Matrix representations of linear transformations on bicomplex space

Anjali [email protected] Department of Applied Mathematics, Gautam Buddha University, Greater Noida, Uttar Pradesh 201312, India , Fahed Zulfeqarr [email protected] Department of Applied Mathematics, Gautam Buddha University, Greater Noida, Uttar Pradesh 201312, India , Akhil Prakash [email protected] Department of Mathematics, Aligarh Muslim University, Aligarh, Uttar Pradesh 202002, India and Prabhat Kumar [email protected] Department of Applied Mathematics, Gautam Buddha University, Greater Noida, Uttar Pradesh 201312, India

Abstract.

An algebraic investigation on bicomplex numbers is carried out here. Particularly matrices and linear maps defined on them are discussed. A new kind of cartesian product, referred to as an idempotent product, is introduced and studied. The elements of this space are linear maps of a special form. These linear maps are examined with respect to usual notions like kernel, range, and singularity. Their matrix representation is also discussed.

Key words and phrases:

Bicomplex Numbers, Conjugation, Vector Space and Linear transformation.

IMS Mathematics Subject Classification:

Primary 15A04, 15A30; Secondary 30G35

Introduction

The theory of bicomplex numbers has been a thrust area of current research in mathematics. It has evolved a lot in the recent past. Several researchers (see [1, 2, 3, 5, 6, 9, 10, 11, 12, 13]) have contributed a lot to the field. They have been working in different directions to analyze their properties and to create concepts consistent with a unified approach to the multivariate theory of complex numbers. In this process, some concepts like bicomplex topology [14], differentiability and analyticity of bicomplex function [8], power series, bicomplex matrices, bicomplex Riemann zeta function and Dirichlet series, integrability, Cauchy’s theory, etc., have been established with necessary modifications. Also, we have followed some of the results and symbols from the existing literature on the fundamental concepts from the book [8] on bicomplex functions.

In section 1, bicomplex numbers and some basic algebraic structures built on them are introduced and discussed. The section also stresses the role of the idempotent representation of bicomplex numbers in their study. A new kind of cartesian product, known as an idempotent product, is defined and discussed. The space becomes a central object of investigation. In section 2, we have revised bicomplex complex linear operator and also we have defined bicomplex linear transformation in this section.

Section 3 deals with matrices and linear maps defined on bicomplex space. It connects bicomplex matrices with elements of an idempotent product. The last section focuses on how the elements of the idempotent product behave for kernel, range, invertibility, and singularity. It renders some results of their matrix representation.

1. Preliminaries and Notations

This section introduces bicomplex numbers. It deals with basic notions such as idempotent representation and the cartesian product of bicomplex space. It presents some fundamental results on bicomplex numbers.

Bicomplex numbers: A bicomplex number is an element of the form :

\displaystyle\xi={u}_{1}+{i}_{1}{u}_{2}+{i}_{2}{u}_{3}+{i}_{1}{i}_{2}{u}_{4},\;\;{u}_{k}\in\mathbb{R},\;\;1\leqslant k\leqslant 4,\;\;\mbox{with}\;\;\;{i}_{1}{i}_{2}\ =\ {i}_{2}{i}_{1},\;\;{i}_{1}^{2}={i}_{2}^{2}=-1.

The set of all bicomplex numbers is denoted by $\displaystyle\mathbb{C}_{2}$ and is referred to as the bicomplex space. The symbols $\displaystyle\mathbb{C}_{1},\;\mathbb{C}_{0}$ , for convenience, denote the set of all complex numbers and the set of all real numbers respectively. The bicomplex space $\displaystyle\mathbb{C}_{2}$ can be described in the following two ways:

	$\displaystyle\displaystyle\mathbb{C}_{2}$	$\displaystyle\displaystyle\coloneqq$	$\displaystyle\displaystyle\{{u}_{1}+{i}_{1}{u}_{2}+{i}_{2}{u}_{3}+{i}_{1}{i}_{2}{u}_{4}\;:\;{u}_{1},{u}_{2},{u}_{3},{u}_{4}\in\mathbb{C}_{0}\},and$
	$\displaystyle\displaystyle\mathbb{C}_{2}$	$\displaystyle\displaystyle\coloneqq$	$\displaystyle\displaystyle\{{z}_{1}+{i}_{2}{z}_{2}\;:\;{z}_{1},{z}_{2}\in\mathbb{C}_{1}\}.$

$\displaystyle\mathbb{C}_{2}$ contains zero-divisors and hence it is not a field but an algebra over $\displaystyle\mathbb{C}_{1}$ . There are exactly four idempotent elements in $\displaystyle\mathbb{C}_{2}$ , viz., $\displaystyle 0,1,e_{1},e_{2}$ where $\displaystyle e_{1},e_{2}$ represent two nontrivial idempotent elements defined as follows:

\displaystyle e_{1}\coloneqq\frac{(1+i_{1}i_{2})}{2}\quad\mbox{and}\quad e_{2}\coloneqq\frac{(1-i_{1}i_{2})}{2}.

Notice that $\displaystyle e_{1}+e_{2}=1$ and $\displaystyle e_{1}e_{2}=e_{2}e_{1}=0$ . Linear independence of these elements over $\displaystyle\mathbb{C}_{1}$ gives rise to a new representation of all bicomplex numbers known as idempotent representation.

Idempotent representation of bicomplex numbers: Every bicomplex number $\displaystyle\xi={z}_{1}+{i}_{2}{z}_{2}$ can be uniquely represented as the complex combination of elements $\displaystyle e_{1}$ and $\displaystyle e_{2}$ in the following form

\displaystyle\xi=(z_{1}-i_{1}z_{2})e_{1}+(z_{1}+i_{1}z_{2})e_{2},

where complex numbers $\displaystyle(z_{1}-i_{1}z_{2})$ and $\displaystyle(z_{1}+i_{1}z_{2})$ are called idempotent components of $\displaystyle\xi$ and will be denoted by $\displaystyle\xi^{-}$ and $\displaystyle\xi^{+}$ respectively. Therefore the number $\displaystyle\xi={z}_{1}+{i}_{2}{z}_{2}$ can also be written as $\displaystyle\xi=\xi^{-}e_{1}+\xi^{+}e_{2}$ , where $\displaystyle\xi^{-}=z_{1}-i_{1}z_{2}$ and $\displaystyle\xi^{+}=z_{1}+i_{1}z_{2}$ .

Remark 1.1.

The idempotent representation of product of elements $\displaystyle\xi,\eta\in\mathbb{C}_{2}$ can be seen easily to be as:

\displaystyle\xi\cdot\eta=\left(\xi^{-}\eta^{-}\right)e_{1}+\left(\xi^{+}\eta^{+}\right)e_{2}.

Cartesian product: The $\displaystyle n$ -times cartesian product of $\displaystyle\mathbb{C}_{2}$ is represented by $\displaystyle\mathbb{C}_{2}^{n}$ and consisting of all $\displaystyle n$ -tuples of bicomplex numbers of the form $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})$ , where $\displaystyle\xi_{i}\in\mathbb{C}_{2};\;i=1,2,\ldots,n$ . That is,

\displaystyle{\mathbb{C}}_{2}^{n}\eqqcolon\{(\xi_{1},\xi_{2},\ldots,\xi_{n}):\;\xi_{i}\in\mathbb{C}_{2};\;i=1,2,\ldots,n\}.

Furthermore, using idempotent representation, we can also express every element of $\displaystyle\mathbb{C}_{2}^{n}$ uniquely as:

\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})=(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})e_{1}+(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})e_{2}

such that $\displaystyle(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})$ and $\displaystyle(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})$ are n-tuples of complex numbers from the space $\displaystyle\mathbb{C}_{1}^{n}$ .

Remark 1.2.

Some remarks can be easily made here.

(1)

Comparison between elements of $\displaystyle\mathbb{C}_{2}^{n}$ can be done by the following rules :

$\displaystyle\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})=(\eta_{1},\eta_{2},\ldots,\eta_{n})$	$\displaystyle\displaystyle\Longleftrightarrow$	$\displaystyle\displaystyle\xi_{i}=\eta_{i}\quad\forall\;i$
	$\displaystyle\displaystyle\Longleftrightarrow$	$\displaystyle\displaystyle\xi^{-}_{i}=\eta^{-}_{i}\;\&\;\;\xi^{+}_{i}=\eta^{+}_{i}\;\;\forall\;i$
$\displaystyle\displaystyle i.e.,\;\;(\xi_{1},\xi_{2},\ldots,\xi_{n})=(\eta_{1},\eta_{2},\ldots,\eta_{n})$	$\displaystyle\displaystyle\Longleftrightarrow$	$\displaystyle\displaystyle\xi^{-}_{i}=\eta^{-}_{i}\;\&\;\;\xi^{+}_{i}=\eta^{+}_{i}\;\;\forall\;i.$

(2)

Generally, we know that if $\displaystyle F^{\prime}$ is a sub-field of a field F then $\displaystyle F^{n}(F^{\prime})$ forms a vector space. We have the result that $\displaystyle\mathbb{C}_{2}$ is not a field so the above result does not give us any proof that the $\displaystyle\mathbb{C}_{2}^{n}(\mathbb{C}_{1})$ is a vector space. It is not hard to see that $\displaystyle\mathbb{C}_{2}^{n}$ forms a vector space over $\displaystyle\mathbb{C}_{1}$ with respect to usual addition and scalar multiplication. This immediately yields that the dimension of $\displaystyle\mathbb{C}_{2}^{n}$ over $\displaystyle C_{1}$ is $\displaystyle 2n$ , i.e., we have

$\displaystyle\dim(\mathbb{C}_{2}^{n})=2n=2\cdot\dim(\mathbb{C}_{1}^{n}).$

(3)

Furthermore for an element $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})\in\mathbb{C}_{2}^{n}$ , we define a scalar product in $\displaystyle\mathbb{C}_{2}^{n}$ with elements $\displaystyle e_{1},\;e_{2}$ by the rules:

	$\displaystyle\displaystyle e_{1}\cdot(\xi_{1},\xi_{2},\ldots,\xi_{n})=(\xi_{1},\xi_{2},\ldots,\xi_{n})e_{1}$	$\displaystyle\displaystyle\eqqcolon$	$\displaystyle\displaystyle(\xi_{1}e_{1},\xi_{2}e_{1},\ldots,\xi_{n}e_{1})=(\xi_{1}^{-}e_{1},\xi_{2}^{-}e_{1},\ldots,\xi_{n}^{-}e_{1}),and$
	$\displaystyle\displaystyle e_{2}\cdot(\xi_{1},\xi_{2},\ldots,\xi_{n})=(\xi_{1},\xi_{2},\ldots,\xi_{n})e_{2}$	$\displaystyle\displaystyle\eqqcolon$	$\displaystyle\displaystyle(\xi_{1}e_{2},\xi_{2}e_{2},\ldots,\xi_{n}e_{2})=(\xi_{1}^{+}e_{2},\xi_{2}^{+}e_{2},\ldots,\xi_{n}^{+}e_{2}).$

It follows that naturally extends to a scalar product in $\displaystyle\mathbb{C}_{2}^{n}$ over $\displaystyle\mathbb{C}_{2}$ and hence to a product in $\displaystyle\mathbb{C}_{2}^{n}$ as follows:

	$\displaystyle\displaystyle\eta\cdot(\xi_{1},\xi_{2},\ldots,\xi_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle(\eta\xi_{1},\eta\xi_{2},\ldots,\eta\xi_{n}),and$
	$\displaystyle\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})\cdot(\eta_{1},\eta_{2},\ldots,\eta_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle(\xi_{1}\eta_{1},\cdots,\xi_{n}\eta_{n}).$

It makes $\displaystyle\mathbb{C}_{2}^{n}$ not just to be a $\displaystyle\mathbb{C}_{2}$ -module but also to be a $\displaystyle\mathbb{C}_{1}$ -algebra.

Cartesian idempotent product: A new kind of cartesian product in bicomplex spaces is hereby introduced which will be referred to as idempotent product which will be dealt and analysed in later part of this article. This helps us to switch from complex case to bicomplex case.
Formally speaking, for any subsets $\displaystyle S_{1},S_{2}\subseteq\mathbb{C}_{1}^{n}$ , their idempotent product, denoted as $\displaystyle S_{1}\times_{e}S_{2}$ is defined to be a subset of $\displaystyle\mathbb{C}_{2}^{n}$ given as

(1)

\displaystyle\displaystyle S_{1}\times_{e}S_{2}\eqqcolon\{e_{1}x+e_{2}y:x\in S_{1},y\in S_{2}\}.

In the same manner, for any subsets $\displaystyle H_{1},H_{2}\subseteq\mathbb{C}_{2}^{m\times n}$ , their idempotent product, denoted as $\displaystyle H_{1}\times_{e}H_{2}$ is defined to be a subset of $\displaystyle\mathbb{C}_{2}^{m\times n}$ given as

(2)

\displaystyle\displaystyle H_{1}\times_{e}H_{2}\eqqcolon\{e_{1}A+e_{2}B:A\in H_{1},B\in H_{2}\}.

Similarly for any non-empty subsets $\displaystyle A_{1},A_{2}$ of the space of $\displaystyle\mathbb{C}_{1}$ -linear maps, $\displaystyle\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})$ , their idempotent product, written as $\displaystyle A_{1}\times_{e}A_{2}$ , is defined to be a subset of $\displaystyle\text{Hom}(\mathbb{C}_{2}^{n},\mathbb{C}_{2}^{m})$ given as

(3)

\displaystyle\displaystyle A_{1}\times_{e}A_{2}\eqqcolon\{e_{1}T_{1}+e_{2}T_{2}:\;T_{1}\in A_{1},T_{2}\in A_{2}\}.

This induces a new formulation of bicomplex space $\displaystyle\mathbb{C}_{2}^{n}$ as $\displaystyle\mathbb{C}_{2}^{n}=\mathbb{C}_{1}^{n}\times_{e}\mathbb{C}_{1}^{n}$ , for all $\displaystyle n\geqslant 1$ . Also we get $\displaystyle\mathbb{C}_{2}^{m\times n}=\mathbb{C}_{1}^{m\times n}\times_{e}\mathbb{C}_{1}^{m\times n}$ . We will later give a meaning to the elements of $\displaystyle A_{1}\times_{e}A_{2}$ .

2. Bicomplex linear operator and bicomplex linear transformation

In this section, we summarize bicomplex linear operator and we refer the reader to [4, 7] for further details. Also, we have defined bicomplex linear transformation. Let $\displaystyle V$ be a module over bicomplex space and Let $\displaystyle T\colon V\to V$ be a map such that

\displaystyle\displaystyle T(u+v)=T(u)+T(v)\quad\mbox{and}\quad T(\alpha v)=\alpha T(v)\quad\forall\alpha\in\mathbb{C}_{2}\quad\mbox{and}\quad u,v\in V.

Then we say that $\displaystyle T$ is a bicomplex linear operator on $\displaystyle V$ . The set $\displaystyle End(V)$ of linear operator on $\displaystyle V$ forms a bicomplex - module by defining

	$\displaystyle\displaystyle(T+S)(v)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T(v)+S(v)\quad\forall\quad T,S\in End(V)$
	$\displaystyle\displaystyle(\alpha T)(v)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\alpha(T(v))\quad\forall\quad T\in End(V),\alpha\in\mathbb{C}_{2}$

Let us set $\displaystyle V_{1}=e_{1}V=\{e_{1}v:v\in v\}$ . Any element $\displaystyle v\in V$ can be written as

\displaystyle\displaystyle v=(e_{1}+e_{2})v=e_{1}v+e_{2}v=v_{1}+v_{2}=e_{1}v_{1}+e_{2}v_{2},\quad\mbox{where}\quad v_{1}=e_{1}v,v_{2}=e_{2}v

It is evident that $\displaystyle V=V_{1}\oplus V_{2}$ . We can define the operators $\displaystyle T_{1},T_{2}$ as

\displaystyle\displaystyle T_{1}(v)=e_{1}T(v),T_{2}(v)=e_{2}T(v)

where $\displaystyle T_{1}\colon V\to V_{1}$ and $\displaystyle T_{2}\colon V\to V_{2}$

Remark 2.1.

The following properties hold

(1)

The operators $\displaystyle T_{1}$ and $\displaystyle T_{2}$ are bicomlex linear
(2)

we get the decomposition $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$
(3)

The action $\displaystyle T$ on $\displaystyle V$ can be decomposed as fellows
$\displaystyle T(v)=e_{1}T_{1}(v_{1})+e_{2}T_{2}(v_{2})$ , where $\displaystyle v=e_{1}v_{1}+e_{2}v_{2}\in V$
(4)

The operator $\displaystyle T_{k}\colon V_{k}\to V_{k};k=1,2$

Definition 2.2.

(Bicomplex linear transformation) If $\displaystyle V$ and $\displaystyle V^{\prime}$ is a module over $\displaystyle\mathbb{C}_{2}$ and let $\displaystyle T\colon V\to V^{\prime}$ be a map such that

	$\displaystyle\displaystyle T(u+v)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T(u)+T(v)$
	$\displaystyle\displaystyle T(\alpha u)$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\alpha T(u)\quad\forall\quad\alpha\in\mathbb{C}_{2}\quad\mbox{and}\quad u,v\in V$

then we say that $\displaystyle T$ is a bicomplex linear transformation from V to V’

As previously, we can define the sets $\displaystyle V_{1},V_{2},V^{\prime}_{1},V^{\prime}_{2}$ and the linear transformation $\displaystyle T_{1}$ and $\displaystyle T_{2}$ here as well

\displaystyle V_{1}=e_{1}V,\quad V_{2}=e_{2}V,\quad V^{\prime}_{1}=e_{1}V^{\prime}\quad\mbox{and}\quad V^{\prime}_{2}=e_{2}V^{\prime}

Any element $\displaystyle v\in V$ and $\displaystyle v^{\prime}\in V^{\prime}$ can be written as

\displaystyle v=e_{1}v_{1}+e_{2}v_{2},\quad v^{\prime}=e_{1}v^{\prime}_{1}+e_{2}v^{\prime}_{2}\quad\mbox{where}\quad v_{i}=e_{i}v,v^{\prime}_{i}=e_{i}v^{\prime}\forall i=1,2

Also it is evident that $\displaystyle V=V_{1}\oplus V_{2}$ and $\displaystyle V^{\prime}=V^{\prime}_{1}\oplus V^{\prime}_{2}$ . The linear transformation $\displaystyle T_{1}\colon V\to V^{\prime}_{1}$ and $\displaystyle T_{2}\colon V\to V^{\prime}_{2}$ will be

\displaystyle T_{1}(v)=e_{1}T(v)\quad\mbox{and}\quad T_{2}(v)=e_{2}T(v)

Remark 2.3.

We can easily prove the following properties

(1)

The linear transformation $\displaystyle T_{1}$ and $\displaystyle T_{2}$ are bicomplex linear.
(2)

We get the decomposition $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$
(3)

The action $\displaystyle T$ on $\displaystyle V$ can be decomposed as fellows

$\displaystyle T(v)=e_{1}T_{1}(v_{1})+e_{2}T_{2}(v_{2});\quad v=e_{1}v_{1}+e_{2}v_{2}\in V$
(4)

The linear transformation $\displaystyle T_{k}\colon V_{k}\to V^{\prime}_{k};k=1,2$

particularly if we set $\displaystyle V=\mathbb{C}_{2}^{n}$ and $\displaystyle V^{\prime}=\mathbb{C}_{2}^{m}$ then $\displaystyle V_{1}=e_{1}\mathbb{C}_{1}^{n},V_{2}=e_{2}\mathbb{C}_{1}^{n}$ and $\displaystyle V^{\prime}_{1}=e_{1}\mathbb{C}_{1}^{m},V^{\prime}_{2}=e_{2}\mathbb{C}_{1}^{m}$ . In this case $\displaystyle T\colon\mathbb{C}_{2}^{n}\to\mathbb{C}_{2}^{m}$ is a bicomplex linear transformation. Also, $\displaystyle T_{1}\colon\mathbb{C}_{2}^{n}\to e_{1}\mathbb{C}_{1}^{m}$ and $\displaystyle T_{2}\colon\mathbb{C}_{2}^{n}\to e_{2}\mathbb{C}_{1}^{m}$ specially, $\displaystyle T_{1}\colon e_{1}\mathbb{C}_{1}^{n}\to e_{1}\mathbb{C}_{1}^{m}$ and $\displaystyle T_{2}\colon e_{2}\mathbb{C}_{1}^{n}\to e_{2}\mathbb{C}_{1}^{m}$ are bicomplex linear transformation.

Remark 2.4.

If $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})\in\mathbb{C}_{2}^{n}$ and

(4)

\displaystyle\displaystyle T(\xi_{1},\xi_{2},\ldots,\xi_{n})=(\eta_{1},\eta_{2},\ldots,\eta_{m})

As $\displaystyle T$ is linear with respect to bicomplex numbers then

\displaystyle\displaystyle e_{1}T(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})+e_{2}T(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})=e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})+e_{2}(\eta_{1}^{+},\eta_{2}^{+},\ldots,\eta_{m}^{+})

(5)

\displaystyle\displaystyle\Rightarrow T(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})=(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})

That is

(6)		$\displaystyle\displaystyle e_{1}T(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$
(7)		$\displaystyle\displaystyle\mbox{and}\quad T[e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})]$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$

Also, by using definition 2.2, $\displaystyle(6)$ and $\displaystyle(7)$ , we have

	$\displaystyle\displaystyle T_{1}[e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})]$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}T[e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})]$
		$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$
(8)		$\displaystyle\displaystyle\therefore\quad T_{1}[e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})]$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$
			$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}T(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})$
			$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})$
			$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$

Now

	$\displaystyle\displaystyle T_{1}(\xi_{1},\xi_{2},\ldots,\xi_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}T(\xi_{1},\xi_{2},\ldots,\xi_{n})$
		$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1},\eta_{2},\ldots,\eta_{m})$
(9)		$\displaystyle\displaystyle\therefore\quad T_{1}(\xi_{1},\xi_{2},\ldots,\xi_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$

$\displaystyle\therefore$ from $\displaystyle(6),(7),(8)$ and (9)

$\displaystyle\displaystyle T_{1}(\xi_{1},\xi_{2},\ldots,\xi_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}(\eta_{1}^{-},\eta_{2}^{-},\ldots,\eta_{m}^{-})$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}T(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T[e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})]$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T_{1}[e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})]$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})$

Similarly

$\displaystyle\displaystyle T_{2}(\xi_{1},\xi_{2},\ldots,\xi_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{2}(\eta_{1}^{+},\eta_{2}^{+},\ldots,\eta_{m}^{+})$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{2}T(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T[e_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})]$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T_{2}[e_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})]$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle T_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})$

It is worth noting here that if $\displaystyle T\colon\mathbb{C}_{2}^{n}\to\mathbb{C}_{2}^{m}$ is a bicomplex linear transformation then we get the bicomplex linear transformation $\displaystyle T_{1}\colon e_{1}\mathbb{C}_{1}^{n}\to e_{1}\mathbb{C}_{1}^{m}$ and $\displaystyle T_{2}\colon e_{2}\mathbb{C}_{1}^{n}\to e_{2}\mathbb{C}_{1}^{m}$ and vice versa. This notation is possible because $\displaystyle T$ is linear with respect to bicomplex numbers. Furthermore, we have defined another map in definition $\displaystyle 3.2$ for which this notion is not true because the map define in there is not linear with respect to bicomplex numbers.

3. Bicomplex Matrices and Linear transformations

This section focuses on bicomplex matrices and linear maps defined on the space $\displaystyle\mathbb{C}_{2}^{n}$ . It connects both these notions like the ordinary ones on complex numbers.

A bicomplex matrix of order $\displaystyle m\times n$ is denoted by $\displaystyle[\xi_{ij}]_{m\times n}$ , $\displaystyle\xi_{ij}\in\mathbb{C}_{2}$ or simply by $\displaystyle[\xi_{ij}]$ if there is no confusion. We let the set of all bicomplex matrices of order $\displaystyle m\times n$ be denoted by $\displaystyle\mathbb{C}_{2}^{m\times n}$ , i.e., we have

(10)

\displaystyle\displaystyle\mathbb{C}_{2}^{m\times n}\eqqcolon\Big{\{}[\xi_{ij}]:\;\xi_{ij}\in\mathbb{C}_{2};\;i=1,2,\ldots,m,\;j=1,2,\ldots,n\Big{\}}.

The usual matrix addition and scalar multiplication makes the space $\displaystyle\mathbb{C}_{2}^{m\times n}$ to be a vector space over the field $\displaystyle\mathbb{C}_{1}$ . Further each bicomplex matrix $\displaystyle[\xi_{ij}]$ can be uniquely written as

(11)

\displaystyle\displaystyle[\xi_{ij}]=e_{1}\ [\xi^{-}_{ij}]\ +\ e_{2}\ [\xi^{+}_{ij}],

where $\displaystyle[\xi^{-}_{ij}],\;[\xi^{+}_{ij}]$ are complex matrices of order $\displaystyle m\times n$ . From this, it follows easily that

(12)

\displaystyle\displaystyle\dim(\mathbb{C}_{2}^{m\times n})=2\cdot\dim(\mathbb{C}_{1}^{m\times n})=2mn.

We now come to linear maps, i.e., linear maps on bicomplex spaces. As we know that $\displaystyle\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})$ represents the set of all $\displaystyle\mathbb{C}_{1}$ -linear maps from $\displaystyle\mathbb{C}_{1}^{n}$ to $\displaystyle\mathbb{C}_{1}^{m}$ so does $\displaystyle\text{Hom}(\mathbb{C}_{2}^{n},\mathbb{C}_{2}^{m})$ . However we simplify these notations to our comfort.

Remark 3.1.

For brevity and convenience, we denote the set of all $\displaystyle\mathbb{C}_{1}$ -linear maps from $\displaystyle\mathbb{C}_{1}^{n}$ to $\displaystyle\mathbb{C}_{1}^{m}$ by $\displaystyle L_{1}^{nm}$ , instead of $\displaystyle\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})$ and set of all $\displaystyle\mathbb{C}_{1}$ -linear maps from $\displaystyle\mathbb{C}_{2}^{n}$ to $\displaystyle\mathbb{C}_{2}^{m}$ by $\displaystyle L_{2}^{nm}$ . Clearly both $\displaystyle L_{1}^{nm}$ and $\displaystyle L_{2}^{nm}$ are vector spaces over $\displaystyle\mathbb{C}_{1}$ . Hence, we can see that

(13)

\displaystyle\displaystyle\dim(L_{1}^{nm})=mn\;\;\mbox{and}\;\;\dim(L_{2}^{nm})=\dim C_{2}^{n}\cdot\dim C_{2}^{m}=2n\cdot 2m=4mn.

It is evident that $\displaystyle L_{1}^{nm}$ = $\displaystyle\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})$ is isomorphic to $\displaystyle\mathbb{C}_{1}^{m\times n}$ as $\displaystyle\mathbb{C}_{1}$ is a field but here we have $\displaystyle\mathbb{C}_{2}$ is not a field so we cannot say that $\displaystyle L_{2}^{nm}$ = $\displaystyle\text{Hom}(\mathbb{C}_{2}^{n},\mathbb{C}_{2}^{m})$ and $\displaystyle\mathbb{C}_{2}^{m\times n}$ are isomorphic. (12) and (13) follow that $\displaystyle L_{2}^{nm}$ is not isomorphic to $\displaystyle\mathbb{C}_{2}^{m\times n}$ . Moreover $\displaystyle\mathbb{C}_{2}^{m\times n}$ is a proper subspace of $\displaystyle L_{2}^{nm}$ in isomorphic sense. So we try to find a subset of $\displaystyle L_{2}^{nm}$ which is isomorphic to $\displaystyle\mathbb{C}_{2}^{m\times n}$ . Thus the following definition.

Definition 3.2.

For any given $\displaystyle T_{1},T_{2}\in L_{1}^{nm}$ , we define a map $\displaystyle f\colon\mathbb{C}_{2}^{n}\to\mathbb{C}_{2}^{m}$ by the following rule:

\displaystyle f(\xi_{1},\xi_{2},\ldots,\xi_{n})\eqqcolon e_{1}\cdot{T}_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})+e_{2}\cdot{T}_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+}).

One can notice that $\displaystyle f$ is a $\displaystyle\mathbb{C}_{1}$ -linear map. This $\displaystyle f$ is defined as $\displaystyle e_{1}T_{1}+e_{2}T_{2}$ . Therefore, the set of all such linear maps is the idempotent product $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ , i.e., we have

(14)

\displaystyle\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}

\displaystyle\displaystyle\eqqcolon

\displaystyle\displaystyle\left\{\;e_{1}T_{1}+e_{2}T_{2}\in L_{2}^{nm}:\;\;T_{1},T_{2}\in L_{1}^{nm}\right\}.

Figure 1. Diagram showing the idempotent components of

\displaystyle T_{1}

and

\displaystyle T_{2}

which are compatible with the idempotent components

\displaystyle(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})

and

\displaystyle(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})

It is a routine matter to check that the idempotent product $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ is a subspace of $\displaystyle L_{2}^{nm}$ . This follows easily from the following proposition.

Proposition 3.3.

Let $\displaystyle T,S\in L_{1}^{nm}\times_{e}L_{1}^{nm}$ be any elements such that $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ and $\displaystyle S=e_{1}S_{1}+e_{2}S_{2}$ . Then, we have

(1)

$\displaystyle T+S=e_{1}(T_{1}+S_{1})+e_{2}(T_{2}+S_{2})$
(2)

$\displaystyle\alpha T=e_{1}(\alpha T_{1})+e_{2}(\alpha T_{2});\quad\forall\alpha\in\mathbb{C}_{1}$

The following theorem contains some basic properties of the elements of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ .

Theorem 3.4.

(Properties) Let $\displaystyle T=e_{1}T_{1}+e_{2}T_{2},\;S=e_{1}S_{1}+e_{2}S_{2}$ be any two elements of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ . Then, we have

(1)

$\displaystyle T=0$ if and only if $\displaystyle T_{1}=0,T_{2}=0$
(2)

$\displaystyle T=S$ if and only if $\displaystyle T_{1}=S_{1},T_{2}=S_{2}$
(3)

$\displaystyle S\circ T=e_{1}(S_{1}\circ T_{1})+e_{2}(S_{2}\circ T_{2})$ , wherever composition defined.

The next theorem deals with the dimension of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ .

Theorem 3.5.

The dimension of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ is equal to $\displaystyle 2\cdot\dim L_{1}^{nm}$

Proof.

As we know that $\displaystyle\dim L_{1}^{nm}=mn$ . So we take $\displaystyle\mathcal{B}_{1}=\{T_{j}:\;j=1,2,\cdots,mn\}$ as a basis for $\displaystyle L_{1}^{nm}$ . This yields a collection $\displaystyle\mathcal{B}_{2}=\{e_{i}T_{j}:i=1,2\;\&\;j=1,2,\cdots,mn\}$ of elements of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ . We assert that $\displaystyle\mathcal{B}_{2}$ is a basis for $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ . For any $\displaystyle S\in L_{1}^{nm}\times_{e}L_{1}^{nm}$ there are $\displaystyle S_{1},S_{2}\in L_{1}^{nm}$ such that $\displaystyle S=e_{1}S_{1}+e_{2}S_{2}$ . Since $\displaystyle<{\mathcal{B}_{1}}>=L_{1}^{nm}$ , there exist $\displaystyle\alpha_{j},\beta_{j}\in\mathbb{C}_{1};\;j=1,2,\ldots,mn$ such that

$\displaystyle\displaystyle S_{1}=\sum_{j=1}^{mn}\alpha_{j}T_{j}$	and	$\displaystyle\displaystyle S_{2}=\sum_{j=1}^{mn}\beta_{j}T_{j}$
$\displaystyle\displaystyle\Rightarrow\quad S=\sum_{j=1}^{mn}e_{1}(\alpha_{j}T_{j})+\sum_{j=1}^{mn}e_{2}(\beta_{j}T_{j})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\sum_{j=1}^{mn}\alpha_{j}(e_{1}T_{j})+\sum_{j=1}^{mn}\beta_{j}(e_{2}T_{j})$
$\displaystyle\displaystyle\Rightarrow\quad S\in<\mathcal{B}_{2}>\;\;$	$\displaystyle\displaystyle\Rightarrow$	$\displaystyle\displaystyle\;<\mathcal{B}_{2}>\;=L_{2}\times_{e}L_{2}.$

Furthermore, if we consider

\displaystyle\sum_{j=1}^{mn}\alpha_{j}(e_{1}T_{j})+\sum_{j=1}^{mn}\beta_{1}(e_{2}T_{i})=0\quad\Rightarrow\quad e_{1}\sum_{j=1}^{mn}(\alpha_{j}T_{j})+e_{2}\sum_{j=1}^{mn}(\beta_{j}T_{j})=0.

Using property (1) of theorem 3.4, we get

\displaystyle\sum_{j=1}^{mn}(\alpha_{j}T_{j})=0\quad\&\quad\sum_{j=1}^{mn}(\beta_{1}T_{j})=0\;\Rightarrow\;\alpha_{j}=0=\beta_{j}\;\;\forall\;j.

Hence, $\displaystyle\mathcal{B}_{2}$ is linearly independent set. Hence the result follows. ∎

Remark 3.6.

(13) and theorem 3.5 show that $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ is a proper subspace of $\displaystyle L_{2}^{nm}$ . It is clear that $\displaystyle\mathbb{C}_{2}^{m\times n}=\mathbb{C}_{1}^{m\times n}\times_{e}\mathbb{C}_{1}^{m\times n}$ and $\displaystyle\mathbb{C}_{2}^{n}=\mathbb{C}_{1}^{n}\times_{e}\mathbb{C}_{1}^{n}$ . But $\displaystyle\text{Hom}(\mathbb{C}_{2}^{n},\mathbb{C}_{2}^{m})\neq\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})\times_{e}\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})$ . In fact, $\displaystyle\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})\times_{e}\text{Hom}(\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m})\subset\text{Hom}(\mathbb{C}_{2}^{n},\mathbb{C}_{2}^{m})$ . This will help us to defining one - one correspondence between bicomplex matrix and linear map.

4. Relationship between bicomplex linear map and matrix

The studying of $\displaystyle\mathbb{C}_{1}$ -linear maps is equivalent to that of complex matrices. Is the same true for a bicomplex matrix? In this section we work on it in the context of bicomplex matrix. Let us suppose that $\displaystyle V$ and $\displaystyle V^{\prime}$ be two vector space of dimension $\displaystyle n$ and $\displaystyle m$ respectively, over the field $\displaystyle F$ . Also we have $\displaystyle\mathcal{B}$ and $\displaystyle\mathcal{B^{\prime}}$ are two ordered bases of vector space $\displaystyle V$ and $\displaystyle V^{\prime}$ respectively. Then the vector spaces $\displaystyle{\text{Hom}({V,V^{\prime}})}(F)$ and $\displaystyle F^{m\times n}(F)$ are isomorphic two each other. If $\displaystyle T$ is any linear map in $\displaystyle\text{Hom}({V,V^{\prime}})$ then the matrix representation of $\displaystyle T$ with respect two ordered bases $\displaystyle\mathcal{B}$ and $\displaystyle\mathcal{B^{\prime}}$ corresponds to a unique matrix in $\displaystyle F^{m\times n}(F)$ and vice-versa. This unique correspondence help us to solving system of equation and to find rank, nullity , eigen value ,eigen vector , characteristic polynomial and minimal polynomial.

If we assume that the field is a set of complex number then for any complex matrix of order $\displaystyle m\times n$ there will be a unique linear map from vector spaces $\displaystyle V$ to $\displaystyle V^{\prime}$ . The vector space $\displaystyle V$ and $\displaystyle V^{\prime}$ can be substitute by vector space $\displaystyle\mathbb{C}_{1}^{n}$ and $\displaystyle\mathbb{C}_{1}^{m}$ because of $\displaystyle V$ and $\displaystyle\mathbb{C}_{1}^{n}$ , $\displaystyle V^{\prime}$ and $\displaystyle\mathbb{C}_{1}^{m}$ have same dimension. So whenever we want to study on complex matrix of order $\displaystyle m\times n$ then we can study its corresponding linear map $\displaystyle T\colon\mathbb{C}_{1}^{n}\to\mathbb{C}_{1}^{m}$ to get the information about concern matrix. This analogous concept can not be define for a bicomplex matrix of order $\displaystyle m\times n$ and the linear map $\displaystyle T\colon\mathbb{C}_{2}^{n}\to\mathbb{C}_{2}^{m}$ . As we know $\displaystyle\mathbb{C}_{2}$ is not a field so a matrix representation of any linear map can not give a bicomplex matrix and 3.1 shows that $\displaystyle L_{2}^{nm}$ is not isomorphic to $\displaystyle\mathbb{C}_{2}^{m\times n}$ . In the same manner the matrix representation of the linear map $\displaystyle T\colon\mathbb{C}_{2}^{n}\to\mathbb{C}_{2}^{m}$ will be a complex matrix of order $\displaystyle 2m\times 2n$ . So using the traditional approach of a matrix representation of a linear map we can not develop the analogous concept for a bicomplex matrix and the linear map $\displaystyle T$ .

To do away this problem we define a new approach called ”Idempotent method” for matrix representation of a linear map.

Definition 4.1.

(Idempotent method) Let $\displaystyle\mathcal{B}_{1},\mathcal{B}_{2}$ be the ordered bases for $\displaystyle\mathbb{C}_{1}^{n}$ and $\displaystyle\mathbb{C}_{1}^{m}$ respectively. Then for any linear map $\displaystyle T\in L_{1}^{nm}\times_{e}L_{1}^{nm}$ so that $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ for some $\displaystyle T_{1},T_{2}\in L_{1}^{nm}$ , the matrix representation of $\displaystyle T$ with respect to these bases is defined as

(15)

\displaystyle\displaystyle[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\eqqcolon e_{1}[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}.

This new approach will help us to find the solution of a system of equations of bicomplex variables and we can define the rank, nullity, eigenvalue, eigenvector, characteristic polynomial, and minimal polynomial for a bicomplex matrix. Here we have not used $\displaystyle T$ as defined in definition $\displaystyle 2.2$ . The reason behind this is that the $\displaystyle T_{1}$ and $\displaystyle T_{2}$ corresponding to $\displaystyle T$ defined in definition $\displaystyle 2.2$ are bicomplex linear transformations from $\displaystyle e_{1}\mathbb{C}_{1}^{n}\to e_{1}\mathbb{C}_{1}^{m}$ and $\displaystyle e_{2}\mathbb{C}_{1}^{n}\to e_{2}\mathbb{C}_{1}^{m}$ respectively. So, the definition $\displaystyle 2.2$ is not more useful for our study.

Remark 4.2.

In the special case when $\displaystyle\mathbb{C}_{1}^{n}=\mathbb{C}_{1}^{m}$ , the matrix representation of the operator $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ with respect to basis $\displaystyle\mathcal{B}$ for $\displaystyle\mathbb{C}_{1}^{n}$ is simplified to $\displaystyle[T]_{\mathcal{B}}$ from $\displaystyle[T]^{\mathcal{B}}_{\mathcal{B}}$ . Thus, we have

(16)

\displaystyle\displaystyle[T]_{\mathcal{B}}=e_{1}[T_{1}]_{\mathcal{B}}+e_{2}[T_{2}]_{\mathcal{B}}.

Example 4.3.

Let $\displaystyle T_{1},T_{2}\colon\mathbb{C}_{1}^{2}\to\mathbb{C}_{1}^{3}$ be linear maps defined as

\displaystyle T_{1}(a,b)=(a,a+b,b)\quad\mbox{and}\quad T_{2}(a,b)=(a-b,b,a)\quad\forall\;(a,b)\in\mathbb{C}_{1}^{2}.

Let $\displaystyle\mathcal{B}_{1}=\{(1,1),(1,0)\}$ and $\displaystyle\mathcal{B}_{2}=\{(1,0,1),(1,1,0),(0,0,1)\}$ be the ordered bases for $\displaystyle\mathbb{C}_{1}^{2}$ and $\displaystyle\mathbb{C}_{1}^{3}$ respectively. This gives

	$\displaystyle\displaystyle\left[T_{1}\right]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}=\begin{bmatrix}-1&0\\ 2&1\\ 2&0\end{bmatrix}$	and	$\displaystyle\displaystyle\left[T_{2}\right]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}=\begin{bmatrix}-1&1\\ 1&0\\ 2&0\end{bmatrix}$
	$\displaystyle\displaystyle\Rightarrow\quad e_{1}[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}\begin{bmatrix}-1&0\\ 2&1\\ 2&0\end{bmatrix}+e_{2}\begin{bmatrix}-1&1\\ 1&0\\ 2&0\end{bmatrix}=\begin{bmatrix}-1&e_{2}\\ 2e_{1}+e_{2}&e_{1}\\ 2&0\end{bmatrix}.$

Therefore, the matrix representation of linear transformation $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ with respect to $\displaystyle\mathcal{B}_{1},\;\mathcal{B}_{2}$ is given by

$\displaystyle[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}=\begin{bmatrix}-1&e_{2}\\ 2e_{1}+e_{2}&e_{1}\\ 2&0\end{bmatrix}$ .

Next we show that the matrix representation as defined in (4.1) behaves well with respect to both the operations in $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ .

Theorem 4.4.

Let $\displaystyle T=e_{1}T_{1}+e_{2}T_{2},\ S=e_{1}S_{1}+e_{2}S_{2}$ be any two elements of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ and let $\displaystyle\mathcal{B}_{1},\mathcal{B}_{2}$ be the ordered bases of $\displaystyle\mathbb{C}_{1}^{n}$ and $\displaystyle\mathbb{C}_{1}^{m}$ respectively. Then

(1)

$\displaystyle[T+S]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}=[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+[S]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$
(2)

$\displaystyle[\alpha T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}=\alpha[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\forall\ \alpha\in\mathbb{C}_{1}$

Proof.

By definition 4.1 and using proposition 3.3, it follows that

$\displaystyle\displaystyle[T+S]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}[T_{1}+S_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[T_{2}+S_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}{\left([T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+[S_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)+e_{2}\left([T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+[S_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\left(e_{1}[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)+\left(e_{1}[S_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[S_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+[S]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}.$

For the second part, proposition 3.3 gives that

$\displaystyle\displaystyle[\alpha T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}[\alpha T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[\alpha T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}\left(\alpha[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)+e_{2}\left(\alpha[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\alpha\left(e_{1}[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}\right)$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\alpha[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}.\hskip 216.81pt$

This completes the proof. ∎

Theorem 4.5.

The space $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ and $\displaystyle\mathbb{C}_{2}^{m\times n}$ are isomorphic.

Proof.

let $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}\in L_{1}^{nm}\times_{e}L_{1}^{nm}$ . Define a function $\displaystyle F\colon L_{1}^{nm}\times_{e}L_{1}^{nm}\to\mathbb{C}_{2}^{m\times n}$ such that

\displaystyle F(T)=[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}.

It is an easy exercise to show that $\displaystyle F$ is one -one onto linear map. Moreover in view of (12) and theorem 3.5, the vector spaces $\displaystyle\mathbb{C}_{2}^{m\times n}$ and $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ have the same dimension over the field $\displaystyle\mathbb{C}_{1}$ . So they are isomorphic. ∎

As per the definition of idempotent product given in (1), we have

(17)		$\displaystyle\displaystyle\ker T_{1}\times_{e}\ker T_{2}$	$\displaystyle\displaystyle\eqqcolon$	$\displaystyle\displaystyle\{e_{1}z+e_{2}w\in\mathbb{C}_{2}^{n}\;:\;z\in\ker T_{1},\;w\in\ker T_{2}\},$
(18)		$\displaystyle\displaystyle\text{Im }T_{1}\times_{e}\text{Im }T_{2}$	$\displaystyle\displaystyle\eqqcolon$	$\displaystyle\displaystyle\{e_{1}z+e_{2}w\in\mathbb{C}_{2}^{m}\;:\;z\in\text{Im }T_{1}\;w\in\text{Im }T_{2}\}.$

The next theorem connect both kernel and range of of the element of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ with the kernels and ranges of their components.

Theorem 4.6.

Let $\displaystyle T\in L_{1}^{nm}\times_{e}L_{1}^{nm}$ so that $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ for some $\displaystyle T_{1},T_{2}\in L_{1}^{nm}$ . Then, we have

(1)

$\displaystyle\ker(e_{1}T_{1}+e_{2}T_{2})=\ker T_{1}\times_{e}\ker T_{2}$
(2)

$\displaystyle\text{Im }(e_{1}T_{1}+e_{2}T_{2})=\text{Im }T_{1}\times_{e}\text{Im }T_{2}$

Proof.

Let $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})\in\ker(e_{1}T_{1}+e_{2}T_{2})$ . Then $\displaystyle\left(e_{1}T_{1}+e_{2}T_{2}\right)(\xi_{1},\xi_{2},\ldots,\xi_{n})=0$ . Using definition 3.2 and from the part (1) of proposition 3.4, it will follows that

$\displaystyle\displaystyle\left(e_{1}T_{1}+e_{2}T_{2}\right)(\xi_{1},\xi_{2},\ldots,\xi_{n})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}\cdot{T}_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})+e_{2}\cdot{T}_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})=0$
$\displaystyle\displaystyle\Leftrightarrow\quad{T}_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})=0$	$\displaystyle\displaystyle{\&}$	$\displaystyle\displaystyle{T}_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})=0$
$\displaystyle\displaystyle\Leftrightarrow\quad(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})\in\ker{T}_{1}$	$\displaystyle\displaystyle{\&}$	$\displaystyle\displaystyle(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})\in\ker{T}_{2}$
$\displaystyle\displaystyle\Leftrightarrow\quad(\xi_{1},\xi_{2},\ldots,\xi_{n})\in$		$\displaystyle\displaystyle\ker T_{1}\times_{e}\ker T_{2}.$

For the secondly part, consider $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{m})\in\text{Im }T_{1}\times_{e}\text{Im }T_{2}$ , then we have

$\displaystyle\displaystyle(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{m}^{-})\in\text{Im }T_{1}$	$\displaystyle\displaystyle and$	$\displaystyle\displaystyle(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{m}^{+})\in\text{Im }T_{2}$
$\displaystyle\displaystyle\Rightarrow\quad\exists\quad(z_{1},z_{2},\ldots,z_{n})$	,	$\displaystyle\displaystyle(w_{1},w_{2},\ldots,w_{n})\in\mathbb{C}_{1}^{n}$
$\displaystyle\displaystyle\mbox{such that}\quad T_{1}(z_{1},z_{2},\ldots,z_{n})=(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{m}^{-})$	$\displaystyle\displaystyle and$	$\displaystyle\displaystyle T_{2}(w_{1},w_{2},\ldots,w_{n})=(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{m}^{+})$
$\displaystyle\displaystyle\mbox{So},\quad e_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{m}^{-})+e_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{m}^{+})$	$\displaystyle\displaystyle\in$	$\displaystyle\displaystyle\text{Im }(e_{1}T_{1}+e_{2}T_{2})$
$\displaystyle\displaystyle\mbox{Hence}\quad\text{Im }T_{1}\times_{e}\text{Im }T_{2}$	$\displaystyle\displaystyle\subseteq$	$\displaystyle\displaystyle\text{Im }(e_{1}T_{1}+e_{2}T_{2}).$

Conversely suppose $\displaystyle e_{1}T_{1}(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{n}^{-})+e_{2}T_{2}(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{n}^{+})\in\text{Im }(e_{1}T_{1}+e_{2}T_{2})$ , for some $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{n})\in\mathbb{C}_{2}^{n}$ . This follows that $\displaystyle\text{Im }(e_{1}T_{1}+e_{2}T_{2})\subseteq\text{Im }T_{1}\times_{e}\text{Im }T_{2}$ . Hence, the equality holds. ∎

Next result exhibits a relation between the rank of $\displaystyle T\in L_{1}^{nm}\times L_{1}^{nm}$ with the ranks of its components.

Theorem 4.7.

Let $\displaystyle T\in L_{1}^{nm}\times L_{1}^{nm}$ so that $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ for some $\displaystyle T_{1},T_{2}\in L_{1}^{nm}$ . Then,

\displaystyle\displaystyle\dim\text{Im }{T}=\dim\text{Im }{T_{1}}+\dim\text{Im }{T_{2}}.

In other words, that rank of $\displaystyle T$ is sum of the ranks of its own components.

Proof.

Let $\displaystyle\mathcal{B}_{1}=\{z_{i}:i==1,2,\ldots,p\},\mathcal{B}_{2}=\{w_{j}:j=1,2,\ldots,q\}$ be the bases for $\displaystyle\text{Im }{T_{1}}$ , $\displaystyle\text{Im }{T_{2}}$ respectively. Consider the collection $\displaystyle\mathcal{B}=\{e_{1}z_{i},e_{2}w_{j}\}$ of $\displaystyle p+q$ elements of $\displaystyle\mathbb{C}_{2}$ . Notice that $\displaystyle<\mathcal{B}>\quad\subseteq\text{Im }{T}$ . If we take $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{m})\in\text{Im }{T}$ , then from theorem 4.6, it follows that

$\displaystyle\displaystyle(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{m}^{-})\in\text{Im }T_{1}$	and	$\displaystyle\displaystyle(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{m}^{+})\in\text{Im }T_{2}$
$\displaystyle\displaystyle\Rightarrow\quad(\xi_{1}^{-},\xi_{2}^{-},\ldots,\xi_{m}^{-})=\sum_{i=1}^{p}\alpha_{i}z_{i}$	and	$\displaystyle\displaystyle(\xi_{1}^{+},\xi_{2}^{+},\ldots,\xi_{m}^{+})=\sum_{j=1}^{q}\beta_{j}w_{j},\quad\mbox{for some}\quad\alpha_{i},\beta_{j}\in\mathbb{C}_{1}$
$\displaystyle\displaystyle\Rightarrow\quad(\xi_{1},\xi_{2},\ldots,\xi_{m})$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}\sum_{i=1}^{p}\alpha_{i}z_{i}+e_{2}\sum_{j=1}^{q}\beta_{j}w_{j}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\sum_{i=1}^{p}\alpha_{i}(e_{1}z_{i})+\sum_{j=1}^{q}\beta_{j}(e_{2}w_{j}).$

This implies $\displaystyle(\xi_{1},\xi_{2},\ldots,\xi_{m})\in\;<\mathcal{B}>$ . We thus have $\displaystyle\text{Im }{T}=<\mathcal{B}>.$ Furthermore, if we consider

$\displaystyle\displaystyle\sum_{i=1}^{p}\alpha_{i}(e_{1}z_{i})$	$\displaystyle\displaystyle+$	$\displaystyle\displaystyle\sum_{j=1}^{q}\beta_{j}(e_{2}w_{j})=0$
$\displaystyle\displaystyle\Rightarrow\quad e_{1}\left(\sum_{i=1}^{p}\alpha_{i}z_{i}\right)$	$\displaystyle\displaystyle+$	$\displaystyle\displaystyle e_{2}\left(\sum_{j=1}^{q}\beta_{j}w_{j}\right)=0$
$\displaystyle\displaystyle\Rightarrow\quad\sum_{i=1}^{p}\alpha_{i}z_{i}=0$	and	$\displaystyle\displaystyle\sum_{j=1}^{q}\beta_{j}w_{j}=0$
$\displaystyle\displaystyle\Rightarrow\quad\alpha_{i}=0=\beta_{j},$		$\displaystyle\displaystyle\forall\ i,j.$

This forces $\displaystyle\mathcal{B}$ to be linearly independent and hence $\displaystyle\mathcal{B}$ becomes a basis for $\displaystyle\text{Im }{T}$ . As we know $\displaystyle\text{rank }T=\dim\text{Im }{T}$ , this implies

$\displaystyle\displaystyle\text{rank }{T}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\mbox{no. of elements of set}\ \mathcal{B}=p+q$
$\displaystyle\displaystyle\Rightarrow\quad\text{rank }{T}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\dim\text{Im }T_{1}\ +\ \dim\text{Im }T_{2}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle\text{rank }\ T_{1}\ +\ \text{rank }\ T_{2}.$

∎

Using rank-nullity theorem, one can deduce the same analogous result for nullity of linear map $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ ..

Corollary 4.8.

For any $\displaystyle T=T=e_{1}T_{1}+e_{2}T_{2}\in L_{1}^{nm}\times L_{1}^{nm}$ , we have

\displaystyle\displaystyle\dim\ker T

\displaystyle\displaystyle=

\displaystyle\displaystyle\dim\ker T_{1}+\dim\ker T_{2}.

Hence, in other words, nullity of $\displaystyle T$ is sum of nullity of its components.

Invertible and non-singular linear maps: We now examine the invertibility and non-singularity of linear maps of the space $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ . For article to be self-contained, these notions are first defined here.

Definition 4.9.

Let $\displaystyle V,W$ be any two finite dimensional vector spaces over the same field. A linear map $\displaystyle T\colon V\to W$ is said to be

•

invertible if there exists a linear map $\displaystyle S\colon W\to V$ such that $\displaystyle S\circ T=I_{V}$ and $\displaystyle T\circ S=I_{W}$ . In other words, $\displaystyle T$ is invertible if and only if $\displaystyle T$ is bijective.
•

non-singular if $\displaystyle\ker T=\{0\}$ , equivalently if $\displaystyle T$ is injective.

Remark 4.10.

For invertible linear map $\displaystyle T$ , the linear map $\displaystyle S$ given in the above definition is called the inverse of $\displaystyle T$ . In case the inverse exists it is unique, so we can denote it by $\displaystyle T^{-1}$ . Also notice that an invertible linear map is an isomorphism so that the dimensions of $\displaystyle V$ and $\displaystyle W$ must be same, i.e., $\displaystyle\dim V=\dim W$ .

The following theorem asserts that idempotent product behaves well with respect to these notions.

Theorem 4.11.

Let $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ be an element of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ . Then, we have

(1)

$\displaystyle T$ is a invertible if and only if both $\displaystyle T_{1},T_{2}$ are invertible. Further in this case, we have following

$\displaystyle(e_{1}T_{1}+e_{2}T_{2})^{-1}=e_{1}T_{1}^{-1}+e_{2}T_{2}^{-1}.$
(2)

$\displaystyle T$ is non-singular if and only if both $\displaystyle T_{1},T_{2}$ are non-singular.

Proof.

Suppose $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ is invertible. Then there exists a linear map $\displaystyle S=e_{1}S_{1}+e_{2}S_{2}$ such that

$\displaystyle\displaystyle S\circ T=T\circ S$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle I_{\mathbb{C}_{2}^{n}}$
$\displaystyle\displaystyle\Leftrightarrow\quad S_{1}\circ T_{1}=T_{1}\circ S_{1}=I_{\mathbb{C}_{1}^{n}}$	and	$\displaystyle\displaystyle S_{2}\circ T_{2}=T_{2}\circ S_{2}=I_{\mathbb{C}_{1}^{n}}$
$\displaystyle\displaystyle\Leftrightarrow\quad S_{1}\ \mbox{is the inverse of}\ T_{1}\$	and	$\displaystyle\displaystyle\ S_{2}\ \mbox{is the inverse of}\ T_{2}$
$\displaystyle\displaystyle\Leftrightarrow\quad T_{1}\;\mbox{and}\;T_{2}\;\mbox{are invertible}.$

Clearly, $\displaystyle S=T^{-1}$ if and only if $\displaystyle S_{1}=T_{1}^{-1}$ and $\displaystyle S_{2}=T_{2}^{-1}$ . This implies that $\displaystyle(e_{1}T_{1}+e_{2}T_{2})^{-1}=e_{1}T_{1}^{-1}+e_{2}T_{2}^{-1}.$ Now for the second part, we first assume that $\displaystyle e_{1}T_{1}+e_{2}T_{2}$ is non-singular. Then, by using theorem 4.6, we have

		$\displaystyle\displaystyle\Leftrightarrow$	$\displaystyle\displaystyle\ker(e_{1}T_{1}+e_{2}T_{2})=\{0\}$
		$\displaystyle\displaystyle\Leftrightarrow$	$\displaystyle\displaystyle\ker(T_{1})=\{0\},\;\ker(T_{2})=\{0\}$
		$\displaystyle\displaystyle\Leftrightarrow$	$\displaystyle\displaystyle T_{1},T_{2}\;\;\mbox{are non-singular}.$

This completes the proof. ∎

The following theorem generalizes the result of $\displaystyle C_{1}$ -linear maps to the elements of $\displaystyle L_{1}^{nm}\times_{e}L_{1}^{nm}$ .

Theorem 4.12.

Let $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}\colon\mathbb{C}_{2}^{n}\to\mathbb{C}_{2}^{m},\ S=e_{1}S_{1}+e_{2}S_{2}\colon\mathbb{C}_{2}^{m}\to\mathbb{C}_{2}^{k}$ be the linear maps. Suppose that $\displaystyle{\mathcal{B}_{1}},{\mathcal{B}_{2}}$ and $\displaystyle{\mathcal{B}_{3}}$ are the ordered bases for $\displaystyle\mathbb{C}_{1}^{n},\mathbb{C}_{1}^{m}$ and $\displaystyle\mathbb{C}_{1}^{k}$ respectively. Then, we have

\displaystyle\left[S\circ T\right]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}=[S]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}}\cdot[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}.

Proof.

By using theorem 3.4 and definition 4.1, it follows

$\displaystyle\displaystyle\left[S\circ T\right]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle[(e_{1}S_{1}+e_{2}S_{2})\circ(e_{1}T_{1}+e_{2}T_{2})]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle[e_{1}(S_{1}\circ T_{1})+e_{2}(S_{2}\circ T_{2})]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}[S_{1}\circ T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}+e_{2}[S_{2}\circ T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle e_{1}[S_{1}]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}}\cdot[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[S_{2}]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}}\cdot[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{3}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle(e_{1}[S_{1}]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}}+e_{2}[S_{2}]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}})\cdot(e_{1}[T_{1}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}+e_{2}[T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}})$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle[e_{1}S_{1}+e_{2}S_{2}]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}}\ \cdot[e_{1}T_{1}+e_{2}T_{2}]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}$
	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle[S]^{\mathcal{B}_{2}}_{\mathcal{B}_{3}}\cdot[T]^{\mathcal{B}_{1}}_{\mathcal{B}_{2}}.$

This completes the theorem. ∎

Remark 4.13.

As a special case in above theorem, if we take $\displaystyle\mathbb{C}_{2}^{m}=\mathbb{C}_{2}^{n}=\mathbb{C}_{2}^{k}$ and $\displaystyle\mathcal{B}$ as a basis, then we get

(19)

\displaystyle\displaystyle[S\circ T]_{\mathcal{B}}=[S]_{\mathcal{B}}\cdot[T]_{\mathcal{B}}.

Invertible and non-singular bicomplex matrices: We now discuss the invertibility and non-singularity of bicomplex matrix of the space $\displaystyle\mathbb{C}_{2}^{n\times n}$ . First we define them here.

Definition 4.14.

A bicomplex square matrix $\displaystyle A\in\mathbb{C}_{2}^{n\times n}$ is said to be

•

invertible if there exists a matrix $\displaystyle B\in\mathbb{C}_{2}^{n\times n}$ such that $\displaystyle A\cdot B=B\cdot A=I_{n}$ .
•

non-singular if $\displaystyle\det A$ is a non-singular element of $\displaystyle\mathbb{C}_{2}$ .

Remark 4.15.

Every square matrix $\displaystyle A\in\mathbb{C}_{2}^{n\times n}$ , like bicomplex number, can also be written uniquely as $\displaystyle A=e_{1}A_{1}+e_{2}A_{2}$ , where $\displaystyle A_{1},A_{2}\in\mathbb{C}_{1}^{n\times n}$ are complex square matrices. With this representation of bicomplex matrices, we have (cf. [8, exercise 6.8])

(1)

$\displaystyle A$ is invertible if and only if $\displaystyle A_{1},A_{2}$ are invertible.
(2)

$\displaystyle A$ is non-singular if and only if $\displaystyle A_{1},A_{2}$ are non-singular.

Theorem 4.16.

Let $\displaystyle T\in L_{1}^{nm}\times_{e}L_{1}^{nm}$ and $\displaystyle\mathcal{B}$ be a basis for $\displaystyle\mathbb{C}_{1}^{n}$ . Then, the linear map $\displaystyle T$ is invertible if and only if the matrix $\displaystyle[T]_{\mathcal{B}}$ is invertible.

Proof.

Let $\displaystyle T=e_{1}T_{1}+e_{2}T_{2}$ is invertible. This implies that $\displaystyle T^{-1}$ exists. From Part (1) of theorem 4.11, it follows that $\displaystyle T^{-1}=e_{1}T_{1}^{-1}+e_{2}T_{2}^{-1}$ . We thus have

$\displaystyle\displaystyle T\circ T^{-1}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle I_{\mathbb{C}_{2}^{n}}$
$\displaystyle\displaystyle\therefore\quad[T\circ T^{-1}]_{\mathcal{B}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle I_{n}$
$\displaystyle\displaystyle{[T]}_{\mathcal{B}}\cdot[T^{-1}]_{\mathcal{B}}$	$\displaystyle\displaystyle=$	$\displaystyle\displaystyle I_{n}\quad\mbox{(from remark \ref{specialcase2})}.$

Hence, $\displaystyle[T]_{\mathcal{B}}$ is invertible.
Conversely suppose $\displaystyle[T]_{\mathcal{B}}$ is invertible. By using part (1) of remark 4.15, we have

			$\displaystyle\displaystyle[T_{1}]_{\mathcal{B}},[T_{2}]_{\mathcal{B}}\ \mbox{are invertible}$
		$\displaystyle\displaystyle\Rightarrow$	$\displaystyle\displaystyle\quad T_{1},T_{2}\ \mbox{are invertible}$
		$\displaystyle\displaystyle\Rightarrow$	$\displaystyle\displaystyle\quad T\ \mbox{is invertible}.$

This completes the theorem. ∎

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