On Cartesian product of matrices

TL;DR

This paper explores the Cartesian product of matrices, providing formulas for traces, establishing identities, and applying these concepts to analyze graph properties.

Contribution

It introduces trace formulas and identities for the Cartesian product of matrices, extending its application to graph theory.

Findings

Derived trace expressions for multiple matrices

Established identities involving Cartesian products

Applied Cartesian product to graph property analysis

Abstract

Recently, Bapat and Kurata [\textit{Linear Algebra Appl.}, 562(2019), 135-153] defined the Cartesian product of two square matrices $A$ and $B$ as $A\oslash B=A\otimes \J+\J\otimes B$, where $\J$ is the all one matrix of appropriate order and $\otimes$ is the Kronecker product. In this article, we find the expression for the trace of the Cartesian product of any finite number of square matrices in terms of traces of the individual matrices. Also, we establish some identities involving the Cartesian product of matrices. Finally, we apply the Cartesian product to study some graph-theoretic properties.

Equations

$$A\otimes B=\left(\begin{array}{ccccc}& a_{1,1}B& a_{1,2}B& \cdots & a_{1,n}B\\ & a_{2,1}B& a_{2,2}B& \cdots & a_{2,n}B\\ & ⋮& ⋮& \ddots & ⋮\\ & a_{m,1}B& a_{m,2}B& \cdots & a_{m,n}B\end{array}\right).$$

$$x^T\mathcal{D}(G)x=\sum _{1\le i<j\le n}{d}_{ij}{x}_i{x}_j.$$

$$(A\otimes B)(C\otimes D)=(AC)\otimes (BD).$$

$$P^{-1}(A\otimes B)P=B\otimes A.$$

$$\text{mbox}tr(A\oslash B)$$

$$=\text{mbox}tr(A\otimes {\mathbb{J}}_n)+\text{mbox}tr({\mathbb{J}}_m\otimes B)$$

$$=\text{mbox}tr(A)\times \text{mbox}tr({\mathbb{J}}_n)+\text{mbox}tr({\mathbb{J}}_m)\times \text{mbox}tr(B)[using\text{lx@cref}creftyperefnumtr-kproduct]$$

$$=n.\text{mbox}tr(A)+m.\text{mbox}tr(B).$$

$$tr(k_1{A}_1\oslash k_2{A}_2\oslash \cdots \oslash k_n{A}_n)=({\Pi }_{i=1}^n{n}_i)\sum _{i=1}^n\frac{k_{i}tr(A_i)}{n_i}.$$

$$\text{mbox}tr(k_1{A}_1\oslash k_2{A}_2\oslash \cdots \oslash k_{\ell }{A}_{\ell })=({\Pi }_{i=1}^{\ell }{n}_i)\sum _{i=1}^{\ell }\frac{k_i\text{mbox}tr(A_i)}{n_i}.$$

$$\text{mbox}tr(k_1{A}_1\oslash k_2{A}_2\oslash \cdots \oslash k_{\ell }{A}_{\ell }\oslash k_{\ell +1}{A}_{\ell +1})$$

$$=$$

$$tr\big{(}(A+B)\oslash(A-B)\big{)}=2n.tr(A).$$

$$\displaystyle\mbox{tr}\big{(}(A+B)\oslash(A-B)\big{)}$$

$$=n[\text{mbox}tr(A)-\text{mbox}tr(B)+\text{mbox}tr(A)+\text{mbox}tr(B)]$$

$$=2n.\text{mbox}tr(A).$$

$$tr(A\otimes (B_1\oslash B_2\oslash \cdots \oslash B_k))=n^{k-1}tr(A)\sum _{i=1}^{k}tr(B_i).$$

$$\text{mbox}tr(A\otimes (B_1\oslash B_2\oslash \cdots \oslash B_k))$$

$$=\text{mbox}tr(A).\text{mbox}tr(B_1\oslash B_2\oslash \cdots \oslash B_k)[\text{mbox}using\text{lx@cref}creftyperefnumtr-kproduct]$$

$$=\text{mbox}tr(A).n^{k-1}\sum _{i=1}^k\text{mbox}tr(B_i)[\text{mbox}using\text{lx@cref}creftyperefnumtracemain]$$

$$=n^{k-1}\text{mbox}tr(A)\sum _{i=1}^k\text{mbox}tr(B_i).$$

$$\displaystyle tr\big{[}(A_{1}\oslash A_{2}\oslash\cdots\oslash A_{\ell})\otimes(A_{\ell+1}\oslash A_{\ell+2}\oslash\cdots\oslash A_{m})\otimes\cdots\otimes(A_{r}\oslash A_{r+1}\cdots A_{t})\big{]}$$

$$={\Pi }_{p=1}^t{n}_p\sum _{i=1}^{\ell }\frac{tr(A_i)}{n_i}\sum _{j=\ell +1}^m\frac{tr(A_j)}{n_j}\cdots \sum _{k=r}^t\frac{tr(A_k)}{n_k}.$$

$$\displaystyle~{}~{}~{}\mbox{tr}\big{[}(A_{1}\oslash A_{2}\oslash\cdots\oslash A_{\ell})\otimes(A_{\ell+1}\oslash A_{\ell+2}\oslash\cdots\oslash A_{m})\otimes\cdots\otimes(A_{r}\oslash A_{r+1}\cdots A_{t})\big{]}$$

$$\displaystyle=\big{[}\mbox{tr}(A_{1}\oslash A_{2}\cdots\oslash A_{\ell})\big{]}\big{[}\mbox{tr}(A_{\ell+1}\oslash A_{\ell+2}\oslash\cdots\oslash A_{m})\big{]}\cdots\big{[}\mbox{tr}(A_{r}\oslash A_{r+1}\oslash\cdots\oslash A_{t})\big{]}$$

$$\displaystyle=\Big{(}\Pi_{i=1}^{\ell}\sum_{i=1}^{\ell}\frac{\mbox{tr}(A_{i})}{n_{i}}\Big{)}\Big{(}\Pi_{\ell+1}^{m}n_{j}\sum_{j=\ell+1}^{m}\frac{\mbox{tr}(A_{j})}{n_{j}}\Big{)}\cdots\Big{(}\Pi_{k=r}^{t}n_{k}\sum_{k=r}^{t}\frac{\mbox{tr}(A_{k})}{n_{k}}\Big{)}\quad[\mbox{using }\lx@cref{creftype~refnum}{tracemain}]$$

$$={\Pi }_{p=1}^t{n}_p\sum _{i=1}^{\ell }\frac{\text{mbox}tr(A_i)}{n_i}\sum _{j=\ell +1}^m\frac{\text{mbox}tr(A_j)}{n_j}\cdots \sum _{k=r}^t\frac{\text{mbox}tr(A_k)}{n_k}.$$

$$\displaystyle tr\big{[}(A_{1}\otimes A_{2}\otimes\cdots\otimes A_{\ell})\oslash(A_{\ell+1}\otimes A_{\ell+2}\otimes\cdots\otimes A_{m})\oslash\cdots\oslash(A_{r}\otimes A_{r+1}\otimes\cdots\otimes A_{t})\big{]}$$

$$\displaystyle=\Pi_{p=1}^{t}n_{p}\Big{[}\Pi_{i=1}^{\ell}\frac{tr(A_{i})}{n_{i}}+\Pi_{j=\ell+1}^{m}\frac{tr(A_{i})}{n_{j}}+\cdots+\Pi_{k=r}^{t}\frac{tr(A_{k})}{n_{k}}\Big{]}.$$

$$\displaystyle~{}~{}~{}~{}\mbox{tr}\big{[}(A_{1}\otimes A_{2}\otimes\cdots\otimes A_{\ell})\oslash(A_{\ell+1}\otimes A_{\ell+2}\otimes\cdots\otimes A_{m})\oslash\cdots\oslash(A_{r}\otimes A_{r+1}\otimes\cdots\otimes A_{t})\big{]}$$

$$\displaystyle=\Pi_{p=1}^{t}n_{p}\Big{[}\frac{\mbox{tr}(A_{1}\otimes A_{2}\otimes\cdots A_{\ell})}{\Pi_{i=1}^{\ell}n_{i}}+\frac{\mbox{tr}(A_{\ell+1}\otimes A_{\ell+2}\cdots\otimes A_{m})}{\Pi_{j=\ell+1}^{m}n_{j}}+\cdots+\frac{\mbox{tr}(A_{r}\otimes\cdots\otimes A_{t})}{\Pi_{k=r}^{t}n_{k}}\Big{]}$$

$$\displaystyle=\Pi_{p=1}^{t}n_{p}\Big{[}\frac{\Pi_{i=1}^{\ell}\mbox{tr}(A_{i})}{\Pi_{i=1}^{\ell}n_{i}}+\frac{\Pi_{j=\ell+1}^{m}\mbox{tr}(A_{i})}{\Pi_{j=1}^{m}n_{j}}+\cdots+\frac{\Pi_{k=r}^{t}\mbox{tr}(A_{k})}{\Pi_{k=r}^{t}n_{k}}\Big{]}$$

$$\displaystyle=\Pi_{p=1}^{t}n_{p}\Big{[}\Pi_{i=1}^{\ell}\frac{\mbox{tr}(A_{i})}{n_{i}}+\Pi_{j=\ell+1}^{m}\frac{\mbox{tr}(A_{i})}{n_{j}}+\cdots+\Pi_{k=r}^{t}\frac{\mbox{tr}(A_{k})}{n_{k}}\Big{]}.$$

$$A\oslash B=A\otimes {\mathbb{J}}_n+J_m\otimes B$$

$$(A\oslash B{)}^T$$

$$=(A\otimes {\mathbb{J}}_n{)}^T+({\mathbb{J}}_m\otimes B{)}^T$$

$$=A^T\otimes {\mathbb{J}}_n^T+{\mathbb{J}}_m^T\otimes B^T[\text{mbox}using\text{lx@cref}creftyperefnumlem-transpose]$$

$$=A^T\otimes {\mathbb{J}}_n+{\mathbb{J}}_m\otimes B^T$$

$$=A^T\oslash B^T.$$

$$(A_1\oslash A_2\oslash \cdots \oslash A_n{)}^T=A_1^T\oslash A_2^T\oslash \cdots \oslash A_n^T.$$

Full text

Keywords: Cartesian product, Kronecker product, Hadamard product, Trace.

AMS Subject Classification: 05C50, 05C12.

1 Introduction and terminology

By $\mathbb{M}_{m,n}$, we denote the class of all matrices of size $m\times n.$ Also, by $\mathbb{M}_{n}$, we denote the class of all square matrices of order $n.$ For $M\in\mathbb{M}_{n}$ we write $m_{ij}$ or $M_{ij}$ to denote the $ij-$th element of $M.$ By $\mathbb{J}$ and $1\!\!1$, we mean the matrix of all one’s and vector of all one’s, respectively of suitable order. Similarly $\large 0$ denotes the zero matrix or the vector. We will mention their order wherever its necessary. Throughout this article, we denote the sum of all entries of a matrix $A$ by $S_{A}$ and the sum of the entries of $i-$th row of $A$ by $A_{i}.$ The inertia of a square matrix $M$ with real eigenvalues is the triplet $(n_{+}(M),n_{0}(M),n_{-}(M)),$ where $n_{+}(M)$ and $n_{-}(M)$ denote the number of positive and negative eigenvalues of $M$, respectively, and $n_{0}(M)$ is the algebraic multiplicity of 0 as an eigenvalue of $M.$

The Kronecker product of two matrices $A$ and $B$ of sizes $m\times n$ and $p\times q,$ respectively, denoted by $A\otimes B$ is defined to be the $mp\times nq$ block matrix

[TABLE]

The Hadamard product of two matrices $A$ and $B$ of the same size, denoted by $A\circ B$ is defined to be the entrywise product $A\circ B=[a_{i,j}b_{i,j}].$

Bapat and Kurata [2] defined the Cartesian product of two square matrices $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n}$ as $A\oslash B=A\otimes\mathbb{J}_{n}+\mathbb{J}_{m}\otimes B.$ The authors proved the Cartesian product to be associative. We use $A^{[k]}$ to mean $\underbrace{A\oslash A\oslash\cdots\oslash A}_{k~{}times}.$

If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $A\oslash B$ can be considered as a block matrix with $i,j-$th block $a_{ii}\mathbb{J}_{n}+B,~{}i=1,2,\ldots,m,$ in other words $A\oslash B$ is the matrix obtained from $A$ by replacing $a_{i,j}$ by $a_{ii}\mathbb{J}_{n}+B$. It can be observed that $a_{i,j}+b_{p,q}$ is the $p,q-$th entry of the $i,j-$th block of $A\oslash B.$

All graphs considered here are finite, undirected, connected and simple. The distance between two vertices $u,v\in V(G)$ is denoted by $d_{uv}$ and is defined as the length of a shortest path between $u$ and $v$ in $G.$ The distance matrix of $G$ is denoted by $\mathcal{D}(G)$ and is defined by $\mathcal{D}(G)=(d_{uv})_{u,v\in V(G)}.$ Since $\mathcal{D}(G)$ is a real symmetric matrix, all its eigenvalues are real. For a column vector $x=(x_{1},\ldots,x_{n})^{T}\in\mathbb{R}^{n},$ we have

[TABLE]

The Wiener index $W(G)$ of a graph is the sum of the distances between all unordered pairs of vertices of $G$, in other words $W(G)=\frac{S_{\mathcal{D}(G)}}{2}.$ The $\textit{distance spectral radius}~{}\rho^{\mathcal{D}}(G)$ of $G$ is the largest eigenvalue of its distance matrix $\mathcal{D}(G).$ The transmission, denoted by $Tr(v)$ of a vertex $v$ is the sum of the distances from $v$ to all other vertices in $G$.

The Cartesian product $G_{1}\Box G_{2}$ of two graphs $G_{1}$ and $G_{2}$ is the graph whose vertex set is the Cartesian product $V(G_{1})\times V(G_{2})$ and in which two vertices $(u,u^{\prime})$ and $(v,v^{\prime})$ are adjacent if and only if either $u=v$ and $u^{\prime}$ is adjacent to $v^{\prime}$ in $G_{2}$, or $u^{\prime}=v^{\prime}$ and $u$ is adjacent to $v$ in $G_{1}.$ Let $Gu*Hv$ denote the graph obtained from two graphs $G$ and $H$ by identifying a vertex $u$ from $G$ with a vertex $v$ from $H.$

The article have been organized as follows. In Section 2, we discuss some existing results involving Kronecker product of matrices and Cartesian product of graphs. In Section 3, we find trace of various compositions of matrices involving Cartesian product. Again in Section 4, we obtain some identities involving Cartesian product of matrices and find some applications in graph theory.

2 Preliminaries

Kronecker product has been extensively studied in the literature. Some of the interesting properties of the Kronecker product are given below.

Lemma 2.1**.**

[5]** If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $tr(A\otimes B)=tr(A)\times tr(B).$

Lemma 2.2**.**

[5]** If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $(A\otimes B)^{T}=A^{T}\times B^{T}.$

Lemma 2.3**.**

[5]** If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $(A\otimes B)^{*}=A^{*}\times B^{*}.$

Lemma 2.4**.**

[5]** For $A\in\mathbb{M}_{m},~{}B\in\mathbb{M}_{n},$ and $a,b\in\mathbb{C},$ $aA\otimes bB=abA\otimes B.$

Lemma 2.5**.**

[5]** For matrices $A,B,C$ and $D$ of appropriate sizes

[TABLE]

Lemma 2.6**.**

[5]** For any $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ there exist a permutation matrix $P$ such that

[TABLE]

For more results on Kronecker product, we refer [3]. The Cartesian product of two graphs have been studied by many researchers. Here we are interested in Cartesian product of two matrices because for any two connected graphs $G_{1}$ and $G_{2},$ the distance matrix of $G_{1}\Box G_{2}$ equals to the Cartesian product of the distance matrices of $G_{1}$ and $G_{2},$ i.e. $\mathcal{D}(G_{1}\Box G_{2})=\mathcal{D}(G_{1})\oslash\mathcal{D}(G_{2}).$ Zhang and Godsil [6] found the distance inertia of the Cartesian product of two graphs.

Theorem 2.7**.**

[6]** If $G$ and $H$ are two connected graphs, where $V(G)=\{u_{1},\ldots,u_{m}\}$ and $V(H)=\{v_{1},\ldots,v_{n}\},$ then, the inertia of distance matrix of $G\Box H$ is $(n_{+}(Gu_{m}*Hu_{n}),(m-1)(n-1)+n_{0}(Gu_{m}*Hu_{n}),n_{-}(Gu_{m}*Hu_{n})).$

Corollary 2.8**.**

[6]** Let $T_{1}$ and $T_{2}$ be two trees on $m$ and $n$ vertices, respectively. Then the distance inertia of $T_{1}\Box T_{2}$ is $(1,(m-1)(n-1),m+n-2).$

3 Trace of Cartesian product

Here we consider different compositions and products involving Cartesian product of matrices and evaluate their trace.

Lemma 3.1**.**

If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $tr(A\oslash B)=n.tr(A)+m.tr(B).$

**Proof. **We have

[TABLE]

Theorem 3.2**.**

If $A_{i}\in\mathbb{M}_{n_{i}}$ and $k_{i}\in\mathbb{C}$ for $i=1,2,\ldots,n,$ then

[TABLE]

**Proof. **We prove the result by induction on $n.$ For $n=1,$ there is nothing to prove. For $n=2,$ the result follows from Lemma 3.1. Suppose the result holds for $n=\ell\leq n-1.$ That is

[TABLE]

Now

[TABLE]

Hence the result follows by induction.

As immediate corollary of the above theorem we get the following result.

Corollary 3.3**.**

For $A\in\mathbb{M}_{n},~{}tr(A^{[k]})=k.n^{k-1}tr(A).$

Proposition 3.4**.**

If $A,B\in\mathbb{M}_{n},$ then

[TABLE]

**Proof. **From Lemma 3.1, we have

[TABLE]

Proposition 3.5**.**

For $A\in\mathbb{M}_{m},B_{i}\in\mathbb{M}_{n};i=1,2,\ldots,k,$ then

[TABLE]

**Proof. **We have

[TABLE]

Theorem 3.6**.**

If $A_{i}\in\mathbb{M}_{n_{i}}$ for $i=1,2,\ldots,t,$ then

[TABLE]

**Proof. **By repeated application of Lemma 2.1 we get

[TABLE]

Theorem 3.7**.**

If $A_{i}\in\mathbb{M}_{n_{i}};i=1,2,\ldots,t,$ then

[TABLE]

**Proof. **Using Theorem 3.2 and then Lemma 2.1, we get

[TABLE]

4 Some identities and applications

From the definition of Cartesian product of two matrices, we get following remarks.

Remark 4.1**.**

If $A$ and $B$ are square matrices and $k\in\mathbb{C},$ then $kA\oslash kB=k(A\oslash B).$

Remark 4.2**.**

For $A\in\mathbb{M}_{n}$ and any $k\in\mathbb{C},$ $k\oslash A=A+k\mathbb{J}_{n}=A\oslash k.$

For any square matrices $A$ and $B,$ from the definitions of Kronecker product and Cartesian product, it can be observed that if $a_{i,j}b_{p,q}$ is an entry of $A\otimes B$ then the corresponding entry of $A\oslash B$ is $a_{i,j}+b_{p,q}.$ Thus from Lemma 2.6, we see that if $P^{-1}(A\otimes B)P=B\otimes A,$ then for the same $P,$ we get $P^{-1}(A\oslash B)P=B\oslash A.$ Thus we get the following result.

Remark 4.3**.**

If $A$ and $B$ are square matrices, then $A\oslash B$ is permutation similar to $B\oslash A.$

Proposition 4.4**.**

For $A\in\mathbb{M}_{m},B\in\mathbb{M}_{n},~{}(A\oslash B)^{T}=A^{T}\oslash B^{T}.$

**Proof. **By definition we have

[TABLE]

which implies

[TABLE]

Hence the result.

By repeated application of Proposition 4.4, we get the following result as a corollary.

Corollary 4.5**.**

For square matrices $A_{i}$ for $i=1,2,\ldots,n,$

[TABLE]

Proceeding as in Proposition 4.4 and using Lemma 2.3, we get the following result.

Proposition 4.6**.**

For $A\in\mathbb{M}_{m},~{}B\in\mathbb{M}_{n},~{}(A\oslash B)^{*}=A^{*}\oslash B^{*}.$

By repeated application of Proposition 4.6, we get the following result as a corollary.

Corollary 4.7**.**

For square matrices $A_{i}$ for $i=1,2,\ldots,n,$

[TABLE]

Theorem 4.8**.**

If $A\in\mathbb{M}_{m},B\in\mathbb{M}_{n},$ then $A\oslash B$ is symmetric if and only if $A$ and $B$ are both symmetric.

**Proof. **If $A$ and $B$ are both symmetric, then $A^{T}=A$ and $B^{T}=B.$ Now

[TABLE]

Therefore $A\oslash B$ is symmetric.

Conversely, suppose that $A\oslash B$ is symmetric. Then $1,1$ block of $A\oslash B$ must be symmetric. But $1,1$ block of $A\oslash B$ is $a_{1,1}\mathbb{J}_{n}+B$ which is symmetric if and only if $B$ is symmetric. Again since $A\oslash B$ is symmetric, the $1,1$ entry of any $i,j-$th block of $A\oslash B$ must be same as $1,1$ entry of $j,i-$th block of $A\oslash B.$ That is $a_{i,j}+b_{1,1}=a_{j,i}+b_{1,1}\hbox{ for all }i,j=1,2\ldots,n.$ Which implies that A is symmetric.

Theorem 4.9**.**

If $A\in\mathbb{M}_{m},B\in\mathbb{M}_{n},$ then $A\oslash B$ is skew-symmetric if and only if $A$ and $B$ are both skew-symmetric.

**Proof. **If $A$ and $B$ are both skew-symmetric, then $A^{T}=-A$ and $B^{T}=-B.$ Now

[TABLE]

Therefore $A\oslash B$ is skew-symmetric.

The other direction is similar to that of the proof of Theorem 4.8.

Theorem 4.10**.**

If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ the $A\oslash B$ is a diagonal matrix if and only if $A=k\mathbb{J}_{m}$ and $B=-k\mathbb{J}_{n}$ for some $k\in\mathbb{C}.$ Furthermore in that case $A\oslash B=\large 0.$

**Proof. **If $A=k\mathbb{J}_{m}$ and $B=-k\mathbb{J}_{n}$ for some $k\in\mathbb{C},$ then

[TABLE]

Again if $A\oslash B$ is a diagonal matrix, then we must have

[TABLE]

Solving all those equations we see that all entries of $A$ are equal (say $k$) and all entries of $B$ are also equal ($-k$). Thus we get our required result.

Corollary 4.11**.**

There exist no square matrices $A,B$ such that $A\oslash B=I.$

Theorem 4.12**.**

If $A,C\in\mathbb{M}_{m},~{}B,D\in\mathbb{M}_{n},$ then $A\oslash B=C\oslash D$ if and only if $C=A-k\mathbb{J}_{m}$ and $D=B+k\mathbb{J}_{n}$ for some $k\in\mathbb{C}.$

**Proof. **If $C=A-k\mathbb{J}_{m}$ and $D=B+k\mathbb{J}_{n}$ for some $k\in\mathbb{C},$ then

[TABLE]

Conversely, suppose that $A\oslash B=C\oslash D.$ Then every block of $A\oslash B$ equals to the corresponding block of $C\oslash D,$

[TABLE]

Which implies that $a_{i,j}+b_{p,q}=c_{i,j}+d_{p,q}$ for any $i,j=1,2.\ldots,m$ and $p,q=1,2,\ldots,n.$ That is $a_{i,j}-c_{i,j}=d_{p,q}-b_{p,q}$ for any $i,j=1,2.\ldots,m$ and $p,q=1,2,\ldots,n.$ Therefore we must have $A-C=\lambda\mathbb{J}_{m}$ and $D-B=\lambda\mathbb{J}_{n}$ for some $\lambda\in\mathbb{C}.$ Hence the theorem follows.

Theorem 4.13**.**

If $A,B\in\mathbb{M}_{n},$ then $A\oslash B=B\oslash A$ if and only if $B=A+k\mathbb{J}_{n}$ for some $k\in\mathbb{C}.$

**Proof. **If $B=A+k\mathbb{J}_{n},$ then by direct calculation we have

[TABLE]

Now suppose $A\oslash B=B\oslash A.$ Then $a_{i,j}+b_{p,q}=b_{i,j}+a_{p,q}\mbox{ for all }i,j,p,q=1,2\ldots,n.$ Therefore

[TABLE]

Thus $B=A+k\mathbb{J}_{n}$ for $k=\frac{S_{B}-S_{A}}{n^{2}}.$

Theorem 4.14**.**

If $A,B,C,D\in\mathbb{M}_{n},$ then

1. (i)

$(A\oslash B)(C\oslash D)=AC\oslash BD+A\mathbb{J}_{n}\otimes\mathbb{J}_{n}B+\mathbb{J}_{n}C\otimes D\mathbb{J}_{n}.$** 2. (ii)

$(A\oslash B)\circ(C\oslash D)=(A\circ C)\oslash(B\circ D)+A\otimes D+C\otimes B.$**

**Proof. **(i) We have

[TABLE]

(ii) Here

[TABLE]

Proposition 4.15**.**

For matrices $A,B,C$ of suitable orders,

[TABLE]

and

[TABLE]

**Proof. **We prove only the first result as the second one can be proved similarly. If $A$ and $B$ are matrices of same order (say $m$) and matrix $C$ is of order $n,$ then

[TABLE]

From Section 4, we have

[TABLE]

and

[TABLE]

Now adding 4.3 and 4.4, we get

[TABLE]

Hence the result follows.

Theorem 4.16**.**

If $A_{i}\in\mathbb{M}_{m},~{}B_{i}\in\mathbb{M}_{n}$ for $i=1,2,\ldots,k,$ then

[TABLE]

**Proof. **We prove the result by induction on $k.$ For $k=1,$ the result is trivial. For $k=2,$ we have

[TABLE]

Thus the result holds for $k=2.$ Suppose the identity holds for $k=1,2,\ldots,\ell<k,$ then

[TABLE]

Hence the result follows.

Using Theorem 4.16 repeatedly, we get the following general result.

Theorem 4.17**.**

For $A_{i}\in\mathbb{M}_{m},B_{i}\in\mathbb{M}_{n},\ldots,C_{i}\in\mathbb{M}_{\ell},$ for $i=1,2\ldots,k,$ then

[TABLE]

Lemma 4.18**.**

If $A$ and $B$ are any square matrices, then

[TABLE]

**Proof. **If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then the $i,j-$th block of $A\otimes B$ is $a_{i,j}B$ and $S_{a_{i,j}B}=a_{i,j}S_{B}.$ Therefore we get

[TABLE]

Hence the result follows.

Theorem 4.19**.**

If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then

[TABLE]

**Proof. **We have

[TABLE]

Hence the theorem holds.

As a corollary of Theorem 4.19, we get the expression for the Wiener index of Cartesian product of two connected graphs.

Corollary 4.20**.**

If $G_{1}$ and $G_{2}$ are two connected graphs of order $m$ and $n$ respectively, then

[TABLE]

As an application of above corollary we get the the following result.

Corollary 4.21**.**

If $H$ is any fixed connected graph and $G_{1},G_{2}$ are connected graphs of same order with $W(G_{1})\geq W(G_{2}),$ then

[TABLE]

with equality if and only if $W(G_{1})=W(G_{2}).$

Theorem 4.22**.**

If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $A\oslash B$ has constant row sum if and only if $A$ and $B$ both have constant row sums.

**Proof. **Let us consider any arbitrary row of $A\oslash B.$ If the first entry of that row is $a_{i,1}+b_{j,i},$ then the row sum of that row of $A\oslash B$ equals to

[TABLE]

Now if $A$ and $B$ have constant row sums, then $A_{i}=\frac{S_{A}}{m}$ and $B_{i}=\frac{S_{B}}{n}.$ Therefore, by 4.6, $A\oslash B$ has constant row sum equal to $\frac{n}{m}S_{A}+\frac{m}{n}S_{B}.$

Again if $A\oslash B$ has constant row sum (say $k$), then from 4.6 we get

[TABLE]

Keeping $i$ fixed, we see that $B_{j}$ is constant for $j=1,2,\ldots,n.$ Similarly, keeping $j$ fixed we get $A_{i}$ is constant for $i=1,2,\ldots,m.$ Hence, the theorem holds.

The following result is a reformulation of Theorem 4.22. Therefore, the proof is omitted.

Theorem 4.23**.**

If $A\in\mathbb{M}_{m}$ and $B\in\mathbb{M}_{n},$ then $1\!\!1_{mn}$ is an eigenvector of $A\oslash B$ if and only if $1\!\!1_{m}$ and $1\!\!1_{n}$ are eigenvectors of $A$ and $B$ respectively.

As an application of Theorem 4.22, we get the following result as a corollary.

Corollary 4.24**.**

The Cartesian product $G_{1}\Box G_{2}$ of two connected graphs $G_{1}$ and $G_{2}$ is transmission regular if and only if $G_{1}$ and $G_{2}$ are both transmission regular.

From the proof of Theorem 4.22, we get a lower bound for the distance spectral radius of the Cartesian product of two connected graphs.

Corollary 4.25**.**

If $G_{1}$ and $G_{2}$ are two connected graphs of order $m$ and $n$ respectively, then

[TABLE]

with equality if and only if $G_{1}$ and $G_{2}$ are both transmission regular.