Some results on multi vector space

TL;DR

This paper introduces the concepts of M-basis and multi dimension in multi vector spaces, exploring their properties to advance understanding of their structure.

Contribution

It presents new definitions of M-basis and multi dimension, and investigates their properties within multi vector spaces.

Findings

Defined M-basis and multi dimension for multi vector spaces

Analyzed properties of these new concepts

Provided foundational results for future research

Abstract

In the present paper, a notion of M-basis and multi dimension of a multi vector space is introduced and some of its properties are studied.

Equations

$$C_{f(M)}(y)=\{\begin{array}{ll}\underset{f(x)=y}{\vee }& C_M(x)if{f}^{-1}(y)\ne \varphi \\ 0& otherwise\end{array}$$

Full text

Department of Mathematics, Visva-Bharati, Santiniketan-731235.

Email: 1moumi.chiney@gmail.com, 2 syamal.samanta@visva-bharati.ac.in, syamal_123@yahoo.co.in

1 Introduction

Theory of Multisets is an important generalization of classical set theory which has emerged by violating a basic property of classical sets that an element can belong to a set only once. Synonymous terms of multisets viz. list, heap, bunch, bag, sample, weighted set, occurrence set and fireset (finitely repeated element set) are used in different contexts but conveying the same idea. It is a set where an element can occur more than once. Many authors like Yager [23], Miyamoto [16, 17], Hickman [9], Blizard [1], Girish and John [7, 8], Chakraborty [2] etc. have studied the properties of multisets. Multisets are very useful structures arising in many areas of mathematics and computer science such as database queries, multicriteria decision making, knowledge representation in data based systems, biological systems and membrane computing etc. [5, 6, 11, 12, 13, 17, 18, 20]. Again the theory of vector space is one of the most important algebraic structures in modern mathematics and this has been extended in different setting such as fuzzy vector space [10, 14, 22], intuitionistic fuzzy vector space , soft vector space [4] etc. In [3], we introduced a notion of multi vector space and studied some of its basic properties. As a continuation of our earlier paper [3], here we have attempted to formulate the concept of basis and dimension of multi vector space and to study their properties.

2 Preliminaries

In this section definition of a multiset (mset in short) and some of its properties are given. Unless otherwise stated, $X$ will be assumed to be an initial universal set and $MS(X)$ denotes the set of all mset over $X.$

$\mathbf{Definition\>2.1}$ [8] An mset $M$ drawn from the set $X$ is represented by a count function $C_{M}:X\rightarrow N$ where $N$ represents the set of non negative integers.

Here $C_{M}(x)$ is the number of occurrence of the element $x$ in the mset $M.$ The presentation of the mset $M$ drawn from $X=\{x_{1},x_{2},....,x_{n}\}$ will be as $M=\{x_{1}/m_{1},x_{2}/m_{2},....,x_{n}/m_{n}\}$ where $m_{i}$ is the number of occurrences of the element $x_{i},$ $i=1,2,...,n$ in the mset $M.$

Also here for any positive integer $\omega,$ $[X]^{\omega}$ is the set of all msets whose elements are in $X$ such that no element in the mset occurs more than $\omega$ times and it will be referred to as mset spaces.

The algebraic operations of msets are considered as in [8].

$\mathbf{Definition\>2.2}$ [19] Let $M$ be a mset over a set $X$. Then a set $M_{n}=\{x\in X:C_{M}(x)\geq n\},$ where $n$ is a natural number, is called $n-$ level set of $M.$

$\mathbf{Proposition\>2.3}$ [19] Let $A,B$ be msets over $X$ and $m,n\in\mathbb{N}$.

$(1)$ If $A\subseteq B,$ then $A_{n}\subseteq B_{n};$

$(2)$ If $m\leq n,$ then $A_{m}\supseteq A_{n};$

$(3)$ $(A\cap B)_{n}=A_{n}\cap B_{n};$

$(4)$ $(A\cup B)_{n}=A_{n}\cup B_{n};$

$(5)$ $A=B$ iff $A_{n}=B_{n},\forall n\in\mathbb{N}.$

$\mathbf{Definition\>2.4}$ [19] Let $P\subseteq X.$ Then for each $n\in\mathbb{N},$ we define a mset $nP$ over $X,$ where $C_{nP}(x)=n,\forall x\in P.$

$\mathbf{Definition\>2.5}$ [19] Let $X$ and $Y$ be two nonempty sets and $f:X\rightarrow Y$ be a mapping. Then

$(1)$ the image of a mset $M\in[X]^{\omega}$ under the mapping $f$ is denoted by $f(M)$ or $f[M]$, where

[TABLE]

$(2)$ the inverse image of a mset $N\in[Y]^{\omega}$ under the mapping $f$ is denoted by $f^{-1}(N)$ or $f^{-1}[N]$ where $C_{f^{-1}(N)}(x)=C_{N}[f(x)]$.

The properties of functions, which are used in this paper, are as in [19].

$\mathbf{Definition\>2.6}$ [3] Let $A_{1},A_{2},...,A_{n}$ $\in\left[X\right]^{\omega}.$ Then we define $A_{1}+A_{2}+...+A_{n}$ as follows:

$C_{A_{1}+A_{2}+...+A_{n}}(x)=\vee\{C_{A_{1}}(x_{1})\wedge C_{A_{2}}(x_{2})\wedge...\wedge C_{A_{n}}(x_{n}):x_{1},x_{2},...,x_{n}\in X\>and\;x_{1}+x_{2}+...+x_{n}=x\}.$

Let $\lambda\in K$ and $B\in\left[X\right]^{\omega}.$ Then $\lambda B$ is defined as follows:

$C_{\lambda B}(y)=\vee\{C_{B}(x):\lambda x=y\}.$

$\mathbf{Lemma\>2.7}$ [3] Let $\lambda\in K$ and $B\in\left[X\right]^{\omega}.$ Then

$(a)$ For $\lambda\neq 0,$ $C_{\lambda B}(y)=C_{B}(\lambda^{-1}y),\forall y\in X.$

For $\lambda=0,$

$C_{\lambda B}(y)=\begin{cases}0,&y\neq 0,\\ \underset{x\in X}{sup}\>C_{B}(x),&y=0\end{cases}$.

$(b)$ For all scalars $\lambda\in K$ and for all $x\in X,$ we have $C_{\lambda B}(\lambda x)\geq C_{B}(x).$

$\mathbf{Definition\>2.8}$ [3] A multiset $V$ in $[X]^{\omega}$ is said to be a multi vector space or multi linear space(in short mvector space) over the linear space $X$ if

$(i)$ $V+V\subseteq V;$

$(ii)$ $\lambda V\subseteq V,$ for every scalar $\lambda.$

We denote the set of all multi vector space over $X$ by $MV(X).$

$\mathbf{Remark\>2.9}$ [3] For a multi vector space $V$ in $[X]^{\omega}$ , $V+V+.....n\>times=V,$ i. e., $nV=V.$

$\mathbf{Remark\>2.10}$ [3] If $V\in MV(X)$ with $dim\>X=m,$ then $\mid C_{V}(X)\mid\leq m+1,$ where $\mid C_{V}(X)\mid$ represents the cardinality of $C_{V}(X).$

$\mathbf{Proposition\>2.11}$ [3] (Representation theorem) Let $V\in MV(X)$ with $dim\>X=m$ and range of $C_{V}=\{n_{0},n_{1},....,n_{k}\}\subseteq\{0,1,2,...,\omega\},$ $k\leq m,$ $n_{0}=C_{V}(\theta)$ and $\omega\geq n_{0}>n_{1}>...>n_{k}\geq 0.$ Then there exists a nested collection of subspaces of $X$ as

$\{\theta\}\subseteq V_{n_{0}}\subsetneqq V_{n_{1}}\subsetneqq V_{n_{2}}\subsetneqq....\subsetneqq V_{n_{k}}=X$ such that $V=n_{0}V_{n_{0}}\cup n_{1}V_{n_{1}}\cup.....\cup n_{k}V_{n_{k}}.$ Also

$(1)$ If $n,m\in(n_{i+1},n_{i}],$ then $V_{n}=V_{m}=V_{n_{i}}.$

$(2)$ If $n\in(n_{i+1},n_{i}]$ and $m\in(n_{i},n_{i-1}]$, then $V_{n}\supsetneqq V_{m}.$

$\mathbf{Definition\>2.12}$ [3] Let $X$ be a finite dimensional vector space with $dim\>X=m$ and $V\in MV(X).$ Consider $Proposition\>2.11.$ Let $B_{n_{i}}$ be a basis on $V_{n_{i}},\>i=0,1,...,k$ such that

$B_{n_{0}}\subsetneqq B_{n_{1}}\subsetneqq B_{n_{2}}\subsetneqq...\subsetneqq B_{n_{k}}$…..$(iii)$

Define a multi subset $\beta$ of $X$ by

$C_{\beta}(x)=\begin{cases}\vee\{n_{i}:x\in B_{n_{i}}\}\\ 0,\>otherwise\end{cases}$

Then $\beta$ is called a multi basis of $V$ corresponding to $(iii).$ We denote the set of all multi bases of $V$ by $\mathcal{B}_{M}(V).$

$\mathbf{Corollary\>2.13}$ [3] Let $\beta$ be a multi basis of $V$ obtained by $(iii).$ Then

$(1)$ If $n,m\in(n_{i+1},n_{i}]$, then $\beta_{n}=\beta_{m}=B_{n_{i}}.$

$(2)$ If $n\in(n_{i+1},n_{i}]$ and $m\in(n_{i},n_{i-1}]$, then $\beta_{n}\supsetneqq\beta_{m}.$

$(3)$ $\beta_{n}$ is a basis of $V_{n}$, for all $n\in\{1,2,....,\omega\}$.

3 Some results on multi vector space

$\mathbf{Lemma\>3.1}$ Let $s,t\in\mathbb{R}$ and $A,\>A_{1}$ and $A_{2}$ be multisets on a vector space $X.$ Then

$(1)$ $s.(t.A)=t.(s.A)=(st).A$ and

$(2)$ $A_{1}\leq A_{2}\Rightarrow t.A_{1}\leq t.A_{2}.$

$\mathbf{Proposition\>3.2}$ Let $V\in MV(X).$ Then $x\in X,\;a\neq 0\Rightarrow C_{V}(ax)=C_{V}(x).$

$\mathbf{Proposition\>3.3}$ Let $V\in MV(X)$ and $u,v\in X$ such that $C_{V}(u)>C_{V}(v)$. Then $C_{V}(u+v)=C_{V}(v).$

$\mathbf{Proposition\>3.4}$ Let $V\in MV(X)$ and $v,w\in X$ with $C_{V}(v)\neq C_{V}(w)$. Then $C_{V}(v+w)=C_{V}(v)\wedge C_{V}(w).$

4 Multi linear independence

$\mathbf{Definition\>4.1}$ Let $V\in MV(X)$ and $dim\>X=n.$ We say that a finite set of vectors $\{x_{i}\}_{i=1}^{n}$ is multi linearly independent in $V$ if and only if $\{x_{i}\}_{i=1}^{n}$ is linearly independent in $X$ and for all $\{a_{i}\}_{i=1}^{n}\subset\mathbb{R}$ with $a_{i}\neq 0$ , $C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{n}{\sum}a_{i}x_{i})$

$=\stackrel{{\scriptstyle[}}{{i}}=1]{n}{\wedge}C_{V}(a_{i}x_{i}).$

The following example shows that every linearly independent set is not multi linearly independent.

$\mathbf{Example\;4.2}$ Let $X=\mathbb{R}^{2}$ and $\omega=4.$ We define a multi vector space $C_{V}:X\rightarrow N$ by

$C_{V}(x)=\begin{cases}4,\>if\>x=(0,0)\\ 2,\;if\;x=(0,a),a\neq 0\\ 1,\>otherwise.\end{cases}$

If we take the vectors $x=(1,0)$ and $y=(-1,1),$ then they are linearly independent but not multi linearly independent. As here $C_{V}(x)=C_{V}(y)=1$, but $C_{V}(x+y)=2>(C_{V}(x)\wedge C_{V}(y))=1.$

$\mathbf{Proposition\>4.3}$ Let $V\in MV(X)$ and $dim\>X=m.$ Then any set of vectors $\{x_{i}\}_{i=1}^{N}(N\leq m),$ which have distinct counts is linearly and multi linearly independent.

$Proof.$ The proof follows by method of induction.

$\mathbf{Note\;4.4}$ Converse of the above proposition is not true. Let $X=\mathbb{R}^{2}$ and $\omega=6.$ We define a multi vector space $C_{V}:X\rightarrow N$ by

$C_{V}(x)=\begin{cases}6,\>if\>x=(0,0)\\ 1,\>otherwise.\end{cases}$

Then we have $\{\theta\}=V_{6}\subset V_{1}=\mathbb{R}^{2}.$ Let $e_{1}=(1,0),$ $e_{2}=(0,1).$ Then $\{e_{1},e_{2}\}$ are multi linearly independent in $V$ , although, $C_{V}(e_{1})=C_{V}(e_{2})$.

5 M-basis

$\mathbf{Definition\>5.1}$ A M-basis for a multi vector space $V\in MV(X)$ is a basis of $X$ which is multi linearly independent in $V$.

We denote the set of all M-bases of $V$ by \text{\mathscr{B}}(V).

$\mathbf{Lemma\>5.2}$ If $V\in MV(X)$ and $Y$ is a proper subspace of $X,$ then for any $t\in X\setminus Y$ with $C_{V}(t)=sup[C_{V}(X\setminus Y)]$, $C_{V}(t+y)=C_{V}(t)\wedge C_{V}(y),$ for all $y\in Y.$

$Proof.$ Since $\omega$ is finite, such a $t$ exists. Let $y\in Y.$ If $C_{V}(y)\neq C_{V}(t)$ then by $Proposition\>3.4,$ $C_{V}(t+y)=C_{V}(t)\wedge C_{V}(y).$ If $C_{V}(y)=C_{V}(t)$ then by $Definition\>2.8,$ $C_{V}(t+y)\geq C_{V}(t)\wedge C_{V}(y).$ Since $t+y\in X\setminus Y$ and $C_{V}(t)=sup[C_{V}(X\setminus Y)],$ we must have $C_{V}(t+y)\leq C_{V}(t)=C_{V}(y)$ and thus $C_{V}(t+y)=C_{V}(t)\wedge C_{V}(y).$

$\mathbf{Lemma\>5.3}$ Let $V\in MV(X)$, $Y$ be a proper subspace of $X$ and $C_{V}\mid_{Y}=C_{V^{\prime}}.$ If $B^{*}$ is a M-basis for $V^{\prime},$ then there exists $t\in X\setminus Y$ such that $B^{+}=B^{*}\cup\{t\}$ is a M-basis for $W,$ where C_{W}=C_{V}\mid_{\prec B^{+}\text{\succ}} and \text{\prec}B^{+}\text{\succ} is the vector space spanned by $B^{+}.$

$Proof.$ Pick $t\in X\setminus Y$ such that $C_{V}(t)=sup[C_{V}(X\setminus Y)].$ Then by $Lemma\;5.2,$ $B^{+}=B^{*}\cup\{t\}$ is a multi linearly independent and hence a M-basis for $W,$ where C_{W}=C_{V}\mid_{\prec B^{+}\text{\succ}}.

$\mathbf{Proposition\>5.4}$ All multi vector spaces $V\in MV(X)$ with $dim\>X=m$ have M-basis.

$Proof.$ The proof follows by mathematical induction.

$\mathbf{Proposition\>5.5}$ Let $V\in MV(X)$ where $dim\>X=m$ and $C_{V}(X\setminus\{\theta\})=\{n_{0},n_{1},n_{2},...,n_{k}\},\;k\leq m$. Then a basis $B=\{e_{1},e_{2},...,e_{m}\}$ of $X$ is a M-basis for $V$ if and only if $B\cap V_{n_{i}}$ is a basis of $V_{n_{i}}$ for any $i=0,1,...,k.$

$\mathbf{Proof.}$ Let $\omega\geq n_{0}>n_{1}>....>n_{k}\geq 0.$ Then $\{\theta\}\subsetneqq V_{n_{0}}\subsetneqq V_{n_{1}}\subsetneqq V_{n_{2}}\subsetneqq....\subsetneqq V_{n_{k}}=X$ . Let $B_{n_{i}}=B\cap V_{n_{i}},\;i=0,1,...,k.$

First suppose that $B\cap V_{n_{i}}=B_{n_{i}}$ is a basis of $V_{n_{i}}$ for any $i=0,1,...,k.$ Then $B_{n_{0}}\subsetneqq B_{n_{1}}\subsetneqq......\subsetneqq B_{n_{k}}=B.$ Let $B_{n_{0}}=\{e_{n_{0}}^{1},e_{n_{0}}^{2},...,e_{n_{0}}^{j}\},j\leq m.$ Then $C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}a_{i}e_{n_{0}}^{i})\geq\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\wedge}C_{V}(e_{n_{0}}^{i})=n_{0}$. Since $n_{0}$ is the highest count,

$C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}a_{i}e_{n_{0}}^{i})=n_{0}=\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\wedge}C_{V}(e_{n_{0}}^{i}).$ Hence $B_{n_{0}}$ is multi linearly independent.

Next let $B_{n_{1}}=B_{n_{0}}\cup\{e_{n_{1}}^{1},e_{n_{1}}^{2},...,e_{n_{1}}^{s}\},j+s\leq m.$ Consider the sum

$\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}b_{i}e_{n_{0}}^{i}+\stackrel{{\scriptstyle[}}{{i}}=1]{s}{\sum}c_{i}e_{n_{1}}^{i}$, where some $c_{i}\neq 0.$ Then $C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}b_{i}e_{n_{0}}^{i}+\stackrel{{\scriptstyle[}}{{i}}=1]{s}{\sum}c_{i}e_{n_{1}}^{i})$

$\geq\left(\underset{i\in J_{1}}{\wedge}C_{V}(e_{n_{0}}^{i})\right)\wedge\left(\underset{i\in J_{2}}{\wedge}C_{V}(e_{n_{1}}^{i})\right)$, [where $J_{1}=\{i:b_{i}\neq 0\},\;J_{2}=\{i:c_{i}\neq 0\}$] $=n_{1}.$

If $C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}b_{i}e_{n_{0}}^{i}+\stackrel{{\scriptstyle[}}{{i}}=1]{s}{\sum}c_{i}e_{n_{1}}^{i})>n_{1},$ then

$C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}b_{i}e_{n_{0}}^{i}+\stackrel{{\scriptstyle[}}{{i}}=1]{s}{\sum}c_{i}e_{n_{1}}^{i})=n_{0}\Rightarrow\stackrel{{\scriptstyle[}}{{i}}=1]{j}{\sum}b_{i}e_{n_{0}}^{i}+\stackrel{{\scriptstyle[}}{{i}}=1]{s}{\sum}c_{i}e_{n_{1}}^{i}\in V_{n_{0}}\Rightarrow c_{i}=0,$ for all $i=1,2,..,s,$ a contradiction. Thus $B_{n_{1}}$ is multi linearly independent. Proceeding in the similar way it can be proved that $B_{n_{k}}=B$ is multi linearly independent and hence a M-basis for $V.$

Conversely, let $B$ be a M-basis for $V.$ Then either $B_{n_{i}}=\phi$ or $B_{n_{i}}\neq\phi.$

Let $B_{n_{i}}=\phi$ and $x\in V_{n_{i}}.$ Then obviously $B_{n_{j}}=\phi,j<i.$ Since $B$ is a basis of $X,$ there exists some $B^{\prime}\subseteq B$ such that $x=\underset{e_{j}\in B}{\sum}b_{j}e_{j},$ $b_{j}\neq 0.$ Then $C_{V}(x)=\underset{e_{j}\in B}{\wedge}C_{V}(e_{j})\leq n_{i+1},$ a contradiction. So, $B_{n_{i}}\neq\phi.$

Then $B_{n_{0}}\subsetneqq B_{n_{1}}\subsetneqq......\subsetneqq B_{n_{k}}=B.$ Let $x\in V_{n_{i}}$ and $B_{n_{i}}$ is not a basis of $V_{n_{i}}$. Choose $x=\underset{e_{i}\in B_{n_{i}}}{\sum}a_{i}e_{i}+\underset{e_{i}^{\prime}\notin B_{n_{i}}}{\sum}b_{i}e_{i}^{\prime}$, for all $b_{i}\neq 0$.

Now, $C_{V}(x)=C_{V}(\underset{e_{i}\in B_{n_{i}}}{\sum}a_{i}e_{i}+\underset{e_{i}^{\prime}\notin B_{n_{i}}}{\sum}b_{i}e_{i}^{\prime})$

$=\left(\underset{e_{i}\in B_{n_{i}}^{\prime}}{\wedge}C_{V}(e_{i})\right)\wedge\left(\underset{e_{i}^{\prime}\notin B_{n_{i}}}{\wedge}C_{V}(e_{i}^{\prime})\right)$, [where $B_{n_{i}}^{\prime}=\{e_{i}\in B_{n_{i}}:a_{i}\neq 0\}$] $=\left(\underset{e_{i}^{\prime}\notin B_{n_{i}}}{\wedge}C_{V}(e_{i}^{\prime})\right)<n_{i},$ a contradiction to the fact that $x\in V_{n_{i}}.$ Thus $x=\underset{e_{i}\in B_{n_{i}}}{\sum}a_{i}e_{i}$ and $B_{n_{i}}$ is a basis of $V_{n_{i}}.$

Hence proved.

$\mathbf{Proposition\>5.6}$ Let $V$ be a multi vector space over $X$ where $dim\>X=m.$ Then there is an one-to-one correspondence between $\mathcal{B}_{M}(V)$ and \text{\mathscr{B}}(V).

$\mathbf{Proposition\;5.7}$ Let $V\in MV(X)$ with $dim\>X=m$ and range of $C_{V}(X\setminus\{\theta\})=\{n_{0},n_{1},....,n_{k}\}\subseteq\{0,1,2,...,\omega\},$ $k\leq m.$ If a basis $B=\{e_{1},e_{2},...,e_{m}\}$ of $X$ is a M-basis, then $C_{V}(B)=\{n_{0},n_{1},....,n_{k}\}.$

$\mathbf{Remark\;5.8}$ Converse of the above $Proposition$ is not true. For example, suppose $X=\mathbb{R}^{4},\;\omega=5.$ Define multi vector space $V$ with count functions $C_{V}$ as follows:

$C_{V}((0,0,0,0))=5$; $C_{V}((0,0,0,\mathbb{R}\setminus\{0\}))=5;$ $C_{V}((0,0,\mathbb{R}\setminus\{0\},\mathbb{R}))=5$, $C_{V}((0,\mathbb{R}\setminus\{0\},\mathbb{R},\mathbb{R}))=2,C_{V}(\mathbb{R}^{4}\setminus(0,\mathbb{R},\mathbb{R},\mathbb{R}))=2.$

Then $B=\{(0,0,0,1),(-1,1,1,1),(1,-1,1,1),(1,1,-1,1)\}$ is a basis of $\mathbb{R}^{4}$ and $C_{V}(B)=\{2,5\}=C_{V}(\mathbb{R}^{4})$. But $B$ is not a M-basis as $B$ is not multi linearly independent. In fact, $C_{V}((-1,1,1,1))=C_{V}(1,-1,1,1))=2.$ But $C_{V}((-1,1,1,1)+(1,-1,1,1))=C_{V}((0,0,1,1))=5>[C_{V}((-1,1,1,1))\wedge C_{V}((-1,1,1,1))]=2.$

$\mathbf{Definition\;5.9}$ Let $V\in MV(X)$ with $dim\>X=m$, range of C_{V}(X\setminus\{\theta\})=\{n_{0},n_{1},$$...,n_{k}\}$$\subseteq\{0,1,2,...,\omega\}, $k\leq m$ and $B_{0}$ be any M-basis for $V.$ Then $C_{V}(B_{0})=\{n_{0},n_{1},....,n_{k}\}.$ We define multi index of a multi M-basis $B_{0}$ with respect to $V$ by $[B_{0}]_{M}=\{r_{i}:$ $r_{i}$ is the number of element of $B_{0}$ taking the value n_{i}$$\}.

$\mathbf{Proposition\;5.10}$ For a multi vector space $V$, multi index of M-basis with respect to $V$ is independent of M-basis.

$\mathbf{Proof.}$ Let $V\in MV(X)$ with $dim\>X=m$, range of $C_{V}(X\setminus\{\theta\})=\{n_{0},n_{1},....,n_{k}\}$

$\subseteq\{0,1,2,...,\omega\},$ $k\leq m$ and $\omega\geq n_{0}>n_{1}>...>n_{k}\geq 0.$ Then for any two M-bases $B_{0},B_{0}^{\prime}$ of $V,$ $C_{V}(B_{0})=C_{V}(B_{0}^{\prime})=\{n_{0},n_{1},....,n_{k}\}.$ Let $[B_{0}]_{M}=\{r_{i}\}$ and $[B_{0}^{\prime}]_{M}=\{r_{i}^{\prime}\}$. Now, $\mid B_{0}\cap V_{n_{i}}\mid=\stackrel{{\scriptstyle[}}{{j}}=0]{i}{\sum}r_{j}$ and $\mid B_{0}^{\prime}\cap V_{n_{i}}\mid=\stackrel{{\scriptstyle[}}{{j}}=0]{i}{\sum}r_{j}^{\prime}$, for $i=0,1,2,...,k.$ As $B_{0}\cap V_{n_{i}}$ and $B_{0}^{\prime}\cap V_{n_{i}}$ are both basis of $V_{n_{i}}$, $\mid B_{0}\cap V_{n_{i}}\mid$

$=\mid B_{0}^{\prime}\cap V_{n_{i}}\mid,$ for all $i=0,1,2,...,k.$ Thus $[B_{0}]_{M}=[B_{0}^{\prime}]_{M}$.

$\mathbf{Note\;5.11}$ As multi index of M-basis with respect to a multi vector space $V$ is independent of M-basis, we can use only the term multi index of a multi vector space $V.$

$\mathbf{Definition\;5.12}$ Let $V\in MV(X)$ with $dim\>X=m$, $C_{V}(X)=\{n_{0},n_{1},....,n_{k}\}$

$\subseteq\{0,1,2,...,\omega\},$ $k\leq m$ and $B$ be any basis for $X.$ We define index of a basis $B$ with respect to $V$ by $[B]=\{r_{i}:\;n_{i}$ $r_{i}$ is the number of element of $B$ taking the value n_{i}$$\}.

$\mathbf{Proposition\;5.13}$ Let $V\in MV(X)$ with $dim\>X=m$, $C_{V}(X\setminus\{\theta\})=\{n_{0},n_{1},....$

,n_{k}\}$$\subseteq\{0,1,2,...,\omega\}, $k\leq m$ and $B$ be any basis of $X$ with $C_{V}(B)=\{n_{0},n_{1},....$

$,n_{k}\}.$ If index $[B]$ of $B$ with respect to $V$ is equal to the multi index of $V$, then $B$ becomes a M-basis.

$\mathbf{Proof.}$ Let us assume that $\omega\geq n_{0}>n_{1}>...>n_{k}\geq 0.$ Then $\{\theta\}\subsetneqq V_{n_{0}}\subsetneqq V_{n_{1}}\subsetneqq V_{n_{2}}\subsetneqq....\subsetneqq V_{n_{k}}=X$. Suppose that $[B]_{M}=\{r_{i}:i=0,1,2,...k\}.$ Then $dim\;V_{n_{i}}=\stackrel{{\scriptstyle[}}{{j}}=0]{i}{\sum}r_{j}=\mid B\cap V_{n_{i}}\mid,$ for all $i=0,1,2,...,k.$ Hence, $B\cap V_{n_{i}}$ becomes a basis for $V_{n_{i}}$ for each $i=0,1,2,..,k.$ Thus by $Proposition\;5.5,$ $B$ is a M-basis for $V.$

6 Dimension of multi vector space

$\mathbf{Definition\>6.1}$ We define the dimension of a multi vector space $V$ over $X$ by $dim(V)=\underset{B\>a\>base\>for\>X}{sup}\left(\underset{x\in B}{\sum}C_{V}(x)\right)$.

Clearly $dim$ is a function from the set of all multi vector spaces to $\mathbb{N}.$

$\mathbf{Proposition\>6.2}$ Let $V\in MV(X)$ where $dim\>X=m<\infty.$ Then if $B$ is a M-basis for $V$ and $B^{*}$ is any basis for $X$ then

$\underset{x\in B^{*}}{\sum}C_{V}(x)\leq\underset{x\in B}{\sum}C_{V}(x).$

$\mathbf{Proposition\>6.3}$ If $V$ is a multi vector space over a finite dimensional vector space $X$, then $dim(V)=\underset{x\in B}{\sum}C_{V}(x),$ where $B$ is any M-basis for $V.$

$\mathbf{Note\;6.4}$ If $V$ is a multi vector space over a finite dimensional vector space $X$, then $dim(V)$ is independent of M-basis for $V.$ It follows from $Proposition\;5.5$ and $Proposition\;5.7.$

$\mathbf{Proposition\;6.5}$ Let $X$ be any finite dimensional vector space and $V,W\in MV(X)$ such that $C_{V}(\theta)\geq sup[C_{W}(X\setminus\{\theta\})]$ and $C_{W}(\theta)\geq sup[C_{V}(X\setminus\{\theta\})]$. Then there exists a basis $B$ for $X$ which is also a M-basis for $V,$ $W$, $V\cap W$ and $V+W.$ In addition, if $A_{1}=\{x\in X:C_{V}(x)<C_{W}(x)\},$ $A_{2}=X\setminus A_{1},$ then for all $v\in B\cap A_{1},$

$(C_{V\cap W})(v)=C_{V}(v)$ and $C_{V+W}(v)=C_{W}(v)$

and for all $v\in B\cap A_{2},$

$(C_{V\cap W})(v)=C_{W}(v)$ and $C_{V+W}(v)=C_{V}(v).$

$\mathbf{Proof.}$ We prove this by induction on $dim\;X.$ In case $dim\;X=1$ the statement is clearly true.

Now suppose that the theorem is true for all the multi vector space with dimension of the underlying vector space equal to n.

Let $V$ and $W$ be two multi vector spaces over $X$ with $dim\;X=n+1>1.$ Let $B_{1}=\{v_{i}\}_{i=1}^{n+1}$ be any M-basis for $V.$ We may assume that $C_{V}(v_{1})\leq C_{V}(v_{i})$ for all $i=\{2,3,...,n+1\}.$ Let $H=\prec\{v_{i}\}_{i=2}^{n+1}\succ.$ Since $n+1>1,$ $H\neq\{\theta\}.$ Clearly $dim\;H=n.$ Define the following two multi vector spaces: $V_{1}$ with count function $C_{V_{1}}=C_{V}\mid_{H}$ and $W_{1}$ with the count function $C_{W_{1}}=C_{W}\mid_{H}$. By inductive hypothesis the exists a basis $B^{*}$ for $H$ which is also a M-basis for $V_{1},$ $W_{1},$ $V_{1}\cap W_{1}$ and $V_{1}+W_{1}.$ Also for all $v\in B^{*}\cap A_{1},$

$(C_{V_{1}\cap W_{1}})(v)=C_{V_{1}}(v)$ and $C_{V_{1}+W_{1}}(v)=C_{W_{1}}(v)$

and for all $v\in B^{*}\cap A_{2},$

$(C_{V_{1}\cap W_{1}})(v)=C_{W_{1}}(v)$ and $C_{V_{1}+W_{1}}(v)=C_{V_{1}}(v).$

We shall now show that $B^{*}$ can be extended to $B$ such that $B$ is a M-basis for $V,$ $W$, $V\cap W$ and $V+W.$ Furthermore, for all $v\in B\cap A_{1},$

$(C_{V\cap W})(v)=C_{V}(v)$ and $C_{V+W}(v)=C_{W}(v)$

and for all $v\in B\cap A_{2},$

$(C_{V\cap W})(v)=C_{W}(v)$ and $C_{V+W}(v)=C_{V}(v).$

Step - 1: First we have to show that for all $x\in H,$

$C_{(V+W)}\mid_{H}(x)=C_{V_{1}+W_{1}}(x).$……….(1)

Let $x\in H.$ Then we have

$C_{(V+W)}\mid_{H}(x)=sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in X\}$

$=sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):x_{1}\in H\}\vee sup\{C_{V}(x_{2})\wedge C_{W}(x-x_{2}):\;x_{2}\in X\setminus H\}.$……..(2)

If $x\in H\setminus\{\theta\},$ we have

$C_{V}(x)\wedge C_{W}(x-x)=C_{V}(x)\wedge C_{W}(\theta)\leq sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\},$

$C_{V}(\theta)\wedge C_{W}(x-\theta)=C_{V}(\theta)\wedge C_{W}(x)\leq sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}.$

Since $C_{V}(\theta)\geq sup[C_{W}(H\setminus\{\theta\})]$ and $C_{W}(\theta)\geq sup[C_{V}(H\setminus\{\theta\})],$

$C_{V}(x)\wedge C_{W}(\theta)=C_{V}(x)$ and $C_{V}(\theta)\wedge C_{W}(x)=C_{W}(x),$

and this leads to the following inequality:

$C_{V}(x)\vee C_{W}(x)\leq sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}.$………(3)

Suppose that $sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}<sup\{C_{V}(x_{2})\wedge C_{W}(x-x_{2}):\;x_{2}\in X\setminus H\}.$………..(4)

This means that there exists $x^{\prime}\in X\setminus H$ such that

$sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}<C_{V}(x^{\prime})\wedge C_{W}(x-x^{\prime}).$

In view of (2) we must have

$C_{V}(x)\vee C_{W}(x)<C_{V}(x^{\prime})\wedge C_{W}(x-x^{\prime}).$…….(5)

Since $x^{\prime}\in X\setminus H$ and $C_{V}(X\setminus H)=C_{V}(v_{1})\leq C_{V}(v_{i})$ for all $i\in\{2,3,...,n+1\},$ we must have $C_{V}(x)\geq C_{V}(x^{\prime}).$ Thus $(4)$ becomes $C_{V}(x)\vee C_{W}(x)<C_{V}(x)\wedge C_{W}(x-x^{\prime}).$ It is not possible (Using the properties of $\vee,\wedge$ and $<$). This means that our assumption (3) is false. Therefore we must have

$sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}\geq sup\{C_{V}(x_{2})\wedge C_{W}(x-x_{2}):\;x_{2}\in X\setminus H\}......(6)$

If $x=\theta,$ $C_{(V+W)}\mid_{H}(\theta)=C_{V}(\theta)\wedge C_{W}(\theta)=sup\{C_{V}(x_{1})\wedge C_{W}(\theta-x_{1}):x_{1}\in H\}\geq sup\{C_{V}(x_{2})\wedge C_{W}(x-x_{2}):\;x_{2}\in X\setminus H\}....(7)$.

Using equations $(6)$ and $(7)$,we have for all $x\in H,$

$sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}\geq sup\{C_{V}(x_{2})\wedge C_{W}(x-x_{2}):\;x_{2}\in X\setminus H\},$

From $(1),$ we have for all $x\in H,$ $C_{(V+W)}\mid_{H}(x)=sup\{C_{V}(x_{1})\wedge C_{W}(x-x_{1}):\;x_{1}\in H\}$

$=sup\{C_{V}\mid_{H}(x_{1})\wedge C_{W}\mid_{H}(x-x_{1}):\;x_{1}\in H\}$

$=C_{V_{1}+W_{1}}(x).$

This establishes (1).

Since $B^{*}$ is a M-basis of V_{1}+W_{1},$$(1) implies that $B^{*}$ is multi linearly independent in $V+W.$

Step - 2: Let $v^{*}\in X\setminus H$ such that $C_{W}(v^{*})=sup[C_{W}(X\setminus H)].$ By $Lemma\;5.2$ and $Lemma\;5.3$, $v^{*}$ is an extension of M-basis $B^{*}$ for $W_{1}$ to $B=B^{*}\cup\{v^{*}\}$ a M-basis for $W.$

Step - 3: Since $C_{V}(X\setminus H)=C_{V}(v_{1}),$ $C_{V}(v_{1})=C_{V}(v^{*})$ and then $v^{*}$is also an extension of M-basis $B^{*}$ for $V_{1}$ to $B$ a M-basis for $V.$

Step - 4: Now we shall show that $v^{*}$ is an extension of M-basis $B^{*}$ for $V_{1}\cap W_{1}$ to $B$ a M-basis for $V\cap W.$ If $v^{*}\in A_{1}$, then $(C_{V}\wedge C_{W})(A_{1}\cap(X\setminus H))=C_{V}(v^{*})$ and for all $z\in A_{2}\cap(X\setminus H),$ $(C_{V}\wedge C_{W})(z)\leq C_{V}(v^{*})$, by definition of $A_{2}.$

From this we may conclude that if $v^{*}\in A_{1}$ then

$(C_{V}\wedge C_{W})(v^{*})=sup[(C_{V}\wedge C_{W})(X\setminus H)].$

If $v^{*}\in A_{2}$ then $C_{W}(v^{*})\leq C_{V}(v^{*}).$ Since $C_{W}(v^{*})=sup[C_{W}(X\setminus H)]$ and $C_{V}$ is constant on $X\setminus H$ we must have $A_{1}\cap(X\setminus H)=\phi.$ Therefore we have that if $v^{*}\in A_{2}$, then $(C_{V}\wedge C_{W})(v^{*})=sup[(C_{V}\wedge C_{W})(X\setminus H)].$ By $Lemma\;5.3,$ we may now conclude that $v^{*}$ extends M-basis $B^{*}$ for $V_{1}\cap W_{1}$ to $B$ a M-basis for $V\cap W.$

Step - 5: Now we shall show that $v^{*}$ is also an extension of $B^{*}$ a M-basis for $V_{1}+W_{1}$ to $B$ a M-basis for $V+W.$ Suppose that there exists $z\in X\setminus H$ such that $C_{(V+W)}(v^{*})<C_{(V+W)}(z).$ Clearly vector $z$ can be written in the form $z=a(v^{*}+v)$ where $a\neq 0$ and $v\in H.$ Therefore we have

$C_{(V+W)}(v^{*})<C_{(V+W)}(z)=C_{(V+W)}(a(v^{*}+v))=C_{(V+W)}(v^{*}+v).$

This means that there exists $x_{1}\in X$ such that for all $x^{\prime}\in X,$

$C_{V}(x^{\prime})\wedge C_{W}(v^{*}-x^{\prime})<C_{V}(x_{1})\wedge C_{W}(v^{*}+v-x_{1}).$ …………(8)

In particular this is true for $x^{\prime}=\theta,$ i.e.

$C_{V}(\theta)\wedge C_{W}(v^{*})<C_{V}(x_{1})\wedge C_{W}(v^{*}+v-x_{1}).$

But since $C_{V}(\theta)\geq sup[C_{W}(X\setminus\{\theta\})]$ we have

$C_{W}(v^{*})<C_{V}(x_{1})\wedge C_{W}(v^{*}+v-x_{1}).$……..(9)

If $x_{1}\in H$ then since $v\in H$ we must have $v-x_{1}\in H.$ Again $v^{*}\in X\setminus H.$ So, by $Lemma\;5.2,$ $C_{W}(v^{*}+v-x_{1})=C_{W}(v^{*})\wedge C_{W}(v-x_{1})$ and so (9) becomes $C_{W}(v^{*})<C_{V}(x_{1})\wedge C_{W}(v^{*})\wedge C_{W}(v-x_{1}),$ which is impossible. Thus $x_{1}\in X\setminus H.$ Let $x^{\prime}=v^{*}$ in (5). Since $C_{W}(\theta)\geq sup[C_{V}(X\setminus\{\theta\})]$ we have

$C_{V}(v^{*})<C_{V}(x_{1})\wedge C_{W}(v^{*}+v-x_{1}).....(10)$

Recall that $C_{V}(X\setminus H)=C_{V}(v_{1})$ and thus $C_{V}(v_{1})=C_{V}(v^{*})=C_{V}(x_{1})$, as $x_{1}\in X\setminus H.$ This again means that the inequality (10) is false. This means that for all $z\in X\setminus H,$ $C_{(V+W)}(v^{*})\geq C_{(V+W)}(z)$. Therefore by $Lemma\;5.3,$ $v^{*}$ is an extension of $B^{*}$ a M-basis for $V_{1}+W_{1}$ to $B$ a M-basis for $V+W.$

Step - 6: Now we shall show that if $v^{*}\in A_{1}$ then $C_{V+W}(v^{*})=C_{W}(v^{*})$ and if $v^{*}\in A_{2}$ then $C_{V+W}(v^{*})=C_{V}(v^{*}).$ From the definition we have:

$C_{(V+W)}(v^{*})=sup\{C_{V}(x_{1})\wedge C_{W}(v^{*}-x_{1}):\;x_{1}\in X\}.$

Let $x^{\prime}$ be such that

$sup\{C_{V}(x_{1})\wedge C_{W}(v^{*}-x_{1}):\;x_{1}\in X\}=C_{V}(x^{\prime})\wedge C_{W}(v^{*}-x^{\prime}).$

By substituting $x_{1}=\theta$ and then $x_{1}=v^{*}$ and recalling that $C_{V}(\theta)\geq sup[C_{W}(X\setminus\{\theta\})]$ and $C_{W}(\theta)\geq sup[C_{V}(X\setminus\{\theta\})]$, we obtain

$C_{V}(v^{*})\vee C_{W}(v^{*})\leq C_{V}(x^{\prime})\wedge C_{W}(v^{*}-x^{\prime}).$

Suppose that

$C_{V}(v^{*})\vee C_{W}(v^{*})<C_{V}(x^{\prime})\wedge C_{W}(v^{*}-x^{\prime}).$ ………(11)

If $x^{\prime}\in H$ then by $Lemma\;5.2$ (as $B=B^{*}\cup\{v^{*}\}$ is a M-basis for $W$), (11) becomes

$C_{V}(v^{*})\vee C_{W}(v^{*})<C_{V}(x^{\prime})\wedge C_{W}(v^{*})\wedge C_{W}(x^{\prime}).$

This is never true, and thus $x^{\prime}\in X\setminus H.$ But now since $C_{V}(v^{*})=C_{V}(x^{\prime})$ the inequality (11) never holds, and so,

$C_{V}(v^{*})\vee C_{W}(v^{*})=C_{V}(x^{\prime})\wedge C_{W}(v^{*}-x^{\prime})=C_{(V+W)}(v^{*}).$ ……..(12)

From equation (12) , we have $v^{*}\in A_{1}$ then $C_{V+W}(v^{*})=C_{W}(v^{*})$ and if $v^{*}\in A_{2}$ then $C_{V+W}(v^{*})=C_{V}(v^{*}).$

This completes the proof.

$\mathbf{Corollary\;6.6}$ If $V$ and $W$ are two multi vector spaces over $X$ such that the dimension of $X$ is finite and $C_{V}(\theta)\geq sup[C_{W}(X\setminus\{\theta\})]$ and $C_{W}(\theta)\geq sup[C_{V}(X\setminus\{\theta\})]$, then

$dim(V+W)=dim\;V+dim\;W-dim\;(V\cap W)$.

$\mathbf{Example\;6.7}$ Suppose $X=\mathbb{R}^{2},\;\omega=6.$ Define two multi vector spaces $V$ and $W$ with count functions $C_{V}$ and $C_{W}$ respectively as follows:

$C_{V}((0,0))=5$; $C_{V}((0,\mathbb{R}\setminus\{0\}))=3;$ $C_{V}(X\setminus\mathbb{R})=1$,

$C_{W}((0,0))=6;$ $C_{W}(\{(x,x):x\in\mathbb{R}\setminus\{0\}\})=2;$ $C_{W}(X\setminus\{(x,x):x\in\mathbb{R}\})=1.$

It is easily checked that $V$ and $W$ are multi vector spaces and $C_{V}(\theta)\geq$

$sup[C_{W}(X\setminus\{\theta\})]$ and $C_{W}(\theta)\geq sup[C_{V}(X\setminus\{\theta\})]$. It is also easy to check that

$C_{V\cap W}((0,0))=5$, $C_{V\cap W}(\{(x,x):x\in\mathbb{R}\setminus\{0\}\})=1$, $C_{V\cap W}(X\setminus\{(x,x):x\in\mathbb{R}\})=1,$ $C_{V+W}((0,0))=5$; $C_{V+W}((0,\mathbb{R}\setminus\{0\}))=3;$ $C_{V+W}(X\setminus(0,\mathbb{R}))=2$ and $B=\{(0,1),(1,1)\}$ is a M-basis for $V,$ $W,$ $V\cap W$ and $V+W.$ Thus

$dim\;(V+W)=3+2=5,$ $dim(V\cap W)=1+1=2,$

$dim\;V=3+1=4$, $dim\;W=2+1=3$.

So, $dim\;V+dim\;W-dim\;(V\cap W)=4+3-2=5=dim\;(V+W).$

$\mathbf{Definition\>6.8}$ Let $V$ be a multi vector space over $X$ and $f:X\rightarrow Y$ be a linear map. Then we define $f(V)$ as

$C_{f(V)}(x)=\begin{cases}sup\{C_{V}(z):z\in f^{-1}(x)\}&\>if\>f^{-1}(x)\neq\phi\\ 0&otherwise\end{cases}$

and $\tilde{kerf}=(kerf,C_{V}\mid_{kerf}),\>\tilde{imf}=(imf,C_{V}\mid_{imf})$.

$\mathbf{Proposition\>6.9}$ If $V$ be a multi vector space over $X$ where $dim\;X$ is finite and $f:X\rightarrow Y$ is a linear map, then

$dim(\tilde{kerf})+dim(\tilde{imf})=dim(V).$

$Proof.$ Suppose that $kerf\neq\{\theta\}.$ If $kerf=\{\theta\}$ then the proof is similar. Now let $B_{Kerf}$ be a M-basis for $\tilde{kerf}$ and $B_{Ex}$ be an extension of $B_{Ker}$ to a M-basis for $V$ (this is clearly possible by repeated application of $Lemma\>5.3$). Then $B_{Kerf}\cup B_{Ex}=B$ is M-basis for $V$ and $B_{Kerf}\cap B_{Ex}=\phi.$

We first show that $f(B_{Ex})=B_{Im}$ is a M-basis for $\tilde{imf}.$ Clearly $B_{Im}$ is a basis for $imf.$ Let $v_{1},v_{2},..,v_{k}\in B_{Ex}$ and $a_{1},...,a_{k}\in\mathbb{R}$ not all zero. By definition we have

$C_{f(V)}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))$

$=\begin{cases}sup\{C_{V}(x):x\in f^{-1}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))\}&if\>f^{-1}\left(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i})\right)\neq\phi\\ 0&otherwise\end{cases}$

Since $\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i})\in imf$ we have

$C_{f(V)}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))=sup\{C_{V}(x):x\in f^{-1}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))\}.$

By linearity of $f$ and by the property of $f^{-1}$ we get

$C_{f(V)}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))=sup\{C_{V}(x):x\in kerf+\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i}\}.$

If $x\in kerf$ then $x=\theta$ or $x=\stackrel{{\scriptstyle[}}{{i}}=1]{p}{\sum}b_{i}u_{i},u_{i}\in B_{kerf}$ where not all $b_{i}$ are zero; so if $x\in kerf+\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i}$ then either $C_{V}(x)=C_{V}(\theta+\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i})$ or $C_{V}(x)=C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{p}{\sum}b_{i}u_{i}+\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i})$ and thus

$C_{V}(x)=min\left(\stackrel{{\scriptstyle[}}{{i}}=1]{p}{\wedge}C_{V}(b_{i}u_{i}),\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\wedge}C_{V}(a_{i}v_{i})\right),$ [ As $u_{i}$and $v_{i}$ are M-basis element of $V$]

which is clearly smaller than or equal to $C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i}).$ Thus

$C_{f(V)}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))=sup\{C_{V}(x):x\in kerf+\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i}\}=C_{V}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}v_{i})=\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\wedge}C_{V}(a_{i}v_{i}).$

By the same argument we get that $C_{f(V)}(f(v_{i}))=C_{V}(v_{i}).$ Thus $C_{f(V)}(\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\sum}a_{i}f(v_{i}))=\stackrel{{\scriptstyle[}}{{i}}=1]{k}{\wedge}C_{f(V)}(a_{i}v_{i}).$

and therefore $B_{Im}$ is a M-basis for $\tilde{imf}.$

Now by definition of multi dimension we get

$dim(V)=\underset{v\in B_{Ker}\cup B_{Ex}}{\sum}C_{V}(v)=\underset{v\in B_{Ker}}{\sum}C_{V}(v)+\underset{v\in B_{Ex}}{\sum}C_{V}(v)$.

But by the above we have if z\in\text{\prec}B_{Ex}\text{\succ}, then $C_{f(V)}(f(z))=C_{V}(z),$ and thus

$dim(V)=\underset{v\in B_{Ker}}{\sum}C_{V}(v)+\underset{v\in B_{Ex}}{\sum}C_{f(V)}(f(v))$

$=\underset{v\in B_{Ker}}{\sum}C_{V}(v)+\underset{v\in B_{Im}}{\sum}C_{f(V)}(v)$

$=dim(\tilde{kerf})+dim(\tilde{imf}).$

7 Conclusion

There is a future scope of study of infinite dimensional multi vector space and behavior of linear operators in multi vector space context.