A comment on some recent results concerning the reverse order law for {1, 3, 4}-inverses

Abstract

This paper has been motivated by the one of Liu and Yang [D. Liu, H. Yang, The reverse order law for {1, 3, 4}-inverse of the product of two matrices, Appl. Math. Comp. 215 (12) (2010) 4293–4303] in which the authors consider separately the cases when $(\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$ and $(\mathit{AB})\{1\text{,}3\text{,}4\}=B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$, where $A\in {\mathbb{C}}^{n\times m}$ and $B\in {\mathbb{C}}^{m\times n}$. Here we prove that $(\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$ is actually equivalent to $(\mathit{AB})\{1\text{,}3\text{,}4\}=B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$. We show that $(\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$ can only be possible if $n⩽m$ and in this case, we present purely algebraic necessary and sufficient conditions for this inclusion to hold. Also we give some new characterizations of $B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}\subseteq (\mathit{AB})\{1\text{,}3\text{,}4\}$.

Introduction

Let $A\in {\mathbb{C}}^{n\times n}$. By $\mathcal{R}(A)\text{,}\mathcal{N}(A)$ and rank(A), we denote the range, the null space and the rank of the matrix A, respectively. The Moore–Penrose inverse of $A\in {\mathbb{C}}^{m\times n}$, is the unique matrix $A^{†}\in {\mathbb{C}}^{n\times m}$ satisfying the four Penrose equations [11]:

$$(1){\mathit{AA}}^{†}A=A\text{,}(2)A^{†}{\mathit{AA}}^{†}=A^{†}\text{,}(3)({\mathit{AA}}^{†}{)}^{\ast }={\mathit{AA}}^{†}\text{,}(4)(A^{†}A{)}^{\ast }=A^{†}A\text{.}$$

It is well-known that each matrix A has its Moore–Penrose inverse.

If $K\subseteq \{1\text{,}2\text{,}3\text{,}4\}$ is arbitrary we shall say that $B\in {\mathbb{C}}^{n\times m}$ is a K-inverse of $A\in {\mathbb{C}}^{m\times n}$ if B satisfies the Penrose equation (j) for each $j\in K$. We shall write AK for the collection of all K inverses of A, and $A^K$ for an unspecified element $X\in \mathit{AK}$. For an arbitrary matrix A, we will denote $E_A=I-A^{†}A$ and $F_A=I-{\mathit{AA}}^{†}$. Remark that $E_A$ is the projection on $\mathcal{N}(A)$, while $F_A$ is the projection on $\mathcal{N}(A^{\ast })$.

The reverse order law for the Moore–Penrose inverse seems first to have been studied by Greville [5], in the 60’s, giving a necessary and sufficient condition for the reverse order law

$$(\mathit{AB}{)}^{†}=B^{†}{A}^{†}\text{,}$$

for matrices A and B. This was followed (see [4]) by further equivalent conditions for the same thing. Sun and Wei [8] extended the reverse order law conditions to the weighted Moore–Penrose inverse, and Hartwig [6] and Tian [9], [10] to the product of three and more matrices, respectively.

The next step was to consider the reverse order law for K-inverses, where $K\subseteq \{1\text{,}2\text{,}3\text{,}4\}$.

The motivation for this paper has been the paper of Liu and Yang [7], in which they give necessary and sufficient conditions for

$$\begin{array}{cc}\hfill & B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}\subseteq (\mathit{AB})\{1\text{,}3\text{,}4\}\text{,}\hfill \\ \hfill & (\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}\text{,}\hfill \\ \hfill & (\mathit{AB})\{1\text{,}3\text{,}4\}=B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}\text{.}\hfill \end{array}$$

What they did not realize was that $(\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$ is actually equivalent to $(\mathit{AB})\{1\text{,}3\text{,}4\}=B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$. So, in Theorem 3 (2) [7], they have two extra conditions.

Here, we will give a very short proof that $(\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$ is equivalent to $(\mathit{AB})\{1\text{,}3\text{,}4\}=B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$. We show that $(\mathit{AB})\{1\text{,}3\text{,}4\}\subseteq B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}$ can only be possible if $n⩽m$ and in this case, we present purely algebraic necessary and sufficient conditions for this inclusion to hold. Also we give some new characterizations of $B\{1\text{,}3\text{,}4\}·A\{1\text{,}3\text{,}4\}\subseteq (\mathit{AB})\{1\text{,}3\text{,}4\}$.

We begin by recalling some well-known results.

Lemma 1.1 [2]

Let $A\in {\mathbb{C}}^{n\times m}$ and $B\in {\mathbb{C}}^{m\times n}$. Then the following statements are equivalent:

• $\mathit{ABA}=A$ and $(\mathit{AB}{)}^{\ast }=\mathit{AB}$,

• $A^{†}\mathit{AB}=A^{†}$,

• There exists $Y\in {\mathbb{C}}^{m\times n}$ such that $B=A^{†}+\left(I-A^{†}A\right)Y$.

Hence, Lemma 1.1 characterizes the set $A\{1\text{,}3\}$. Analogously, we have the following result which characterizes $A\{1\text{,}4\}$:

Lemma 1.2 [2]

Let $A\in {\mathbb{C}}^{n\times m}$ and $B\in {\mathbb{C}}^{m\times n}$. Then the following statements are equivalent:

• $\mathit{ABA}=A$ and $(\mathit{BA}{)}^{\ast }=\mathit{BA}$,

• ${\mathit{BAA}}^{†}=A^{†}$,

• There exists $Y\in {\mathbb{C}}^{m\times n}$ such that $B=A^{†}+Y\left(1-{\mathit{AA}}^{†}\right)$.

We can prove the analogous result for A{1, 3, 4}:

Lemma 1.3

Let $A\in {\mathbb{C}}^{n\times m}$ and $B\in {\mathbb{C}}^{m\times n}$. Then the following statements are equivalent:

• $B\in A\{1\text{,}3\text{,}4\}$,

• $A^{\ast }\mathit{AB}=A^{\ast }$ and ${\mathit{BAA}}^{\ast }=A^{\ast }$,

• There exists $Y\in {\mathbb{C}}^{m\times n}$ such that $B=A^{†}+\left(I-A^{†}A\right)Y\left(I-{\mathit{AA}}^{†}\right)$.

Proof

• If $B\in A\{1\text{,}3\text{,}4\}$, then

  $$A^{\ast }\mathit{AB}=A^{\ast }(\mathit{AB}{)}^{\ast }=(\mathit{ABA}{)}^{\ast }=A^{\ast }\text{and}{\mathit{BAA}}^{\ast }=(\mathit{ABA}{)}^{\ast }=A^{\ast }\text{.}$$

• If $A^{\ast }\mathit{AB}=A^{\ast }$ and ${\mathit{BAA}}^{\ast }=A^{\ast }$, then

  $$\begin{array}{cc}\hfill & \mathit{ABA}={\mathit{AA}}^{†}\mathit{ABA}={\left(A^{†}\right)}^{\ast }{A}^{\ast }\mathit{ABA}={\left(A^{†}\right)}^{\ast }{A}^{\ast }A=A\text{,}\hfill \\ \hfill & \mathit{AB}=B^{\ast }{A}^{\ast }\mathit{AB}=(\mathit{AB}{)}^{\ast }(\mathit{AB})\text{,}\hfill \\ \hfill & \mathit{BA}={\mathit{BAA}}^{\ast }{B}^{\ast }=(\mathit{BA})(\mathit{BA}{)}^{\ast }\text{,}\hfill \end{array}$$

  so, $B\in A\{1\text{,}3\text{,}4\}$.

• If $B\in A\{1\text{,}3\text{,}4\}$, then $B\in A\{1\text{,}3\}$, so by Lemma 1.1, $B=A^{†}+\left(I-A^{†}A\right)T$, for some $T\in {\mathbb{C}}^{m\times n}$. Also, $\mathit{BA}=A^{†}A$, so $\left(I-A^{†}A\right)\mathit{TA}=0$. Set $Z=\left(I-A^{†}A\right)T$. We have by Lemma 1.2 that Z is a solution of equation $\mathit{ZA}=0$, so $Z=Y\left(1-{\mathit{AA}}^{†}\right)$, for some $Y\in {\mathbb{C}}^{m\times n}$. Now, $B=A^{†}+\left(I-A^{†}A\right)Y\left(I-{\mathit{AA}}^{†}\right)$.

• This is evident.  □

Hence,

$$A\{1\text{,}3\text{,}4\}=\left\{A^{†}+\left(I-A^{†}A\right)Y\left(I-{\mathit{AA}}^{†}\right):Y\in {\mathbb{C}}^{m\times n}\right\}\text{.}$$

The following lemma is an auxiliary result which can easily be checked:

Lemma 1.4

Let $A\in {\mathbb{C}}^{n\times m}\text{,}P\in {\mathbb{C}}^{n\times n}$ and $Q\in {\mathbb{C}}^{m\times m}$. If P and Q are idempotents, then

$$(\mathit{PAQ}{)}^{†}=Q(\mathit{PAQ}{)}^{†}P\text{.}$$

The following result is proved in [3] for bounded linear operators on Hilbert spaces, so it is also valid in the matrix case.

Lemma 1.5

Let $A\in {\mathbb{C}}^{n\times m}$ and $B\in {\mathbb{C}}^{m\times n}$. The following statements are equivalent:

• $(\mathit{AB}{)}^{†}=B^{†}{A}^{†}$,

• $(\mathit{AB})(\mathit{AB}{)}^{†}A={\mathit{ABB}}^{†}$ and $B(\mathit{AB}{)}^{†}\mathit{AB}=A^{†}\mathit{AB}$.