Null Spaces of Sums and Differences of Inner Regular Elements in Unital Rings

Abstract

In this paper, we investigate the right and left null spaces of sums, differences, and modified sums of inner regular elements in unital rings. We establish sufficient conditions under which these null spaces coincide with intersections of the corresponding kernel ideals, and we derive criteria for their triviality. For commuting $\{1,5\}$-invertible elements, we obtain decompositions of the null spaces of differences of associated idempotent-like elements. We also establish equivalence conditions for the invertibility of sums and differences of special classes of generalized invertible elements. Applications to bounded linear operators on Banach and Hilbert spaces, as well as to matrices, are presented.

1. Introduction

Let $\mathcal{R}$ be an associative ring with unity $1\ne 0$.

We say that $a\in \mathcal{R}$ is inner regular, inner generalized invertible, or $\left\{1\right\}$-invertible element if there exists $a^{\left(1\right)}\in \mathcal{R}$ such that $a{a}^{\left(1\right)}a=a$. Such an element $a^{\left(1\right)}\in \mathcal{R}$ is called an inner generalized inverse of a or a $\left\{1\right\}$-inverse of a.

In the theory of generalized inverses, we use the standard numbering for the following conditions. Namely, an element x may satisfy one or more of the following relations:

$$\begin{array}{c}\hfill \left(1\right)axa=a,\left(2\right)xax=x,\left(5\right)ax=xa.\end{array}$$

An element satisfying conditions $\left(1\right)$ and $\left(5\right)$ is called a $\{1,5\}$-inverse of $a\in \mathcal{R}$ and is denoted by $a^{(1,5)}$. In this case, the element a is called $\{1,5\}$-invertible.

The group inverse of $a\in \mathcal{R}$, or $\{1,2,5\}$-inverse, when it exists, is denoted by $a^{♯}$ and is the unique solution to

$$\begin{array}{c}\hfill \left(1\right)a{a}^{♯}a=a,\left(2\right)a^{♯}a{a}^{♯}=a^{♯},\left(5\right)a{a}^{♯}=a^{♯}a.\end{array}$$

Such an element a is called group invertible, or $\{1,2,5\}$-invertible.

We denote by ${\mathcal{R}}^{-1}$, ${\mathcal{R}}^{\left\{1\right\}}$, ${\mathcal{R}}^{\{1,5\}}$, ${\mathcal{R}}^{♯}$ the subsets of invertible, inner regular, $\{1,5\}$-invertible, group invertible elements of $\mathcal{R}$, respectively.

For convenience, the notation for the generalized inverses used throughout the paper is summarized in Table 1.

Table 1. Notation for the generalized inverses used throughout the paper.

Type of Generalized Inverse	Notation	Defining Conditions	Corresponding Class
Inner inverse ($\left\{1\right\}$-inverse)	$a^{\left(1\right)}$	$a{a}^{\left(1\right)}a=a$	$a\in {\mathcal{R}}^{\left\{1\right\}}$
{1,5}-inverse	$a^{(1,5)}$	$\begin{array}{c}a{a}^{(1,5)}a=a,\\ a{a}^{(1,5)}=a^{(1,5)}a\end{array}$	$a\in {\mathcal{R}}^{\{1,5\}}$
Group inverse ($\{1,2,5\}$-inverse)	$a^{♯}$	$\begin{array}{c}a{a}^{♯}a=a,\\ a^{♯}a{a}^{♯}=a^{♯},\\ a{a}^{♯}=a^{♯}a\end{array}$	$a\in {\mathcal{R}}^{♯}$

Elements $a,b\in \mathcal{R}$ are mutually orthogonal, written $a\perp b$, if $ab=0=ba$. An element $a\in \mathcal{R}$ is called $k$-potent if $a^k=a$ for some natural $k>1$.

With each element $a\in \mathcal{R}$ we associate two image ideals

$$\begin{array}{c}\hfill a\mathcal{R}=\{ax:x\in \mathcal{R}\},\mathcal{R}a=\{xa:x\in \mathcal{R}\}\end{array}$$

(1)

and two kernel ideals, which are also referred to as the right and left annihilators of a, respectively,

$$\begin{array}{c}\hfill a^0=\{x\in \mathcal{R}:ax=0\},{}^{0}a=\{x\in \mathcal{R}:xa=0\}.\end{array}$$

(2)

The symbol ⊕ denotes a direct sum; in the ring setting it refers to a direct sum of right or left ideals, while in the operator setting it refers to a direct sum of subspaces.

The behavior of these ideals provides useful information about algebraic properties of elements in rings, in particular about sums and differences of elements possessing generalized inverses.

Inner regular elements and their generalized inverses have been widely studied in ring theory and operator theory (see [1,2]). General background on generalized inverses can be found in [3,4,5,6]. Various algebraic aspects of generalized inverses in rings and semigroups have been investigated in numerous papers; see, for example, [7,8].

Motivated by earlier results concerning the invertibility of sums and differences of special classes of elements, such as idempotents and related operators (see [9,10,11,12]), we study the structure of the null spaces of sums and differences of inner regular elements. In particular, we establish several relations between the null spaces of sums and differences of inner regular elements and the intersections of the corresponding kernel ideals.

The obtained results lead to various consequences concerning the triviality of null spaces and, in certain situations, the invertibility of sums and differences of $\{1,5\}$-invertible elements and $k$-potents. As applications, we derive corresponding statements for bounded linear operators on Banach and Hilbert spaces as well as for matrices.

In contrast to the classical results concerning idempotents and projectors, our approach is formulated for the broader class of inner regular and generalized invertible elements in a unital ring. Rather than considering only the invertibility of sums and differences, we first describe the corresponding right and left null spaces in terms of intersections and direct sums of image and kernel ideals. The invertibility criteria obtained later in the paper arise as consequences of these null-space relations and, in particular cases, extend analogous results known for idempotents and projectors.

The paper is organized as follows. In Section 2, we investigate relations between the null spaces of sums and differences of inner regular elements and obtain several sufficient conditions under which these null spaces coincide with intersections of the corresponding kernel ideals. In Section 3, we present applications of the obtained results to bounded linear operators on Banach and Hilbert spaces and to matrices.

2. Main Results

Throughout this section, whenever $a,b\in {\mathcal{R}}^{\left\{1\right\}}$, $a^{\left(1\right)}$ and $b^{\left(1\right)}$ denote fixed inner inverses of a and b, respectively. Similarly, whenever $\{1,5\}$-inverses are considered, $a^{(1,5)}$ and $b^{(1,5)}$ denote fixed chosen $\{1,5\}$-inverses.

2.1. Null Spaces of Sums and Modified Sums of Inner Regular Elements

The following theorem provides a sufficient condition under which the null space of the sum of three inner regular elements coincides with the intersection of their null spaces.

Theorem 1.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and $c\in \mathcal{R}$, and let $a^{\left(1\right)}$ and $b^{\left(1\right)}$ be fixed inner inverses of a and b, respectively. If $a{a}^{\left(1\right)}(b+c)=0\mathit{and}b{b}^{\left(1\right)}c=0,$ then

$${(a+b+c)}^0=a^0\cap b^0\cap c^0.$$

Proof. 

First, we prove that $a^0\cap b^0\cap c^0\subseteq {(a+b+c)}^0.$ Let $x\in a^0\cap b^0\cap c^0$. Then $ax=bx=cx=0,$ and hence $(a+b+c)x=0.$ Therefore, $x\in {(a+b+c)}^0$.

Conversely, let $x\in {(a+b+c)}^0$. Then $(a+b+c)x=0,$ that is,

$$ax+bx+cx=0.$$

(3)

Multiplying (3) from the left by $a{a}^{\left(1\right)}$, we obtain $a{a}^{\left(1\right)}ax+a{a}^{\left(1\right)}bx+a{a}^{\left(1\right)}cx=0.$ Since $a^{\left(1\right)}$ is an inner inverse of a, we have $a{a}^{\left(1\right)}a=a$. Moreover, by the assumption $a{a}^{\left(1\right)}(b+c)=0$, it follows that $ax=0.$ Thus, (3) reduces to $bx+cx=0.$ Multiplying this equality from the left by $b{b}^{\left(1\right)}$, we get $b{b}^{\left(1\right)}bx+b{b}^{\left(1\right)}cx=0.$ Since $b{b}^{\left(1\right)}b=b$ and $b{b}^{\left(1\right)}c=0$, we obtain $bx=0.$ Consequently, from (3) we also have $cx=0.$ Hence, $x\in a^0\cap b^0\cap c^0.$ Therefore, ${(a+b+c)}^0=a^0\cap b^0\cap c^0.$ □

Example 1.

Let $\mathcal{R}=M_2\left(\mathbb{Q}\right)$ and

$$a=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),b=\left(\begin{array}{cc}0& 1\\ 0& 1\end{array}\right),c=\left(\begin{array}{cc}0& -1\\ 0& 0\end{array}\right).$$

Since $a^2=a$ and $b^2=b$, we may choose $a^{\left(1\right)}=a$ and $b^{\left(1\right)}=b$. A direct computation gives

$$a{a}^{\left(1\right)}(b+c)=0,b{b}^{\left(1\right)}c=0.$$

Hence, the hypotheses of Theorem 1 are satisfied. Moreover,

$$a+b+c=I_2,$$

and therefore

$${(a+b+c)}^0=\left\{0\right\}=a^0\cap b^0\cap c^0.$$

Notice also that

$$a{a}^{\left(1\right)}b=ab\ne 0,a{a}^{\left(1\right)}c=ac\ne 0,$$

showing that the condition $a{a}^{\left(1\right)}(b+c)=0$ does not require the two terms to vanish separately.

Replacing right null spaces by left null spaces, we obtain the following analogue of the previous theorem.

Theorem 2.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and $c\in \mathcal{R}$, and let $a^{\left(1\right)}$ and $b^{\left(1\right)}$ be fixed inner inverses of a and b, respectively. If $(b+c)a^{\left(1\right)}a=0\mathit{and}c{b}^{\left(1\right)}b=0,$ then

$${}^0(a+b+c)={}^{0}a\cap {}^{0}b\cap {}^{0}c.$$

Proof. 

The inclusion ${}^{0}a\cap {}^{0}b\cap {}^{0}c\subseteq {}^0(a+b+c)$ is immediate. Conversely, let $x\in {}^0(a+b+c)$. Then $xa+xb+xc=0.$ Multiplying this equality from the right by $a^{\left(1\right)}a$ and using $a{a}^{\left(1\right)}a=a$ and $(b+c)a^{\left(1\right)}a=0$, we obtain $xa=0$. Hence, $xb+xc=0.$ Multiplying the latter equality from the right by $b^{\left(1\right)}b$ and using $b{b}^{\left(1\right)}b=b$ and $c{b}^{\left(1\right)}b=0$, we get $xb=0$. Consequently, $xc=0$. Thus, $x\in {}^{0}a\cap {}^{0}b\cap {}^{0}c,$ which proves the assertion. □

Example 2.

Let $\mathcal{R}=M_2\left(\mathbb{Q}\right)$ and

$$A=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),B=\left(\begin{array}{cc}0& 0\\ 1& 1\end{array}\right),C=\left(\begin{array}{cc}0& 0\\ -1& 0\end{array}\right).$$

Since $A^2=A$ and $B^2=B$, we may choose $A^{\left(1\right)}=A$ and $B^{\left(1\right)}=B$. A direct computation gives

$$(B+C)A^{\left(1\right)}A=0,C{B}^{\left(1\right)}B=0.$$

Hence, the hypotheses of Theorem 2 are satisfied. Moreover,

$$A+B+C=I_2,$$

and therefore

$${}^0(A+B+C)=\left\{0\right\}={}^{0}A\cap {}^{0}B\cap {}^{0}C.$$

Notice also that

$$B{A}^{\left(1\right)}A=BA\ne 0,C{A}^{\left(1\right)}A=CA\ne 0,$$

showing that the condition $(B+C)A^{\left(1\right)}A=0$ does not require the two terms to vanish separately.

Complementing Theorem 1, we provide several alternative sufficient conditions under which the null space of the sum of two inner regular elements coincides with the intersection of their null spaces.

Theorem 3.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$. If one of the following conditions holds

(i)

$a+b+a{a}^{\left(1\right)}b+b{b}^{\left(1\right)}a+a{a}^{\left(1\right)}b{b}^{\left(1\right)}a=0$,

(ii)

$a+b+a{a}^{\left(1\right)}b+b{b}^{\left(1\right)}a+b{b}^{\left(1\right)}a{a}^{\left(1\right)}b=0$,

(iii)

$a+b+a{a}^{\left(1\right)}b{b}^{\left(1\right)}a=0$,

(iv)

$a+b+b{b}^{\left(1\right)}a{a}^{\left(1\right)}b=0$,

then

$${(a+b)}^0=a^0\cap b^0.$$

Proof. 

Let condition $\left(i\right)$ be satisfied. It is obvious that $a^0\cap b^0\subseteq {(a+b)}^0$. Conversely, we prove the reverse inclusion. For any $x\in {(a+b)}^0$, we obtain

$$\begin{array}{c}\hfill ax=-bx.\end{array}$$

(4)

Premultiplying both sides of (4) on the left first by $a{a}^{\left(1\right)}$ and then by $b{b}^{\left(1\right)}$, leads to

$$\begin{array}{c}\hfill ax=b{b}^{\left(1\right)}ax\end{array}$$

(5)

and

$$\begin{array}{c}\hfill bx=a{a}^{\left(1\right)}bx.\end{array}$$

(6)

From (5) and (6), it follows that

$$\begin{array}{c}\hfill ax=a{a}^{\left(1\right)}b{b}^{\left(1\right)}ax.\end{array}$$

(7)

and

$$\begin{array}{c}\hfill bx=b{b}^{\left(1\right)}a{a}^{\left(1\right)}bx,\end{array}$$

(8)

respectively. Multiplying condition $\left(i\right)$ on the right by x and using (4)–(7), we obtain

$$\begin{array}{cc}\hfill 0=& ax+bx+a{a}^{\left(1\right)}bx+b{b}^{\left(1\right)}ax+a{a}^{\left(1\right)}b{b}^{\left(1\right)}ax\hfill \\ \hfill =& ax.\hfill \end{array}$$

Thus, $ax=0$, i.e., $x\in a^0$. Since (4) holds, we conclude that $bx=0$, i.e., $x\in b^0$. Hence, $x\in a^0\cap b^0$. This establishes the equality ${(a+b)}^0=a^0\cap b^0$.

For condition $\left(ii\right)$, using (4)–(8), we obtain

$$\begin{array}{cc}\hfill 0=& ax+bx+a{a}^{\left(1\right)}bx+b{b}^{\left(1\right)}ax+b{b}^{\left(1\right)}a{a}^{\left(1\right)}bx\hfill \\ \hfill =& ax-ax-ax+ax-ax\hfill \\ \hfill =& -ax.\hfill \end{array}$$

Therefore, $ax=0$, and by (4), also $bx=0$. In case $\left(iii\right)$,

$$0=ax+bx+a{a}^{\left(1\right)}b{b}^{\left(1\right)}ax=ax,$$

whereas in case $\left(iv\right)$,

$$0=ax+bx+b{b}^{\left(1\right)}a{a}^{\left(1\right)}bx=bx.$$

Thus, in each case, $ax=bx=0$, and the desired equality follows. □

The corresponding result for the left null space is obtained by applying the same argument with multiplication on the right, while preserving the order of all factors.

Theorem 4.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$. If one of the following conditions holds

(i)

$a+b+b{a}^{\left(1\right)}a+a{b}^{\left(1\right)}b+a{b}^{\left(1\right)}b{a}^{\left(1\right)}a=0$,

(ii)

$a+b+b{a}^{\left(1\right)}a+a{b}^{\left(1\right)}b+b{a}^{\left(1\right)}a{b}^{\left(1\right)}b=0$,

(iii)

$a+b+a{b}^{\left(1\right)}b{a}^{\left(1\right)}a=0$,

(iv)

$a+b+b{a}^{\left(1\right)}a{b}^{\left(1\right)}b=0$,

then

$${}^0(a+b){=}^{0}a{\cap }^{0}b.$$

The following theorem provides sufficient conditions under which the right null space of the modified sum $a+b+a{a}^{\left(1\right)}b$ coincides with the intersection of the right null spaces of a and b.

Theorem 5.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and let $3\in {\mathcal{R}}^{-1}$. If $a{a}^{\left(1\right)}b=b{b}^{\left(1\right)}a$, then

$${(a+b+a{a}^{\left(1\right)}b)}^0=a^0\cap b^0.$$

Proof. 

Suppose that condition $a{a}^{\left(1\right)}b=b{b}^{\left(1\right)}a$ holds. Let $x\in {(a+b+a{a}^{\left(1\right)}b)}^0$. Premultiplying the equality

$$\begin{array}{c}\hfill ax+bx+a{a}^{\left(1\right)}bx=0\end{array}$$

(9)

by $a{a}^{\left(1\right)}$ and $b{b}^{\left(1\right)}$, respectively, leads to $ax+a{a}^{\left(1\right)}bx+a{a}^{\left(1\right)}bx=0$ and $b{b}^{\left(1\right)}ax+bx+b{b}^{\left(1\right)}a{a}^{\left(1\right)}bx=0$, i.e., $b{b}^{\left(1\right)}ax+bx+b{b}^{\left(1\right)}ax=0$. Adding the last two equalities and using (9), we get $3a{a}^{\left(1\right)}bx=0$. Since $3\in {\mathcal{R}}^{-1}$, it follows that $a{a}^{\left(1\right)}bx=0$, which, together with (9), implies $ax+bx=0$. Premultiplying the last equality by $a{a}^{\left(1\right)}$ yields $ax=0$, and hence $bx=0$. This proves that $x\in a^0\cap b^0$. Thus, ${(a+b+a{a}^{\left(1\right)}b)}^0\subseteq a^0\cap b^0$. The reverse inclusion is trivial, and therefore ${(a+b+a{a}^{\left(1\right)}b)}^0=a^0\cap b^0$. □

The following example shows that the assumption $3\in {\mathcal{R}}^{-1}$ in Theorem 5 cannot be omitted.

Example 3.

The assumption $3\in {\mathcal{R}}^{-1}$ in Theorem 5 is essential. Indeed, let $\mathcal{R}={\mathbb{F}}_3$ and take

$$a=b=1,a^{\left(1\right)}=b^{\left(1\right)}=1.$$

Then $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and

$$a{a}^{\left(1\right)}b=b{b}^{\left(1\right)}a=1.$$

However, since $char{\mathbb{F}}_3=3$,

$$a+b+a{a}^{\left(1\right)}b=1+1+1=0.$$

Consequently,

$${(a+b+a{a}^{\left(1\right)}b)}^0={\mathbb{F}}_3,$$

whereas

$$a^0\cap b^0=\left\{0\right\}.$$

Thus, the conclusion of Theorem 5 may fail when 3 is not invertible.

By applying the same argument with multiplication on the right while preserving the order of all factors, we obtain sufficient conditions under which the left null space of the modified sum $a+b+b{a}^{\left(1\right)}a$ coincides with the intersection of the left null spaces of a and b.

Theorem 6.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and let $3\in {\mathcal{R}}^{-1}$. If $b{a}^{\left(1\right)}a=a{b}^{\left(1\right)}b$, then

$${}^0(a+b+b{a}^{\left(1\right)}a){=}^{0}a{\cap }^{0}b.$$

2.2. Triviality and Decomposition of Null Spaces

Motivated by the well-known necessary and sufficient conditions for the invertibility of the sum and the difference of two idempotent matrices (see [13]), we investigate analogous questions for inner regular elements. We first provide a sufficient condition under which the null space of the sum of inner regular elements is trivial.

Theorem 7.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and let $2,3\in {\mathcal{R}}^{-1}$. If $a(1-a^{\left(1\right)}b)\mathcal{R}\cap b(1-b^{\left(1\right)}a)\mathcal{R}=\left\{0\right\}$ and $a^0\cap b^0=\left\{0\right\},$ then

(i)

${(a+b)}^0=\left\{0\right\}$,

(ii)

${(a+b+a{a}^{\left(1\right)}b)}^0=\left\{0\right\}$,

(iii)

${(a+b+b{b}^{\left(1\right)}a)}^0=\left\{0\right\}$.

Proof. 

$\left(i\right)$ Let $x\in {(a+b)}^0$, i.e., $ax=-bx$. Premultiplying both sides of $ax=-bx$ on the left first by $a{a}^{\left(1\right)}$ and then by $b{b}^{\left(1\right)}$, we get $ax=-a{a}^{\left(1\right)}bx=-bx=b{b}^{\left(1\right)}ax$. Using $ax=-a{a}^{\left(1\right)}bx$, we have $2ax=ax-a{a}^{\left(1\right)}bx=a(1-a^{\left(1\right)}b)x$. Since $2\in {\mathcal{R}}^{-1}$, we obtain

$$\begin{array}{c}\hfill ax=a(1-a^{\left(1\right)}b)\left(2^{-1}x\right)\in a(1-a^{\left(1\right)}b)\mathcal{R}.\end{array}$$

(10)

Similarly, using $ax=b{b}^{\left(1\right)}ax$, we have $2ax=ax+ax=-bx+b{b}^{\left(1\right)}ax$. Since $2\in {\mathcal{R}}^{-1}$, we obtain

$$\begin{array}{c}\hfill ax=-b(1-b^{\left(1\right)}a)\left(2^{-1}x\right)\in b(1-b^{\left(1\right)}a)\mathcal{R}.\end{array}$$

(11)

Expressions (10) and (11) imply $ax\in a(1-a^{\left(1\right)}b)\mathcal{R}\cap b(1-b^{\left(1\right)}a)\mathcal{R}$. In view of the assumption that $a(1-a^{\left(1\right)}b)\mathcal{R}\cap b(1-b^{\left(1\right)}a)\mathcal{R}=\left\{0\right\}$, it follows that $ax=0$. Combining this with the equation $ax=-bx$ leads to $x\in a^0\cap b^0$. The second assumption implies $x=0$.

$\left(ii\right)$ For any $x\in {(a+b+a{a}^{\left(1\right)}b)}^0$, we have

$$\begin{array}{c}\hfill ax+bx+a{a}^{\left(1\right)}bx=0.\end{array}$$

(12)

Premultiplying (12) on the left by $a{a}^{\left(1\right)}$ gives

$$\begin{array}{c}\hfill ax+a{a}^{\left(1\right)}bx+a{a}^{\left(1\right)}bx=0,\end{array}$$

(13)

i.e.,

$$\begin{array}{c}\hfill ax=-2a{a}^{\left(1\right)}bx.\end{array}$$

(14)

From (12) and (13), it follows that

$$\begin{array}{c}\hfill bx=a{a}^{\left(1\right)}bx.\end{array}$$

(15)

Substituting (15) into (14) yields

$$\begin{array}{c}\hfill ax=-2bx.\end{array}$$

(16)

Similarly, premultiplying (12) on the left by $b{b}^{\left(1\right)}$, we obtain

$$\begin{array}{c}\hfill b{b}^{\left(1\right)}ax+2bx=0.\end{array}$$

(17)

Equalities (16) and (17) imply

$$\begin{array}{c}\hfill ax=b{b}^{\left(1\right)}ax.\end{array}$$

(18)

Further, using (16) and (18), we have $3ax=ax+2ax=-2bx+2b{b}^{\left(1\right)}ax=-2b(1-b^{\left(1\right)}a)x$. Since $3\in {\mathcal{R}}^{-1}$, it follows that

$$ax=-2b(1-b^{\left(1\right)}a)\left(3^{-1}x\right)\in b(1-b^{\left(1\right)}a)\mathcal{R}.$$

Similarly, the equality (14) yields $3ax=2ax+ax=2ax-2a{a}^{\left(1\right)}bx=2a(1-a^{\left(1\right)}b)x$. Since $3\in {\mathcal{R}}^{-1}$, it follows that

$$ax=2a(1-a^{\left(1\right)}b)\left(3^{-1}x\right)\in a(1-a^{\left(1\right)}b)\mathcal{R}.$$

Consequently, $ax\in a(1-a^{\left(1\right)}b)\mathcal{R}\cap b(1-b^{\left(1\right)}a)\mathcal{R}=\left\{0\right\}$ and $ax=0$. Since the equality (16) holds, we also have $bx=0$. Thus, $x\in a^0\cap b^0=\left\{0\right\}$ and $x=0$.

$\left(iii\right)$ For any $x\in {(a+b+b{b}^{\left(1\right)}a)}^0$, we have

$$ax+bx+b{b}^{\left(1\right)}ax=0.$$

Premultiplying this equality on the left by $b{b}^{\left(1\right)}$ and then by $a{a}^{\left(1\right)}$, similarly as in $\left(ii\right)$, we obtain

$$ax=b{b}^{\left(1\right)}ax,bx=-2ax,bx=a{a}^{\left(1\right)}bx.$$

Therefore,

$$3ax=ax-bx=ax-a{a}^{\left(1\right)}bx=a(1-a^{\left(1\right)}b)x$$

and

$$-3ax=bx-ax=bx-b{b}^{\left(1\right)}ax=b(1-b^{\left(1\right)}a)x.$$

Since $3\in {\mathcal{R}}^{-1}$, it follows that

$$ax\in a(1-a^{\left(1\right)}b)\mathcal{R}\cap b(1-b^{\left(1\right)}a)\mathcal{R}=\left\{0\right\}.$$

Thus, $ax=0$, and consequently $bx=0$. Hence, $x\in a^0\cap b^0=\left\{0\right\}$ and $x=0$. □

The following result is the left-sided analogue of the previous theorem.

Theorem 8.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$ and let $2,3\in {\mathcal{R}}^{-1}$. If $\mathcal{R}(1-b{a}^{\left(1\right)})a\cap \mathcal{R}(1-a{b}^{\left(1\right)})b=\left\{0\right\}$ and ${}^{0}a{\cap }^{0}b=\left\{0\right\},$ then

(i)

${}^0(a+b)=\left\{0\right\}$,

(ii)

${}^0(a+b+b{a}^{\left(1\right)}a)=\left\{0\right\}$,

(iii)

${}^0(a+b+a{b}^{\left(1\right)}b)=\left\{0\right\}$.

The following theorem characterizes the triviality of the null space of the difference of two inner regular elements under suitable conditions.

Theorem 9.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$. If $a+b$ and $1-a{a}^{\left(1\right)}b{b}^{\left(1\right)}$ are invertible, then

$${(a-b)}^0=\left\{0\right\}.$$

Furthermore, suppose that $2\in {\mathcal{R}}^{-1}$, $a{a}^{\left(1\right)}b=b{b}^{\left(1\right)}a$, and

$$a^2{a}^{\left(1\right)}b{b}^{\left(1\right)}=ba{a}^{\left(1\right)}b{b}^{\left(1\right)}.$$

If ${(a-b)}^0=\left\{0\right\}$ and the elements $a+b$ and $1-a{a}^{\left(1\right)}b{b}^{\left(1\right)}$ are regular, then both $a+b$ and $1-a{a}^{\left(1\right)}b{b}^{\left(1\right)}$ are left invertible.

Proof. 

Let $x\in {(a-b)}^0$, i.e., $ax=bx$. Premultiplying $ax=bx$ by $a{a}^{\left(1\right)}$ on the left, we obtain $ax=a{a}^{\left(1\right)}bx$. Similarly, we obtain $bx=b{b}^{\left(1\right)}ax$. Using $bx=b{b}^{\left(1\right)}ax$ and $ax=a{a}^{\left(1\right)}bx$, we obtain $a{a}^{\left(1\right)}b{b}^{\left(1\right)}ax=a{a}^{\left(1\right)}\left(b{b}^{\left(1\right)}ax\right)=a{a}^{\left(1\right)}bx$ and $a{a}^{\left(1\right)}b{b}^{\left(1\right)}bx=a{a}^{\left(1\right)}\left(b{b}^{\left(1\right)}bx\right)=a{a}^{\left(1\right)}bx=ax$. Further,

$$\begin{array}{c}\hfill (1-a{a}^{\left(1\right)}b{b}^{\left(1\right)})(a+b)x=ax+bx-a{a}^{\left(1\right)}bx-ax=bx-ax=0.\end{array}$$

(19)

Since $1-a{a}^{\left(1\right)}b{b}^{\left(1\right)}$ and $a+b$ are invertible, from (19) follows that $x=0$. Now, we prove the converse implication. Let $x\in {(a+b)}^0$. Then

$$ax=-bx.$$

Premultiplying this equality by $a{a}^{\left(1\right)}$ and $b{b}^{\left(1\right)}$, respectively, we obtain

$$ax=-a{a}^{\left(1\right)}bx$$

and

$$b{b}^{\left(1\right)}ax=-bx.$$

Since $a{a}^{\left(1\right)}b=b{b}^{\left(1\right)}a$, it follows that

$$ax=-b{b}^{\left(1\right)}ax=bx.$$

Hence, $2ax=0$. Since $2\in {\mathcal{R}}^{-1}$, we obtain $ax=bx=0$, and therefore

$$x\in {(a-b)}^0=\left\{0\right\}.$$

Thus, ${(a+b)}^0=\left\{0\right\}$. Since $a+b$ is regular, it follows from Lemma 2.2 in [10] that $a+b$ is left invertible. Moreover, let

$$x\in {(1-a{a}^{\left(1\right)}b{b}^{\left(1\right)})}^0.$$

Then

$$x=a{a}^{\left(1\right)}b{b}^{\left(1\right)}x.$$

Using the condition $a^2{a}^{\left(1\right)}b{b}^{\left(1\right)}=ba{a}^{\left(1\right)}b{b}^{\left(1\right)}$, we obtain

$$ax=a^2{a}^{\left(1\right)}b{b}^{\left(1\right)}x=ba{a}^{\left(1\right)}b{b}^{\left(1\right)}x=bx.$$

Hence, $x\in {(a-b)}^0=\left\{0\right\}$, and consequently

$${(1-a{a}^{\left(1\right)}b{b}^{\left(1\right)})}^0=\left\{0\right\}.$$

Since $1-a{a}^{\left(1\right)}b{b}^{\left(1\right)}$ is regular, Lemma 2.2 in [10] implies that it is left invertible. This completes the proof. □

The following example shows that, in general, the conclusion “left invertible” in the converse part of Theorem 9 cannot be replaced by “invertible”.

Example 4.

Let $H={\ell }^2\left(\mathbb{N}\right)$ and let $S\in B\left(H\right)$ be the unilateral shift,

$$S(x_1,x_2,\dots )=(0,x_1,x_2,\dots ).$$

Take

$$a=S,b=0,a^{\left(1\right)}=S^{*},b^{\left(1\right)}=0.$$

Since $S{S}^{*}S=S$, we have $a\in B{\left(H\right)}^{\left\{1\right\}}$. Moreover,

$$a{a}^{\left(1\right)}b=b{b}^{\left(1\right)}a=0$$

and

$$a^2{a}^{\left(1\right)}b{b}^{\left(1\right)}=ba{a}^{\left(1\right)}b{b}^{\left(1\right)}=0.$$

Since $N\left(S\right)=\left\{0\right\}$, the equality $ST=0$, for $T\in B\left(H\right)$, implies $T=0$. Hence,

$${(a-b)}^0=S^0=\left\{0\right\}.$$

Furthermore,

$$a+b=S,I-a{a}^{\left(1\right)}b{b}^{\left(1\right)}=I,$$

and both operators are relatively regular. Indeed,

$$S^{*}S=I,$$

so S is left invertible. However,

$$S{S}^{*}\ne I,$$

and hence S is not invertible. Thus, in the converse part of Theorem 9, left invertibility cannot, in general, be replaced by invertibility.

The corresponding result for the left null spaces is obtained analogously by multiplying on the right and preserving the order of all products.

Theorem 10.

Let $a,b\in {\mathcal{R}}^{\left\{1\right\}}$. If $a+b$ and $1-a^{\left(1\right)}a{b}^{\left(1\right)}b$ are invertible, then

$${}^0(a-b)=\left\{0\right\}.$$

Furthermore, suppose that $2\in {\mathcal{R}}^{-1}$, $b{a}^{\left(1\right)}a=a{b}^{\left(1\right)}b$, and

$$a^{\left(1\right)}a{b}^{\left(1\right)}ba=a^{\left(1\right)}a{b}^{\left(1\right)}{b}^2.$$

If ${}^0(a-b)=\left\{0\right\}$ and the elements $a+b$ and $1-a^{\left(1\right)}a{b}^{\left(1\right)}b$ are regular, then both $a+b$ and $1-a^{\left(1\right)}a{b}^{\left(1\right)}b$ are right invertible.

In the case of $\{1,5\}$-invertible elements, a related result holds.

Theorem 11.

Let $a,b\in {\mathcal{R}}^{\{1,5\}}$ such that a commutes with b and with $b^{(1,5)}$, and b commutes with $a^{(1,5)}$, and let $4\in {\mathcal{R}}^{-1}$. If $a+b+a{a}^{(1,5)}b+b{b}^{(1,5)}a$ and $1-a{a}^{(1,5)}b{b}^{(1,5)}$ are invertible, then ${(a-b)}^0=\left\{0\right\}$ and ${}^0(a-b)=\left\{0\right\}$.

Proof. 

Let $x\in {(a-b)}^0$, i.e., $ax=bx$. Then $a{a}^{(1,5)}bx=a{a}^{(1,5)}ax=ax$, and $b{b}^{(1,5)}ax=b{b}^{(1,5)}bx=bx$. Therefore, $ax+bx+a{a}^{(1,5)}bx+b{b}^{(1,5)}ax=4ax$. Further, $(1-a{a}^{(1,5)}b{b}^{(1,5)})(a+b+a{a}^{(1,5)}b+b{b}^{(1,5)}a)x=(1-a{a}^{(1,5)}b{b}^{(1,5)})4ax=4(ax-ax)=0$. Since $4\in {\mathcal{R}}^{-1}$ and the elements $a+b+a{a}^{(1,5)}b+b{b}^{(1,5)}a$ and $1-a{a}^{(1,5)}b{b}^{(1,5)}$ are invertible, it follows that $x=0$, i.e., ${(a-b)}^0=\left\{0\right\}$.

For the left null space, let $x\in {}^0(a-b)$, so that $xa=xb$. Using the commutativity assumptions, we obtain $xa{a}^{(1,5)}b=xba{a}^{(1,5)}=x{a}^2{a}^{(1,5)}=xa$ and, analogously, $xb{b}^{(1,5)}a=xa.$ Hence, $x(a+b+a{a}^{(1,5)}b+b{b}^{(1,5)}a)=4xa.$ Moreover, $x{a}^2{a}^{(1,5)}b{b}^{(1,5)}=xab{b}^{(1,5)}=x{b}^2{b}^{(1,5)}=xb=xa.$ Therefore, $x(a+b+a{a}^{(1,5)}b+b{b}^{(1,5)}a)(1-a{a}^{(1,5)}b{b}^{(1,5)})=0.$ The invertibility assumptions and $4\in {\mathcal{R}}^{-1}$ now imply $x=0$. Thus, ${}^0(a-b)=\left\{0\right\}$. □

Remark 1.

Note that the following result for $\{1,5\}$-invertible elements can be proved analogously to Theorem 11: Let $a,b\in {\mathcal{R}}^{\{1,5\}}$ such that a commutes with b and with $b^{(1,5)}$, and b commutes with $a^{(1,5)}$, and let $2\in {\mathcal{R}}^{-1}$. If $a+b$ and $1-a{a}^{(1,5)}b{b}^{(1,5)}$ are invertible, then ${(a-b)}^0=\left\{0\right\}$ and ${}^0(a-b)=\left\{0\right\}$. Moreover, observe that a more general proposition of this type is a corollary of Theorem 9.

Before proving the following decomposition results, we first establish some elementary properties of $\{1,5\}$-invertible elements.

Lemma 1.

Let $a\in {\mathcal{R}}^{\{1,5\}}$. Then

$$a\mathcal{R}=a{a}^{(1,5)}\mathcal{R},a^0={\left(a{a}^{(1,5)}\right)}^0,$$

and

$$a\mathcal{R}\cap a^0=\left\{0\right\}.$$

Analogously,

$$\mathcal{R}a=\mathcal{R}a{a}^{(1,5)},{}^{0}a={}^0\left(a{a}^{(1,5)}\right),\mathcal{R}a\cap {}^{0}a=\left\{0\right\}.$$

Proof. 

Since $a{a}^{(1,5)}a=a,$ we have $a\mathcal{R}\subseteq a{a}^{(1,5)}\mathcal{R}.$ On the other hand, $a{a}^{(1,5)}\mathcal{R}\subseteq a\mathcal{R},$ and therefore $a\mathcal{R}=a{a}^{(1,5)}\mathcal{R}.$ Since $a{a}^{(1,5)}=a^{(1,5)}a,$ if $x\in a^0$, then $a{a}^{(1,5)}x=a^{(1,5)}ax=0.$ Hence, $a^0\subseteq {\left(a{a}^{(1,5)}\right)}^0.$

Conversely, let $x\in {\left(a{a}^{(1,5)}\right)}^0.$ Using $a{a}^{(1,5)}=a^{(1,5)}a$ and $a{a}^{(1,5)}a=a,$ we obtain $a^2{a}^{(1,5)}=a\left(a{a}^{(1,5)}\right)=a\left(a^{(1,5)}a\right)=a{a}^{(1,5)}a=a.$ Therefore, $ax=a^2{a}^{(1,5)}x=a\left(a{a}^{(1,5)}x\right)=0.$ Thus, $a^0={\left(a{a}^{(1,5)}\right)}^0.$

Finally, let $x\in a\mathcal{R}\cap a^0$. Then $x=au$ for some $u\in \mathcal{R}$ and $ax=0$. Moreover,

$$a^{(1,5)}{a}^2=\left(a^{(1,5)}a\right)a=\left(a{a}^{(1,5)}\right)a=a{a}^{(1,5)}a=a.$$

Hence,

$$x=au=a^{(1,5)}{a}^{2}u=a^{(1,5)}ax=0.$$

Therefore, $a\mathcal{R}\cap a^0=\left\{0\right\}.$

The left-sided identities follow similarly by multiplication on the right. □

In the following theorem, we investigate a decomposition of the right null space associated with the difference of $\{1,5\}$-invertible elements.

Theorem 12.

Let $a,b\in {\mathcal{R}}^{\{1,5\}}$. Then ${(a{a}^{(1,5)}-b{b}^{(1,5)})}^0=(a\mathcal{R}\cap b\mathcal{R})\oplus (a^0\cap b^0)$.

Proof. 

We begin by proving the equality ${(a{a}^{(1,5)}-b{b}^{(1,5)})}^0=(a\mathcal{R}\cap b\mathcal{R})+(a^0\cap b^0)$. Note that $a^0\cap b^0\subseteq {(a{a}^{(1,5)}-b{b}^{(1,5)})}^0$. Further, suppose that $x\in a\mathcal{R}\cap b\mathcal{R}$. Then there exist $u,v\in \mathcal{R}$ such that $x=au=bv$. Premultiplying the equalities $x=au=bv$ on the left by $a{a}^{(1,5)}$ and $b{b}^{(1,5)}$, respectively, yields $x=a{a}^{(1,5)}x=b{b}^{(1,5)}x$. Thus, $x\in {(a{a}^{(1,5)}-b{b}^{(1,5)})}^0$. This proves the inclusion $(a\mathcal{R}\cap b\mathcal{R})+(a^0\cap b^0)\subseteq {(a{a}^{(1,5)}-b{b}^{(1,5)})}^0$. We now prove the reverse inclusion. For any $x\in {(a{a}^{(1,5)}-b{b}^{(1,5)})}^0$, we obtain $a{a}^{(1,5)}x=b{b}^{(1,5)}x\in a\mathcal{R}\cap b\mathcal{R}$. Premultiplying both sides of the equality $a{a}^{(1,5)}x=b{b}^{(1,5)}x$ on the left by a, it follows that $ax=ab{b}^{(1,5)}x$, i.e., $x-b{b}^{(1,5)}x\in a^0$. It is evident that $x-b{b}^{(1,5)}x\in b^0$. The decomposition $x=b{b}^{(1,5)}x+(x-b{b}^{(1,5)}x)\in (a\mathcal{R}\cap b\mathcal{R})+(a^0\cap b^0)$ establishes the reverse inclusion of the spaces.

Furthermore, we prove that the sum $(a\mathcal{R}\cap b\mathcal{R})+(a^0\cap b^0)$ is direct. By Lemma 1, we have $a\mathcal{R}\cap a^0=\left\{0\right\}.$ Therefore,

$$(a\mathcal{R}\cap b\mathcal{R})\cap (a^0\cap b^0)\subseteq a\mathcal{R}\cap a^0=\left\{0\right\}.$$

Hence, ${(a{a}^{(1,5)}-b{b}^{(1,5)})}^0=(a\mathcal{R}\cap b\mathcal{R})\oplus (a^0\cap b^0).$ □

Example 5.

Let $R=M_3\left(\mathbb{F}\right)$, where $\mathbb{F}$ is either $\mathbb{R}$ or $\mathbb{C}$, and let

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

Since $a^2=a$ and $b^2=b$, we may take

$$a^{(1,5)}=a,b^{(1,5)}=b.$$

Thus, $a,b\in R^{\{1,5\}}$.

We have

$$aR=\left\{\left(\begin{array}{ccc}{x}_1& x_2& x_3\\ 0& 0& 0\\ 0& 0& 0\end{array}\right):x_1,x_2,x_3\in \mathbb{F}\right\},$$

and

$$bR=\left\{\left(\begin{array}{ccc}0& 0& 0\\ y_1& y_2& y_3\\ 0& 0& 0\end{array}\right):y_1,y_2,y_3\in \mathbb{F}\right\}.$$

Hence,

$$aR\cap bR=\left\{0\right\}.$$

On the other hand,

$$a^0=\left\{\left(\begin{array}{ccc}0& 0& 0\\ x_1& x_2& x_3\\ y_1& y_2& y_3\end{array}\right):x_i,y_i\in \mathbb{F}\right\},$$

whereas

$$b^0=\left\{\left(\begin{array}{ccc}{x}_1& x_2& x_3\\ 0& 0& 0\\ y_1& y_2& y_3\end{array}\right):x_i,y_i\in \mathbb{F}\right\}.$$

Therefore,

$$a^0\cap b^0=\left\{\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ y_1& y_2& y_3\end{array}\right):y_1,y_2,y_3\in \mathbb{F}\right\}\ne \left\{0\right\}.$$

Moreover,

$$a{a}^{(1,5)}-b{b}^{(1,5)}=a-b=\left(\begin{array}{ccc}1& 0& 0\\ 0& -1& 0\\ 0& 0& 0\end{array}\right),$$

and therefore

$${(a{a}^{(1,5)}-b{b}^{(1,5)})}^0=\left\{\left(\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ y_1& y_2& y_3\end{array}\right):y_1,y_2,y_3\in \mathbb{F}\right\}.$$

Consequently,

$${(a{a}^{(1,5)}-b{b}^{(1,5)})}^0=(aR\cap bR)\oplus (a^0\cap b^0)=a^0\cap b^0\ne \left\{0\right\},$$

which illustrates Theorem 12.

Using the left-sided identities from Lemma 1 and arguing analogously, with multiplication on the right, we obtain the following result.

Theorem 13.

Let $a,b\in {\mathcal{R}}^{\{1,5\}}$. Then ${}^0(a{a}^{(1,5)}-b{b}^{(1,5)})=(\mathcal{R}a\cap \mathcal{R}b)\oplus {(}^{0}a{\cap }^{0}b)$.

2.3. Invertibility of Sums and Differences

The following theorem presents necessary and sufficient conditions for the invertibility of the sum and the difference of two $\{1,5\}$-invertible elements whose $\{1,5\}$-inverses are mutually orthogonal.

Theorem 14.

Let $a,b\in {\mathcal{R}}^{\{1,5\}}$ such that $a^{(1,5)}\perp b^{(1,5)}$. Then the following conditions are equivalent:

(i)

$a+b\in {\mathcal{R}}^{-1}$,

(ii)

$a-b\in {\mathcal{R}}^{-1}$,

(iii)

$a{a}^{(1,5)}+b{b}^{(1,5)}=1$,

(iv)

$a\mathcal{R}\oplus b\mathcal{R}=\mathcal{R}$,

(v)

$a^0\oplus b^0=\mathcal{R}$.

Proof. 

First, we note that from the assumption $a^{(1,5)}\perp b^{(1,5)}$, i.e., $a^{(1,5)}{b}^{(1,5)}=b^{(1,5)}{a}^{(1,5)}=0$ it follows that $ab=ba=0$, $a^{(1,5)}b=b{a}^{(1,5)}=0$ and $a{b}^{(1,5)}=b^{(1,5)}a=0$. Indeed, since $a=a^2{a}^{(1,5)}=a^{(1,5)}{a}^2$ and $b=b^2{b}^{(1,5)}=b^{(1,5)}{b}^2,$ we have

$$a^{(1,5)}b=a^{(1,5)}{b}^{(1,5)}{b}^2=0,b{a}^{(1,5)}=b^2{b}^{(1,5)}{a}^{(1,5)}=0,$$

and, similarly,

$$a{b}^{(1,5)}=a^2{a}^{(1,5)}{b}^{(1,5)}=0,b^{(1,5)}a=b^{(1,5)}{a}^{(1,5)}{a}^2=0.$$

Therefore,

$$ab=a^2{a}^{(1,5)}b=0$$

and

$$ba=b^2{b}^{(1,5)}a=0.$$

$\left(i\right)\iff \left(ii\right)$ When we use the equality $ab=0=ba$, we obtain ${(a+b)}^2={(a-b)}^2=a^2+b^2$. Therefore, $\left(i\right)$ is equivalent to $\left(ii\right)$.

$\left(i\right)\Rightarrow \left(iii\right)$ Using $a^2{a}^{(1,5)}=a$, $b^2{b}^{(1,5)}=b$ and $ab=ba=0$, we get

$$(a+b)(a{a}^{(1,5)}+b{b}^{(1,5)})=aa{a}^{(1,5)}+bb{b}^{(1,5)}=a+b.$$

Since $a+b$ is invertible, then $a{a}^{(1,5)}+b{b}^{(1,5)}=1$.

$\left(iii\right)\Rightarrow \left(i\right)$ Using the equalities $a{b}^{(1,5)}=0$, $b{a}^{(1,5)}=0$ and $a{a}^{(1,5)}+b{b}^{(1,5)}=1$, we obtain

$$(a+b)(a^{(1,5)}+b^{(1,5)})=a{a}^{(1,5)}+b{b}^{(1,5)}=1.$$

Thus, $a+b$ is right invertible.

Moreover, using $a^{(1,5)}b=0$, $b^{(1,5)}a=0$ and $a^{(1,5)}a=a{a}^{(1,5)}$, $b^{(1,5)}b=b{b}^{(1,5)}$, we have

$$(a^{(1,5)}+b^{(1,5)})(a+b)=a^{(1,5)}a+b^{(1,5)}b=a{a}^{(1,5)}+b{b}^{(1,5)}=1.$$

Hence, $a+b$ is also left invertible, and therefore $a+b$ is invertible.

$\left(iii\right)\Rightarrow \left(iv\right)$ First, we prove that $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$.

Let $x\in a\mathcal{R}\cap b\mathcal{R}$. Then there exist $u,v\in \mathcal{R}$ such that $x=au=bv$. Premultiplying $au=bv$ by $a{a}^{(1,5)}$ leads to $au=0$, i.e., $x=0$. Hence, $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$.

To prove $a\mathcal{R}+b\mathcal{R}=\mathcal{R}$, it is sufficient to prove $1\in a\mathcal{R}+b\mathcal{R}$. Indeed, $1=a{a}^{(1,5)}+b{b}^{(1,5)}\in a\mathcal{R}+b\mathcal{R}$.

$\left(iv\right)\Rightarrow \left(iii\right)$ Since $a\mathcal{R}+b\mathcal{R}=\mathcal{R}$ and $1\in \mathcal{R}$, there exist $u,v\in \mathcal{R}$ such that

$$1=au+bv.$$

(20)

Premultiplying (20) by $a{a}^{(1,5)}$ and using $a^{(1,5)}b=0$, we obtain $a{a}^{(1,5)}=au$. Analogously, the equality $b{b}^{(1,5)}=bv$ holds. Substituting in (20) gives $a{a}^{(1,5)}+b{b}^{(1,5)}=1$.

$\left(iii\right)\Rightarrow \left(v\right)$ In order to prove that $a^0\oplus b^0=\mathcal{R}$, we prove that $a^0+b^0=\mathcal{R}$. The condition $\left(iii\right)$ implies $1=1-a{a}^{(1,5)}+1-b{b}^{(1,5)}\in a^0+b^0$. Hence, $a^0+b^0=\mathcal{R}$.

Next, we prove $a^0\cap b^0=\left\{0\right\}$. Let $x\in a^0\cap b^0$. Then $ax=0=bx$, i.e., $a^{(1,5)}ax=0=b^{(1,5)}bx$. Now, applying the condition $\left(iii\right)$, we obtain $0=a^{(1,5)}ax+b^{(1,5)}bx=(a^{(1,5)}a+b^{(1,5)}b)x=x$. Thus, $a^0\cap b^0=\left\{0\right\}$.

$\left(v\right)\Rightarrow \left(iii\right)$ Since $a^0+b^0=\mathcal{R}$ and $1\in \mathcal{R}$, there exist $u\in a^0$ and $v\in b^0$ such that

$$1=u+v.$$

(21)

The previous equality leads to $a=av$ and $b=bu$. Now, from $0=a-a=a{a}^{(1,5)}a-av=a(a^{(1,5)}a-v)$, it follows that $a^{(1,5)}a-v\in a^0$. Using $b{a}^{(1,5)}=0$ and $bv=0$, we deduce $b(a^{(1,5)}a-v)=0$, i.e., $a^{(1,5)}a-v\in b^0$. Hence, $a^{(1,5)}a-v\in a^0\cap b^0=\left\{0\right\}$, i.e., $a^{(1,5)}a=v$. Similarly, we have $b^{(1,5)}b=u$. Finally, $a^{(1,5)}a=v$, $b^{(1,5)}b=u$ and (21) give that the equality $\left(iii\right)$ holds. □

Remark 2.

From the proof of Theorem 14 we conclude that the condition $\left(iv\right)$ can be replaced by the condition $a\mathcal{R}+b\mathcal{R}=\mathcal{R}$.

Remark 3.

Under the assumptions of Theorem 14, it holds $a{a}^{(1,5)}+b{b}^{(1,5)}=1\iff {(a{a}^{(1,5)}+b{b}^{(1,5)})}^n=1\iff {(a{a}^{(1,5)}-b{b}^{(1,5)})}^{2n}=1,$ $n\in \mathbb{N}$.

Example 6.

Let $\mathcal{R}=M_2\left(\mathbb{C}\right)$ and consider

$$a=\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right),b=\left(\begin{array}{cc}0& 0\\ 0& 3\end{array}\right).$$

Then $a,b\in {\mathcal{R}}^{\{1,5\}}$, with

$$a^{(1,5)}=\left(\begin{array}{cc}{\displaystyle \frac{1}{2}}& 0\\ 0& 0\end{array}\right),b^{(1,5)}=\left(\begin{array}{cc}0& 0\\ 0& {\displaystyle \frac{1}{3}}\end{array}\right).$$

Clearly,

$$a^{(1,5)}{b}^{(1,5)}=b^{(1,5)}{a}^{(1,5)}=0,$$

and hence $a^{(1,5)}\perp b^{(1,5)}$. Moreover,

$$a{a}^{(1,5)}+b{b}^{(1,5)}=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I_2.$$

Also,

$$a+b=\left(\begin{array}{cc}2& 0\\ 0& 3\end{array}\right),a-b=\left(\begin{array}{cc}2& 0\\ 0& -3\end{array}\right),$$

are invertible.

Furthermore,

$$a\mathcal{R}=\left\{\left(\begin{array}{cc}x& y\\ 0& 0\end{array}\right):x,y\in \mathbb{C}\right\},b\mathcal{R}=\left\{\left(\begin{array}{cc}0& 0\\ x& y\end{array}\right):x,y\in \mathbb{C}\right\},$$

and therefore

$$a\mathcal{R}\oplus b\mathcal{R}=M_2\left(\mathbb{C}\right).$$

Similarly,

$$a^0=\left\{\left(\begin{array}{cc}0& 0\\ x& y\end{array}\right):x,y\in \mathbb{C}\right\},b^0=\left\{\left(\begin{array}{cc}x& y\\ 0& 0\end{array}\right):x,y\in \mathbb{C}\right\},$$

so that

$$a^0\oplus b^0=M_2\left(\mathbb{C}\right).$$

Thus, all the equivalent conditions of Theorem 14 are satisfied in this example.

Some formulae are given in the following corollary.

Corollary 1.

Let $a,b\in {\mathcal{R}}^{\{1,5\}}$ such that $a^{(1,5)}\perp b^{(1,5)}$. If one of the conditions $\left(i\right)–\left(v\right)$ of the preceding theorem holds, then

(i)

${(a\pm b)}^{-1}=a^{(1,5)}\pm b^{(1,5)}$.

(ii)

$a^{(1,5)}={(a+b)}^{-1}a{(a+b)}^{-1}={(a+b)}^{-1}a{(a-b)}^{-1}={(a-b)}^{-1}a{(a-b)}^{-1}$;

$b^{(1,5)}={(a+b)}^{-1}b{(a+b)}^{-1}=-{(a+b)}^{-1}b{(a-b)}^{-1}={(a-b)}^{-1}b{(a-b)}^{-1}$.

(iii)

$a=a{(a+b)}^{-1}a=a{(a-b)}^{-1}a$; $b=b{(a+b)}^{-1}b=-b{(a-b)}^{-1}b$.

(iv)

${(a{a}^{(1,5)}-b{b}^{(1,5)})}^{-1}={(a+b)}^{-1}{(a^{(1,5)}-b^{(1,5)})}^{-1}={(a-b)}^{-1}{(a^{(1,5)}+b^{(1,5)})}^{-1}$.

Proof. 

$\left(i\right)$ The equality ${(a+b)}^{-1}=a^{(1,5)}+b^{(1,5)}$ follows from the proof of Theorem 14. Similarly, condition $\left(iii\right)$ of Theorem 14 implies

$$(a-b)(a^{(1,5)}-b^{(1,5)})=a{a}^{(1,5)}+b{b}^{(1,5)}=1.$$

Moreover, using $a^{(1,5)}b=b^{(1,5)}a=0$ and $a^{(1,5)}a=a{a}^{(1,5)}$, $b^{(1,5)}b=b{b}^{(1,5)}$, we obtain

$$(a^{(1,5)}-b^{(1,5)})(a-b)=a^{(1,5)}a+b^{(1,5)}b=a{a}^{(1,5)}+b{b}^{(1,5)}=1.$$

Thus, the equality

$${(a-b)}^{-1}=a^{(1,5)}-b^{(1,5)}$$

holds.

$\left(ii\right)$ From $(a+b)a^{(1,5)}(a+b)=(a+b)a^{(1,5)}(a-b)=(a-b)a^{(1,5)}(a-b)=a$ follows the first part of item $\left(ii\right)$.

Similarly to the previous equality, we prove the second part of item $\left(ii\right)$.

$\left(iii\right)$ The identity $a{a}^{(1,5)}(a+b)=a$ implies $a{a}^{(1,5)}=a{(a+b)}^{-1}$, i.e., $a=a{(a+b)}^{-1}a$.

The other equality in (iii) follows analogously.

$\left(iv\right)$ Using the identities established above, we have

$$\begin{array}{cc}\hfill (a+b)(a^{(1,5)}-b^{(1,5)})& =(a^{(1,5)}-b^{(1,5)})(a+b)\hfill \\ & =a{a}^{(1,5)}-b{b}^{(1,5)}\hfill \\ & =(a-b)(a^{(1,5)}+b^{(1,5)})\hfill \\ & =(a^{(1,5)}+b^{(1,5)})(a-b).\hfill \end{array}$$

Since all the factors are invertible by $\left(i\right)$, it follows that

$${(a{a}^{(1,5)}-b{b}^{(1,5)})}^{-1}={(a+b)}^{-1}{(a^{(1,5)}-b^{(1,5)})}^{-1}={(a-b)}^{-1}{(a^{(1,5)}+b^{(1,5)})}^{-1}.$$

□

Proposition 1.

Let $\mathcal{R}$ be a unital algebra with unity 1 and null 0. Let $a,b\in {\mathcal{R}}^{\{1,5\}}$ and $a\ne 0$.

(i)

If $ba=a^2,$ then $a-b$ is noninvertible.

(ii)

If $ba=-a^2$, then $a+b$ is noninvertible.

Proof. 

$\left(i\right)$ Suppose that $a-b$ is invertible. Using $ba=a^2$ and, by the commutativity condition $a{a}^{(1,5)}=a^{(1,5)}a$, $a^2{a}^{(1,5)}=a{a}^{(1,5)}a=a,$ we have

$$(a-b)(1-a{a}^{(1,5)})=a-b.$$

Since $a-b$ is invertible, then $1-a{a}^{(1,5)}=1$, i.e., $a{a}^{(1,5)}=0$. Hence, $a=0$, which contradicts the assumption $a\ne 0$. Thus, $a-b$ is noninvertible.

$\left(ii\right)$ The proof is analogous to $\left(i\right)$. □

Let us have a look at arbitrary $a\in {\mathcal{R}}^{\{1,5\}}$ with $a^{(1,5)}=a^m$ for some natural $m>1$. From $a^{♯}=a^{(1,5)}=a^m$, note that a is $(m+2)$-potent. Indeed,

$$a^{m+2}=a{a}^{m}a=a{a}^{♯}a=a.$$

Now, as $ax=a^{m+2}x=a^m\left(a^{2}x\right)$ for every $x\in \mathcal{R}$, it easily follows $a\mathcal{R}=a^m\mathcal{R}$ and $a^0={\left(a^m\right)}^0$.

Before stating the next result, we note that, for commuting elements $a,b\in \mathcal{R}$,

$$a^{m+1}-b^{m+1}=(a-b){\displaystyle \sum _{i=0}^m}{a}^{m-i}{b}^i.$$

If $a^{♯}=a^m$ and $b^{♯}=b^m$, then

$${\displaystyle \sum _{i=0}^m}{a}^{m-i}{b}^i=a^{♯}+{\displaystyle \sum _{i=1}^{m-1}}{a}^{m-i}{b}^i+b^{♯}.$$

Thus, the latter expression arises naturally as the second factor in the above factorization.

Theorem 15.

Let $a,b\in {\mathcal{R}}^{♯}$ such that $a^{♯}=a^m$ and $b^{♯}=b^m$ for some natural $m>1$. If $ab=ba$, then the following conditions are equivalent:

(i)

$a-b$ and $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ are invertible,

(ii)

$a\mathcal{R}\oplus b\mathcal{R}=\mathcal{R}$,

(iii)

$a^0\oplus b^0=\mathcal{R}$,

(iv)

$a-b$ and $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ are regular, $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$ and $a^0\cap b^0=\left\{0\right\}$.

Proof. 

$\left(iv\right)\Rightarrow \left(i\right)$ We first note that the corresponding left-sided conditions also follow from the assumptions in (iv). Indeed, since $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$ and $ab=ba$, we have $ab=ba=0$. If $x\in \mathcal{R}a\cap \mathcal{R}b$, then $x=ua=vb$ for some $u,v\in \mathcal{R}$, and hence $x=x{a}^{m+1}=x{b}^{m+1}=x{a}^{m+1}{b}^{m+1}=0$. Thus, $\mathcal{R}a\cap \mathcal{R}b=\left\{0\right\}$.

Further, put $p=a^{m+1}$ and $q=b^{m+1}$. Then p and q are commuting idempotents, $a^0=p^0$, $b^0=q^0$, and $pq=0$. Hence, $(1-p)(1-q)\in p^0\cap q^0=\left\{0\right\}$. If $x\in {}^{0}a\cap {}^{0}b$, then $xp=xq=0$, and therefore $x=x(1-p)(1-q)=0$. Thus, ${}^{0}a\cap {}^{0}b=\left\{0\right\}$.

First, we show that ${(a-b)}^0=\left\{0\right\}$. Let $(a-b)x=0$. Then $ax=bx\in a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$, and $x\in a^0\cap b^0=\left\{0\right\}$. Hence, $x=0$. From the regularity of $a-b$ and ${(a-b)}^0=\left\{0\right\}$ it follows by Lemma 2.2. [10] that $a-b$ is left invertible. Using the left-sided conditions established above, the analogous argument applied to the left null space shows that $a-b$ is right invertible. Thus, $a-b$ is invertible.

Now, we prove that $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ is invertible. Let $x\in {(a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯})}^0$, i.e., $(a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯})x=0$. Then $a^{m-i}{b}^i=b^i{a}^{m-i}\in a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$. Hence, $(a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯})x=0\iff a^{♯}x+b^{♯}x=0$. Since $a^{♯}=a^m$ and $b^{♯}=b^m$, we obtain $a^{m}x=-b^{m}x\in a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$. Hence, $x\in {\left(a^m\right)}^0\cap {\left(b^m\right)}^0$. Since $a^{m+2}=a$ and $b^{m+2}=b$, it follows that $ax=a^2\left(a^{m}x\right)=0$ and $bx=b^2\left(b^{m}x\right)=0$. Therefore, $x\in a^0\cap b^0=\left\{0\right\}$. By Lemma 2.2. [10] it follows that $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ is left invertible. Using the corresponding left-sided conditions established above and applying the analogous argument to the left null space, we conclude that $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ is right invertible. Thus, $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ is invertible.

$\left(i\right)\Rightarrow \left(iv\right)$ If $a-b$ and $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ are invertible, then $a-b$ and $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ are regular.

Let $x\in a\mathcal{R}\cap b\mathcal{R}$. Then $x=au=bv$ for some $u,v\in \mathcal{R}$. Also, we have $x=a^{m+1}x=b^{m+1}x$ and $(a^{m+1}-b^{m+1})x=0$. Since $a^{m+1}-b^{m+1}=(a-b){\sum }_{i=0}^m{a}^{m-i}{b}^i=(a-b)(a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯})$ is invertible, then $(a^{m+1}-b^{m+1})x=0\Rightarrow x=0$. Therefore, the condition $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$ holds.

Let $u\in \mathcal{R}$ be the inverse of the sum $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$, i.e., of the sum ${\sum }_{i=0}^m{a}^{m-i}{b}^i$. From $1=u\left({\sum }_{i=0}^m{a}^{m-i}{b}^i\right)=u\left({\sum }_{i=0}^{m-1}{a}^{m-1-i}{b}^i\right)a+u{b}^m$ we obtain that $a^0\cap b^0=\left\{0\right\}$. Indeed, if $x\in a^0\cap b^0$, then $ax=0$ and $b^{m}x=0$, and hence $x=u\left({\sum }_{i=0}^{m-1}{a}^{m-1-i}{b}^i\right)ax+u{b}^{m}x=0.$

$\left(i\right)\Rightarrow \left(ii\right)$ Since the item $\left(i\right)$ is equivalent to the item $\left(iv\right)$, we prove only that $a\mathcal{R}+b\mathcal{R}=\mathcal{R}$ is satisfied. From the invertibility of the sum $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$, i.e., of the sum ${\sum }_{i=0}^m{a}^{m-i}{b}^i$, we conclude that exists $u\in \mathcal{R}$ such that $1=\left({\sum }_{i=0}^m{a}^{m-i}{b}^i\right)u=a\left({\sum }_{i=0}^{m-1}{a}^{m-1-i}{b}^i\right)u+b{b}^{m-1}u\in a\mathcal{R}+b\mathcal{R}$. Thus, the condition $a\mathcal{R}+b\mathcal{R}=\mathcal{R}$ is satisfied.

$\left(ii\right)\Rightarrow \left(i\right)$ Suppose that $a\mathcal{R}\oplus b\mathcal{R}=\mathcal{R}$. Since $ab=ba\in a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$, we have $ab=ba=0$. Let $1=au+bv$ for some $u,v\in \mathcal{R}$. Then $a^{m+1}=a^{m+1}(au+bv)=au$ and, similarly, $b^{m+1}=bv$. Hence, $a^{m+1}+b^{m+1}=1$. Since $ab=ba=0$, all mixed terms vanish, and therefore $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}=a^m+b^m$. Now, $(a-b)(a^m-b^m)=(a^m-b^m)(a-b)=a^{m+1}+b^{m+1}=1$, so $a-b$ is invertible. Similarly, $(a^m+b^m)(a+b)=(a+b)(a^m+b^m)=a^{m+1}+b^{m+1}=1$. Thus, $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ is invertible.

$\left(ii\right)\Rightarrow \left(iii\right)$ Since the item $\left(ii\right)$ is equivalent to the item $\left(iv\right)$, the condition $a^0\cap b^0=\left\{0\right\}$ holds. To prove $a^0+b^0=\mathcal{R}$ it is sufficient to prove $1\in a^0+b^0$. From $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$ follows that $1=au+bv$ for some $u,v\in \mathcal{R}$. Then $abv\in a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$, and $bv\in a^0$. Similarly, $bau\in a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$, and $au\in b^0$. Now, we get $1=au+bv\in a^0+b^0$. Hence, $a^0+b^0=\mathcal{R}$, and therefore item $\left(iii\right)$ holds.

$\left(iii\right)\Rightarrow \left(ii\right)$ Put $p=a^{m+1}$ and $q=b^{m+1}$. Since $a^{m+2}=a$ and $b^{m+2}=b$, the elements p and q are commuting idempotents, and $a\mathcal{R}=p\mathcal{R}$, $b\mathcal{R}=q\mathcal{R}$, $a^0=p^0$, and $b^0=q^0$. Hence, by (iii), $p^0\oplus q^0=\mathcal{R}$. Since $p^0=(1-p)\mathcal{R}$ and $q^0=(1-q)\mathcal{R}$, we have $(1-p)(1-q)\in p^0\cap q^0=\left\{0\right\}$, and therefore $1-p-q+pq=0$. Moreover, $p^0+q^0=\mathcal{R}$, so there exist $u,v\in \mathcal{R}$ such that $1=u+v$, $pu=0$, and $qv=0$. Since $pq=qp$, $pq=pq(u+v)=qpu+pqv=0$. Thus, $p+q=1$, and hence $p\mathcal{R}+q\mathcal{R}=\mathcal{R}$. If $x\in p\mathcal{R}\cap q\mathcal{R}$, then $x=pr=qs$ for some $r,s\in \mathcal{R}$, and $qx=qpr=0$, whereas $qx=x$. Thus, $x=0$, and consequently $p\mathcal{R}\oplus q\mathcal{R}=\mathcal{R}$. Since $a\mathcal{R}=p\mathcal{R}$ and $b\mathcal{R}=q\mathcal{R}$, it follows that $a\mathcal{R}\oplus b\mathcal{R}=\mathcal{R}$. □

Example 7.

Let $\mathcal{R}=M_2\left(\mathbb{C}\right)$ and let $\omega =e^{2\pi i/3}$. Consider

$$a=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right),b=\left(\begin{array}{cc}0& 0\\ 0& \omega \end{array}\right).$$

Then $ab=ba=0$ and

$$a^{♯}=a=a^2,b^{♯}=\left(\begin{array}{cc}0& 0\\ 0& {\omega }^2\end{array}\right)=b^2.$$

Thus, the assumptions of Theorem 15 are satisfied for $m=2$. Moreover,

$$a-b=\left(\begin{array}{cc}1& 0\\ 0& -\omega \end{array}\right)$$

and

$$a^{♯}+ab+b^{♯}=\left(\begin{array}{cc}1& 0\\ 0& {\omega }^2\end{array}\right)$$

are invertible. Hence, by Theorem 15,

$$a\mathcal{R}\oplus b\mathcal{R}=M_2\left(\mathbb{C}\right)\mathit{and}{a}^0\oplus b^0=M_2\left(\mathbb{C}\right).$$

Indeed,

$$a\mathcal{R}=\left\{\left(\begin{array}{cc}x& y\\ 0& 0\end{array}\right):x,y\in \mathbb{C}\right\},b\mathcal{R}=\left\{\left(\begin{array}{cc}0& 0\\ x& y\end{array}\right):x,y\in \mathbb{C}\right\},$$

and hence $a\mathcal{R}\oplus b\mathcal{R}=M_2\left(\mathbb{C}\right)$. Moreover,

$$a^0=\left\{\left(\begin{array}{cc}0& 0\\ x& y\end{array}\right):x,y\in \mathbb{C}\right\},b^0=\left\{\left(\begin{array}{cc}x& y\\ 0& 0\end{array}\right):x,y\in \mathbb{C}\right\},$$

so that $a^0\oplus b^0=M_2\left(\mathbb{C}\right)$.

In the following corollary, we give the form of the inverse of the difference of two commuting elements $a,b\in {\mathcal{R}}^{\{1,5\}}$ such that $a^{♯}=a^m$ and $b^{♯}=b^m$ for some natural $m>1$.

Corollary 2.

Let $a,b\in {\mathcal{R}}^{♯}$ such that $ab=ba$, $a^{♯}=a^m$ and $b^{♯}=b^m$ for some natural $m>1$. If one of the conditions $\left(i\right)–\left(iv\right)$ of the preceding theorem holds, then ${(a-b)}^{-1}=a^{♯}-b^{♯}$.

Proof. 

From the proof of $\left(iii\right)\Rightarrow \left(ii\right)$ of Theorem 15, we get $1=a^{m+1}+b^{m+1}$, $a^{m+1}\in b^0$ and $b^{m+1}\in a^0$. Since $a^{m+2}=a$ and $b^{m+2}=b$, we have

$$a^{2m+1}=a^{m-1}{a}^{m+2}=a^m=a^{♯}and{b}^{2m+1}=b^{m-1}{b}^{m+2}=b^m=b^{♯}.$$

Moreover, since $ab=ba$, every mixed term containing positive powers of both a and b belongs to $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$ and therefore vanishes. Hence,

$$\begin{array}{cc}\hfill 1& =(a-b){\displaystyle \sum _{i=0}^m}{a}^{m-i}{b}^i(a^{m+1}-b^{m+1})\hfill \\ & =(a-b)(a^{2m+1}-b^{2m+1})\hfill \\ & =(a-b)(a^{♯}-b^{♯}).\hfill \end{array}$$

Therefore, ${(a-b)}^{-1}=a^{♯}-b^{♯}$. □

The next result shows that if one of the conditions $\left(i\right)–\left(iv\right)$ of Theorem 15 holds, then the sum is also invertible.

Corollary 3.

Let $a,b\in {\mathcal{R}}^{♯}$ such that $ab=ba$, $a^{♯}=a^m$ and $b^{♯}=b^m$ for some natural $m>1$. If one of the conditions $\left(i\right)–\left(iv\right)$ of the preceding theorem holds, then $(a+b)\in {\mathcal{R}}^{-1}$.

Proof. 

As established in the proof of $\left(iii\right)\Rightarrow \left(ii\right)$ of Theorem 15, we have $1=a^{m+1}+b^{m+1}$ and $a\mathcal{R}\cap b\mathcal{R}=\left\{0\right\}$. Moreover, the commutativity of a and b implies that all mixed terms occurring below lie in $a\mathcal{R}\cap b\mathcal{R}$ and hence are zero. Therefore,

$$\begin{array}{cc}& (a+b)\left(a^{♯}+{\displaystyle \sum _{i=1}^{m-1}}{a}^{m-i}{b}^i+b^{♯}\right)(a^{m+1}+b^{m+1})\hfill \\ & =(a+b)(a^m+b^m)(a^{m+1}+b^{m+1})\hfill \\ & =a^{2m+2}+b^{2m+2}=a^{m+1}+b^{m+1}=1.\hfill \end{array}$$

Since a and b commute, the element $a^{♯}+{\sum }_{i=1}^{m-1}{a}^{m-i}{b}^i+b^{♯}$ commutes with $a+b$. Hence, it is also a left inverse of $a+b$. Therefore, $a+b$ is invertible. □

3. Applications to Banach and Hilbert Space Operators

3.1. Applications to Banach Space Operators

In this subsection, we turn our attention to the case when $\mathcal{A}$ is the Banach algebra $B\left(X\right)$ of all bounded linear operators on a Banach space X. The characterization of when an operator $A\in B\left(X\right)$ has a generalized inverse in $B\left(X\right)$ and methods of the construction of a generalized inverse are well-known ([1,2,5,6]). Concepts of inner generalized and group inverses of Banach space operators are similar to those in rings: we say that $A\in B\left(X\right)$ is relatively regular, or g-invertible, if there exists an operator $A^{\left(1\right)}\in B\left(X\right)$ satisfying $A{A}^{\left(1\right)}A=A$; for $A\in B\left(X\right)$, the group inverse of A is the unique operator (if it exists) $A^{♯}\in B\left(X\right)$ such that

$$\begin{array}{c}\hfill \left(1\right)A{A}^{♯}A=A,\left(2\right)A^{♯}A{A}^{♯}=A^{♯},\left(5\right)A{A}^{♯}=A^{♯}A.\end{array}$$

If $A{A}^{\left(1\right)}=A^{\left(1\right)}A$, then $A\in B\left(X\right)$ is $\{1,5\}$-invertible operator and $B^{\{1,5\}}\left(X\right)$ consists of all $\{1,5\}$-invertible operators. As usual, the identity operator is represented by I.

Moreover, for a relatively regular operator A,

$$AB\left(X\right)=\{T\in B(X):R(T)\subseteq R(A\left)\right\},$$

whereas

$$A^0=\{T\in B\left(X\right):R\left(T\right)\subseteq N\left(A\right)\}.$$

Indeed, if $T=AS$ for some $S\in B\left(X\right)$, then $R\left(T\right)\subseteq R\left(A\right)$. Conversely, if $R\left(T\right)\subseteq R\left(A\right)$, then $A{A}^{\left(1\right)}T=T$, since $A{A}^{\left(1\right)}$ acts as the identity on $R\left(A\right)$. Hence, $T=A\left(A^{\left(1\right)}T\right)\in AB\left(X\right)$. Moreover, $AT=0$ if and only if $R\left(T\right)\subseteq N\left(A\right)$, which proves the second identity. Thus, the right annihilator $A^0$ is determined by the operator null space $N\left(A\right)$ through the above relation.

The following results represent operator versions of Theorems 3, 5, 7, 9, 11 and 12 in the setting of bounded linear operators on a Banach space.

Theorem 16.

Let $F,G\in B^{\left\{1\right\}}\left(X\right)$. If one of the following conditions holds

(i)

$F+G+F{F}^{\left(1\right)}G+G{G}^{\left(1\right)}F+F{F}^{\left(1\right)}G{G}^{\left(1\right)}F=0$,

(ii)

$F+G+F{F}^{\left(1\right)}G+G{G}^{\left(1\right)}F+G{G}^{\left(1\right)}F{F}^{\left(1\right)}G=0$,

(iii)

$F+G+F{F}^{\left(1\right)}G{G}^{\left(1\right)}F=0$,

(iv)

$F+G+G{G}^{\left(1\right)}F{F}^{\left(1\right)}G=0$,

then

$$N(F+G)=N\left(F\right)\cap N\left(G\right).$$

Theorem 17.

Let $F,G\in B^{\left\{1\right\}}\left(X\right)$ and let $3I\in B{\left(X\right)}^{-1}$. If $F{F}^{\left(1\right)}G=G{G}^{\left(1\right)}F$, then

$$N(F+G+F{F}^{\left(1\right)}G)=N\left(F\right)\cap N\left(G\right).$$

Theorem 18.

Let $F,G\in B^{\left\{1\right\}}\left(X\right)$. If $\mathcal{R}\left(F(I-F^{\left(1\right)}G)\right)\cap \mathcal{R}\left(G(I-G^{\left(1\right)}F)\right)=\left\{0\right\}$ and $N\left(F\right)\cap N\left(G\right)=\left\{0\right\},$ then

(i)

$N(F+G)=\left\{0\right\}$,

(ii)

$N(F+G+F{F}^{\left(1\right)}G)=\left\{0\right\}$,

(iii)

$N(F+G+G{G}^{\left(1\right)}F)=\left\{0\right\}$.

Theorem 19.

Let $F,G\in B^{\left\{1\right\}}\left(X\right)$. If $F+G$ and $I-F{F}^{\left(1\right)}G{G}^{\left(1\right)}$ are invertible, then

$$N(F-G)=\left\{0\right\}.$$

Furthermore, suppose that

$$F{F}^{\left(1\right)}G=G{G}^{\left(1\right)}F$$

and

$$F^2{F}^{\left(1\right)}G{G}^{\left(1\right)}=GF{F}^{\left(1\right)}G{G}^{\left(1\right)}.$$

If $N(F-G)=\left\{0\right\}$ and the operators $F+G$ and $I-F{F}^{\left(1\right)}G{G}^{\left(1\right)}$ are relatively regular, then both $F+G$ and $I-F{F}^{\left(1\right)}G{G}^{\left(1\right)}$ are left invertible.

Theorem 20.

Let $F,G\in B^{\{1,5\}}\left(X\right)$ be operators such that F commutes with G and with $G^{(1,5)}$. If $F+G+F{F}^{(1,5)}G+G{G}^{(1,5)}F$ and $I-F{F}^{(1,5)}G{G}^{(1,5)}$ are invertible, then $N(F-G)=\left\{0\right\}$.

Theorem 21.

Let $F,G\in B^{\{1,5\}}\left(X\right)$. Then $N(F{F}^{(1,5)}-G{G}^{(1,5)})=(\mathcal{R}\left(F\right)\cap \mathcal{R}\left(G\right))\oplus (N\left(F\right)\cap N\left(G\right))$.

The following lemma will be used in the sequel to relate algebraic ideals to operator ranges and null spaces. Note that $AB\left(X\right)$ and $A^0$ denote the image and kernel ideals of an operator A in $B\left(X\right)$.

Lemma 2.

Let $F,G\in B\left(X\right)$ be relatively regular operators on a Banach space X. Then

(i)

$FB\left(X\right)\cap GB\left(X\right)=\left\{0\right\}\iff R\left(F\right)\cap R\left(G\right)=\left\{0\right\}$,

(ii)

$B\left(X\right)=FB\left(X\right)\oplus GB\left(X\right)\iff X=R\left(F\right)\oplus R\left(G\right)$,

(iii)

$F^0\cap G^0=\left\{0\right\}\iff N\left(F\right)\cap N\left(G\right)=\left\{0\right\}$,

(iv)

$B\left(X\right)=F^0\oplus G^0\iff X=N\left(F\right)\oplus N\left(G\right)$.

Proof. 

It follows from Lemma 5.1. [10] and equalities $FB\left(X\right)=F{F}^{\left(1\right)}B\left(X\right)$, $F^0={\left(F{F}^{\left(1\right)}\right)}^0$, $R\left(F\right)=R\left(F{F}^{\left(1\right)}\right)$ and $N\left(F\right)=N\left(F{F}^{\left(1\right)}\right)$ for relatively regular operators. □

For bounded linear operators on a Banach space, Theorem 14 provides an operator-theoretic characterization of the invertibility of the sum and the difference of two $\{1,5\}$-invertible operators with mutually orthogonal $\{1,5\}$-inverses. In particular, it relates the invertibility of both the sum and the difference to direct-sum decompositions of the underlying space in terms of the ranges and null spaces of the operators, while the corresponding classical results concern idempotents and projections, the present result replaces these restrictive assumptions by $\{1,5\}$-invertibility together with the orthogonality of the corresponding $\{1,5\}$-inverses. Thus, Theorem 22 provides an analogous range-null-space characterization for this broader class of operators.

Theorem 22.

Let $F,G\in B^{\{1,5\}}\left(X\right)$ be operators such that $F^{(1,5)}\perp G^{(1,5)}$. Then the following conditions are equivalent:

(i)

$F+G$ is invertible,

(ii)

$F-G$ is invertible,

(iii)

$F{F}^{(1,5)}+G{G}^{(1,5)}=I$,

(iv)

$X=R\left(F\right)\oplus R\left(G\right)$,

(v)

$X=N\left(F\right)\oplus N\left(G\right)$.

Proof. 

By Theorem 14, conditions (i)–(v) are equivalent to the corresponding algebraic conditions in $B\left(X\right)$. Since $F,G\in B^{\{1,5\}}\left(X\right)$ are relatively regular, Lemma 2 gives

$$FB\left(X\right)\oplus GB\left(X\right)=B\left(X\right)⟺X=R\left(F\right)\oplus R\left(G\right),$$

and

$$F^0\oplus G^0=B\left(X\right)⟺X=N\left(F\right)\oplus N\left(G\right).$$

Thus, the stated conditions are equivalent. □

Theorem 15 yields a related characterization for a class of commuting group-invertible operators. We obtain the following result.

Theorem 23.

Let $F,G\in B^{♯}\left(X\right)$ be operators such that $F^{♯}=F^m$ and $G^{♯}=G^m$ for some natural $m>1$. If $FG=GF$, then the following conditions are equivalent:

(i)

$F-G$ and $F^{♯}+{\sum }_{i=1}^{m-1}{F}^{m-i}{G}^i+G^{♯}$ are invertible,

(ii)

$R\left(F\right)\oplus R\left(G\right)=X$,

(iii)

$N\left(F\right)\oplus N\left(G\right)=X$,

(iv)

$F-G$ and $F^{♯}+{\sum }_{i=1}^{m-1}{F}^{m-i}{G}^i+G^{♯}$ are relatively regular, $R\left(F\right)\cap R\left(G\right)=\left\{0\right\}$ and $N\left(F\right)\cap N\left(G\right)=\left\{0\right\}$.

Proof. 

Applying Theorem 15 in the algebra $B\left(X\right)$ and using Lemma 2, we obtain

$$FB\left(X\right)\cap GB\left(X\right)=\left\{0\right\}⟺R\left(F\right)\cap R\left(G\right)=\left\{0\right\},$$

and

$$F^0\cap G^0=\left\{0\right\}⟺N\left(F\right)\cap N\left(G\right)=\left\{0\right\}.$$

The direct-sum conditions are translated in the same way as in the proof of the preceding theorem. Hence, the stated conditions are equivalent. □

3.2. Applications to Hilbert Space Operators

In this subsection, we consider $\{1,5\}$-invertible operators of $B\left(H\right)$, the $C^{*}$-algebra of all bounded linear operators on a Hilbert space H. Here, we use the concepts and notations from previous subsections.

Theorems 14 and 15 yield the following results for Hilbert space operators when we recall that an operator $A\in B\left(H\right)$ is regular if and only if it has a closed range. No additional geometric properties specific to Hilbert spaces are required in these results; they follow directly from the corresponding Banach space results by taking $X=H$.

Theorem 24.

Let H be a Hilbert space and $F,G\in B^{\{1,5\}}\left(H\right)$ such that $F^{(1,5)}\perp G^{(1,5)}$. Then the following conditions are equivalent:

(i)

$F+G$ is invertible,

(ii)

$F-G$ is invertible,

(iii)

$F{F}^{(1,5)}+G{G}^{(1,5)}=I$,

(iv)

$R\left(F\right)\oplus R\left(G\right)=H$,

(v)

$N\left(F\right)\oplus N\left(G\right)=H$.

Proof. 

The result follows directly from Theorem 22 by taking $X=H$. □

Theorem 25.

Let H be a Hilbert space and $F,G\in B^{♯}\left(H\right)$ such that $F^{♯}=F^m$ and $G^{♯}=G^m$ for some natural $m>1$. If $FG=GF$, then the following conditions are equivalent:

(i)

$F-G$ and $F^{♯}+{\sum }_{i=1}^{m-1}{F}^{m-i}{G}^i+G^{♯}$ are invertible,

(ii)

$R\left(F\right)\oplus R\left(G\right)=H$,

(iii)

$N\left(F\right)\oplus N\left(G\right)=H$,

(iv)

$F-G$ and $F^{♯}+{\sum }_{i=1}^{m-1}{F}^{m-i}{G}^i+G^{♯}$ have closed range, $R\left(F\right)\cap R\left(G\right)=\left\{0\right\}$ and $N\left(F\right)\cap N\left(G\right)=\left\{0\right\}$.

Proof. 

Applying Theorem 23 to the Hilbert space H yields the desired equivalences. □

Finally, we state the corresponding finite-dimensional matrix result. Let $\mathbb{F}$ denote either $\mathbb{R}$ or $\mathbb{C}$. Then $M_n\left(\mathbb{F}\right)$ denotes the algebra of all $n\times n$ matrices over $\mathbb{F}$. In the finite-dimensional setting, all subspaces are closed and complemented, so the closedness and complementedness assumptions required in the infinite-dimensional operator setting are automatic. However, the algebraic assumptions on the matrices, such as group invertibility, the power conditions on the group inverses, and commutativity, remain essential. As a consequence of Theorem 25, we obtain the following result.

Corollary 4.

Let $F,G\in M_n\left(\mathbb{F}\right)$ be group invertible matrices such that $F^{♯}=F^m$ and $G^{♯}=G^m$ for some natural $m>1$. If $FG=GF$, then the following conditions are equivalent:

(i)

$F-G$ and $F^{♯}+{\sum }_{i=1}^{m-1}{F}^{m-i}{G}^i+G^{♯}$ are invertible,

(ii)

$R\left(F\right)\oplus R\left(G\right)={\mathbb{F}}^n$,

(iii)

$N\left(F\right)\oplus N\left(G\right)={\mathbb{F}}^n$,

(iv)

$R\left(F\right)\cap R\left(G\right)=\left\{0\right\}$ and $N\left(F\right)\cap N\left(G\right)=\left\{0\right\}$.

4. Conclusions

In this paper, we investigated the structure of the null spaces of sums and differences of inner regular elements in unital rings. We established several relations between these null spaces and the intersections of the corresponding annihilators and provided sufficient conditions under which they coincide or are trivial. These results yield further insights into the interplay between algebraic properties of elements and the behavior of their associated null spaces.

As applications, we obtained corresponding results for bounded linear operators on Banach and Hilbert spaces, as well as for matrices, thereby demonstrating the applicability of the developed theory in different settings. The obtained conditions also lead to consequences concerning the invertibility of sums and differences of special classes of elements.

The presented approach provides a unified framework for analyzing such problems and suggests possible directions for further research, including extensions to other classes of generalized inverses and more general algebraic structures.