New Results on Idempotent Operators in Hilbert Spaces

Abstract

This paper provides a new proof of the operator norm identity $\parallel Q\parallel = \parallel I-Q\parallel$, where Q is a bounded idempotent operator on a complex Hilbert space, and I is the identity operator. We also derive explicit lower and upper bounds for the distance from an arbitrary idempotent operator to the set of orthogonal projections. Our approach simplifies existing proofs.

1. Introduction and Notations

The study of idempotent operators and orthogonal projections on Hilbert spaces is a fundamental part of operator theory, with applications in numerical analysis, $C^{*}$-algebra theory, quantum mechanics, and approximation theory [1,2,3,4]. A bounded linear operator Q on a complex Hilbert space $\mathcal{H}$ is said to be idempotent if it satisfies $Q^2=Q$. Such operators are fundamental in decomposition theorems, spectral analysis, and the study of operator algebras [5,6,7,8]. Notably, idempotence does not imply boundedness; in every infinite-dimensional Hilbert space, unbounded idempotent operators exist [7].

A bounded linear operator P on $\mathcal{H}$ is called an orthogonal projection if it is both idempotent ($P^2=P$) and self-adjoint ($P=P^{*}$). Orthogonal projections play an important role in functional analysis, and their properties have been widely studied in [9,10,11,12,13]. Although every orthogonal projection is idempotent, the reverse is not always true [7,14]. Many researchers have focused on the norm behavior of idempotent operators and their distance from orthogonal projections, as discussed in [15,16,17,18,19].

A fundamental result for bounded idempotent operators is the following norm equality. Let Q be a bounded, idempotent, linear operator, and let I denote the identity operator, both defined on a complex Hilbert space $\mathcal{H}$. Then,

$$\parallel Q\parallel = \parallel I-Q\parallel ,$$

(1)

except in the trivial cases where $Q=0$ or $Q\left(\mathcal{H}\right)=\mathcal{H}$. Here, $\parallel ·\parallel$ denotes the operator norm on the space of bounded linear operators defined on $\mathcal{H}$.

The identity (1), often mentioned in standard textbooks, has been carefully proven by several authors using different methods. For example, Ando [15], Kato [3], Koliha and Rakočević [14], Szyld [17], and Andreev [16] gave elegant proofs based on spectral theory, operator inequalities, and geometric ideas in Hilbert spaces. More norm inequalities related to idempotents and projections are discussed in [20,21,22,23] and the references cited therein.

When Q is an orthogonal projection, the equality reduces to $\parallel Q\parallel = \parallel I-Q\parallel = 1$, due to its spectral properties, since the eigenvalues of Q and $I-Q$ lie in $\{0,1\}$. In contrast, for non-self-adjoint idempotent operators, the norms of Q and $I-Q$ require more detailed analysis, as they are not fully determined by spectral arguments [11,19,24].

Next, we introduce notations and preliminary results essential for proving our main results in this work. Let $(\mathcal{H},⟨·,·⟩)$ be a complex inner product space, where $\mathbb{K}$ denotes the field of complex numbers.As is customary, 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 $\parallel ·\parallel$. For any $T\in \mathcal{B}\left(\mathcal{H}\right)$, let $T^{*}$ denote its adjoint. An operator T is said to be self-adjoint if $T=T^{*}$. For any $T\in \mathcal{B}\left(\mathcal{H}\right)$, we denote by $\mathcal{R}\left(T\right)$ and $\mathcal{N}\left(T\right)$ the range and null space of T, respectively. The symbol I represents the identity operator.

An operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ is called positive if it is self-adjoint and satisfies $⟨Tx,x⟩\ge 0$ for all $x\in \mathcal{H}$. In this case, there exists a unique positive operator $S\in \mathcal{B}\left(\mathcal{H}\right)$ such that $S^2=T$; this operator S is called the square root of T and is denoted by $T^{\frac{1}{2}}$.

For any operator $T\in \mathcal{B}\left(\mathcal{H}\right)$, the spectrum of T, denoted $\sigma \left(T\right)$, is defined as

$$\sigma \left(T\right)=\{\lambda \in \mathbb{C}:T-\lambda I\mathrm{is}\mathrm{not}\mathrm{invertible}\},$$

and its spectral radius is

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

Another subset of the complex plane associated with T is its numerical range (or field of values),

$$W\left(T\right)=\{⟨Tx,x⟩:x\in \mathcal{H},\parallel x\parallel = 1\}.$$

It is well known that

$$r\left(T\right) \le \parallel T\parallel .$$

Moreover, if T is normal (i.e., $T^{*}T=T{T}^{*}$), then equality holds. More generally, an operator T is called normaloid if

$$r\left(T\right) = \parallel T\parallel .$$

Let $\mathrm{Gr}\left(\mathcal{H}\right)$ denote the Grassmannian manifold of $\mathcal{H}$, consisting of all closed subspaces $\mathcal{M}\subseteq \mathcal{H}$. For each $\mathcal{S}\in \mathrm{Gr}\left(\mathcal{H}\right)$, the orthogonal projection onto $\mathcal{S}$ is denoted by $P_{\mathcal{S}}$. To facilitate our discussion, we define two important subsets of $\mathcal{B}\left(\mathcal{H}\right)$: $\mathcal{OP}$, the set of all oblique projections, and $\mathcal{P}$, the set of orthogonal projections. Specifically, these sets are given by

$$\mathcal{P}=\{P\in \mathcal{B}\left(\mathcal{H}\right)\mid P^2=P=P^{*}\},$$

and

$$\mathcal{OP}=\{Q\in \mathcal{B}\left(\mathcal{H}\right)\mid Q^2=Q\}.$$

It is straightforward to see that the inclusion $\mathcal{P}⊊\mathcal{OP}$ is strict, since not every idempotent operator is self-adjoint. For example, the operator

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

is idempotent but not self-adjoint, showing that $\mathcal{P}⊊\mathcal{OP}$. Moreover, it is well known and will be used throughout the manuscript that, for any $Q\in \mathcal{OP}$, one has $\parallel Q\parallel \ge 1$ and in addition, $\parallel Q\parallel = 1$ if and only if $Q\in \mathcal{P}$ (see [1] Theorem 7.72).

It is well known that if $Q\in \mathcal{OP}$, then its adjoint $Q^{*}$, its complement $I-Q$, and the adjoint of its complement ${(I-Q)}^{*}$ all belong to $\mathcal{OP}$. By convention, elements of $\mathcal{P}$ are called orthogonal projections, while those of $\mathcal{OP}$ are called oblique projections. Moreover, for any $P_{{\mathcal{S}}_1},P_{{\mathcal{S}}_2}\in \mathcal{P}$, one has

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

This identity, due to Duncan and Taylor [22], admits an alternative proof by one of the authors in [20].

This manuscript has two main goals. First, we give a new proof of the norm equality (1) for general idempotent operators on Hilbert spaces, bringing together existing results in a unified way. Second, we provide explicit bounds for the distance between an idempotent operator and the set of orthogonal projections, extending recent developments [18,19] and offering new insights into the structure of idempotent operators.

2. Main Results

2.1. Norm Identity in $\mathcal{OP}$ and Its Consequences

In this subsection, we present our main results. Our first main result provides a new proof of the equality (1), which relies primarily on two intrinsic properties of orthogonal projections: idempotence and positivity. To establish this proof, we require the following well-known lemmas.

The first lemma, stated in [23] (Theorem 1.7 (b)), characterizes the norm of the sum of two operators whose ranges and adjoint ranges are mutually orthogonal.

Lemma 1.

Let $S,T\in \mathcal{B}\left(\mathcal{H}\right)$ such that the ranges of S and T, and the ranges of $S^{*}$ and $T^{*}$, are orthogonal to each other, respectively. Then

$$\parallel S+T\parallel = max\{\parallel S\parallel ,\parallel T\parallel \}.$$

The second lemma, see [5,6], asserts that the spectra of the products $Q_1{Q}_2$ and $(I-Q_1)(I-Q_2)$ coincide outside of the points 0 and 1, when $Q_1$ and $Q_2$ are orthogonal projections. This fact plays a crucial role in spectral comparisons arising in our analysis.

Lemma 2

([5] (Lemma 3) or [6] (Equation (2.5)). Let $Q_1,Q_2\in \mathcal{OP}$. Then

$$\sigma \left(Q_1{Q}_2\right)\setminus \{0,1\}=\sigma \left((I-Q_1)(I-Q_2)\right)\setminus \{0,1\}.$$

We are now in a position to present our first result in this work.

Theorem 1.

Let $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$. Then it follows that

$$\parallel Q\parallel = \parallel I-Q\parallel = \parallel Q+Q^{*}-I\parallel .$$

Proof. 

Let us begin by noting that the self-adjoint operator $Q+Q^{*}-I$ can be decomposed as follows:

$$Q+Q^{*}-I=Q-{(I-Q)}^{*}=S+T,$$

where $S=Q$ and $T=-{(I-Q)}^{*}$.

Then, from the fact that Q is an idempotent operator, it follows that

$$T^{*}S=-(I-Q)Q=-Q+Q^2=0,$$

and

$$T{S}^{*}=-{(I-Q)}^{*}{Q}^{*}=-Q^{*}+{\left(Q^{*}\right)}^2=0.$$

From the previous equalities, we can conclude that the ranges of S and T, as well as the ranges of $S^{*}$ and $T^{*}$, are orthogonal to each other, respectively. Then by Lemma 1, we conclude that

$$\begin{array}{cc}\hfill \parallel Q+Q^{*}-I\parallel & = \parallel S+T\parallel \hfill \\ \hfill & = max\{\parallel S\parallel ,\parallel T\parallel \}\hfill \\ \hfill & = max\{\parallel Q\parallel ,\parallel I-Q\parallel \}.\hfill \end{array}$$

Since $Q\in \mathcal{OP}$, it follows immediately that $\parallel Q\parallel \ge 1$, with equality holding only in the special case that $Q\in \mathcal{P}$. We now examine the case where $\parallel Q\parallel \ge 1$ and $\parallel I-Q\parallel \ge 1$

To prove the equality between the norms of Q and $I-Q$, we proceed by contradiction.

Assume, without loss of generality, that

$$\parallel I-Q\parallel \ge \parallel Q\parallel \ge 1.$$

Since $Q\in \mathcal{B}\left(\mathcal{H}\right)$, both Q and $I-Q$ are bounded linear operators, and their products with their adjoints are positive and self-adjoint. In particular, the operators

$$Q^{*}Q\mathrm{and}{(I-Q)}^{*}(I-Q)$$

are positive and normal, so their norms coincide with their spectral radii:

$${\parallel Q\parallel }^2= \parallel Q^{*}Q\parallel =r\left(Q^{*}Q\right){,\parallel I-Q\parallel }^2=\parallel {(I-Q)}^{*}(I-Q)\parallel =r\left({(I-Q)}^{*}(I-Q)\right).$$

Therefore, from our assumption,

$$r\left({(I-Q)}^{*}(I-Q)\right)={\parallel I-Q\parallel }^2>{\parallel Q\parallel }^2=r\left(Q^{*}Q\right)>1.$$

Now, consider the spectral properties of the operators $Q^{*}Q$ and ${(I-Q)}^{*}(I-Q)$. By Lemma 2, applied to the pair $Q_1=Q^{*}$, $Q_2=Q$, we know that the nonzero spectral values of $Q^{*}Q$ and ${(I-Q)}^{*}(I-Q)$, other than possibly 0 and 1, coincide:

$$\sigma \left(Q^{*}Q\right)\setminus \{0,1\}=\sigma \left({(I-Q)}^{*}(I-Q)\right)\setminus \{0,1\}.$$

But since the spectrum of a positive operator lies in $[0,\parallel T{\parallel }^2]$, the inequality

$$r\left({(I-Q)}^{*}(I-Q)\right)>r\left(Q^{*}Q\right)$$

would imply that a spectral value of ${(I-Q)}^{*}(I-Q)$, outside of $\{0,1\}$, exceeds all the spectral values of $Q^{*}Q$, contradicting the equality of spectra stated above.

Therefore, our assumption must be false. We conclude that

$$\parallel I-Q\parallel = \parallel Q\parallel .$$

The case $\parallel Q\parallel \ge \parallel I-Q\parallel$ is entirely analogous; it leads to the reversed inequality $r\left(Q^{*}Q\right)>r\left({(I-Q)}^{*}(I-Q)\right)$, and the same contradiction with the spectral correspondence given by Lemma 2 arises.

In conclusion, we have proven that

$$\parallel Q\parallel = \parallel I-Q\parallel = \parallel Q+Q^{*}-I\parallel .$$

This completes the proof. □

Remark 1.

We clarify why we must assume that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$ for the equality

$$\parallel I-Q\parallel = \parallel Q\parallel ,$$

to hold. If $\mathcal{N}\left(Q\right)=\mathcal{H}$, then $Q=0$, and hence $\parallel Q\parallel =0$ while $\parallel I-Q\parallel =1$. On the other hand, if $\mathcal{R}\left(Q\right)=\mathcal{H}$, then, since Q is idempotent, we have that $Q\left(h\right)=h$ for every $h\in \mathcal{H}$. Indeed, if $h\in \mathcal{R}\left(Q\right)=\mathcal{H}$, there exists $x\in \mathcal{H}$ such that $h=Q\left(x\right)$, and thus

$$Q\left(h\right)=Q^2\left(x\right)=Q\left(x\right)=h.$$

In this case, $\parallel Q\parallel =1$ and $\parallel I-Q\parallel =0$.

Before proceeding, let us illustrate with an example that the norm equality (1) is not a property exclusive to bounded idempotent operators.

Remark 2.

Let $\mathcal{H}={\ell }^2\left(\mathbb{N}\right)$ be the Hilbert space of square-summable sequences. We aim to construct a bounded linear operator P defined on $\mathcal{H}$ such that $P\notin \mathcal{OP}$ and satisfies $\parallel P\parallel = \parallel I-P\parallel$.

Define the operator P as follows:

$$P(x_1,x_2,x_3,\dots )=\left(x_1,{\displaystyle \frac{1}{2}}{x}_2,{\displaystyle \frac{1}{3}}{x}_3,\dots \right).$$

This operator is bounded and linear. However, it is not idempotent, since $P^2\ne P$. For example, consider the sequence $(0,1,0,\dots )$:

$$P^2(0,1,0,\dots )=P\left(0,{\displaystyle \frac{1}{2}},0,\dots \right)=\left(0,\frac{1}{4},0,\dots \right)\ne \left(0,{\displaystyle \frac{1}{2}},0,\dots \right)=P(0,1,0,\dots ).$$

Furthermore, it is trivial to see that $\mathcal{N}\left(P\right)=\left\{\right(0,0,0,\dots \left)\right\}$. Now, we will demonstrate that the range of P does not coincide with $\mathcal{H}$. Consider the sequence $y:={\left\{y_k\right\}}_{k=1}^{\infty }$ with $y_k:={\displaystyle \frac{1}{k}}$. Clearly, $y\in \mathcal{H}$. However, if $y\in \mathcal{R}\left(P\right)$, then $x_k=k{y}_k=1$ for all $k\ge 1$, which implies that

$$\sum _{k=1}^{\infty }{\left|x_k\right|}^2=\sum _{k=1}^{\infty }1=\infty .$$

Thus, $x\notin \mathcal{H}$, and $y={\left({\displaystyle \frac{1}{k}}\right)}_{k=1}^{\infty }$ does not belong to the range of P.

Finally, it is immediate to verify that

$$\parallel P\parallel =1= \parallel I-P\parallel .$$

Corollary 1.

Let $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$. Then it follows that

$$\parallel Q\parallel = \parallel I-Q\parallel = \parallel Q^{*}\parallel = \parallel I-Q^{*}\parallel .$$

Proof. 

Let us first verify that if $Q\in \mathcal{OP}$, then $Q^{*}\in \mathcal{OP}$ as well, and that it satisfies the hypotheses of Theorem 1.

Suppose, for contradiction, that $\mathcal{N}\left(Q^{*}\right)=\mathcal{H}$. Then $Q^{*}y=0$ for all $y\in \mathcal{H}$. By the definition of the adjoint, we have

$$⟨Qx,y⟩=⟨x,Q^{*}y⟩=0\mathrm{for}\mathrm{all}x,y\in \mathcal{H},$$

which implies that $Qx=0$ for all $x\in \mathcal{H}$, and hence $Q=0$, a contradiction.

On the other hand, if $\mathcal{R}\left(Q^{*}\right)=\mathcal{H}$, then, as noted in Remark 1, it follows that $Q^{*}=I$, and consequently $Q=I$, which contradicts the assumption that $\mathcal{R}\left(Q\right)\ne \mathcal{H}$.

We thus conclude that $Q^{*}\in \mathcal{OP}$, and in particular, neither $\mathcal{N}\left(Q^{*}\right)$ nor $\mathcal{R}\left(Q^{*}\right)$ coincides with $\mathcal{H}$. Therefore, by Theorem 1, we obtain

$$\parallel Q^{*}\parallel = \parallel I-Q^{*}\parallel = \parallel Q^{*}+{\left(Q^{*}\right)}^{*}-I\parallel = \parallel Q+Q^{*}-I\parallel = \parallel Q\parallel = \parallel I-Q\parallel ,$$

the desired conclusion follows. □

As a consequence of Lemma 1, we have the following norm equalitites.

Corollary 2.

Let $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$. Then it follows that

$$\parallel Q\parallel = \parallel I-Q\parallel = \parallel Q^{*}-Q+I\parallel .$$

Proof. 

Let us note that it is sufficient to prove the last equality. Let us consider $S=Q^{*}$ and $T=(I-Q)$, then

$$S+T=Q^{*}-Q+I,$$

where the ranges of S and T, as well as the ranges of $S^{*}$ and $T^{*}$, are orthogonal to each other, respectively. Thus, applying Lemma 1, we deduce that

$$\begin{array}{cc}\hfill \parallel Q^{*}-Q+I\parallel & = \parallel S+T\parallel \hfill \\ \hfill & = max\{\parallel S\parallel ,\parallel T\parallel \}\hfill \\ \hfill & = max\{\parallel Q\parallel ,\parallel I-Q\parallel \}.\hfill \end{array}$$

Hence, the proof is complete. □

Combining Theorem 1 and Corollary 2, we can assert that if $Q\in \mathcal{OP}$ is such that neither the null space $\mathcal{N}\left(Q\right)$ nor the range $\mathcal{R}\left(Q\right)$ coincides with the whole space $\mathcal{H}$, then the following identities hold:

$$\parallel Q\parallel = \parallel I-Q\parallel = \parallel I-(Q+Q^{*})\parallel = \parallel I-2\Re \left(Q\right)\parallel ,$$

where $\Re \left(Q\right)={\displaystyle \frac{Q+Q^{*}}{2}}$ denotes the real part of the operator $Q\in \mathcal{B}\left(\mathcal{H}\right)$.

Analogously, considering the imaginary part $\Im \left(Q\right)={\displaystyle \frac{Q-Q^{*}}{2i}}$, we also obtain:

$$\parallel I-2\Re \left(Q\right)\parallel = \parallel I-2i\Im \left(Q\right)\parallel = \parallel Q\parallel = \parallel I-Q\parallel .$$

(2)

Moreover, we may conclude that

$$\begin{array}{cc}\hfill 2\parallel Q\parallel = 2\parallel I-Q\parallel = \parallel 2I-2Q\parallel & = \parallel 2I-2(\Re (Q)+i\Im (Q\left)\right)\parallel \hfill \\ \hfill & \le \parallel I-2\Re \left(Q\right)\parallel +\parallel I-2i\Im \left(Q\right)\parallel \hfill \\ \hfill & = \parallel Q\parallel +\parallel Q\parallel =2\parallel Q\parallel ,\hfill \end{array}$$

i.e., for any $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$, it holds that

$$\begin{array}{c}\hfill \parallel 2I-2(\Re (Q)+i\Im (Q\left)\right)\parallel = \parallel I-2\Re \left(Q\right)\parallel +\parallel I-2i\Im \left(Q\right)\parallel .\end{array}$$

This means that the norm of the sum of the two operators $I-2\Re \left(Q\right)$ and $I-2i\Im \left(Q\right)$ is equal to the sum of their norms. In [24], the authors provide a characterization of when such an equality holds for arbitrary pairs of operators in $\mathcal{B}\left(\mathcal{H}\right)$, and we therefore recommend their work for further reading.

Let $\mathcal{R},\mathcal{N}\in Gr\left(\mathcal{H}\right)$, and let $P_{\mathcal{R}}$ and $P_{\mathcal{N}}$ denote the orthogonal projections onto these subspaces. In [7], Buckholtz established necessary and sufficient conditions to guarantee the existence of an operator $Q\in \mathcal{OP}$ with $\mathcal{R}\left(Q\right)=\mathcal{R}$ and $\mathcal{N}\left(Q\right)=\mathcal{N}$, respectively. Furthermore, in the same article, Buckholtz derived an explicit formula for $\parallel Q\parallel$, with $Q\ne 0$, in terms of the orthogonal projections $P_{\mathcal{R}}$ and $P_{\mathcal{N}}$. More precisely,

$$\parallel Q\parallel ={\displaystyle \frac{1}{\sqrt{1-\parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}{\parallel }^2}}}.$$

(3)

It is worth mentioning that the proof presented by Buckholtz of the identity (3) was not the first. For this reason, in [17], Szyld provides a historical overview of the various proofs for such equality, particularly in Sections 6 and 7.

We now present an identity for the norm of Q in terms of the norm of the sum and the anticommutator of the orthogonal projections $P_{\mathcal{R}\left(Q\right)}$ and $P_{\mathcal{N}\left(Q\right)}$.

Let us recall that if $S,T\in \mathcal{B}\left(\mathcal{H}\right)$, the anticommutator of S and T is defined as $ST+TS$. Recently, Walters, in [13], derived a formula for the norm of the anticommutator in the context of orthogonal projections. In particular, if ${\mathcal{S}}_1,{\mathcal{S}}_2\in Gr\left(\mathcal{H}\right)$, then the norm of the anticommutator $P_{{\mathcal{S}}_1}{P}_{{\mathcal{S}}_2}+P_{{\mathcal{S}}_2}{P}_{{\mathcal{S}}_1}$ is expressed as a simple quadratic function of $\parallel P_{{\mathcal{S}}_1}{P}_{{\mathcal{S}}_2}\parallel$, more precisely

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

(4)

Theorem 2.

Let $Q\in \mathcal{OP}$ with range $\mathcal{R}\left(Q\right)$ and kernel $\mathcal{N}\left(Q\right)$. Then, if $Q\ne 0$, we have

$${\displaystyle \parallel Q\parallel =\frac{1}{\sqrt{\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel - \parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}{P}_{\mathcal{R}\left(Q\right)}\parallel }}.}$$

Proof. 

Since, by assumption, we are excluding the trivial case $Q=0$, we may assert that the orthogonal projection onto the range of Q, namely $P_{\mathcal{R}\left(Q\right)}$, is nonzero. In this setting, we invoke the Duncan–Taylor inequality, which establishes a sharp norm identity for the sum of two orthogonal projections (see [22]). Specifically, applying this result to the pair $P_{\mathcal{R}\left(Q\right)}$ and $P_{\mathcal{N}\left(Q\right)}$, we obtain

$$\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel =1 + \parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}\parallel .$$

(5)

Now, by combining this expression with the identity given in (4), we can derive the following chain of equalities:

$$\begin{array}{cc}\hfill 1- \parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}{\parallel }^2& =\left(1+ \parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}\parallel \right)-\left(\parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}\parallel + \parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}{\parallel }^2\right)\hfill \\ \hfill & = \parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel - \parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}{P}_{\mathcal{R}\left(Q\right)}\parallel .\hfill \end{array}$$

In conclusion, this final identity, in conjunction with the previously established norm formula in (3), allows us to complete the proof directly and without further elaboration. □

Using the recently obtained identity for the norm of Q, we can derive the following characterization for Q to be an orthogonal projection.

Corollary 3.

Let $Q\in \mathcal{OP}$, with $Q\ne 0$. Then, the following statements are equivalent:

• 1.   $Q\in \mathcal{P}$, i.e., $Q=Q^{*}$.
• 2.   $\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel = \parallel P_{\mathcal{R}\left(Q\right)}-P_{\mathcal{N}\left(Q\right)}\parallel = 1.$
• 3.   $P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}=0$.

Proof. 

Suppose that $Q\in \mathcal{P}$, with $Q\ne 0$. In particular, this ensures that either $P_{\mathcal{R}\left(Q\right)}\ne 0$ or $P_{\mathcal{N}\left(Q\right)}\ne 0$. By (3), we have

$$\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel -\parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}{P}_{\mathcal{R}\left(Q\right)}\parallel =1=1-\parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}{\parallel }^2,$$

which leads to $\parallel P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}{\parallel }^2=0$. Consequently, using Duncan-Taylor’s identity for the norm of the sum of orthogonal projections, we deduce that $\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel =1$. Furthermore, by [21] (Proposition 2.9), it follows that

$$\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel = \parallel P_{\mathcal{R}\left(Q\right)}-P_{\mathcal{N}\left(Q\right)}\parallel .$$

Observe that if $\parallel P_{\mathcal{R}\left(Q\right)}+P_{\mathcal{N}\left(Q\right)}\parallel =1$, then by Duncan-Taylor’s formula, $P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}=0$.

Finally, when $P_{\mathcal{R}\left(Q\right)}{P}_{\mathcal{N}\left(Q\right)}=0$, we conclude that $Q\in \mathcal{OP}$ with $\parallel Q\parallel =1$, i.e., $Q\in \mathcal{P}$. □

In [4], the renowned Krein–Krasnoselskii–Milman equality (KKME) was derived. It establishes a precise relationship for the norm of the difference between two orthogonal projections in terms of their interactions with complementary subspaces. More precisely, if $P_{{\mathcal{S}}_1},P_{{\mathcal{S}}_2}\in \mathcal{P}$, then it holds

$$\parallel P_{{\mathcal{S}}_1}-P_{{\mathcal{S}}_2}\parallel =max\{\parallel P_{{\mathcal{S}}_1}(I-P_{{\mathcal{S}}_2})\parallel ,\parallel P_{{\mathcal{S}}_2}(I-P_{{\mathcal{S}}_1})\parallel \}.$$

We now present a generalization of KKME within the framework of oblique projections. Unlike the original KKME, which is restricted to orthogonal projections, this extended result captures analogous properties for oblique projections, thereby broadening its applicability to a wider class of operators.

Theorem 3.

Let $Q_1,Q_2\in \mathcal{OP}$. Then, we have

$$∥Q_2-Q_1^{*}-[Q_2{Q}_1-Q_2^{*}{Q}_1^{*}]∥=max\{\parallel Q_2(I-Q_1)\parallel ,\parallel Q_1(I-Q_2)\parallel \}.$$

Proof. 

Define $T=Q_2(I-Q_1)$ and $S=-(I-Q_2^{*})Q_1^{*}$. Notice that

$$S^{*}T=-Q_1(I-Q_2)Q_2(I-Q_1)=0,$$

and

$$S{T}^{*}=-(I-Q_2^{*})Q_1^{*}(I-Q_1^{*})Q_2^{*}=0.$$

Thus, the ranges of T and S, as well as those of $T^{*}$ and $S^{*}$, are pairwise orthogonal. By Lemma 1, we conclude that

$$\begin{array}{ccc}\hfill ∥Q_2-Q_1^{*}-[Q_2{Q}_1-Q_2^{*}{Q}_1^{*}]∥& =& \parallel Q_2(I-Q_1)-(I-Q_2^{*})Q_1^{*}\parallel \hfill \\ & =& max\{\parallel Q_2(I-Q_1)\parallel ,\parallel Q_1(I-Q_2)\parallel \}.\hfill \end{array}$$

So, the proof is finished. □

Corollary 4.

Let $Q_1,Q_2\in \mathcal{OP}$ such that $Q_2{Q}_1=Q_2^{*}{Q}_1^{*}$. Then, the following equality holds:

$$∥Q_2-Q_1^{*}∥=max\{\parallel Q_2(I-Q_1)\parallel ,\parallel Q_1(I-Q_2)\parallel \}.$$

Before proceeding, we will first demonstrate, through a simple example, that the result presented above is indeed a generalization of KKME. More specifically, we will prove that there exist operators $Q_1,Q_2\in \mathcal{OP}\setminus \mathcal{P}$ that satisfy the condition $Q_2{Q}_1=Q_2^{*}{Q}_1^{*}.$

Let

$$Q_1=\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right),Q_2=\left(\begin{array}{cc}0& -1\\ 0& 1\end{array}\right).$$

It is trivial to verify that $Q_1^2=Q_1$ and $Q_2^2=Q_2$. Furthermore, since neither $Q_1$ nor $Q_2$ is self-adjoint, we conclude that $Q_1,Q_2\in \mathcal{OP}\setminus \mathcal{P}$.

Now, we compute:

$$Q_2{Q}_1=\left(\begin{array}{cc}0& -1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right),$$

and

$$Q_2^{*}{Q}_1^{*}=\left(\begin{array}{cc}0& 0\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right).$$

In conclusion, the matrices $Q_1$ and $Q_2$ are oblique projections (but not orthogonal projections), and they satisfy the condition: $Q_2{Q}_1=Q_2^{*}{Q}_1^{*}.$

In particular, if $Q_1,Q_2\in \mathcal{P}$, the hypotheses of Corollary 4 are immediately satisfied, and thus, the KKME follows as a consequence.

Proposition 1.

Sean $Q_1,Q_2\in \mathcal{P}$, then it holds

$$\parallel Q_1-Q_2\parallel =max\{\parallel Q_1(I-Q_2)\parallel ,\parallel Q_2(I-Q_1)\parallel \}.$$

2.2. The Operator Distances from Oblique Projections to Orthogonal Projections

We now turn our attention we turn our attention to studying the distance from a given oblique projection Q to the set $\mathcal{P}$ of all orthogonal projections. We recall that Zhang, Tian, and Xu recently provided a complete characterization of the operator distances from an idempotent operator to the set of orthogonal projections, including the minimum, maximum, and intermediate values (see [19]). Before recalling this result, we must review the notion of generalized inverse for bounded linear operators.

Let $T,S\in \mathcal{B}\left(\mathcal{H}\right)$. An operator S is said to be a generalized inverse of T if it satisfies the conditions

$$TST=T\mathrm{and}STS=S.$$

Such operators arise naturally in the study of linear equations, approximation theory, and optimization, especially in the context of solving inconsistent or underdetermined systems. It is important to emphasize that the generalized inverse of a given operator is not unique in general.

A particularlysignificant instance of a generalized inverse is the Moore–Penrose inverse. The operator S is called the Moore–Penrose inverse of T if, in addition to the two identities above, it also satisfies the symmetry conditions

$${\left(TS\right)}^{*}=TS\mathrm{and}{\left(ST\right)}^{*}=ST.$$

When it exists, the Moore–Penrose inverse is unique and is denoted by $T^{†}$. It plays a central role in functional analysis and operator theory, as it generalizes the notion of the inverse to non-invertible or non-square operators in a canonical way.

It is well known (see [25]) that an operator T admits a Moore–Penrose inverse if and only if its range is closed. Moreover, if T is invertible, then the Moore–Penrose inverse coincides with the usual inverse, i.e., $T^{†}=T^{-1}$. For a comprehensive treatment of generalized inverses in Hilbert spaces, we refer the interested reader to [26].

More precisely, they obtained the following results.

Lemma 3

([19] Theorem 2.3 and 2.4). Let $Q\in \mathcal{OP}$. The distance $\parallel m\left(Q\right)-Q\parallel$ is the minimum distance from Q to all orthogonal projections on $\mathcal{H}$, where

$$m\left(Q\right)={\displaystyle \frac{1}{2}}\left(\left(|Q^{*}|+Q^{*}\right){\left|Q^{*}\right|}^{†}{\left(|Q^{*}|+I\right)}^{-1}\left(|Q^{*}|+Q\right)\right)\in \mathcal{P},$$

and $|Q^{*}|={\left(Q{Q}^{*}\right)}^{{\displaystyle \frac{1}{2}}}$. If $P_0\in \mathcal{P}$ is such that $\parallel P_0-Q\parallel$ is the minimum distance from Q to all projections on $\mathcal{H}$, then

$$\parallel I-P_0-Q\parallel = 1 + \parallel P_0-Q\parallel ,$$

(6)

which represents the maximum distance from Q to all projections on $\mathcal{H}$.

Remark 3.

If $Q\in \mathcal{P}$, then from the identities $Q=Q^{*}=|Q^{*}|=|Q^{*}{|}^{†}$ and the fact that Q commutes with $\left(\right|Q^{*}{|+I)}^{-1}$, together with the equality

$$\left(\right|Q^{*}{|+I)}^{-1}=\frac{1}{2}Q+I-Q,$$

it follows almost immediately that

$$m\left(Q\right)=2Q\left(\right|Q^{*}{|+I)}^{-1}=Q.$$

In the more general case where $Q\in \mathcal{OP}\setminus \mathcal{P}$, the explicit computation of the orthogonal projection $m\left(Q\right)$ closest to the oblique projection Q requires deeper insight. For a detailed treatment of this problem, we refer the reader to the recent manuscript [19]. In particular, the motivation behind this approach is clearly articulated in Problem 1.1 of that work, where the authors begin by analyzing the problem explicitly in the setting of ${\mathbb{M}}_2\left(\mathbb{C}\right)$, i.e., $2\times 2$ complex matrices.

In that specific context, the minimizing orthogonal projection is computed in closed form. The general formula for $m\left(Q\right)$, valid for any Hilbert space $\mathcal{H}$, is then obtained by extending the techniques developed for matrices and applying operator-theoretic tools. In particular, the authors rely heavily on the block matrix representation of an operator relative to a projection, which plays a central role in deriving the general result.

As an immediate consequence of Theorem 1 and Lemma 3, we can conclude the following result.

Proposition 2.

Let $Q\in \mathcal{OP}$ be such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$. Let $P_0\in \mathcal{P}$ be an orthogonal projection minimizing the distance to Q, i.e.,

$$\parallel P_0-Q\parallel =\underset{P\in \mathcal{P}}{min}\parallel P-Q\parallel .$$

Then, $P_0\ne 0$ and $P_0\ne I$.

Proof. 

Suppose that there exists Q with the given hypotheses, and by reductio ad absurdum, assume that $P_0=0$ or $P_0=I$. Then, by (6), it follows that

$$\parallel I-Q\parallel =1+\parallel Q\parallel ,$$

if $P_0=0$, or

$$\parallel Q\parallel =1+\parallel I-Q\parallel ,$$

if $P_0=I$. Notice that both equalities contradict the fact that the norms of Q and $I-Q$ coincide. Therefore, we can conclude that $P_0\ne 0$ and $P_0\ne I$. □

From the previous result, we derive the following inequality. It is worth noting that the same inequality was proved in [18] (Corollary 3.13), using a different approach. Furthermore, our inequality is strict for every $Q\in \mathcal{OP}$ whose null space and range are both distinct from the entire Hilbert space.

Corollary 5.

Let $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$, then

$$\parallel m\left(Q\right)-Q\parallel < \parallel Q\parallel = \parallel I-Q\parallel .$$

(7)

Taking into account that, in [18], the authors recently derived an explicit formula for the norm of $m\left(Q\right)-Q$, we can establish the following refinement of (7).

Theorem 4.

Let $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$, then

$$\parallel m\left(Q\right)-Q\parallel \le \sqrt{\parallel Q\parallel \left(\parallel Q\parallel -1\right)}<\parallel Q\parallel -{\displaystyle \frac{1}{2}}.$$

Proof. 

By [18] (Theorem 3.17), we have that

$$\parallel m\left(Q\right)-Q\parallel ={\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right].$$

(8)

We first rewrite the expression:

$$\begin{array}{cc}\hfill \parallel m\left(Q\right)-Q\parallel & ={\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]\hfill \\ \hfill & =\sqrt{\parallel Q\parallel -1}\left({\displaystyle \frac{\sqrt{\parallel Q\parallel -1}+\sqrt{\parallel Q\parallel +1}}{2}}\right).\hfill \end{array}$$

Now, applying the inequality

$${\displaystyle \frac{\sqrt{a}+\sqrt{b}}{2}}\le \sqrt{{\displaystyle \frac{a+b}{2}}},\mathrm{for}\mathrm{all}a,b>0,$$

which follows from the concavity of the square root function, we obtain

$$\begin{array}{cc}\hfill \parallel m\left(Q\right)-Q\parallel & \le \sqrt{\parallel Q\parallel -1}\sqrt{{\displaystyle \frac{\parallel Q\parallel -1+\parallel Q\parallel +1}{2}}}\hfill \\ \hfill & =\sqrt{\parallel Q\parallel \left(\parallel Q\parallel -1\right)}\hfill \\ \hfill & <\sqrt{\parallel Q\parallel \left(\parallel Q\parallel -1\right)+{\displaystyle \frac{1}{4}}}\hfill \\ \hfill & =\parallel Q\parallel -{\displaystyle \frac{1}{2}}.\hfill \end{array}$$

This finishes the proof. □

Remark 4.

We will show that the first inequality established in Theorem 4 is strict whenever $\parallel Q\parallel >1$.

Suppose, by contradiction, that there exists $Q\in \mathcal{OP}$, with neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ equal to $\mathcal{H}$, such that

$$\parallel m\left(Q\right)-Q\parallel =\sqrt{\parallel Q\parallel \left(\parallel Q\parallel -1\right)},\mathrm{with}\parallel Q\parallel >1.$$

By identity (8), this is equivalent to

$${\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]=\sqrt{\parallel Q\parallel \left(\parallel Q\parallel -1\right)}.$$

Squaring both sides and performing standard algebraic manipulations, we obtain

$${(\parallel Q\parallel -1)}^2+{(\parallel Q\parallel }^2-1)+2(\parallel Q\parallel -1)\sqrt{{\parallel Q\parallel }^2-1}={4(\parallel Q\parallel }^2-\parallel Q\parallel ).$$

That is,

$${\parallel Q\parallel }^2-2\parallel Q\parallel +1+{\parallel Q\parallel }^2-1+2(\parallel Q\parallel -1)\sqrt{{\parallel Q\parallel }^2-1}={4(\parallel Q\parallel }^2-\parallel Q\parallel ),$$

or equivalently,

$$2(\parallel Q\parallel -1)\sqrt{{\parallel Q\parallel }^2-1}=2\parallel Q\parallel (\parallel Q\parallel -1).$$

Since we assumed that $\parallel Q\parallel \ne 1$, we may cancel the common factor and deduce that

$$\sqrt{{\parallel Q\parallel }^2-1}=\parallel Q\parallel .$$

Squaring once more, we arrive at ${\parallel Q\parallel }^2-1={\parallel Q\parallel }^2$, a contradiction.

Therefore, we have the following statement

Corollary 6.

Let $Q\in \mathcal{OP}$, with neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ equal to $\mathcal{H}$, and $\parallel Q\parallel >1$, then

$$\parallel m\left(Q\right)-Q\parallel <\sqrt{\parallel Q\parallel \left(\parallel Q\parallel -1\right)}.$$

At this stage, we are in a position to establish a lower bound for the minimum distance between an idempotent operator $Q\in \mathcal{OP}$ and the set of orthogonal projections $\mathcal{P}$.

Proposition 3.

Let $Q\in \mathcal{OP}$. Then

$$\parallel Q\parallel -1\le {\left(\parallel Q\parallel -1\right)}^{\frac{3}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}\le \parallel m\left(Q\right)-Q\parallel .$$

(9)

Furthermore, equality $\parallel Q\parallel -1= \parallel m\left(Q\right)-Q\parallel$ holds if and only if $Q\in \mathcal{P}\setminus \left\{0\right\}$.

Proof. 

If $Q=0$, the inequality is trivial.

Hence, we may assume that $\parallel Q\parallel \ge 1$.

Then, by identity (8) and the AM–GM inequality, we obtain

$$\begin{array}{cc}\hfill \parallel m\left(Q\right)-Q\parallel & ={\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]\hfill \\ \hfill & =\sqrt{\parallel Q\parallel -1}\left({\displaystyle \frac{\sqrt{\parallel Q\parallel -1}+\sqrt{\parallel Q\parallel +1}}{2}}\right)\hfill \\ \hfill & \ge \sqrt{\parallel Q\parallel -1}\sqrt[4]{\left(\parallel Q\parallel -1\right)\left(\parallel Q\parallel +1\right)}\hfill \\ \hfill & ={\left(\parallel Q\parallel -1\right)}^{\frac{3}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}\hfill \\ \hfill & \ge \parallel Q\parallel -1.\hfill \end{array}$$

Suppose from now on that

$$\parallel Q\parallel -1= \parallel m\left(Q\right)-Q\parallel .$$

Then, the identity (8) immediately yields

$$\parallel Q\parallel =1,$$

so that $Q\in \mathcal{P}\setminus \left\{0\right\}$.

Conversely, if $Q\in \mathcal{P}\setminus \left\{0\right\}$, then clearly $m\left(Q\right)=Q$, and hence

$$\parallel Q\parallel -1=0= \parallel m\left(Q\right)-Q\parallel .$$

This finishes the proof. □

Remark 5.

In Proposition 3, we proved that if $Q\in \mathcal{P}$, then

$${\left(\parallel Q\parallel -1\right)}^{\frac{3}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}=\parallel m\left(Q\right)-Q\parallel .$$

(10)

Suppose, for the sake of contradiction, that there exists an operator $Q\in \mathcal{OP}\setminus \mathcal{P}$ satisfying (10), that is,

$${\left(\parallel Q\parallel -1\right)}^{\frac{3}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}={\displaystyle \frac{1}{2}}\left[(\parallel Q\parallel -1)+\sqrt{{\parallel Q\parallel }^2-1}\right],$$

with $\parallel Q\parallel >1$.

Then,

$${\left(\parallel Q\parallel -1\right)}^{\frac{3}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}={\displaystyle \frac{1}{2}}\sqrt{\parallel Q\parallel -1}\left[\sqrt{\parallel Q\parallel -1}+\sqrt{\parallel Q\parallel +1}\right],$$

and since $\parallel Q\parallel \ne 1$, we conclude that

$$2{\left(\parallel Q\parallel -1\right)}^{\frac{1}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}=\sqrt{\parallel Q\parallel -1}+\sqrt{\parallel Q\parallel +1}.$$

Equivalently,

$${\left[{(\parallel Q\parallel -1)}^{1/4}-{(\parallel Q\parallel +1)}^{1/4}\right]}^2=0,$$

which implies that

$$\parallel Q\parallel -1= \parallel Q\parallel +1,$$

a contradiction.

Combining Proposition 3 with the preceding observation, we arrive at the following result.

Corollary 7.

Let $Q\in \mathcal{OP}\setminus \mathcal{P}$. Then,

$$\parallel Q\parallel -1<{\left(\parallel Q\parallel -1\right)}^{\frac{3}{4}}{\left(\parallel Q\parallel +1\right)}^{\frac{1}{4}}<\parallel m\left(Q\right)-Q\parallel .$$

Given that if $Q\in \mathcal{OP}$ with $Q\ne 0$, then $\parallel Q\parallel \ge 1$, it follows that

$$0\le \parallel Q\parallel -1\le \parallel I-Q\parallel .$$

By employing the matched operator and the inequalities established earlier, we can derive the following refinement, which we present in the proposition below.

Proposition 4.

Let $Q\in \mathcal{OP}$ be such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincides with $\mathcal{H}$. Then,

$$\parallel Q\parallel -1\le \parallel m\left(Q\right)-Q\parallel < \parallel Q\parallel -{\displaystyle \frac{1}{2}}< \parallel Q\parallel =\parallel I-Q\parallel < \parallel Q\parallel +1.$$

In particular,

$$\left|\parallel m\left(Q\right)-Q\parallel -\parallel Q\parallel \right|\le 1,$$

and if $Q\in \mathcal{OP}\setminus \mathcal{P}$, the inequality is strict.

We conclude this section by deriving upper bounds for the norm of $Q+I$, where $Q\in \mathcal{OP}$. If $Q\in \mathcal{P}$, it is straightforward to verify that $\parallel Q+I\parallel =2$. Observe that, based on (8), we obtain

$$\begin{array}{ccc}\hfill \parallel Q+I\parallel & =& \parallel Q-m\left(Q\right)+m\left(Q\right)+I\parallel \hfill \\ & \le & \parallel Q-m\left(Q\right)\parallel +\parallel m\left(Q\right)+I\parallel \hfill \\ & =& {\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]+2,\hfill \end{array}$$

for any $Q\in \mathcal{OP}$.

This bound is not sharp since

$$\parallel Q\parallel +1\le {\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]+2.$$

Indeed, suppose, for the sake of contradiction, that

$${\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]+2<\parallel Q\parallel +1,$$

or equivalently,

$${\displaystyle \frac{1}{2}}\left[\parallel Q\parallel -1+\sqrt{{\parallel Q\parallel }^2-1}\right]+1<\parallel Q\parallel .$$

However, this inequality implies that $\parallel m\left(Q\right)-Q\parallel <\parallel Q\parallel -1$, which contradicts (9).

Next, we present an upper bound for the norm of $Q+I$, which is sharper than the one given by the triangle inequality.

Theorem 5.

Let $Q\in \mathcal{OP}$ such that neither $\mathcal{N}\left(Q\right)$ nor $\mathcal{R}\left(Q\right)$ coincide with $\mathcal{H}$. Then,

$$\parallel Q+I\parallel \le \sqrt{{\parallel Q\parallel }^2+\parallel Q\parallel +2}\le \parallel Q\parallel +1.$$

(11)

Furthermore, equality $\parallel Q+I\parallel = \parallel Q\parallel +1$ holds if and only if $Q\in \mathcal{P}\setminus \left\{0\right\}$.

Proof. 

Note that

$$\begin{array}{ccc}\hfill {\parallel Q+I\parallel }^2=\parallel (Q^{*}+I)(Q+I)\parallel & =& \parallel Q^{*}Q+2\Re \left(Q\right)+I\parallel \hfill \\ & =& \parallel Q^{*}Q+2I+2\Re \left(Q\right)-I\parallel \hfill \\ & \le & \parallel Q^{*}Q+2I\parallel +\parallel 2\Re \left(Q\right)-I\parallel .\hfill \end{array}$$

As a consequence of the positivity of $Q^{*}Q$ and I, and using the equality (2), we conclude that $\parallel Q^{*}{Q+2I\parallel = \parallel Q\parallel }^2+2$ and $\parallel 2\Re \left(Q\right)-I\parallel = \parallel Q\parallel$, respectively.

Thus,

$$\begin{array}{ccc}\hfill {\parallel Q+I\parallel }^2=\parallel (Q^{*}+I)(Q+I)\parallel & \le & \parallel Q^{*}Q+2I\parallel +\parallel 2\Re \left(Q\right)-I\parallel \hfill \\ & =& {\parallel Q\parallel }^2+2+\parallel Q\parallel ,\hfill \end{array}$$

We conclude the proof by observing that if we consider the functions

$$g\left(t\right)=\sqrt{t^2+t+2}\mathrm{and}f\left(t\right)=t+1,$$

where $t\ge 1$. Then,

$$\begin{array}{ccc}\hfill f^2\left(t\right)-(t^2+t+2)& =& t^2+2t+1-t^2-t-2\hfill \\ & =& t-1\ge 0,\hfill \end{array}$$

for any $t\ge 1$.

Now suppose that equality holds, namely,

$$\parallel Q+I\parallel = \parallel Q\parallel +1.$$

Then, by (11), we have

$$\sqrt{{\parallel Q\parallel }^2+\parallel Q\parallel +2}=\parallel Q\parallel +1,$$

which implies that $\parallel Q\parallel =1$, and hence $Q\in \mathcal{P}\setminus \left\{0\right\}$. Conversely, if $Q\in \mathcal{P}\setminus \left\{0\right\}$, then Duncan–Taylor’s formula implies

$$\parallel Q+I\parallel = \parallel Q\parallel +1.$$

This finishes the proof. □

It is well known that if $P\in \mathcal{P}$ then P is self–adjoint and hence normal. Every normal operator satisfies $r\left(P\right)=\parallel P\parallel ,$ that is, P is normaloid. By contrast, we will show that oblique non orthogonal projections do not necessarily share this property.

Corollary 8.

Let $Q\in \mathcal{OP}\setminus \mathcal{P}$. Then $r\left(Q\right)< \parallel Q\parallel$. In particular, Q is not normaloid.

Proof. 

Suppose, to the contrary, that Q is normaloid, i.e., $r\left(Q\right)= \parallel Q\parallel$. By the definition of the numerical radius, there exists a sequence of unit vectors $\left\{x_n\right\}\subset \mathcal{H}$ such that

$$\left|⟨Q{x}_n,x_n⟩\right|⟶\parallel Q\parallel .$$

For each n, choose ${\theta }_n\in [0,2\pi )$ so that

$$z_n:=e^{i{\theta }_n}⟨Q{x}_n,x_n⟩=\left|⟨Q{x}_n,x_n⟩\right|\ge 0.$$

Since the unit circle $\{e^{i\theta }:0\le \theta \le 2\pi \}$ is compact, by Bolzano–Weierstrass there is a subsequence $\left\{e^{i{\theta }_{n_k}}\right\}$ converging to some $e^{i{\theta }_0}$. Hence

$$z_{n_k}=e^{i{\theta }_{n_k}}\left|⟨Q{x}_{n_k},x_{n_k}⟩\right|⟶e^{i{\theta }_0}\parallel Q\parallel .$$

It follows that $e^{i{\theta }_0}\parallel Q\parallel \in \overline{W\left(Q\right)}$, and in particular, $\parallel Q\parallel \in \overline{W\left(Q\right)}$. By [24] (Theorem 2.1), this implies

$$\parallel Q+I\parallel = \parallel Q\parallel +1.$$

But then Theorem 5 forces $Q\in \mathcal{P}$, contradicting $Q\notin \mathcal{P}$. Therefore, no oblique projection $Q\in \mathcal{OP}\setminus \mathcal{P}$ can be normaloid. □

Remark 6.

We observe that an alternative proof of Corollary 8 follows from the classical characterization of normaloid operators (see [2]): an operator $T\in \mathcal{B}\left(\mathcal{H}\right)$ is normaloid if and only if ${\parallel T\parallel }^n=\parallel T^n\parallel \mathrm{for}\mathrm{all}n\in \mathbb{N}$. Indeed, if $Q\in \mathcal{OP}\setminus \mathcal{P}$ were normaloid, then for every $n\in \mathbb{N}$ one would have

$${\parallel Q\parallel }^n= \parallel Q^n\parallel = \parallel Q\parallel ,$$

since $Q^n=Q$ by idempotence. It follows that $\parallel Q\parallel \in \{0,1\}$, and hence $Q\in \mathcal{P}$, a contradiction. While this concise argument is elegant, we have chosen to develop and present proofs based on the results established earlier in this manuscript, as they offer deeper insight into the geometric and spectral properties of oblique projections.

In Theorem 5, we show that if $Q\in \mathcal{OP}\setminus \mathcal{P}$, then

$$\parallel Q+I\parallel < \parallel Q\parallel +1.$$

(12)

Remark 7.

Within the framework of bounded linear operators on Hilbert spaces, the identity

$$\parallel T+I\parallel = \parallel T\parallel +1,$$

with $T\in \mathcal{B}\left(\mathcal{H}\right)$, is referred to as the Daugavet equation; see, for example, [27,28] and the references therein. In particular, we have shown that no operator $Q\in \mathcal{OP}\setminus \mathcal{P}$ satisfies the Daugavet equation.

We now show that non-orthogonal oblique projections satisfy an inequality that extends (12). More precisely:

Theorem 6.

Let $Q\in \mathcal{OP}\setminus \mathcal{P}$. Then

$$\parallel Q+\lambda I\parallel < \parallel Q\parallel +1,$$

for any $\lambda \in \mathbb{T}=\{z\in \mathbb{C}\mid |z|=1\}.$

Proof. 

Suppose that there exists ${\lambda }_0\in \mathbb{T}$ such that

$$\parallel Q+{\lambda }_{0}I\parallel = \parallel Q\parallel +1.$$

Then, by [29] (Proposition 4.7), we conclude that Q is normaloid, which contradicts Corollary 8. This completes the proof. □

3. Conclusions

In this paper, we presented a new and streamlined proof of the operator norm identity

$$\parallel Q\parallel =\parallel I-Q\parallel ,$$

for bounded idempotent operators on a complex Hilbert space. This approach offers a clearer perspective compared to existing methods. Additionally, we derived explicit lower and upper bounds for the distance between an idempotent operator and the set of orthogonal projections, providing new insights into their geometric and algebraic connections.

These results enhance our understanding of idempotent operators and have potential applications in various areas, including operator theory, numerical analysis, and quantum mechanics.

We believe this work not only consolidates recent developments in the field but also opens up new avenues for future research. By establishing a firm foundation, we hope to encourage further exploration into the rich structure of Hilbert space theory and related disciplines.