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$, onto closed subspaces ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$, to attain its maximum, namely $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2.$ These conditions are expressed in terms of the geometric relationship and symmetry between the ranges of the projections. We apply these results to orthogonal projections associated with a closed-range operator via its Moore–Penrose inverse. Additionally, for any bounded operator T with closed range in $\mathcal{H}$, we derive sufficient conditions ensuring $\parallel T{T}^{†}+T^{†}T\parallel =2,$ where $T^{†}$ denotes the Moore–Penrose inverse of T. This work highlights how symmetry between operator ranges and their algebraic structure governs norm extremality and extends a recent finite-dimensional result to the general Hilbert space setting.

1. Introduction

Orthogonal projections are fundamental in functional analysis and operator theory, appearing in a wide range of theoretical and applied contexts. As self-adjoint idempotent operators, they are symmetric in the operator-theoretic sense, since each projection coincides with its adjoint. This intrinsic symmetry plays a crucial role in understanding their behavior, especially when analyzing sums of projections and their norm properties. Foundational results on the norm of projection sums are presented in [1,2], while various norm inequalities involving projections and related operators are explored in [3,4,5], often revealing deeper geometric and algebraic symmetries. Further structural insights into projection operators and their ranges are given in [6,7]. In addition, the Moore–Penrose inverse is a key tool in the study of closed-range operators, with its properties—particularly in forming symmetric operator expressions like $T{T}^{†}+T^{†}T$—examined in [8,9,10].

This work studies the operator norm of the sum of two orthogonal projections, $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel$, where $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$ denote the projections onto closed subspaces ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$ of a complex Hilbert space $\mathcal{H}$. Our goal is to find conditions under which this norm reaches its maximum value, $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$, and to apply these results to projections associated with bounded operators and their Moore–Penrose inverses. This extends recent finite-dimensional results [11,12] to infinite-dimensional Hilbert spaces, addressing a significant open problem in operator theory.

A crucial result by Duncan and Taylor [13] shows that

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =1+\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel .$$

(1)

This identity has been studied in finite-dimensional spaces [11,12] and within the framework of two-projection theory [14], but the infinite-dimensional case remains challenging. The Dixmier angle, explored by Deutsch [15], offers a geometric tool for analyzing subspace relationships and guides our characterizations. Works on operator ranges [16] and range inclusion [17] also support our study.

A key motivation is to connect these ideas to bounded operators with closed ranges and their Moore–Penrose inverses, which are fundamental in operator theory [9,10]. The Moore–Penrose inverse, detailed in [9,18], plays a central role in studying projections onto ranges and kernels of operators. Inspired by Conde’s recent work [12] on projection norm maximization, we explore conditions under which

$$\parallel T{T}^{†}+T^{†}T\parallel =2,$$

where T is a bounded operator with closed range and $T^{†}$ denotes its Moore–Penrose inverse. This generalizes finite-dimensional results [11] to infinite-dimensional spaces, despite challenges such as non-compact operators and unbounded inverses [19,20]. We suggest that the reader consult the following manuscripts for further details and background on the topic [21,22,23].

Next, we introduce the key concepts and notations used in this paper.

Let $\mathcal{H}$ be a complex Hilbert space. We denote by $\mathcal{B}\left(\mathcal{H}\right)$ the $C^{*}$-algebra of all bounded linear operators on $\mathcal{H}$, equipped with the operator norm. For $T\in \mathcal{B}\left(\mathcal{H}\right)$, we write $T^{*}$ for its adjoint, $Ker\left(T\right)$ for its kernel, and $Ran\left(T\right)$ for its range.

An operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ is positive if $\langle Tx,x\rangle \ge 0$ for all $x\in \mathcal{H}$. This implies T is self-adjoint, i.e., $T=T^{*}$. The set of positive operators is denoted by $\mathcal{B}{\left(\mathcal{H}\right)}^{+}$.

An operator $P\in \mathcal{B}\left(\mathcal{H}\right)$ is an orthogonal projection if $P^2=P$ and $P^{*}=P$. Such projections are essential for studying subspaces of $\mathcal{H}$ and their geometry.

The subset $CR\left(\mathcal{H}\right)$ consists of operators in $\mathcal{B}\left(\mathcal{H}\right)$ with closed range. An operator T is normal if $T^{*}T=T{T}^{*}$. Every orthogonal projection is normal and has closed range.

The spectral radius of $T\in \mathcal{B}\left(\mathcal{H}\right)$ is

$$r\left(T\right)=sup\left\{\right|\lambda |:\lambda \in \sigma (T\left)\right\},$$

where $\sigma \left(T\right)$ is the spectrum of T. The numerical radius is

$$\omega \left(T\right)=sup\left\{\right|\langle Tx,x\rangle |:\parallel x\parallel =1\}.$$

The Dixmier angle, introduced in [15], measures the relative position of two closed subspaces $\mathcal{M},\mathcal{N}\subseteq \mathcal{H}$. It is defined as

$$c_0(\mathcal{M},\mathcal{N})=sup\left\{\right|\langle x,y\rangle |:x\in \mathcal{M},\parallel x\parallel \le 1,y\in \mathcal{N},\parallel y\parallel \le 1\}.$$

This quantity is crucial in operator theory and spectral analysis.

We work with various classes of operators in $\mathcal{B}\left(\mathcal{H}\right)$. Beyond normality, several classes of non-normal operators have distinct structural and spectral properties. An operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ is defined as follows:

• Normaloid if $\parallel T\parallel =\omega \left(T\right)$.
• Spectraloid if $\omega \left(T\right)=r\left(T\right)$.

These classes are related by the following inclusions:

$$\begin{array}{ccc}\hfill \mathrm{Self-adjoint}& ⊊& \mathrm{Normal}⊊\mathrm{Normaloid}⊊\mathrm{Spectraloid},\hfill \end{array}$$

(2)

For an operator $T\in CR\left(\mathcal{H}\right)$, the Moore–Penrose inverse $T^{†}\in \mathcal{B}\left(\mathcal{H}\right)$ is the unique operator satisfying the Penrose equations:

$$T{T}^{†}T=T,T^{†}T{T}^{†}=T^{†},{\left(T{T}^{†}\right)}^{*}=T{T}^{†},{\left(T^{†}T\right)}^{*}=T^{†}T.$$

For more details, see [9].

Define the operators

$$P:=T{T}^{†}\mathrm{and}Q:=T^{†}T.$$

These are orthogonal projections onto $Ran\left(T\right)$ and $Ran\left(T^{*}\right)$, respectively. Then, we have

$$\parallel P+Q\parallel =\parallel T{T}^{†}+T^{†}T\parallel \le 2.$$

We conclude by recalling a notable subclass within the set of closed-range operators. An operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ is called an EP operator if it has closed range and commutes with its Moore–Penrose inverse, that is, $[T,T^{†}]=0$, where $[A,B]=AB-BA$ denotes the usual commutator in $\mathcal{B}\left(\mathcal{H}\right)$; see [10].

2. Main Results

In this section, we provide necessary and sufficient conditions for the norm of the sum of two orthogonal projections to attain its maximum value. This characterization will be formulated in terms of the geometry of their ranges, or, equivalently, in terms of the closedness of certain sums or products of the associated projections.

To obtain our main result, we rely on two key corollaries. The first derives from a general identity concerning the sum of operator ranges—originally due to Crimmins (unpublished) and later simplified by Fillmore and Williams [16]. The second extends this result specifically to the setting of orthogonal projections.

Lemma 1

(Corollary 3 in [16]). Let $A,B\in \mathcal{B}{\left(\mathcal{H}\right)}^{+}\cap CR\left(\mathcal{H}\right)$. Then $A+B\in CR\left(\mathcal{H}\right)$ if and only if $Ran\left(A\right)+Ran\left(B\right)$ is closed.

Lemma 2

(Corollary X in [2]). $P_{{\mathcal{W}}_1},P_{{\mathcal{W}}_2}$ be orthogonal projections on $\mathcal{H}$. Then, the following conditions are equivalent:

1.

${\mathcal{W}}_1+{\mathcal{W}}_2$ is closed.

2.

$P_{{\mathcal{W}}_1}\pm P_{{\mathcal{W}}_2}\in CR\left(\mathcal{H}\right)$.

3.

$P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\in CR\left(\mathcal{H}\right)$.

4.

$P_{{\mathcal{W}}_2}-P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\in CR\left(\mathcal{H}\right)$.

Using these results, we are ready to prove one of the main theorems of this section.

Theorem 1.

Let $P_{{\mathcal{W}}_1},P_{{\mathcal{W}}_2}$ be orthogonal projections on $\mathcal{H}$. Then, the following conditions are equivalent.

1.

$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$.

2.

${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}$ or ${\mathcal{W}}_1+{\mathcal{W}}_2$ is not closed.

3.

$Ran\left(P_{{\mathcal{W}}_1}\right)\cap Ran\left(P_{{\mathcal{W}}_2}\right)\ne \left\{0\right\}$ or $P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\notin CR\left(\mathcal{H}\right)$.

4.

$Ran\left(P_{{\mathcal{W}}_1}\right)\cap Ran\left(P_{{\mathcal{W}}_2}\right)\ne \left\{0\right\}$ or $P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2}\notin CR\left(\mathcal{H}\right)$.

5.

$Ran\left(P_{{\mathcal{W}}_1}\right)\cap Ran\left(P_{{\mathcal{W}}_2}\right)\ne \left\{0\right\}$ or $P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\notin CR\left(\mathcal{H}\right)$.

6.

$Ran\left(P_{{\mathcal{W}}_1}\right)\cap Ran\left(P_{{\mathcal{W}}_2}\right)\ne \left\{0\right\}$ or $P_{{\mathcal{W}}_2}-P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\notin CR\left(\mathcal{H}\right)$.

Proof. 

Recall from (1) (or a well-known identity for sums of projections) that

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =1+\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel .$$

Therefore,

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2\iff \parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel =1.$$

The quantity $\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel$ corresponds to the cosine of the Dixmier angle $c_0({\mathcal{W}}_1,{\mathcal{W}}_2)$ between the subspaces ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$, as defined in [15]:

$$c_0({\mathcal{W}}_1,{\mathcal{W}}_2):=\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel .$$

From [15], Lemma 10, we know that

$$c_0({\mathcal{W}}_1,{\mathcal{W}}_2)=1$$

if and only if either ${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}$ or ${\mathcal{W}}_1+{\mathcal{W}}_2$ is not closed. This establishes the equivalence between statements (1) and (2).

Statements (3) to (6) all relate the closure of sums and differences of projections and the closedness of ranges of corresponding operators.

By Lemma 1, we know that for positive operators $A,B\in \mathcal{B}{\left(\mathcal{H}\right)}^{+}\cap CR\left(\mathcal{H}\right)$, the sum $A+B$ has closed range if and only if the sum of their ranges is closed. Here, projections $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$ satisfy these conditions.

Furthermore, Lemma 2 states that for orthogonal projections $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$, the condition that the sum of subspaces ${\mathcal{W}}_1+{\mathcal{W}}_2$ is closed is equivalent to the operators $P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}$ and $P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2}$ having closed range, as well as to the operators $P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}-P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}$ having closed range.

Thus, the closedness of ${\mathcal{W}}_1+{\mathcal{W}}_2$ is equivalent to each of the conditions that the operators in (3), (4), (5), and (6) belong to $CR\left(\mathcal{H}\right)$.

Therefore, the negation of the closedness condition in (2), i.e., ${\mathcal{W}}_1+{\mathcal{W}}_2$ is not closed, is equivalent to each of the operators listed in (3) to (6) not having closed range. The intersection condition ${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}$ is equivalent to

$$Ran\left(P_{{\mathcal{W}}_1}\right)\cap Ran\left(P_{{\mathcal{W}}_2}\right)\ne \left\{0\right\},$$

so all statements consistently use the ranges of projections. In conclusion, this proves the equivalence between statements (2) and (3), (4), (5), and (6). □

Remark 1.

Before presenting a new characterization, we recall a result previously obtained by one of the authors (see [12]).

Lemma 3.

Let $P_{{\mathcal{W}}_1},P_{{\mathcal{W}}_2}$ be the orthogonal projections on $\mathcal{H}$. Then, the following conditions are equivalent.

1.

$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =\parallel P_{{\mathcal{W}}_1}\parallel +\parallel P_{{\mathcal{W}}_2}\parallel =2.$

2.

$\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel =\parallel P_{{\mathcal{W}}_1}\parallel \parallel P_{{\mathcal{W}}_2}\parallel =1.$

3.

$r\left(P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\right)=r\left(P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\right)=1.$

4.

$\omega \left(P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\right)=\omega \left(P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\right)=1.$

5.

$\parallel \alpha P_{{\mathcal{W}}_1}+\beta P_{{\mathcal{W}}_2}\parallel =\alpha +\beta$ for any $\alpha ,\beta \ge 0.$

6.

$1\in \overline{W\left(P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\right)}\cap \overline{W\left(P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\right)}.$

Note that the condition that the sum of the ranges is a not closed set, as stated in Theorem 1, is superfluous when these subspaces are finite-dimensional. More precisely, we have the following result:

Corollary 1.

Let $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$ be orthogonal projections on $\mathcal{H}$, where ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$ are finite-dimensional subspaces. Then, the equality $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$ holds if and only if ${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}.$

Proof. 

The result follows immediately from the well-known fact that if ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$ are finite-dimensional subspaces, then their sum ${\mathcal{W}}_1+{\mathcal{W}}_2$ is closed, together with Theorem 1. □

We will now demonstrate that the hypotheses given in the obtained characterization in Theorem 1 are independent in an infinite-dimensional setting. Specifically, we seek three examples that illustrate this independence: (1) ${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}$ but ${\mathcal{W}}_1+{\mathcal{W}}_2$ is closed, (2) ${\mathcal{W}}_1\cap {\mathcal{W}}_2=\left\{0\right\}$ but ${\mathcal{W}}_1+{\mathcal{W}}_2$ is not closed, and (3) both ${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}$ and ${\mathcal{W}}_1+{\mathcal{W}}_2$ are not closed.

Moreover, in each example, we compute the norm of $P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}$, where $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$ denote the orthogonal projections onto ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$, respectively.

Example 1.

1. Let $\mathcal{H}={\ell }^2$ be the space of square-summable sequences, and let ${\left\{e_n\right\}}_{n=0}^{\infty }$ be its canonical basis. Consider the following closed subspaces of $\mathcal{H}$:

$${\mathcal{W}}_1=\overline{span\{e_1,e_2,e_3,\dots \}},\mathrm{and}{\mathcal{W}}_2=\overline{span\{e_2,e_3,e_4,\dots \}}.$$

• It is immediately clear that ${\mathcal{W}}_1+{\mathcal{W}}_2={\mathsf{\ell }}^2$, meaning that the sum of ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$ is closed. Let $P_{{\mathcal{W}}_1}$ and $P_{{\mathcal{W}}_2}$ denote the respective orthogonal projections onto ${\mathcal{W}}_1$ and ${\mathcal{W}}_2$. In this case, the intersection of their ranges is given by

  $${\mathcal{W}}_1\cap {\mathcal{W}}_2=\overline{span\{e_2,e_3,e_4,\dots \}},$$
• which is a proper subspace of $\mathcal{H}$. Therefore, ${\mathcal{W}}_1\cap {\mathcal{W}}_2\ne \left\{0\right\}$.
• Moreover, applying $P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}$ to $e_1$ yields $(P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})e_1=2e_1.$ Thus, we conclude that $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$.

2.

Let $\mathcal{H}$ be a Hilbert space with orthonormal basis ${\left\{e_n\right\}}_{n\in \mathbb{N}}$. For each $n\in \mathbb{N}$, define

$$s_n=e_{2n}\mathrm{and}{t}_n=s_n+{\displaystyle \frac{e_{2n-1}}{n+1}}.$$

Consider the closed subspaces ${\mathcal{W}}_1=\overline{span\{t_n:n\in \mathbb{N}\}}$ and ${\mathcal{W}}_2=\overline{span\{s_n:n\in \mathbb{N}\}}.$

It is evident that every basis vector $e_n$ belongs to ${\mathcal{W}}_2+{\mathcal{W}}_1$, so the closure of ${\mathcal{W}}_2+{\mathcal{W}}_1$ is the entire space $\mathcal{H}$. However, one can show that the vector $z={\sum }_{n\in \mathbb{N}}{\displaystyle \frac{e_{2n-1}}{n+1}}$ does not belong to ${\mathcal{W}}_2+{\mathcal{W}}_1$, and therefore ${\mathcal{W}}_2+{\mathcal{W}}_1\ne \mathcal{H}$. To see this, assume that we can express

$$z=\sum _{n\in \mathbb{N}}{\alpha }_n{s}_n+\sum _{n\in \mathbb{N}}{\beta }_n{t}_n^{\prime },$$

where ${\alpha }_n,{\beta }_n\in \mathbb{C}$ and $t_n^{\prime }={\displaystyle \frac{(n+1)t_n}{\sqrt{1+{(n+1)}^2}}}.$ A careful examination shows that we must have ${\beta }_n=\sqrt{1+{\displaystyle \frac{1}{{(n+1)}^2}}},$ and consequently ${\alpha }_n=-1$ for every $n\in \mathbb{N}$. However, this is a contradiction since both $\left\{{\alpha }_n\right\}$ and $\left\{{\beta }_n\right\}$ must belong to ${\mathsf{\ell }}^2$, implying that

$$z\notin {\mathcal{W}}_2+{\mathcal{W}}_1,$$

and hence the sum is not closed.

Moreover, it is straightforward to verify that ${\mathcal{W}}_2\cap {\mathcal{W}}_1=\left\{0\right\}$. Indeed, if $u\in {\mathcal{W}}_2\cap {\mathcal{W}}_1$, then u can be written both as $u={\sum }_n{\alpha }_n{s}_n={\sum }_n{\beta }_n{t}_n.$ Since the decomposition is unique, we deduce that ${\alpha }_n={\beta }_n$ for all n. This forces

$$\sum _n{\alpha }_n{\displaystyle \frac{e_{2n-1}}{n+1}}=0,$$

and hence all ${\alpha }_n$ must vanish, so that $u=0$.

Finally, note that $P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\left(s_1\right)=s_1\in {\mathcal{W}}_1$. This fact, along with standard arguments on the interplay between projections, shows that $\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel =1$ or equivalently $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$.

3.

Let $\mathcal{H}$ be a Hilbert space with an orthonormal basis ${\left\{e_n\right\}}_{n\in \mathbb{N}}$. Consider the closed subspaces

$${\mathcal{W}}_1=\overline{span\{e_1,t_n:n\in \mathbb{N}\}},{\mathcal{W}}_2=\overline{span\{e_1,s_n:n\in \mathbb{N}\}}.$$

Using a similar argument to the one in the previous example, we can conclude that ${\mathcal{W}}_2+{\mathcal{W}}_1$ is not closed. Moreover, it is trivial that $e_1\in {\mathcal{W}}_2\cap {\mathcal{W}}_1$, which implies that the intersection of these subspaces is nontrivial. In particular, we conclude that $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2.$

Remark 2.

It is worth mentioning that Example 1 (2) was previously presented in [12] as a counterexample to a result discussed in that paper (see Proposition 2.6).

We are now in a position to characterize precisely when the norm of the sum of two orthogonal projections attains its maximal value.

Theorem 2.

Let $P_{{\mathcal{W}}_1},P_{{\mathcal{W}}_2}$ be orthogonal projections on $\mathcal{H}$. Then, the following assertions are equivalent:

1.

$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$;

2.

There exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_2}{x}_n\parallel =1.$$

Proof. 

Assume that $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$. By the definition of the operator norm, there exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel =2.$$

Since the sequences $\{\parallel P_{{\mathcal{W}}_1}{x}_n\parallel \}$ and $\{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel \}$ are bounded in $\mathbb{R}$, we may extract subsequences—still denoted by $\left\{x_n\right\}$—such that both norms converge. Now observe the following:

$$2=\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel \le \underset{n\to \infty }{lim}\left(\parallel P_{{\mathcal{W}}_1}{x}_n\parallel +\parallel P_{{\mathcal{W}}_2}{x}_n\parallel \right)\le \parallel P_{{\mathcal{W}}_1}\parallel +\parallel P_{{\mathcal{W}}_2}\parallel =2.$$

Since the first and last expressions are equal, it follows that the inequalities in between must actually hold as equalities. Therefore, we conclude that

$$\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_2}{x}_n\parallel =1,$$

and

$$\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel =2.$$

Now for the converse direction. Suppose there exists a sequence $\left\{x_n\right\}\subset \mathcal{H}$, with $\parallel x_n\parallel =1$ for all n, such that

$$\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_2}{x}_n\parallel =1.$$

For each n, we have

$$\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel \ge 〈(P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n,x_n〉=\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2+{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2.$$

Taking limits as $n\to \infty$ gives

$$\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel \ge \underset{n\to \infty }{lim}\left(\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2+{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2\right)=2,$$

Therefore

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel \ge \underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel =2.$$

Since it is well known that $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel \le 2,$ and we have exhibited unit vectors $x_n$ for which

$$\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel \to 2,$$

it follows that

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2,$$

as claimed. □

As a consequence of Theorem 2, we obtain the following result, which provides new characterizations for when the norm of the sum of two orthogonal projections attains its maximum value.

Corollary 2.

Let $P_{{\mathcal{W}}_1},P_{{\mathcal{W}}_2}$ be orthogonal projections on $\mathcal{H}$. Then, the following conditions are equivalent.

1.

$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2.$

2.

There exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}\Re \langle P_{{\mathcal{W}}_1}{x}_n,P_{{\mathcal{W}}_2}{x}_n\rangle =1$$

where $\Re \left(z\right)$ denotes the real part of $z\in \mathbb{C}$.

3.

$\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel =2$

4.

$\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel +\parallel P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel =2.$

Proof. 

Assume that (1) holds. By Theorem 2, there exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_2}{x}_n\parallel =1,$$

and

$$\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel =2.$$

Therefore,

$$\begin{array}{cc}\hfill 4& =\underset{n\to \infty }{lim}{\parallel (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\parallel }^2\hfill \\ \hfill & =\underset{n\to \infty }{lim}\langle (P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n,(P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2})x_n\rangle \hfill \\ \hfill & =\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2+{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2+\langle (P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1})x_n,x_n\rangle \hfill \\ \hfill & =2+2\underset{n\to \infty }{lim}\Re \langle P_{{\mathcal{W}}_1}{x}_n,P_{{\mathcal{W}}_2}{x}_n\rangle ,\hfill \end{array}$$

which implies that

$$\underset{n\to \infty }{lim}\Re \langle P_{{\mathcal{W}}_1}{x}_n,P_{{\mathcal{W}}_2}{x}_n\rangle =1.$$

If (2) holds, and noting that the operator $P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}$ is self-adjoint, we obtain

$$\begin{array}{cc}\hfill 2=\underset{n\to \infty }{lim}2\Re \langle P_{{\mathcal{W}}_1}{x}_n,P_{{\mathcal{W}}_2}{x}_n\rangle & =\underset{n\to \infty }{lim}\langle (P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1})x_n,x_n\rangle \hfill \\ \hfill & \le \underset{n\to \infty }{lim}\left|\langle (P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1})x_n,x_n\rangle \right|\hfill \\ \hfill & \le \underset{\parallel x\parallel =1}{sup}\left|\langle (P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1})x,x\rangle \right|\hfill \\ \hfill & =\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel \le 2,\hfill \end{array}$$

and hence,

$$\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel =2.$$

Suppose that the norm of the anticommutator $P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}$ is equal to 2, then we have

$$2=\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}+P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel \le \parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel +\parallel P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel \le 2,$$

which implies

$$\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel +\parallel P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel =2.$$

Finally, if (4) holds, and since $\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel =\parallel P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel$, it follows that $\parallel P_{{\mathcal{W}}_2}{P}_{{\mathcal{W}}_1}\parallel =1$. Then, by (1), we conclude that

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =1+\parallel P_{{\mathcal{W}}_1}{P}_{{\mathcal{W}}_2}\parallel =2.$$

□

In order to obtain a new characterization of when the sum of two orthogonal projections is maximal in norm, we begin by recalling the following lemma. We use the notion of Birkhoff–James orthogonality: for elements $u,v$ in a normed linear space $\mathcal{X}$, we say that u is orthogonal to v in the sense of Birkhoff–James, denoted $u{\perp }_{B}v$, if

$$\parallel u\parallel \le \parallel u+\lambda v\parallel \mathrm{for}\mathrm{all}\lambda \in \mathbb{C}.$$

Lemma 4

(Proposition 4.1 in [24]). Let $\mathcal{X}$ be a normed linear space. For $u_1,u_2\in \mathcal{X}$, the following statements are equivalent:

1.

$\parallel u_1+u_2\parallel =\parallel u_1\parallel +\parallel u_2\parallel$.

2.

$u_1{\perp }_B\left(\parallel u_2\parallel u_1-\parallel u_1\parallel u_2\right)$.

3.

$u_2{\perp }_B\left(\parallel u_2\parallel u_1-\parallel u_1\parallel u_2\right)$.

Theorem 3.

Let $P_{{\mathcal{W}}_1},P_{{\mathcal{W}}_2}$ be orthogonal projections on $\mathcal{H}$. Then $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$ if and only if there exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}(P_{{\mathcal{W}}_1}-I)x_n=0,\underset{n\to \infty }{lim}\left(\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2-{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2\right)=0.$$

Proof. 

Assume that $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$. Then, by the triangle inequality, we have

$$2=\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel \le \parallel P_{{\mathcal{W}}_1}\parallel +\parallel P_{{\mathcal{W}}_2}\parallel \le 2,$$

which implies in particular that

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =\parallel P_{{\mathcal{W}}_1}\parallel +\parallel P_{{\mathcal{W}}_2}\parallel .$$

Using Lemma 4, we conclude that

$$P_{{\mathcal{W}}_1}{\perp }_B{P}_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2}.$$

Since $P_{{\mathcal{W}}_1}$ is a positive operator, by the characterization of orthogonality in $\mathcal{B}\left(\mathcal{H}\right)$ given in [25], Lemma 1, it follows that there exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}(P_{{\mathcal{W}}_1}-I)x_n=0\mathrm{and}\underset{n\to \infty }{lim}\langle (P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2})x_n,x_n\rangle =0.$$

Note that, since orthogonal projections are idempotent, we have

$$\langle (P_{{\mathcal{W}}_1}-P_{{\mathcal{W}}_2})x_n,x_n\rangle =\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2-{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2,$$

and therefore we can conclude that the sequence $\left\{x_n\right\}\subset \mathcal{H}$ satisfies

$$\underset{n\to \infty }{lim}(P_{{\mathcal{W}}_1}-I)x_n=0\mathrm{and}\underset{n\to \infty }{lim}\left(\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2-{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2\right)=0.$$

(3)

Conversely, if there exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that (3) holds, then applying again [25], Lemma 1 together with Lemma 4, we conclude that

$$\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =\parallel P_{{\mathcal{W}}_1}\parallel +\parallel P_{{\mathcal{W}}_2}\parallel =2.$$

□

Remark 3.

It is clear that the conditions obtained in Theorems 2 and 3 are equivalent to each other via the equality $\parallel P_{{\mathcal{W}}_1}+P_{{\mathcal{W}}_2}\parallel =2$. However, for the reader’s better understanding, we present the equivalence explicitly below.

• Let us consider the following statements:

1.

There exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$, $\parallel x_n\parallel =1$, such that

$$\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}-I)x_n\parallel =0\mathrm{and}\underset{n\to \infty }{lim}(\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2-\parallel P_{{\mathcal{W}}_2}{x}_n{\parallel }^2)=0.$$

2.

There exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =1\mathrm{and}\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_2}{x}_n\parallel =1.$$

 Suppose that (1) holds. Since $\parallel x_n\parallel =1$ and

$$\parallel (P_{{\mathcal{W}}_1}-I)x_n{\parallel }^2=\parallel x_n{\parallel }^2-\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2=1-{\parallel P_{{\mathcal{W}}_1}{x}_n\parallel }^2,$$

taking limits gives

$$\underset{n\to \infty }{lim}{\parallel P_{{\mathcal{W}}_1}{x}_n\parallel }^2=1,$$

hence $\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =1.$ Moreover,

$$\parallel P_{{\mathcal{W}}_2}{x}_n{\parallel }^2={\parallel P_{{\mathcal{W}}_1}{x}_n\parallel }^2+\left(\parallel P_{{\mathcal{W}}_2}{x}_n{\parallel }^2-{\parallel P_{{\mathcal{W}}_1}{x}_n\parallel }^2\right),$$

so by passing to the limit,

$$\underset{n\to \infty }{lim}{\parallel P_{{\mathcal{W}}_2}{x}_n\parallel }^2=1,$$

and therefore $\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_2}{x}_n\parallel =1.$

Conversely, if $\underset{n\to \infty }{lim}\parallel P_{{\mathcal{W}}_1}{x}_n\parallel =1$, then we have

$$\parallel (P_{{\mathcal{W}}_1}-I)x_n{\parallel }^2=1-{\parallel P_{{\mathcal{W}}_1}{x}_n\parallel }^2⟶0,$$

so $\underset{n\to \infty }{lim}\parallel (P_{{\mathcal{W}}_1}-I)x_n\parallel =0$. And since both $\parallel P_{{\mathcal{W}}_1}{x}_n\parallel \to 1$ and $\parallel P_{{\mathcal{W}}_2}{x}_n\parallel \to 1$, it follows that

$$\underset{n\to \infty }{lim}(\parallel P_{{\mathcal{W}}_1}{x}_n{\parallel }^2-\parallel P_{{\mathcal{W}}_2}{x}_n{\parallel }^2)=0.$$

For the sequel, we aim to investigate the case when T is a closed-range operator.

Let $T\in CR\left(\mathcal{H}\right)$ with $T\ne 0$. Since $T{T}^{†}$ and $T^{†}T$ are orthogonal projections, it follows from (1) that

$$1\le \parallel T{T}^{†}+T^{†}T\parallel =1+\parallel T{T}^{†}{T}^{†}T\parallel \le \parallel T{T}^{†}\parallel +\parallel T^{†}T\parallel =2.$$

(4)

In this section, we explore conditions on the operator T to ensure that

$$\parallel T{T}^{†}+T^{†}T\parallel =2.$$

(5)

A direct characterization of this equality, derived from (4) (or Proposition 3), is as follows:

$$\parallel T{T}^{†}+T^{†}T\parallel =2\iff \parallel T{T}^{†}{T}^{†}T\parallel =\parallel T^{†}TT{T}^{†}\parallel =1,$$

indicating that the product of the orthogonal projections $T{T}^{†}$ and $T^{†}T$ has norm 1.

An intuitive assumption to explore is the containment relationship between the ranges of the projections, specifically

$$Ran\left(T\right)\subseteq Ran\left(T^{*}\right)\mathrm{or}Ran\left(T^{*}\right)\subseteq Ran\left(T\right).$$

Since these subspaces are closed, we express this condition in terms of their kernels.

Theorem 4.

Let $T\in CR\left(\mathcal{H}\right)$ with $T\ne 0$ satisfy

$$Ker\left(T\right)\subseteq Ker\left(T^{*}\right)\mathrm{or}Ker\left(T^{*}\right)\subseteq Ker\left(T\right).$$

Then,

$$\parallel T{T}^{†}+T^{†}T\parallel =2.$$

Proof. 

Assume $Ker\left(T\right)\subseteq Ker\left(T^{*}\right)$ and suppose, for contradiction, that $Ran\left(T\right)\cap Ran\left(T^{*}\right)=\left\{0\right\}.$ Since T has closed range, it follows that

$$Ran\left(T\right)=Ker{\left(T^{*}\right)}^{\perp }\subseteq Ker{\left(T\right)}^{\perp }=Ran\left(T^{*}\right).$$

Thus, we obtain $Ran\left(T\right)=Ran\left(T\right)\cap Ran\left(T^{*}\right)=\left\{0\right\},$ which contradicts the assumption that $T\ne 0$. Therefore, $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}.$ Applying Theorem 1, we conclude that $\parallel T{T}^{†}+T^{†}T\parallel =2.$ □

The inclusion conditions on the kernels stated in the previous theorem can be expressed as a well-known consequence of Douglas’ theorem on range inclusions (see [17]). Specifically, given $T\in CR\left(\mathcal{H}\right)$, the following equivalences hold:

$$Ker\left(T\right)\subseteq Ker\left(T^{*}\right)\iff \exists S_1\in \mathcal{B}\left(\mathcal{H}\right):T^{*}{S}_1=T,$$

or

$$Ker\left(T^{*}\right)\subseteq Ker\left(T\right)\iff \exists S_2\in \mathcal{B}\left(\mathcal{H}\right):T{S}_2=T^{*}.$$

For operators with closed range that satisfy the equality $Ker\left(T\right)=Ker\left(T^{*}\right)$, the following result follows directly from the preceding theorem.

Corollary 3.

Let $T\in CR\left(\mathcal{H}\right)$ with $T\ne 0$ such that $Ker\left(T\right)=Ker\left(T^{*}\right)$ Then, $\parallel T{T}^{†}+T^{†}T\parallel =2$.

Remark 4.

We can provide an ad hoc proof for closed-range operators T whose kernel coincides with that of their adjoint. More precisely, if T satisfies these conditions, Brock showed in [19] that the equality $Ker\left(T\right)=Ker\left(T^{*}\right)$ is equivalent to the commutativity of T and its Moore–Penrose inverse, namely, $T{T}^{†}=T^{†}T.$ Consequently,

$$\parallel T{T}^{†}+T^{†}T\parallel =2\parallel T{T}^{†}\parallel =2,$$

since $T{T}^{†}$ is an orthogonal projection.

The closed-range operators that satisfy the commutativity condition mentioned above have been extensively studied and are referred to as EP operators (see [10]). More precisely, given $T\in CR\left(\mathcal{H}\right)$, we say that T is an EP operator if and only if its range coincides with that of its adjoint, i.e.,

$$Ran\left(T\right)=Ran\left(T^{*}\right)\iff T{T}^{†}=T^{†}T.$$

To conclude this remark, we mention that in [19], a characterization of the equality of the kernels of T and $T^{*}$, where T is a closed-range operator, was obtained. This result is closely related to the previously discussed theorem of Douglas concerning the inclusion of ranges. Specifically, it is shown that

$$\begin{array}{ccc}\hfill Ker\left(T\right)=Ker\left(T^{*}\right)& \iff & \exists S_1\in \mathcal{B}\left(\mathcal{H}\right),bijective:T^{*}{S}_1=T\hfill \\ \hfill & \iff & \exists S_2\in \mathcal{B}\left(\mathcal{H}\right),bijective:T{S}_2=T^{*}\hfill \end{array}$$

In conclusion, we give the following statement for EP operators.

Proposition 1.

Let T be an EP operator with $T\ne 0$. Then,

$$\parallel T{T}^{†}+T^{†}T\parallel =2.$$

To support the preceding discussion, we now present an example of an infinite-dimensional operator in the EP class and verify the assertions stated above.

Example 2.

Let $\mathcal{H}={\ell }^2\left(\mathbb{N}\right)$, and consider the operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ defined by

$$T(x_1,x_2,x_3,\dots )=(x_1,0,0,\dots ).$$

This operator can be represented by the infinite matrix

$$T=\left[\begin{array}{ccccc}1& 0& 0& 0& \dots \\ 0& 0& 0& 0& \dots \\ 0& 0& 0& 0& \dots \\ ⋮& ⋮& ⋮& ⋮& \ddots \end{array}\right].$$

In this case, the operator T has range $Ran\left(T\right)=\mathrm{span}\left\{e_1\right\}$, which is closed, implying that T belongs to the class $CR\left(\mathcal{H}\right)$. Moreover, T is a self-adjoint and idempotent operator, i.e., an orthogonal projection. As a consequence, its Moore–Penrose inverse coincides with itself, $T^{†}=T$, and thus it commutes with its Moore–Penrose inverse, satisfying $[T,T^{†}]=0$. Therefore, T is an EP operator. However, T is not invertible since $Ker\left(T\right)$ is infinite-dimensional.

Let us now compute the norm $\parallel T{T}^{†}+T^{†}T\parallel$. Since $T=T^{†}$, we have the following:

$$T{T}^{†}+T^{†}T=2T,$$

and thus,

$$\parallel T{T}^{†}+T^{†}T\parallel =\parallel 2T\parallel =2\parallel T\parallel =2.$$

We obtain the following corollary for normal operators.

Corollary 4.

Let $T\in CR\left(\mathcal{H}\right)$ be a normal operator with $T\ne 0$. Then, we obtain $\parallel T{T}^{†}+T^{†}T\parallel =2$.

Proof. 

Since T is normal, it satisfies $T{T}^{*}=T^{*}T$, which implies that $Ker\left(T\right)=Ker\left(T^{*}\right)$. Consequently, by Corollary 3, we conclude that $\parallel T{T}^{†}+T^{†}T\parallel =2.$ □

Recently, in [18], new bounds for the numerical radius were obtained using the Moore–Penrose inverse, refining several previously known estimates. More precisely, if $T\in CR\left(\mathcal{H}\right)$, then we have

$$w^2\left(T\right)\le {\displaystyle \frac{1}{4}}\parallel T{T}^{*}+T^{*}T\parallel ·\parallel T{T}^{†}+T^{†}T\parallel .$$

(6)

This inequality represents a refinement, within the class of closed-range operators, of the classical bound established by Kittaneh in [26],

$${\displaystyle \frac{1}{4}}\parallel T^{*}T+T{T}^{*}\parallel \le w^2\left(T\right)\le {\displaystyle \frac{1}{2}}\parallel T^{*}T+T{T}^{*}\parallel ,$$

which holds for every $T\in \mathcal{B}\left(\mathcal{H}\right)$.

In particular, in [18, Theorem 2.4], the authors established necessary and sufficient conditions for the validity of (5) in the finite-dimensional setting.

Lemma 5

(Proposition 2.1 in [18]). Let $T\in \mathcal{B}\left(\mathcal{H}\right)$, with $\mathcal{H}$ finite-dimensional and $T\ne 0$. Then, $\parallel T{T}^{†}+T^{†}T\parallel =2,$ if and only $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}$.

It is worth mentioning that in [18], the authors provide two different proofs of the preceding result. The first proof uses the fact that $\parallel PQ\parallel = \parallel \left(T^{†}T\right)\left(T{T}^{†}\right)\parallel =1$, while the second is based on the observation that both P and Q have 1 as an eigenvalue and share a common eigenvector. However, the result can also be deduced from Corollary 1 by considering $P_{{\mathcal{W}}_1}=T{T}^{†}$ and $P_{{\mathcal{W}}_2}=T^{†}T$.

In the following result, we establish that if a closed-range operator T attains Kittaneh’s upper bound, then it also satisfies (5).

Theorem 5.

Let $T\in CR\left(\mathcal{H}\right)$, with $T\ne 0$ being such that

$$w^2\left(T\right)={\displaystyle \frac{1}{2}}\parallel T^{*}T+T{T}^{*}\parallel .$$

Then, $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}$ or $Ran\left(T\right)+Ran\left(T^{*}\right)$ is not closed.

Proof. 

Suppose, for the sake of contradiction, that $Ran\left(T\right)\cap Ran\left(T^{*}\right)=\left\{0\right\}$ and $Ran\left(T\right)+Ran\left(T^{*}\right)$ is closed. Equivalently, by Theorem 1, we deduce that $\parallel T{T}^{†}+T^{†}T\parallel <2.$ Then, by inequality (6), it follows that

$$\begin{array}{ccc}\hfill {\displaystyle \frac{1}{2}}\parallel T^{*}T+T{T}^{*}\parallel =w^2\left(T\right)& \le & {\displaystyle \frac{1}{4}}\parallel T{T}^{*}+T^{*}T\parallel \parallel T{T}^{†}+T^{†}T\parallel \hfill \\ & <& {\displaystyle \frac{1}{2}}\parallel T^{*}T+T{T}^{*}\parallel .\hfill \end{array}$$

This is a contradiction, and hence we conclude that $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}$ or $Ran\left(T\right)+Ran\left(T^{*}\right)$ is not closed. □

As an immediate consequence of the previous result, the following holds.

Corollary 5.

Let $T\in CR\left(\mathcal{H}\right)$, $T\ne 0$, and T normaloid. Then, $\parallel T{T}^{†}+T^{†}T\parallel =2.$

Proof. 

The fact that T is normaloid implies that $\omega \left(T\right)=\parallel T\parallel$. Then,

$${\parallel T\parallel }^2={\omega }^2\left(T\right)\le {\displaystyle \frac{1}{2}}\parallel T{T}^{*}+T^{*}T\parallel \le {\displaystyle \frac{1}{2}}\left(\parallel T{T}^{*}\parallel +\parallel T^{*}T\parallel \right)={\parallel T\parallel }^2,$$

which means that ${\omega }^2\left(T\right)={\displaystyle \frac{1}{2}}\parallel T{T}^{*}+T^{*}T\parallel .$ Thus, by Theorem 5, we conclude that $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}$ or $Ran\left(T\right)+Ran\left(T^{*}\right)$ is not closed. By Theorem 1, we conclude that $\parallel T{T}^{†}+T^{†}T\parallel =2.$ □

In the following statement, we see that with the additional condition that the Hilbert space is finite-dimensional, the previous Corollary is more precise regarding its conclusion.

Proposition 2.

Let $T\in \mathcal{B}\left(\mathcal{H}\right)$, where $\mathcal{H}$ is finite-dimensional, with T being normaloid and $T\ne 0$. Then, we conclude that $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\},$ and, in particular, $\parallel T{T}^{†}+T^{†}T\parallel =2.$

Proof. 

Since $\mathcal{H}$ is finite-dimensional, it follows that $T\in CR\left(\mathcal{H}\right)$, and therefore, it satisfies the hypotheses of Corollary 5. Consequently, we can assert that either $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}$ or $Ran\left(T\right)+Ran\left(T^{*}\right)$ is not closed. However, due to finite dimensionality, we know that $Ran\left(T\right)+Ran\left(T^{*}\right)$ is a closed set in $\mathcal{H}$, which implies that necessarily $Ran\left(T\right)\cap Ran\left(T^{*}\right)\ne \left\{0\right\}$. Finally, as a consequence of Lemma 5, the nontrivial intersection of the ranges is equivalent to the operator $T{T}^{†}+T^{†}T$ achieving its maximum norm. □

Observe that Corollary 4 follows directly from Corollary 5, since every normal operator is, specifically, normaloid. Accordingly, if $T\in CR\left(\mathcal{H}\right)$, $T\ne 0$, and T belongs to any of the following subclasses—namely, self-adjoint, normal, or normaloid operators (see (2))—then we have

$$\parallel T{T}^{†}+T^{†}T\parallel =2.$$

A partir de lo anterior, it is natural to wonder whether the fact that a nonzero operator $T\in CR\left(\mathcal{H}\right)$ satisfies the equality $\parallel T{T}^{†}+T^{†}T\parallel =2$ implies that T is normaloid. The following counterexample demonstrates that this is not necessarily the case.

On the other hand, it is well known that every normaloid operator is also spectraloid, i.e., $r\left(T\right)=\omega \left(T\right)$, which naturally raises the question of whether this new class of operators satisfies this equality. We provide two examples: the first shows that there exist spectraloid operators that are not normaloid yet satisfy the norm equality studied in this manuscript, while the second gives a negative answer to the previous question.

Example 3.

Motivated by the manuscript of Goldberg and Zwas [27], we consider, for each $n\ge 3$, the following block matrix

$$A_n=I_{n-2}\oplus \left[\begin{array}{cc}0& 0\\ 2& 0\end{array}\right],$$

where $I_k$ denotes the $k\times k$ identity matrix.

In [27], the authors showed that $A_n$ is a spectraloid matrix but not normaloid (that is, the inclusion between these two subclasses is strict). Moreover, for each $n\ge 3$, it holds that

$$A_n^{†}=I_{n-2}\oplus \left[\begin{array}{cc}0& {\displaystyle \frac{1}{2}}\\ 0& 0\end{array}\right].$$

By performing block-wise multiplication, we obtain

$$A_n{A}_n^{†}+A_n^{†}{A}_n=2I_{n-2}\oplus \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right].$$

Therefore,

$$∥A_n{A}_n^{†}+A_n^{†}{A}_n∥=∥2I_{n-2}\oplus \left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]∥=2,$$

for any $n\ge 3$.

Another potential counterexample, based on Proposition 1, is to exhibit a nonzero operator that is EP but not normaloid.

Example 4.

Consider the matrix

$$T=\left[\begin{array}{cc}1& 2\\ 0& 1\end{array}\right].$$

It is immediate to see that $r\left(T\right)=1$ and that if we take $x_0={\displaystyle \frac{1}{\sqrt{2}}}{(1,1)}^t$, then

$$|\langle T{x}_0,x_0\rangle |=2,$$

hence

$$\omega \left(T\right)\ge 2>1=r\left(T\right),$$

which shows that T is not spectraloid. Furthermore, it can be readily verified that

$$\parallel T\parallel =1+\sqrt{2}>2=\omega \left(T\right),$$

thus concluding that T is not normaloid.

Since T is invertible, its Moore–Penrose pseudoinverse coincides with its usual inverse. In particular, we have $T{T}^{†}=T^{†}T,$ which shows that T is an EP operator, and

$$\parallel T^{†}T+T{T}^{†}\parallel =2.$$

3. Conclusions

In this paper, we investigated the operator norm of the sum of two orthogonal projections on a complex Hilbert space, identifying necessary and sufficient conditions for this norm to attain its maximum value, namely 2. Our analysis was guided by geometric considerations, specifically the Dixmier angle between closed subspaces, and we expressed our conditions directly in terms of the ranges of the involved projections.

As a central application, we extended these results to bounded linear operators with closed range through their Moore–Penrose inverses. In particular, we derived sufficient conditions under which the operator

$$T{T}^{†}+T^{†}T$$

has norm equal to 2. This result generalizes previous findings from the finite-dimensional setting to the broader context of general (possibly infinite-dimensional) Hilbert spaces.

Beyond these core contributions, our approach highlights deep connections between subspace geometry, projection theory, and the theory of generalized inverses. The interplay between these areas opens several potential avenues for future research. For instance, the further exploration of norm-attaining properties for sums or compositions of more than two projections, or for other classes of generalized inverses, could yield new insights.

We believe that the techniques and results developed in this work may serve as a foundation for a broader line of research in operator theory and Hilbert space geometry. In particular, this study may be a starting point for new investigations into norm inequalities, operator angles, and their role in solving structural problems related to projections and closed-range operators.