The Reverse Order Law for the {1,3M,4N}—The Inverse of Two Matrix Products

Abstract

By using the maximal and minimal ranks of some generalized Schur complement, the equivalent conditions for the reverse order law $\left(AB\right)\{1,3M,4K\}=B\{1,3N,4K\}A\{1,3M,4N\}$ are presented.

1. Introduction

At first, we will provide some definitions for the convenience of readers.

Definition 1 

([1,2]).

• $C^{m\times n}$: the set of $m\times n$ complex matrices;
• $R\left(A\right)$, $N\left(A\right)$: the range space and the null space of A, respectively;
• $r\left(A\right)$, $A^{*}$: the rank and the conjugate transpose of A, respectively;
• $A^{⊛(M,N)}=A^{\#}=N^{-1}{A}^{*}M$: the weighted conjugate transpose matrices of A with positive-definite Hermitian matrices M and N.

Definition 2 

([3]). Let M be a positive-definite Hermitian matrix and let $A\in C^{m\times n}$, $b\in C^m$. The aim of the so-called weighted least squares problem [WLS] is to find $x\in C^n$ that satisfies the following:

$$\begin{array}{c}\hfill {\parallel Ax-b\parallel }_M=\parallel M^{{\displaystyle \frac{1}{2}}}{(Ax-b)\parallel }_2^2=\underset{x\in C^n}{min}{\parallel Ax-b\parallel }_M,\end{array}$$

where $M=M^{{\displaystyle \frac{1}{2}}}{M}^{{\displaystyle \frac{1}{2}}}$, ${\parallel .\parallel }_2$ is a spectral norm, and ${\parallel .\parallel }_M$ is a weighted $M-$norm.

Definition 3 

([2]). Let $A\in C^{m\times n}$, and let M and N be two positive-definite Hermitian matrices. X is the weighted Moore–Penrose inverse of A when it satisfies

$$\begin{array}{c}\hfill \left(1\right)AXA=A,\left(2\right)XAX=X,\left(3M\right){\left(MAX\right)}^{*}=MAX,\left(4N\right){\left(NXA\right)}^{*}=NXA,\end{array}$$

(1)

where X is denoted by $X=A^{(1,2,3M,4N)}=A_{M,N}^{†}$. Set’s $A\{1,3M,4N\}$-inverses of A are $A\{1,3M,4N\}=\left\{X\in C^{n\times m}{|AXA=A,\left(MAX\right)}^{*}=MAX,{\left(NXA\right)}^{*}=NXA\right\}$.

The reverse order law for the weighted generalized inverse has been widely applied in the study of the WLS (see [4,5,6]), the weighted perturbation theory (see [7,8,9,10,11,12]), the optimization problems and other related topics (see [13,14,15,16,17,18]).

The reverse order laws for the weighted generalized inverses of matrix products have attracted considerable attentions (see [19,20,21,22,23,24,25,26,27,28]). It is well known that the core problem here concerns the reverse order law, which depends on what conditions the equations

$$\begin{array}{c}\hfill A_n^{(i,j,\dots ,k)}{A}_{n-1}^{(i,j,\dots ,k)}\cdots A_1^{(i,j,\dots ,k)}={(A_1{A}_2\cdots A_n)}^{(i,j,\dots ,k)}\end{array}$$

hold, where $i,j,\dots ,k\in \{1,2,3M,4N\}$.

The purpose of this paper is to show some equivalent conditions for the following reverse order law:

$$\begin{array}{c}\hfill B\{1,3N,4K\}A\{1,3M,4N\}=\left(AB\right)\{1,3M,4K\}.\end{array}$$

(2)

Furthermore, the equivalent conditions for the inclusions

$$\begin{array}{c}\hfill B\{1,3N,4K\}A\{1,3M,4N\}\subseteq \left(AB\right)\{1,3M,4K\}\end{array}$$

(3)

and

$$\begin{array}{c}\hfill B\{1,3N,4K\}A\{1,3M,4N\}\supseteq \left(AB\right)\{1,3M,4K\},\end{array}$$

(4)

are presented.

The following lemmas are essential in the rest of this paper.

Lemma 1 

([2,4]). Let M and N be two positive-definite Hermitian matrices, and let $A\in C^{m\times n}$. Then,

$$\begin{array}{c}\hfill X\in A\{1,3M\}⟺A^{*}MAX=A^{*}M,\end{array}$$

(5)

$$\begin{array}{c}\hfill X\in A\{1,4N\}⟺XA{N}^{-1}{A}^{*}=N^{-1}{A}^{*},\end{array}$$

(6)

$$\begin{array}{c}\hfill X\in A\{1,4N\}⟺X^{*}\in A^{*}\{1,3N^{-1}\}.\end{array}$$

(7)

Lemma 2 

([17]). Suppose that A, B, C and D are four complex matrices, and suppose that M and N are two positive-definite Hermitian matrices. Then,

$$\begin{array}{c}\hfill r(D-C{A}_{M,N}^{†}B)=r\left(\begin{array}{cc}{A}^{*}MA{N}^{-1}{A}^{*}& A^{*}MB\\ C{N}^{-1}{A}^{*}& D\end{array}\right)-r\left(A\right);\end{array}$$

(8)

$$\begin{array}{c}\hfill \underset{A^{(1,3M,4N)}}{max}r(D-C{A}^{(1,3M,4N)}B)=min\{r\left(\begin{array}{cc}{A}^{*}MA& A^{*}MB\\ C& D\end{array}\right)-r\left(A\right),\\ \hfill r\left(\begin{array}{cc}A{N}^{-1}{A}^{*}& B\\ C{N}^{-1}{A}^{*}& D\end{array}\right)-r\left(A\right)\};\end{array}$$

(9)

$$\begin{array}{c}\hfill \underset{A^{(1,3M,4N)}}{min}r(D-C{A}^{(1,3M,4N)}B)=r\left(\begin{array}{cc}{A}^{*}MA& A^{*}MB\\ C& D\end{array}\right)+r\left(\begin{array}{cc}A{N}^{-1}{A}^{*}& B\\ C{N}^{-1}{A}^{*}& D\end{array}\right)\\ \\ \hfill -r\left(\begin{array}{cc}A& B\\ C& D\end{array}\right)-r\left(A\right).\end{array}$$

(10)

Lemma 3 

([29]). Let X and Y be two arbitrary matrices, and let $A_{ij}\in C^{m_i\times n_j}(1\le i,j\le 3)$. Then,

$$\begin{array}{c}\hfill \underset{X,Y}{min}r\left(\begin{array}{ccc}{A}_{11}& A_{12}& X\\ A_{21}& A_{22}& A_{23}\\ Y& A_{32}& A_{33}\end{array}\right)=r\left(\begin{array}{ccc}{A}_{21},& A_{22},& A_{23}\end{array}\right)+r\left(\begin{array}{c}{A}_{12}\\ A_{22}\\ A_{32}\end{array}\right)+max\{r\left(\begin{array}{cc}{A}_{11}& A_{12}\\ A_{21}& A_{22}\end{array}\right)\\ \hfill -r\left(\begin{array}{c}{A}_{12}\\ A_{22}\end{array}\right)-r\left(\begin{array}{cc}{A}_{21},& A_{22}\end{array}\right),r\left(\begin{array}{cc}{A}_{22}& A_{23}\\ A_{32}& A_{33}\end{array}\right)-r\left(\begin{array}{c}{A}_{22}\\ A_{32}\end{array}\right)-r\left(\begin{array}{cc}{A}_{22},& A_{23}\end{array}\right)\}.\end{array}$$

(11)

The contents of this paper are organized as follows: In Section 2, we first present the equivalent conditions for Inclusions (3) and (4) using the maximal and minimal ranks of the generalized Schur complement and completions of partial matrices. Then, applying these results, we study the reverse order law (2). At last, in Theorem 3, we obtain necessary and sufficient conditions for the reverse order law (2) via rank conditions of known matrices.

2. Main Results

To obtain the main results (Theorem 3), we need present the minimal rank of the generalized Schur complement $D-B^{(1,3N,4K)}{A}^{(1,3M,4N)}$.

Lemma 4. 

Let M, N, and K be three positive-definite Hermitian matrices, and let $A\in C^{m\times n}$, $B\in C^{n\times k}$, $D\in C^{k\times m}$. Then,

$$\begin{array}{cc}& \underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{min}r(D-B^{(1,3N,4K)}{A}^{(1,3M,4N)})\\ & =r\left(\begin{array}{cc}{B}^{*}NBD& B^{*}N\\ A^{*}M& A^{*}MA\end{array}\right)+r\left(\begin{array}{cc}{N}^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ DA{N}^{-1}{A}^{*}& K^{-1}{B}^{*}\end{array}\right)-r\left(A\right)-r\left(B\right)& \\ & +max\{r(B^{*}{A}^{*}-B^{*}NBDA{N}^{-1}{A}^{*})-r(BDA{N}^{-1}{A}^{*}-N^{-1}{A}^{*})-r(B^{*}NBDA-B^{*}N),& \\ & r\left(\begin{array}{cc}{K}^{-1}{B}^{*}& D\\ A^{*}MAB{K}^{-1}{B}^{*}& A^{*}M\end{array}\right)-r(DAB{K}^{-1}{B}^{*}-K^{-1}{B}^{*})-r(A^{*}MABD-A^{*}M)-n\}.\end{array}$$

(12)

Proof. 

It is well known that $A^{(1,3M,4N)}=A_{M,N}^{†}+\widetilde{F_A}V\widetilde{E_A}$, where $\widetilde{F_A}=I_n-A_{M,N}^{†}A$, $\widetilde{E_A}=I_m-A{A}_{M,N}^{†}$ and V is an arbitrary matrix (see [17]). Combining this fact, we have

$$\begin{array}{ccc}& & r(D-B^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& r[(B_{N,K}^{†}+\widetilde{F_B}W\widetilde{E_B})(A_{M,N}^{†}+\widetilde{F_A}V\widetilde{E_A})-D]\hfill \\ & =& r(B_{N,K}^{†}{A}_{M,N}^{†}+B_{N,K}^{†}\widetilde{F_A}V\widetilde{E_A}+\widetilde{F_B}W\widetilde{E_B}{A}_{M,N}^{†}+\widetilde{F_B}W\widetilde{E_B}\widetilde{F_A}V\widetilde{E_A}-D)\hfill \\ & =& r\left(\begin{array}{cccccc}O& O& O& O& I_n& V\\ O& O& -\widetilde{E_A}& O& O& I_m\\ O& O& O& I_n& \widetilde{F_A}& O\\ O& \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}& O& O\\ I_n& O& \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}& O& O\\ W& I_k& O& O& O& O\end{array}\right)-k-m-3n.\hfill \end{array}$$

(13)

where $\widetilde{F_A}=I_n-A_{M,N}^{†}A$, $\widetilde{E_A}=I_m-A{A}_{M,N}^{†}$, $\widetilde{F_B}=I_k-B_{N,K}^{†}B$, $\widetilde{E_B}=I_n-B{B}_{N,K}^{†}$ and the matrices V and W are arbitrary. Applying Lemma 3, we have

$$\begin{array}{ccc}& & \underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{min}r(D-B^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& r\left(\begin{array}{ccc}\widetilde{F_B},& B_{N,K}^{†}{A}_{M,N}^{†}-D,& B_{N,K}^{†}\widetilde{F_A}\end{array}\right)\hfill \\ & +& r\left(\begin{array}{c}-\widetilde{E_A}\\ B_{N,K}^{†}{A}_{M,N}^{†}-D\\ \widetilde{E_B}{A}_{M,N}^{†}\end{array}\right)+max\{r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\end{array}\right)-r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\\ O& \widetilde{E_B}{A}_{M,N}^{†}\end{array}\right)\hfill \\ & -& r\left(\begin{array}{ccc}O& -\widetilde{E_A}& O\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\end{array}\right),r\left(\begin{array}{cc}{B}_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)\hfill \\ & -& r\left(\begin{array}{cc}-\widetilde{E_A}& O\\ B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)-r\left(\begin{array}{ccc}\widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ O& \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)\}\hfill \end{array}$$

(14)

The next step is to use block matrix operations to simplify the rank of the right-hand side of (14). For the first one, combining $\widetilde{F_A}=I_n-A_{M,N}^{†}A$ and $\widetilde{F_B}=I_k-B_{N,K}^{†}B$, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{ccc}\widetilde{F_B},& B_{N,K}^{†}{A}_{M,N}^{†}-D,& B_{N,K}^{†}\widetilde{F_A}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{ccc}{I}_k-B_{N,K}^{†}B,& B_{N,K}^{†}{A}_{M,N}^{†}-D,& B_{N,K}^{†}-B_{N,K}^{†}{A}_{M,N}^{†}A\end{array}\right)\hfill \\ & =& r\left(\begin{array}{ccc}{I}_k-B_{N,K}^{†}B,& B_{N,K}^{†}{A}_{M,N}^{†}-D,& B_{N,K}^{†}-DA\end{array}\right)\hfill \\ & =& r\left(\begin{array}{ccccc}{B}_{N,K}^{†}& B_{N,K}^{†}B& -B_{N,K}^{†}{A}_{M,N}^{†}& -B_{N,K}^{†}& O\\ O& I_k-B_{N,K}^{†}B,& B_{N,K}^{†}{A}_{M,N}^{†}-D,& B_{N,K}^{†}-DA& O\\ O& O& A_{M,N}^{†}& O& A_{M,N}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{ccccc}{B}_{N,K}^{†}& B_{N,K}^{†}B& O& -B_{N,K}^{†}& B_{N,K}^{†}{A}_{M,N}^{†}\\ B_{N,K}^{†}& I_k,& -D,& -DA& O\\ O& O& A_{M,N}^{†}& O& A_{M,N}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{ccc}{B}_{N,K}^{†}BD& B_{N,K}^{†}BDA-B_{N,K}^{†}& B_{N,K}^{†}{A}_{M,N}^{†}\\ A_{M,N}^{†}& O& A_{M,N}^{†}\end{array}\right)+k-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cc}{B}_{N,K}^{†}BD& B_{N,K}^{†}\\ A_{M,N}^{†}& A_{M,N}^{†}A\end{array}\right)+k-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(15)

From Definition 1, we have $A^{\#}=N^{-1}{A}^{*}M$, $B^{\#}=K^{-1}{B}^{*}N$ and

$$\begin{array}{c}\hfill \left(\begin{array}{cc}{B}^{\#}B& O\\ O& A^{\#}A\end{array}\right)\left(\begin{array}{cc}{B}_{N,K}^{†}BD& B_{N,K}^{†}\\ A_{M,N}^{†}& A_{M,N}^{†}A\end{array}\right)=\left(\begin{array}{cc}{B}^{\#}BD& B^{\#}\\ A^{\#}& A^{\#}A\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill \left(\begin{array}{cc}{B}_{N,K}^{†}{\left(B_{N,K}^{†}\right)}^{\#}& O\\ O& A_{M,N}^{†}{\left(A_{M,N}^{†}\right)}^{\#}\end{array}\right)\left(\begin{array}{cc}{B}^{\#}BD& B^{\#}\\ A^{\#}& A^{\#}A\end{array}\right)=\left(\begin{array}{cc}{B}_{N,K}^{†}BD& B_{N,K}^{†}\\ A_{M,N}^{†}& A_{M,N}^{†}A\end{array}\right).\end{array}$$

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}{B}_{N,K}^{†}BD& B_{N,K}^{†}\\ A_{M,N}^{†}& A_{M,N}^{†}A\end{array}\right)=r\left(\begin{array}{cc}{B}^{\#}BD& B^{\#}\\ A^{\#}& A^{\#}A\end{array}\right)=r\left(\begin{array}{cc}{B}^{*}NBD& B^{*}N\\ A^{*}M& A^{*}MA\end{array}\right).\end{array}$$

(16)

Substituting Formula (16) into (15), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{ccc}\widetilde{F_B},& B_{N,K}^{†}{A}_{M,N}^{†}-D,& B_{N,K}^{†}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{cc}{B}^{*}NBD& B^{*}N\\ A^{*}M& A^{*}MA\end{array}\right)+k-r\left(A\right)-r\left(B\right).\end{array}$$

(17)

For the second partitioned matrix of the right-hand of (14), we have

$$\begin{array}{ccc}& & r\left(\begin{array}{c}-\widetilde{E_A}\\ B_{N,K}^{†}{A}_{M,N}^{†}-D\\ \widetilde{E_B}{A}_{M,N}^{†}\end{array}\right)=r\left(\begin{array}{cc}{I}_n& A_{M,N}^{†}\\ O& -\widetilde{E_A}\\ -B_{N,K}^{†}& -D\\ -\widetilde{E_B}& O\end{array}\right)-n\hfill \\ & =& r\left(\begin{array}{cccc}{A}_{M,N}^{†}& O& A_{M,N}^{†}& O\\ O& I_n& A_{M,N}^{†}& O\\ O& O& -\widetilde{E_A}& O\\ O& -B_{N,K}^{†}& -D& O\\ O& -\widetilde{E_B}& O& O\\ O& B_{N,K}^{†}& O& B_{N,K}^{†}\end{array}\right)-n-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cccc}{A}_{M,N}^{†}& O& A_{M,N}^{†}& O\\ -A_{M,N}^{†}& I_n& O& O\\ -A{A}_{M,N}^{†}& O& -I_m& O\\ O& O& -D& O\\ O& -I_n& O& -B{B}_{N,K}^{†}\\ O& B_{N,K}^{†}& O& B_{N,K}^{†}\end{array}\right)-n-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cc}{A}_{M,N}^{†}& B{B}_{N,K}^{†}\\ DA{A}_{M,N}^{†}& B_{N,K}^{†}\end{array}\right)+m-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(18)

Similarly to (16), we have

$$\begin{array}{c}\hfill \left(\begin{array}{cc}{A}_{M,N}^{†}& B{B}_{N,K}^{†}\\ DA{A}_{M,N}^{†}& B_{N,K}^{†}\end{array}\right)\left(\begin{array}{cc}A{A}^{\#}& O\\ O& B{B}^{\#}\end{array}\right)=\left(\begin{array}{cc}{A}^{\#}& B{B}^{\#}\\ DA{A}^{\#}& B^{\#}\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill \left(\begin{array}{cc}{A}^{\#}& B{B}^{\#}\\ DA{A}^{\#}& B^{\#}\end{array}\right)\left(\begin{array}{cc}{\left(A_{M,N}^{†}\right)}^{\#}{A}_{M,N}^{†}& O\\ O& {\left(B_{N,K}^{†}\right)}^{\#}{B}_{N,K}^{†}\end{array}\right)=\left(\begin{array}{cc}{A}_{M,N}^{†}& B{B}_{N,K}^{†}\\ DA{A}_{M,N}^{†}& B_{N,K}^{†}\end{array}\right).\end{array}$$

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}{A}_{M,N}^{†}& B{B}_{N,K}^{†}\\ DA{A}_{M,N}^{†}& B_{N,K}^{†}\end{array}\right)=r\left(\begin{array}{cc}{A}^{\#}& B{B}^{\#}\\ DA{A}^{\#}& B^{\#}\end{array}\right)=r\left(\begin{array}{cc}{N}^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ DA{N}^{-1}{A}^{*}& K^{-1}{B}^{*}\end{array}\right).\end{array}$$

(19)

Substituting Formula (19) into (18), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{c}-\widetilde{E_A}\\ B_{N,K}^{†}{A}_{M,N}^{†}-D\\ \widetilde{E_B}{A}_{M,N}^{†}\end{array}\right)=r\left(\begin{array}{cc}{N}^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ DA{N}^{-1}{A}^{*}& K^{-1}{B}^{*}\end{array}\right)+m-r\left(A\right)-r\left(B\right).\end{array}$$

(20)

Substituting $\widetilde{E_A}=I_m-A{A}_{M,N}^{†}$ and $\widetilde{F_B}=I_k-B_{N,K}^{†}B$ into the third block matrix in (14), we have

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\end{array}\right)=r\left(\begin{array}{cccc}{A}_{M,N}^{†}& O& A_{M,N}^{†}& O\\ O& O& -\widetilde{E_A}& O\\ O& \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\\ O& O& O& B_{N,K}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cccc}{A}_{M,N}^{†}& O& A_{M,N}^{†}& O\\ -A{A}_{M,N}^{†}& O& -I_m& O\\ O& I_k& -D& B_{N,K}^{†}\\ B_{N,K}^{†}{A}_{M,N}^{†}& B_{N,K}^{†}B& O& B_{N,K}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{c}{B}_{N,K}^{†}{A}_{M,N}^{†}-B_{N,K}^{†}BDA{A}_{M,N}^{†}\end{array}\right)+k+m-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(21)

This is because

$$\begin{array}{ccc}\hfill r\left(\begin{array}{c}{B}^{\#}{A}^{\#}-B^{\#}BDA{A}^{\#}\end{array}\right)& =& r\left[B^{\#}B\left(\begin{array}{c}{B}_{N,K}^{†}{A}_{M,N}^{†}-B_{N,K}^{†}BDA{A}_{M,N}^{†}\end{array}\right)A{A}^{\#}\right]\hfill \\ & \le & r\left(\begin{array}{c}{B}_{N,K}^{†}{A}_{M,N}^{†}-B_{N,K}^{†}BDA{A}_{M,N}^{†}\end{array}\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}\hfill r\left(\begin{array}{c}{B}_{N,K}^{†}{A}_{M,N}^{†}-B_{N,K}^{†}BDA{A}_{M,N}^{†}\end{array}\right)& =& r\left[B_{N,K}^{†}{\left(B_{N,K}^{†}\right)}^{\#}\left(\begin{array}{c}{B}^{\#}{A}^{\#}-B^{\#}BDA{A}^{\#}\end{array}\right){\left(A_{M,N}^{†}\right)}^{\#}{A}_{M,N}^{†}\right]\hfill \\ & \le & r\left(\begin{array}{c}{B}^{\#}{A}^{\#}-B^{\#}BDA{A}^{\#}\end{array}\right).\hfill \end{array}$$

Then,

$$\begin{array}{ccc}\hfill r\left(\begin{array}{c}{B}_{N,K}^{†}{A}_{M,N}^{†}-B_{N,K}^{†}BDA{A}_{M,N}^{†}\end{array}\right)& =& r\left(\begin{array}{c}{B}^{\#}{A}^{\#}-B^{\#}BDA{A}^{\#}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{c}{B}^{*}{A}^{*}-B^{*}NBDA{N}^{-1}{A}^{*}\end{array}\right)\hfill \end{array}$$

(22)

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\end{array}\right)=r\left(\begin{array}{c}{B}^{*}{A}^{*}-B^{*}NBDA{N}^{-1}{A}^{*}\end{array}\right)+k+m-r\left(A\right)-r\left(B\right).\end{array}$$

(23)

For the fourth partitioned matrix on the right-hand of (14), we have

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\\ O& \widetilde{E_B}{A}_{M,N}^{†}\end{array}\right)=r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\\ O& A_{M,N}^{†}-BD\end{array}\right)=r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}BD-D\\ O& A_{M,N}^{†}-BD\end{array}\right)\hfill \\ & =& r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& O\\ O& A_{M,N}^{†}-BD\end{array}\right)=r\left(\widetilde{F_B}\right)+r\left(\begin{array}{cc}{A}_{M,N}^{†}& A_{M,N}^{†}\\ O& \widetilde{E_A}\\ O& A_{M,N}^{†}-BD\end{array}\right)-r\left(A\right)\hfill \\ & =& k-r\left(B\right)+r\left(\begin{array}{cc}{A}_{M,N}^{†}& A_{M,N}^{†}\\ A{A}_{M,N}^{†}& I_m\\ -A_{M,N}^{†}& -BD\end{array}\right)-r\left(A\right)\hfill \\ & =& r(BDA{A}_{M,N}^{†}-A_{M,N}^{†})+k+m-r\left(B\right)-r\left(A\right).\hfill \end{array}$$

(24)

Using Definition 1, we obtain

$$\begin{array}{c}\hfill r(BDA{A}_{M,N}^{†}-A_{M,N}^{†})=r\left[(BDA{A}^{\#}-A^{\#}){\left(A_{M,N}^{†}\right)}^{\#}{A}_{M,N}^{†}\right]\le r(BDA{A}^{\#}-A^{\#})\end{array}$$

and

$$\begin{array}{c}\hfill r(BDA{A}^{\#}-A^{\#})=r\left[(BDA{A}_{M,N}^{†}-A_{M,N}^{†})A{A}^{\#}\right]\le r(BDA{A}_{M,N}^{†}-A_{M,N}^{†}).\end{array}$$

That is,

$$\begin{array}{c}\hfill r(BDA{A}_{M,N}^{†}-A_{M,N}^{†})=r(BDA{A}^{\#}-A^{\#})=r(BDA{N}^{-1}{A}^{*}-N^{-1}{A}^{*}).\end{array}$$

(25)

Substituting Formula (25) into (24), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}O& -\widetilde{E_A}\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D\\ O& \widetilde{E_B}{A}_{M,N}^{†}\end{array}\right)=r(BDA{N}^{-1}{A}^{*}-N^{-1}{A}^{*})+k+m-r\left(B\right)-r\left(A\right).\end{array}$$

(26)

For the fifth block matrix in (14), via the row or column elementary block matrix operations and $\widetilde{E_A}A=O$, $\widetilde{E_A}=I_m-A{A}_{M,N}^{†}$, $\widetilde{F_A}=I_n-A_{M,N}^{†}A$ and $\widetilde{F_B}=I_k-B_{N,K}^{†}B$, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{ccc}O& -\widetilde{E_A}& O\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{ccc}O& -\widetilde{E_A}& O\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}-DA\end{array}\right)\hfill \\ & =& r\left(\begin{array}{ccccc}{A}_{M,N}^{†}& O& A_{M,N}^{†}& O& O\\ O& O& \widetilde{E_A}& O& O\\ O& \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}-DA& B_{N,K}^{†}\\ O& O& O& O& B_{N,K}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{ccccc}{A}_{M,N}^{†}& O& A_{M,N}^{†}& O& O\\ A{A}_{M,N}^{†}& O& I_m& O& O\\ O& I_k& -D& -DA& B_{N,K}^{†}\\ B_{N,K}^{†}{A}_{M,N}^{†}& B_{N,K}^{†}B& O& -B_{N,K}^{†}& B_{N,K}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{c}{B}_{N,K}^{†}BDA-B_{N,K}^{†}\end{array}\right)+m+k-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(27)

From Definition 1, we have

$$\begin{array}{ccc}\hfill r\left(\begin{array}{c}{B}_{N,K}^{†}BDA-B_{N,K}^{†}\end{array}\right)& =& r\left[B_{N,K}^{†}{\left(B_{N,K}^{†}\right)}^{\#}\left(\begin{array}{c}{B}^{\#}BDA-B^{\#}\end{array}\right)\right]\le r\left(\begin{array}{c}{B}^{\#}BDA-B^{\#}\end{array}\right)\hfill \\ & =& r\left[B^{\#}B\left(\begin{array}{c}{B}_{N,K}^{†}BDA-B_{N,K}^{†}\end{array}\right)\right]\le r\left(\begin{array}{c}{B}_{N,K}^{†}BDA-B_{N,K}^{†}\end{array}\right).\hfill \end{array}$$

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{c}{B}_{N,K}^{†}BDA-B_{N,K}^{†}\end{array}\right)=r\left(\begin{array}{c}{B}^{\#}BDA-B^{\#}\end{array}\right)=r\left(\begin{array}{c}{B}^{*}NBDA-B^{*}N\end{array}\right).\end{array}$$

(28)

Substituting Formula (28) into (27), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{ccc}O& -\widetilde{E_A}& O\\ \widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{c}{B}^{*}NBDA-B^{*}N\end{array}\right)+m+k-r\left(A\right)-r\left(B\right).\end{array}$$

(29)

For the sixth block matrix in (14), using the same method as above, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}{B}_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{cc}{B}_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ A_{M,N}^{†}-BD& \widetilde{F_A}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{cc}{B}_{N,K}^{†}BD-D& O\\ A_{M,N}^{†}-BD& \widetilde{F_A}\end{array}\right)=r\left(\begin{array}{cccc}{B}_{N,K}^{†}& O& O& O\\ B_{N,K}^{†}& B_{N,K}^{†}BD-D& O& O\\ O& A_{M,N}^{†}-BD& \widetilde{F_A}& A_{M,N}^{†}\\ O& O& O& A_{M,N}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cccc}{B}_{N,K}^{†}& -B_{N,K}^{†}BD& O& O\\ B_{N,K}^{†}& -D& O& O\\ O& -BD& I_n& A_{M,N}^{†}\\ O& -A_{M,N}^{†}& A_{M,N}^{†}A& A_{M,N}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cc}{B}_{N,K}^{†}& D\\ A_{M,N}^{†}AB{B}_{N,K}^{†}& A_{M,N}^{†}\end{array}\right)+n-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(30)

From Definition 1, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}{B}_{N,K}^{†}& D\\ A_{M,N}^{†}AB{B}_{N,K}^{†}& A_{M,N}^{†}\end{array}\right)\hfill \\ & =& r\left\{\left(\begin{array}{cc}I& O\\ O& A_{M,N}^{†}{\left(A_{M,N}^{†}\right)}^{\#}\end{array}\right)\left(\begin{array}{cc}{B}^{\#}& D\\ A^{\#}AB{B}^{\#}& A^{\#}\end{array}\right)\left(\begin{array}{cc}{\left(B_{N,K}^{†}\right)}^{\#}{B}_{N,K}^{†}& O\\ O& I\end{array}\right)\right\}\hfill \\ & \le & r\left(\begin{array}{cc}{B}^{\#}& D\\ A^{\#}AB{B}^{\#}& A^{\#}\end{array}\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}{B}^{\#}& D\\ A^{\#}AB{B}^{\#}& A^{\#}\end{array}\right)\hfill \\ & =& r\left\{\left(\begin{array}{cc}I& O\\ O& A^{\#}A\end{array}\right)\left(\begin{array}{cc}{B}_{N,K}^{†}& D\\ A_{M,N}^{†}AB{B}_{N,K}^{†}& A_{M,N}^{†}\end{array}\right)\left(\begin{array}{cc}B{B}^{\#}& O\\ O& I\end{array}\right)\right\}\hfill \\ & \le & r\left(\begin{array}{cc}{B}_{N,K}^{†}& D\\ A_{M,N}^{†}AB{B}_{N,K}^{†}& A_{M,N}^{†}\end{array}\right).\hfill \end{array}$$

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}{B}_{N,K}^{†}& D\\ A_{M,N}^{†}AB{B}_{N,K}^{†}& A_{M,N}^{†}\end{array}\right)=r\left(\begin{array}{cc}{B}^{\#}& D\\ A^{\#}AB{B}^{\#}& A^{\#}\end{array}\right)=r\left(\begin{array}{cc}{K}^{-1}{B}^{*}& D\\ A^{*}MAB{K}^{-1}{B}^{*}& A^{*}M\end{array}\right)\end{array}$$

(31)

Substituting Formula (31) into (30), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}{B}_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{cc}{K}^{-1}{B}^{*}& D\\ A^{*}MAB{K}^{-1}{B}^{*}& A^{*}M\end{array}\right)+n-r\left(A\right)-r\left(B\right).\end{array}$$

(32)

For the seventh partitioned matrix in (14), using the same method as above with $\widetilde{E_A}A=O$, $\widetilde{E_A}=I_m-A{A}_{M,N}^{†}$$,\widetilde{F_A}=I_n-A_{M,N}^{†}A$ and $\widetilde{E_B}=I_n-B{B}_{N,K}^{†}$, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}-\widetilde{E_A}& O\\ B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{cc}-\widetilde{E_A}& O\\ B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}-DA\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{cc}-\widetilde{E_A}& O\\ DA{A}_{M,N}^{†}-D& B_{N,K}^{†}-DA\\ O& \widetilde{E_B}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{cccc}{A}_{M,N}^{†}& A_{M,N}^{†}& O& O\\ O& -\widetilde{E_A}& O& O\\ O& DA{A}_{M,N}^{†}-D& B_{N,K}^{†}-DA& O\\ O& O& \widetilde{E_B}& O\\ O& O& B_{N,K}^{†}& B_{N,K}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{cccc}{A}_{M,N}^{†}& A_{M,N}^{†}& O& O\\ -A{A}_{M,N}^{†}& -I_m& O& O\\ -DA{A}_{M,N}^{†}& -D& -DA& -B_{N,K}^{†}\\ O& O& I_n& B{B}_{N,K}^{†}\\ O& O& B_{N,K}^{†}& B_{N,K}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{c}DAB{B}_{N,K}^{†}-B_{N,K}^{†}\end{array}\right)+m+n-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(33)

From Definition 1, we have

$$\begin{array}{c}\hfill r\left(\begin{array}{c}DAB{B}_{N,K}^{†}-B_{N,K}^{†}\end{array}\right)=r\left[\left(\begin{array}{c}DAB{B}^{\#}-B^{\#}\end{array}\right){\left(B_{N,K}^{†}\right)}^{\#}{B}_{N,K}^{†}\right]\le r\left(\begin{array}{c}DAB{B}^{\#}-B^{\#}\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill r\left(\begin{array}{c}DAB{B}^{\#}-B^{\#}\end{array}\right)=r\left[\left(\begin{array}{c}DAB{B}_{N,K}^{†}-B_{N,K}^{†}\end{array}\right)B{B}^{\#}\right]\le r\left(\begin{array}{c}DAB{B}_{N,K}^{†}-B_{N,K}^{†}\end{array}\right).\end{array}$$

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{c}DAB{B}_{N,K}^{†}-B_{N,K}^{†}\end{array}\right)=r\left(\begin{array}{c}DAB{B}^{\#}-B^{\#}\end{array}\right)=r\left(\begin{array}{c}DAB{K}^{-1}{B}^{*}-K^{-1}{B}^{*}\end{array}\right).\end{array}$$

(34)

Substituting Formula (34) into (33), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{cc}-\widetilde{E_A}& O\\ B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{c}DAB{K}^{-1}{B}^{*}-K^{-1}{B}^{*}\end{array}\right)+n+m-r\left(A\right)-r\left(B\right).\end{array}$$

(35)

For the last partitioned matrix in (14), using the basic block matrix operations for rows or columns and $B\widetilde{F_B}=O$$,\widetilde{F_A}=I_n-A_{M,N}^{†}A$, $\widetilde{F_B}=I_k-B_{N,K}^{†}B$ and $\widetilde{E_B}=I_n-B{B}_{N,K}^{†}$, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{ccc}\widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ O& \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{ccc}\widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ O& A_{M,N}^{†}-BD& \widetilde{F_A}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{ccc}\widetilde{F_B}& B_{N,K}^{†}BD-D& O\\ O& A_{M,N}^{†}-BD& \widetilde{F_A}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{ccccc}{B}_{N,K}^{†}& O& O& O& O\\ B_{N,K}^{†}& \widetilde{F_B}& B_{N,K}^{†}BD-D& O& O\\ O& O& A_{M,N}^{†}-BD& \widetilde{F_A}& A_{M,N}^{†}\\ O& O& O& O& A_{M,N}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{ccccc}{B}_{N,K}^{†}& B_{N,K}^{†}B& -B_{N,K}^{†}BD& O& O\\ B_{N,K}^{†}& I_k& -D& O& O\\ O& O& -BD& I_n& A_{M,N}^{†}\\ O& O& -A_{M,N}^{†}& A_{M,N}^{†}A& A_{M,N}^{†}\end{array}\right)-r\left(A\right)-r\left(B\right)\hfill \\ & =& r\left(\begin{array}{c}{A}_{M,N}^{†}ABD-A_{M,N}^{†}\end{array}\right)+n+k-r\left(A\right)-r\left(B\right).\hfill \end{array}$$

(36)

From Definition 1, we have

$$\begin{array}{c}\hfill r\left(\begin{array}{c}{A}_{M,N}^{†}ABD-A_{M,N}^{†}\end{array}\right)=r\left[A_{M,N}^{†}{\left(A_{M,N}^{†}\right)}^{\#}\left(\begin{array}{c}{A}^{\#}ABD-A^{\#}\end{array}\right)\right]\le r\left(\begin{array}{c}{A}^{\#}ABD-A^{\#}\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill r\left(\begin{array}{c}{A}^{\#}ABD-A^{\#}\end{array}\right)=r\left[A^{\#}A\left(\begin{array}{c}{A}_{M,N}^{†}ABD-A_{M,N}^{†}\end{array}\right)\right]\le r\left(\begin{array}{c}{A}_{M,N}^{†}ABD-A_{M,N}^{†}\end{array}\right).\end{array}$$

That is,

$$\begin{array}{c}\hfill r\left(\begin{array}{c}{A}_{M,N}^{†}ABD-A_{M,N}^{†}\end{array}\right)=r\left(\begin{array}{c}{A}^{\#}ABD-A^{\#}\end{array}\right)=r\left(\begin{array}{c}{A}^{*}MABD-A^{*}M\end{array}\right).\end{array}$$

(37)

Substituting Formula (37) into (36), we have

$$\begin{array}{c}\hfill r\left(\begin{array}{ccc}\widetilde{F_B}& B_{N,K}^{†}{A}_{M,N}^{†}-D& B_{N,K}^{†}\widetilde{F_A}\\ O& \widetilde{E_B}{A}_{M,N}^{†}& \widetilde{E_B}\widetilde{F_A}\end{array}\right)=r\left(\begin{array}{c}{A}^{*}MABD-A^{*}M\end{array}\right)+n+k-r\left(A\right)-r\left(B\right).\end{array}$$

(38)

According to the above proof process, by substituting Formulas (17), (20), (23), (26), (29), (32), (35) and (38) into (14), we have Formula (12). Thus, Lemma 4 is proved. □

Theorem 1. 

Let M, N, and K be three positive-definite Hermitian matrices, and let $A\in C^{m\times n}$, $B\in C^{n\times k}$. Then,

$$\begin{array}{ccc}& & \left(AB\right)\{1,3M,4K\}\subseteq B\{1,3N,4K\}A\{1,3M,4N\}\hfill \\ & ⟺& r\left(AB\right)=max\left\{min\{a_1,a_2\},min\{k-r\left(B\right),m-r\left(A\right)\}+a_3+a_4-n\right\}.\hfill \end{array}$$

(39)

where

$$\begin{array}{c}\hfill a_1=r\left(\begin{array}{cc}{B}^{*}NB& B^{*}{A}^{*}\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right),a_3=r\left(\begin{array}{c}{B}^{*}N\\ B^{*}{A}^{*}MA\end{array}\right)\end{array}$$

(40)

and

$$\begin{array}{c}\hfill a_2=r\left(\begin{array}{cc}A{N}^{-1}{A}^{*}& AB{K}^{-1}{B}^{*}{A}^{*}\\ B^{*}{A}^{*}& B^{*}NB{K}^{-1}{B}^{*}{A}^{*}\end{array}\right),a_4=r\left(\begin{array}{c}A{N}^{-1}\\ AB{K}^{-1}{B}^{*}\end{array}\right).\end{array}$$

(41)

Proof. 

We can see that there are some ${\left(AB\right)}^{(1,3M,4K)}\in \left(AB\right)\{1,3M,4K\}$, $A^{(1,3M,4N)}\in A\{1,3M,4N\}$ and $B^{(1,3N,4K)}\in B\{1,3N,4K\}$, such that the following three formulas are equivalent:

$$\begin{array}{c}\hfill \left(AB\right)\{1,3M,4K\}\subseteq B\{1,3N,4K\}A\{1,3M,4N\},\end{array}$$

(42)

$$\begin{array}{c}\hfill B^{(1,3N,4K)}{A}^{(1,3M,4N)}={\left(AB\right)}^{(1,3M,4K)}\end{array}$$

(43)

and

$$\begin{array}{c}\hfill \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}\underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{min}r({\left(AB\right)}^{(1,3M,4K)}-B^{(1,3N,4K)}{A}^{(1,3M,4N)})=0.\end{array}$$

(44)

It is well known that $AB{\left(AB\right)}_{M,K}^{†}=AB{\left(AB\right)}^{(1,3M,4K)}=P_{R\left(AB\right),N\left(B^{\#}{A}^{\#}\right)}$ and ${\left(AB\right)}_{M,K}^{†}AB={\left(AB\right)}^{(1,3M,4K)}AB=P_{R\left(B^{\#}{A}^{\#}\right),N\left(AB\right)}$ (see [4]). Combining this fact with Formula (12) in Lemma 4, we have

$$\begin{array}{ccc}& & \underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{min}r({\left(AB\right)}^{(1,3M,4K)}-B^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& r\left(\begin{array}{cc}O& B^{*}N\\ A^{*}M-A^{*}MAB{\left(AB\right)}_{M,K}^{†}& A^{*}MA\end{array}\right)+r\left(\begin{array}{cc}{N}^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ O& K^{-1}{B}^{*}-{\left(AB\right)}_{M,K}^{†}AB{K}^{-1}{B}^{*}\end{array}\right)\hfill \\ & -& r\left(A\right)-r\left(B\right)+max\left\{r\right(B^{*}{A}^{*}-B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*})\hfill \\ & -& r(N^{-1}{A}^{*}-B{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*})-r(B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A-B^{*}N),\hfill \\ & & r\left(\begin{array}{cc}{A}^{*}M& A^{*}MAB{K}^{-1}{B}^{*}\\ {\left(AB\right)}^{(1,3M,4K)}& K^{-1}{B}^{*}\end{array}\right)-r({\left(AB\right)}_{M,K}^{†}AB{K}^{-1}{B}^{*}-K^{-1}{B}^{*})\hfill \\ & -& r(A^{*}MAB{\left(AB\right)}_{M,K}^{†}-A^{*}M)-n\}.\hfill \end{array}$$

(45)

Applying (8) of Lemma 2, we have

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}O& B^{*}N\\ A^{*}M-A^{*}MAB{\left(AB\right)}_{M,K}^{†}& A^{*}MA\end{array}\right)\hfill \\ & =& r\left[\left(\begin{array}{cc}O& B^{*}N\\ A^{*}M& A^{*}MA\end{array}\right)-\left(\begin{array}{c}O\\ A^{*}MAB\end{array}\right){\left(AB\right)}_{M,K}^{†}\left(\begin{array}{cc}{I}_m& O\end{array}\right)\right]\hfill \\ & =& r\left(\begin{array}{ccc}{\left(AB\right)}^{*}MAB{K}^{-1}{\left(AB\right)}^{*}& {\left(AB\right)}^{*}M& O\\ O& O& B^{*}N\\ A^{*}MAB{K}^{-1}{\left(AB\right)}^{*}& A^{*}M& A^{*}MA\end{array}\right)-r\left(AB\right)\hfill \\ & =& a_3+r\left(A\right)-r\left(AB\right)\hfill \end{array}$$

(46)

and

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}{N}^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ O& K^{-1}{B}^{*}-{\left(AB\right)}_{M,K}^{†}AB{K}^{-1}{B}^{*}\end{array}\right)\hfill \\ & =& r\left[\left(\begin{array}{cc}{N}^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ O& K^{-1}{B}^{*}\end{array}\right)-\left(\begin{array}{c}O\\ I_k\end{array}\right){\left(AB\right)}_{M,K}^{†}\left(\begin{array}{cc}O& AB{K}^{-1}{B}^{*}\end{array}\right)\right]\hfill \\ & =& r\left(\begin{array}{ccc}{\left(AB\right)}^{*}MAB{K}^{-1}{\left(AB\right)}^{*}& O& {\left(AB\right)}^{*}MAB{K}^{-1}{B}^{*}\\ O& N^{-1}{A}^{*}& B{K}^{-1}{B}^{*}\\ K^{-1}{\left(AB\right)}^{*}& O& K^{-1}{B}^{*}\end{array}\right)-r\left(AB\right)\hfill \\ & =& a_4+r\left(B\right)-r\left(AB\right)\hfill \end{array}$$

(47)

and

$$\begin{array}{ccc}& & r({\left(AB\right)}_{M,K}^{†}AB{K}^{-1}{B}^{*}-K^{-1}{B}^{*})=r(K^{-1}{B}^{*}-{\left(AB\right)}_{M,K}^{†}AB{K}^{-1}{B}^{*})\hfill \\ & =& r\left(\begin{array}{cc}{\left(AB\right)}^{*}MAB{K}^{-1}{\left(AB\right)}^{*}& {\left(AB\right)}^{*}MAB{K}^{-1}{B}^{*}\\ K^{-1}{\left(AB\right)}^{*}& K^{-1}{B}^{*}\end{array}\right)-r\left(AB\right)\hfill \\ & =& r\left(B\right)-r\left(AB\right)\hfill \end{array}$$

(48)

and

$$\begin{array}{ccc}& & r(A^{*}MAB{\left(AB\right)}_{M,K}^{†}-A^{*}M)=r(A^{*}M-A^{*}MAB{\left(AB\right)}_{M,K}^{†})\hfill \\ & =& r\left(\begin{array}{cc}{\left(AB\right)}^{*}MAB{K}^{-1}{\left(AB\right)}^{*}& {\left(AB\right)}^{*}M\\ A^{*}MAB{K}^{-1}{\left(AB\right)}^{*}& A^{*}M\end{array}\right)-r\left(AB\right)\hfill \\ & =& r\left(A\right)-r\left(AB\right).\hfill \end{array}$$

(49)

Applying (9) and (10) of Lemma 2, we have

$$\begin{array}{ccc}& & \underset{{\left(AB\right)}^{(1,3M,4K)}}{min}r(N^{-1}{A}^{*}-B{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*})\hfill \\ & =& r\left(\begin{array}{cc}{\left(AB\right)}^{*}MAB& {\left(AB\right)}^{*}MA{N}^{-1}{A}^{*}\\ B& N^{-1}{A}^{*}\end{array}\right)+r\left(\begin{array}{cc}AB{K}^{-1}{\left(AB\right)}^{*}& A{N}^{-1}{A}^{*}\\ B{K}^{-1}{\left(AB\right)}^{*}& N^{-1}{A}^{*}\end{array}\right)\hfill \\ & -& r\left(\begin{array}{cc}AB& A{N}^{-1}{A}^{*}\\ B& N^{-1}{A}^{*}\end{array}\right)-r\left(AB\right)\hfill \\ & =& a_4-r\left(AB\right)\hfill \\ & =& \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}r(N^{-1}{A}^{*}-B{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*})\hfill \end{array}$$

and

$$\begin{array}{ccc}& & \underset{{\left(AB\right)}^{(1,3M,4K)}}{min}r(B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A-B^{*}N)\hfill \\ & =& \underset{{\left(AB\right)}^{(1,3M,4K)}}{min}r(B^{*}N-B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A)\hfill \\ & =& r\left(\begin{array}{cc}{\left(AB\right)}^{*}MAB& {\left(AB\right)}^{*}MA\\ B^{*}NB& B^{*}N\end{array}\right)+r\left(\begin{array}{cc}AB{K}^{-1}{\left(AB\right)}^{*}& A\\ B^{*}NB{K}^{-1}{\left(AB\right)}^{*}& B^{*}N\end{array}\right)\hfill \\ & -& r\left(\begin{array}{cc}AB& A\\ B^{*}NB& B^{*}N\end{array}\right)-r\left(AB\right)\hfill \\ & =& a_3-r\left(AB\right)\hfill \\ & =& \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}r(B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A-B^{*}N).\hfill \end{array}$$

For any ${\left(AB\right)}^{(1,3M,4K)}\in \left(AB\right)\{1,3M,4K\}$, we have

$$\begin{array}{c}\hfill r(N^{-1}{A}^{*}-B{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*})=a_4-r\left(AB\right)\end{array}$$

(50)

and

$$\begin{array}{c}\hfill r(B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A-B^{*}N)=a_3-r\left(AB\right).\end{array}$$

(51)

Combining Formulas Combining Formulas (45)–(51), we have

$$\begin{array}{ccc}& & \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}\underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{min}r({\left(AB\right)}^{(1,3M,4K)}-B^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& max\{\underset{{\left(AB\right)}^{(1,3M,4K)}}{max}r(B^{*}{A}^{*}-B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*}),\hfill \end{array}$$

$$\begin{array}{ccccc}& & & & {\displaystyle \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}}r\left(\begin{array}{cc}{A}^{*}M& A^{*}MAB{K}^{-1}{B}^{*}\\ {\left(AB\right)}^{(1,3M,4K)}& K^{-1}{B}^{*}\end{array}\right)+a_3+a_4-r\left(A\right)-r\left(B\right)-n\}.\hfill \end{array}$$

(52)

Using (9) of Lemma 2 with $A=AB$, $B=A{N}^{-1}{A}^{*}$, $C=B^{*}NB$ and $D=B^{*}{A}^{*}$, we obtain

$$\begin{array}{ccc}& & \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}r(B^{*}{A}^{*}-B^{*}NB{\left(AB\right)}^{(1,3M,4K)}A{N}^{-1}{A}^{*})\hfill \\ & =& min\{r\left(\begin{array}{cc}{\left(AB\right)}^{*}MAB& {\left(AB\right)}^{*}MA{N}^{-1}{A}^{*}\\ B^{*}NB& B^{*}{A}^{*}\end{array}\right)-r\left(AB\right),\hfill \\ & & r\left(\begin{array}{cc}AB{K}^{-1}{\left(AB\right)}^{*}& A{N}^{-1}{A}^{*}\\ B^{*}NB{K}^{-1}{\left(AB\right)}^{*}& B^{*}{A}^{*}\end{array}\right)-r\left(AB\right)\}\hfill \\ & =& min\left\{a_1-r\left(AB\right),a_2-r\left(AB\right)\right\}.\hfill \end{array}$$

(53)

Using (9) of Lemma 2 with $\tilde{A}=AB$, $\tilde{X}={\left(AB\right)}^{(1,3M,4K)}$, $\tilde{B}=\left(\begin{array}{cc}{I}_m,& O\end{array}\right)$, $\tilde{C}=\left(\begin{array}{c}O\\ -I_k\end{array}\right)$ and $\tilde{D}=\left(\begin{array}{cc}{A}^{*}M& A^{*}MAB{K}^{-1}{B}^{*}\\ O& K^{-1}{B}^{*}\end{array}\right)$, we obtain

$$\begin{array}{ccc}& & \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}r\left(\begin{array}{cc}{A}^{*}M& A^{*}MAB{K}^{-1}{B}^{*}\\ {\left(AB\right)}^{(1,3M,4K)}& K^{-1}{B}^{*}\end{array}\right)=\underset{{\left(AB\right)}^{(1,3M,4K)}}{max}r(\tilde{D}-\tilde{C}\tilde{X}\tilde{B})\hfill \\ & =& min\{r\left(\begin{array}{ccc}{\left(AB\right)}^{*}MAB& {\left(AB\right)}^{*}M& O\\ O& A^{*}M& A^{*}MAB{K}^{-1}{B}^{*}\\ -I_k& O& K^{-1}{B}^{*}\end{array}\right),\hfill \\ & & r\left(\begin{array}{ccc}AB{K}^{-1}{\left(AB\right)}^{*}& I_m& O\\ O& A^{*}M& A^{*}MAB{K}^{-1}{B}^{*}\\ -K^{-1}{\left(AB\right)}^{*}& O& K^{-1}{B}^{*}\end{array}\right)\}-r\left(AB\right)\hfill \\ & =& min\left\{r\left(A\right)-r\left(AB\right)+k,r\left(B\right)-r\left(AB\right)+m\right\}.\hfill \end{array}$$

(54)

Combining Formulas (52), (53) and (54), we have

$$\begin{array}{ccc}& & \underset{{\left(AB\right)}^{(1,3M,4K)}}{max}\underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{min}r({\left(AB\right)}^{(1,3M,4K)}-B^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& max\left\{min\{a_1,a_2\},min\{k-r\left(B\right),m-r\left(A\right)\}+a_3+a_4-n\right\}-r\left(AB\right).\hfill \end{array}$$

(55)

According to Formulas (42), (43) and (44) (i.e., (55)), we obtain Formula (39). □

Theorem 2. 

Let M, N, and K be three positive-definite Hermitian matrices, and let $A\in C^{m\times n}$, $B\in C^{n\times k}$. Then,

$$\begin{array}{ccc}& & B\{1,3N,4K\}A\{1,3M,4N\}\subseteq \left(AB\right)\{1,3M,4K\}\hfill \\ & ⟺& min\left\{a_1+m-r\left(A\right),a_3\right\}=r\left(B\right)andmin\left\{a_2+k-r\left(B\right),a_4\right\}=r\left(A\right),\hfill \end{array}$$

(56)

where $a_1$, $a_2$, $a_3$ and $a_4$ are the same as in Theorem 1.

Proof. 

For the convenience of future research, we simply provide an idea for proving this theorem. For any $A^{(1,3M,4N)}\in A\{1,3M,4N\}$ and $B^{(1,3N,4K)}\in B\{1,3N,4K\}$, according to Lemma 1, Formulas (57)–(59) are equivalent, as follows:

$$\begin{array}{c}\hfill B\{1,3N,4K\}A\{1,3M,4N\}\subseteq \left(AB\right)\{1,3M,4K\}.\end{array}$$

(57)

$$\begin{array}{ccc}& & {\left(AB\right)}^{*}MAB{B}^{(1,3N,4K)}{A}^{(1,3M,4N)}={\left(AB\right)}^{*}M\hfill \\ & and& \\ & & B^{(1,3N,4K)}{A}^{(1,3M,4N)}AB{K}^{-1}{\left(AB\right)}^{*}=K^{-1}{\left(AB\right)}^{*}.\hfill \end{array}$$

(58)

and

$$\begin{array}{ccc}& & \underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{max}r(B^{*}{A}^{*}M-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{A}^{(1,3M,4N)})=0\hfill \\ & and& \\ & & \underset{B^{(1,3N,4K)},A^{(1,3M,4N)}}{max}r(K^{-1}{B}^{*}{A}^{*}-B^{(1,3N,4K)}{A}^{(1,3M,4N)}AB{K}^{-1}{B}^{*}{A}^{*})=0.\hfill \end{array}$$

(59)

From (9) of Lemma 2 with $A=A$, $B=I_m$, $C=B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}$ and $D=B^{*}{A}^{*}M$, we have

$$\begin{array}{ccc}& & \underset{A^{(1,3M,4N)}}{max}r(B^{*}{A}^{*}M-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& min\{r\left(\begin{array}{cc}{A}^{*}MA& A^{*}M\\ B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}& B^{*}{A}^{*}M\end{array}\right)-r\left(A\right),\hfill \\ & & r\left(\begin{array}{cc}A{N}^{-1}{A}^{*}& I_m\\ B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{N}^{-1}{A}^{*}& B^{*}{A}^{*}M\end{array}\right)-r\left(A\right)\}\hfill \\ & =& min\left\{r\right(B^{*}{A}^{*}MA-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}),\hfill \\ & & r(B^{*}{A}^{*}MA{N}^{-1}{A}^{*}-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{N}^{-1}{A}^{*})+m-r\left(A\right)\}.\hfill \end{array}$$

(60)

Using (10) of Lemma 2 with $A=B$, $B=I_n$, $C=B^{*}{A}^{*}MAB$ and $D=B^{*}{A}^{*}MA$, we have

$$\begin{array}{ccc}& & \underset{B^{(1,3N,4K)}}{min}r(B^{*}{A}^{*}MA-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)})\hfill \\ & =& r\left(\begin{array}{cc}{B}^{*}NB& B^{*}N\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA\end{array}\right)+r\left(\begin{array}{cc}B{K}^{-1}{B}^{*}& I_n\\ B^{*}{A}^{*}MAB{K}^{-1}{B}^{*}& B^{*}{A}^{*}MA\end{array}\right)\hfill \\ & -& r\left(\begin{array}{cc}B& I_n\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA\end{array}\right)-r\left(B\right)=a_3-r\left(B\right).\hfill \end{array}$$

(61)

Using (10) of Lemma 2 with $A=B$, $B=N^{-1}{A}^{*}$, $C=B^{*}{A}^{*}MAB$ and $D=B^{*}{A}^{*}MA{N}^{-1}{A}^{*}$, we have

$$\begin{array}{ccc}& & \underset{B^{(1,3N,4K)}}{min}r(B^{*}{A}^{*}MA{N}^{-1}{A}^{*}-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{N}^{-1}{A}^{*})\hfill \\ & =& r\left(\begin{array}{cc}{B}^{*}NB& B^{*}{A}^{*}\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)+r\left(\begin{array}{cc}B{K}^{-1}{B}^{*}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1}{B}^{*}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\hfill \\ & -& r\left(\begin{array}{cc}B& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)-r\left(B\right).\hfill \end{array}$$

(62)

This is because

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}B{K}^{-1}{B}^{*}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1}{B}^{*}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\hfill \\ & =& r\left[\left(\begin{array}{cc}B{K}^{-1/2}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1/2}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\left(\begin{array}{cc}{\left(B{K}^{-1/2}\right)}^{*}& O\\ O& I\end{array}\right)\right]\hfill \\ & \le & r\left(\begin{array}{cc}B{K}^{-1/2}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1/2}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\hfill \end{array}$$

and

$$\begin{array}{ccc}& & r\left(\begin{array}{cc}B{K}^{-1/2}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1/2}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\hfill \\ & =& r\left[\left(\begin{array}{cc}B{K}^{-1}{B}^{*}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1}{B}^{*}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\left(\begin{array}{cc}{\left(B{K}^{-1}{B}^{*}\right)}^{†}B{K}^{-1/2}& O\\ O& I\end{array}\right)\right]\hfill \\ & \le & r\left(\begin{array}{cc}B{K}^{-1}{B}^{*}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1}{B}^{*}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right).\hfill \end{array}$$

(63)

Then, we have

$$\begin{array}{ccc}\hfill r\left(\begin{array}{cc}B{K}^{-1}{B}^{*}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1}{B}^{*}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)& =& r\left(\begin{array}{cc}B{K}^{-1/2}& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB{K}^{-1/2}& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)\hfill \\ & =& r\left(\begin{array}{cc}B& N^{-1}{A}^{*}\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right).\hfill \end{array}$$

(64)

Combining Formulas (62)–(64), we have

$$\begin{array}{ccc}& & \underset{B^{(1,3N,4K)}}{min}r(B^{*}{A}^{*}MA{N}^{-1}{A}^{*}-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{N}^{-1}{A}^{*})\hfill \\ & =& r\left(\begin{array}{cc}{B}^{*}NB& B^{*}{A}^{*}\\ B^{*}{A}^{*}MAB& B^{*}{A}^{*}MA{N}^{-1}{A}^{*}\end{array}\right)-r\left(B\right)\hfill \\ & =& a_1-r\left(B\right).\hfill \end{array}$$

(65)

From Formulas (60), (61) and (65), we have

$$\begin{array}{ccc}& & \underset{A^{(1,3M,4N)},B^{(1,3N,4K)}}{max}r(B^{*}{A}^{*}M-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{A}^{(1,3M,4N)})\hfill \\ & =& min\{\underset{B^{(1,3N,4K)}}{min}r(B^{*}{A}^{*}MA-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}),\hfill \\ & & \left[\underset{B^{(1,3N,4K)}}{min}r(B^{*}{A}^{*}MA{N}^{-1}{A}^{*}-B^{*}{A}^{*}MAB{B}^{(1,3N,4K)}{N}^{-1}{A}^{*})\right]+m-r\left(A\right)\}\hfill \\ & =& min\left\{a_1-r\left(A\right)-r\left(B\right)+m,a_3-r\left(B\right)\right\}.\hfill \end{array}$$

(66)

On the other hand, using Lemma 2 again and similarly to the method in (66), we have

$$\begin{array}{ccc}& & \underset{B^{(1,3N,4K)},A^{(1,3M,4N)}}{max}r(K^{-1}{B}^{*}{A}^{*}-B^{(1,3N,4K)}{A}^{(1,3M,4N)}AB{K}^{-1}{B}^{*}{A}^{*})\hfill \\ & =& min\left\{a_2+k-r\left(A\right)-r\left(B\right),a_4-r\left(A\right)\right\}.\hfill \end{array}$$

(67)

Finally, by combining Formulas (57), (58), (59), (66) and (67), we have Formula (56). □

The reverse order law (2) holds if and only if the two inclusions (3) and (4) hold. Thus, combining Theorems 1 and 2, we immediately obtain the main result of this paper.

Theorem 3. 

Let M, N, and K be three positive-definite Hermitian matrices. Let $A\in C^{m\times n}$ and $B\in C^{n\times k}$. Let $a_1$, $a_2$, $a_3$ and $a_4$ be the same as in Theorem 1. Then, the following statements are equivalent:

$$\begin{array}{ccc}& \left(1\right)& B\{1,3N,4K\}A\{1,3M,4N\}=\left(AB\right)\{1,3M,4K\}\hfill \\ & \left(2\right)& r\left(AB\right)=max\left\{min\{a_1,a_2\},min\{k-r\left(B\right),m-r\left(A\right)\}+a_3+a_4-n\right\}and\hfill \\ & & r\left(B\right)=min\left\{a_1+m-r\left(A\right),a_3\right\}andr\left(A\right)=min\left\{a_2+k-r\left(B\right),a_4\right\}.\hfill \end{array}$$

Corollary 1. 

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

$$\begin{array}{ccc}& \left(1\right)& B\{1,3,4\}A\{1,3,4\}=\left(AB\right)\{1,3,4\}\hfill \\ & \left(2\right)& r\left(AB\right)=max\left\{min\{b_1,b_2\},min\{k-r\left(B\right),m-r\left(A\right)\}+b_3+b_4-n\right\}and\hfill \\ & & r\left(B\right)=min\left\{b_1+m-r\left(A\right),b_3\right\}andr\left(A\right)=min\left\{b_2+k-r\left(B\right),b_4\right\},\hfill \end{array}$$

where

$$\begin{array}{c}\hfill b_1=r\left(\begin{array}{cc}{B}^{*}B& B^{*}{A}^{*}\\ B^{*}{A}^{*}AB& B^{*}{A}^{*}A{A}^{*}\end{array}\right),b_3=r\left(\begin{array}{c}{B}^{*}\\ B^{*}{A}^{*}A\end{array}\right)\end{array}$$

and

$$\begin{array}{c}\hfill b_2=r\left(\begin{array}{cc}A{A}^{*}& AB{B}^{*}{A}^{*}\\ B^{*}{A}^{*}& B^{*}B{B}^{*}{A}^{*}\end{array}\right),b_4=r\left(\begin{array}{c}A\\ AB{B}^{*}\end{array}\right).\end{array}$$

3. Conclusions

The reverse order law for a matrix product’s weighted generalized inverses was studied in this article by using the ranks of generalized Schur complements. The work in this paper can be utilized as a useful tool in many algorithms for the computation of matrix equations’ weighted least squares technique.