Lobachevskii Journal of Mathematics http://ljm.ksu.ru Vol. 26, 2007, 5–15

© H. Azanchiler

Abstract. In [1] the author extended $n$-line splitting from graphs to binary matroids and characterized the circuits of the result matroid, i.e. line-splitting matroid (es-splitting). In this paper, we characterize dependent, independent and base sets in line-splitting matroid $M_X^e$. Moreover, we determine rank function of $M_X^e$.

1. Introduction

Fleischner [2] introduced the idea of splitting a vertex of degree at least three in a connected graph and Raghunathan, Shikare and Waphare [4] extended the splitting operation from graphs to binary matroids. Shikare, Azadi and Waphare [6] further generalized this operation and also in [7] extended the $n$-point splitting operation from graphs to a binary matroid. Moreover, in [5] Shikare and Azadi determined the base of splitting matroids and the author in [1] extended the $n$-line splitting operation [8] from graphs to the binary matroids by the following way.

Let $M_X^e$ be the vector matroid of the matrix $A_X^e$. We say that $M_X^e$ has been obtained from $M$ by splitting $e$ and $X$ in $M$. The transition from $M$ to $M_X^e$ is called splitting of $M$ with respect to $e$ and $X$. For convenience, we say that $M_X^e$ is an element-set splitting (es-splitting) matroid.

The terminology from matroid theory which we use can be obtained from Oxely [3].

2. Independent Set in $M_X^e$

Next theorem characterize the dependent set in es-splitting matroid $M_X^e$.

Theorem 2.1. Let $M$ be a binary matroid on a set $S$ and $D$ be a dependent set in $M$. Then $D$ is dependent in $M_X^e$ if and only if $D$ does not contain precisely one circuit of $M$ containing an odd number of elements of $X$.

Proof. Let $D$ be a dependent set in $M$ and suppose $D$ does not contain precisely one circuit of $M$ containing an odd number of elements of $X$. Then we have the following two cases:

• $D$ contains a circuit $C$ of $M$ with even number of elements of $X$. Then $C$ is a circuit of $M_X^e$ and is contained in $D$. Therefore, $D$ is dependent set in $M_X^e$.
• $D$ contains at least two circuits, say $C$ and $C^{\prime }$, with odd number of elements of $X$. Then $$C\Delta C^{\prime }=C_1\cup C_2\cup \dots \cup C_m.$$

If for any of the $1\le i\le m$, $C_i$ contains an even number of elements of $X$, then it is a circuit in $M_X^e$ and is contained in $D$. Suppose there is no such $C_i$. Let $C_j$ and $C_k$ be two circuits each of which contains an odd number of elements of $X$. If $C_j\cup C_k$ contains a member of ${\mathcal{C}}_0$, say $C^{\prime \prime }$, then $C^{\prime \prime }\subseteq D$ and we are done. Otherwise $C_j\cup C_k$, or a minimal member of ${\mathcal{C}}_1$, contained in it, is a circuit of $M_X^e$ contained in $D$.

Conversely, let $D$ be a dependent set of $M$ which is also dependent in $M_X^e$. Since $D\subseteq S$, $a$ or $\gamma$ or both do not belong to $D$. Suppose $C$ is a circuit of $M_X^e$ contained in $D$. Then $C$ contains an even number of elements of $X$. We have two cases:

(i) $C$ is a circuit of $M$ containing an even number of elements of $X$.

(ii) $C$ is a disjoint union of two circuits of $M$ each of which contains an odd number of elements of $X$.

In both cases $D$ cannot contain precisely one circuit of $M$ containing an odd number of elements of $X$. $\square$

Lemma 2.2. Every independent set in $M$ is independent in $M_X^e$.

Remark 2.3. Converse of the lemma is not true. By Theorem 2.1, every circuit of $M$ containing an odd number of element of $X$ is a independent set of $M_X^e$.

The next theorem characterizes the independent sets of $M_X^e$.

Theorem 2.4. Let $I\subseteq S\cup \left\{a,\gamma \right\}$. Then $I$ is independent in $M_X^e$ if and only if one of the following conditions hold.

• $I=I_1\cup J$, where $I_1$ is an independent set in $M$ and $J\in \left\{\phi ,\left\{a\right\}\right\}$.
• $I=I_1\cup \left\{\gamma \right\}$, where $I_1$ is an independent set in $M$ and no circuit of $M$ is contained in $I_1\cup \left\{e\right\}$.
• $I=\left(I_1\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$, where $I_1$ is an independent set in $M$ containing $e$.
• $I=\left({\cup }_{i=1}^m{C}_i\right)\cup J$, where $J\in \left\{\phi ,\left\{\gamma \right\}\right\}$, each $C_i$ contains an odd number of element of $X,C_i\cap C_j\ne \phi ,i\ne j,i,j=1,2,\dots ,m$, $I$ contains no member of ${\mathcal{C}}_0$, and $I\cup \left\{e\right\}$ contains no circuit of $M$ other than $C_i$ for $i=1,2,\dots ,m$.

Proof. (1) Let $I_1$ be an independent set in $M$. By Lemma 2.2, $I_1$ is independent in $M_X^e$. Thus, $I=I_1\cup J$, where $J=\phi$, is independent in $M_X^e$. Further, by Definition 1.1 of $A_X^e,I_1\cup \left\{a\right\}$ is an independent set in $M_X^e$.

(2) We show that $I_1\cup \left\{\gamma \right\}$ is independent in $M_X^e$, where $I$ and $\gamma$ satisfy conditions in (2). On the contrary, suppose $I_1\cup \left\{\gamma \right\}$ is dependent in $M_X^e$ and let $C^{\prime }$ be a circuit of $M_X^e$ contained in $I_1\cup \left\{\gamma \right\}$. We have the following cases:

(i) $C^{\prime }\in {\mathcal{C}}_0$ or ${\mathcal{C}}_1$. Then $C^{\prime }\subseteq I_1$ and we know $C^{\prime }$ is a circuit or contains a circuit of $M$. This shows that $I_1$ is dependent in $M$, a contradiction.

(ii) $C^{\prime }\in {\mathcal{C}}_2$. Then $C^{\prime }=C\cup \left\{a\right\}$, where $C$ is a cocircuit of $M$. But then $C\subseteq I_1$, a contradiction.

(iii) $C^{\prime }\in {\mathcal{C}}_3$. Then $C^{\prime }=C_1\cup \left\{e,\gamma \right\}$ or $C^{\prime }=\left(C_2\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$ or $C^{\prime }=\left(C_3\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$, where $C_1,C_2,C_3$ are circuits in $M$ and $C_1,C_2$ and $C_3\setminus \left\{e\right\}$ each contain an odd number of elements of $X$. Consequently, $C_1\cup \left\{e,\gamma \right\}\subseteq I_1\cup \left\{\gamma \right\}$ implies that $C_1\subseteq I_1.\left(C_2\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}\subseteq I_1\cup \left\{\gamma \right\}$ implies $C_2\setminus \left\{e\right\}\subseteq I_1$; and $\left(C_3\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}\subseteq I_1\cup \left\{\gamma \right\}$ implies $C_3\setminus \left\{e\right\}\subseteq I_1$, contradictions to hypotheses in (2).

(3) Let $I_1$ be an independent set in $M$ containing $e$. We show that $\left(I_1\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is independent in $M_X^e$. On the contrary, suppose $\left(I_1\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is dependent in $M_X^e$. By similar argument as in (2), we get contradictions.

(4) Let $C_1,C_2,\dots$ and $C_m$ be circuits in $M$, where each $C_i$ contains an odd number of elements of $X$ and $C_i\cap C_j\ne \phi$ for $i\ne j,i,j=1,2,\dots ,m$. Clearly each $C_i$ is independent in $M_X^e$ and $I_1={\cup }_{i=1}^m{C}_i$ is independent in $M_X^e$. Further, $\gamma \notin {\cup }_{i=1}^m{C}_i$ and by hypothesis, $I=\left({\cup }_{i=1}^m{C}_i\right)\cup \left\{\gamma \right\}$ contains no circuit of ${\mathcal{C}}_0$ and $I\cup \left\{e\right\}$ contains no circuit of $M$ other than $C_i$ for $i=1,2,\dots ,m$. Therefore, $I$ is a independent set in $M_X^e$.

Conversely, let $I\subseteq S\cup \left\{a,\gamma \right\}$ be an independent set in $M_X^e$. We have the following cases:

(I) Let $I\cap \left\{a,\gamma \right\}=\phi$. Then $I\subseteq S$ and we have two subcases:

(i) $I$ be independent in $M$. Then $I_1=I$.

(ii) $I$ be dependent set in $M$. Let $C_1,C_2,\dots ,C_m$ be the circuits of $M$, contained in $I$. Then each $C_i$ must contain an odd number of elements of $X$ and $C_i\cap C_j\ne \phi$ for $i\ne j$. If $I-\left(C_1\cup C_2\cup \dots \cup C_m\right)=\phi$, then $I=C_1\cup C_2\cup \dots \cup C_m$ is independent in $M_X^e$ such that $I$ does not contain a member of ${\mathcal{C}}_0$ and $I\cup \left\{e\right\}$ does not contain any circuit of $M$ other than $C_i$ for $i=1,2,\dots ,m$. Thus, $I$ is of type (4), where $J=\phi$. If $I-\left(C_1\cup C_2\cup \dots \cup C_m\right)\ne \phi$ then $I=\left(C_1\cup C_2\cup \dots \cup C_m\right)\cup Y$, where $Y\subseteq S\cup \left\{a,\gamma \right\},Y\cap \left\{a,\gamma \right\}=\phi$, so $Y\subseteq S$. But $Y$ does not contain a circuit of $M$ with even number of elements of $X$, and also $Y$ does not contain any $C_i$ for $i=1,2,\dots ,m$, thus $Y=\phi$; a contradiction.

(II) Suppose $I\cap \left\{a,\gamma \right\}\ne \phi$. We have the following cases:

(i) $a\in I$ and $\gamma \notin I$. Then $I-\left\{a\right\}$ is independent in $M$, if $I-\left\{a\right\}=I_1$, then $I=I_1\cup \left\{a\right\}$.

(ii) Let $a\notin I$ and $\gamma \in I$. We show that $I-\left\{\gamma \right\}$ is independent in $M$. On the contrary, suppose $I-\left\{\gamma \right\}$ contains a circuit say $C$ of $M$. Then $C\subseteq \left(I-\left\{\gamma \right\}\right)\cup \left\{e\right\}$ and $\left(C\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}\subseteq I$. But $\left(C\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$ is a circuit of $M_X^e$ contained in $I$, a contradiction. So, if $I-\left\{\gamma \right\}=I_1$ then $I=I_1\cup \left\{\gamma \right\}$. Now, suppose $I-\left\{\gamma \right\}$ contains more than one circuit of $M$, say $C_1,C_2,\dots ,C_m$, where each $C_i$ contains an odd number of elements of $X$ and $C_i\cap C_j\ne \phi$ for $i\ne j,i,j=1,2,\dots ,m$. Thus $C_1\cup C_2\cup \dots \cup C_m\subseteq I-\left\{\gamma \right\}$ and ${\cup }_{i=1}^m{C}_i\subseteq \left(I-\left\{\gamma \right\}\right)\cup \left\{e\right\}$. Consequently, $\left(\left({\cup }_{i=1}^m{C}_i\right)\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}\subseteq I$. For $1\le j\le m,\left(C_j\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$ is a circuit of $M_X^e$, contained in $\left(\left({\cup }_{i=1}^m{C}_i\right)\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$, that is, in $I$; a contradiction. So $I=\left({\cup }_{i=1}^m{C}_i\right)\cup \left\{\gamma \right\}$ is a independent set in $M_X^e$. Moreover, $I\cup \left\{e\right\}$ contains no circuit of $M$ other than $C_i,i=1,2,\dots ,m$. Thus $I$ is of type (4), where $j=\left\{\gamma \right\}$.

(iii) Let $a,\gamma \in I$. Then we show that $I-\left\{a,\gamma \right\}$ is independent in $M$. On the contrary, suppose $I-\left\{a,\gamma \right\}$ contains a circuit say $C$ of $M$. Thus $C\subseteq \left(I-\left\{a,\gamma \right\}\right)\cup \left\{e\right\}$ and $\left(C\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}\subseteq I$. But $\left(C\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is a circuit of $M_X^e$, where $C$ contains an odd number of elements of $X$. If $C$ contains an even number of elements of $X$, then $C\subseteq I$; a contradiction. We conclude that $I_1=\left(I\mid \left\{a,\gamma \right\}\right)\cup \left\{e\right\}$ is independent in $M$ and $I=\left(I_1\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$. $I$ is of type (3). This completes the proof of the theorem. $\square$

3. Bases in $M_X^e$

In the next theorem, we characterize the bases of the matroid $M_X^e$ in terms of the bases of $M$.

Theorem 3.1. Let $\mathcal{ℬ}$ be a collection of bases of $M$. A subset $B^{\prime }$ of $S\cup \left\{a,\gamma \right\}$ is a base of $M_X^e$ if and only if one of the following conditions hold:

• $B^{\prime }=B\cup \left\{a\right\}$
• $B^{\prime }=B\cup \left\{\gamma \right\}$, where $B\in \mathcal{ℬ}$ and no circuit $C$ of $M$ containing $e$ contains an odd number of elements of $X$ such that $C\setminus \left\{e\right\}\subseteq B$.
• $B^{\prime }=\left(B\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$, where $B\in \mathcal{ℬ}$, no circuit $C$ of $M$ containing $e$ contains an odd number of elements of $X$ such that $C\setminus \left\{e\right\}\subseteq B$.
• $B^{\prime }=B\cup \left\{z\right\}$, where $B\in \mathcal{ℬ}$, $z\in S-B$ and the fundamental circuit of $M$ contained in $B\cup \left\{z\right\}$ contains an odd number of elements of $X$.

Proof. (1) Let $B$ be a base of $M$. Then $B$ is independent in $M$ and, by Lemma 2.2, $B$ is independent in $M_X^e$. Further, by Theorem 2.4, $B\cup \left\{a\right\}$ is independent in $M_X^e$. Then

Thus, $B^{\prime }=B\cup \left\{a\right\}$ is a base of $M_X^e$.

(2) Let $B\cup \left\{\gamma \right\}$ satisfies the conditions in (2). We show that $B\cup \left\{\gamma \right\}$ is independent in $M_X^e$. On the contrary, suppose $B\cup \left\{\gamma \right\}$ is dependent in $M_X^e$ and $C^{\prime }$ is a circuit of $M_X^e$ contained in $B\cup \left\{\gamma \right\}$. If $C^{\prime }\in {\mathcal{C}}_0$ or ${\mathcal{C}}_1$, then $C^{\prime }\subseteq B$ and this leads to a contradiction. If $C^{\prime }=C\cup \left\{a\right\}$, where $C$ is a circuit in $M$ and $a\notin C$, then $C\cup \left\{a\right\}\subseteq B\cup \left\{\gamma \right\}$ implies $C\cup \left\{a\right\}\subseteq B$ and again $C\subseteq B$; a contradiction. If $C^{\prime }\in {\mathcal{C}}_2$, then $C^{\prime }=C_1\cup \left\{e,\gamma \right\},C^{\prime }=\left(C_2\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$ or $C^{\prime }=\left(C_3\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$. If $C_1\cup \left\{e,\gamma \right\}\subseteq B\cup \left\{\gamma \right\}$, then $C_1\cup \left\{e\right\}\subseteq B$, that is, $C_1\subseteq B$. If $\left(C_2\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}\subseteq B\cup \left\{\gamma \right\}$, then $C_2\setminus \left\{e\right\}\subseteq B$, that is, $C_2\subseteq B$. If $\left(C_3\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}\subseteq B\cup \left\{\gamma \right\}$, then $\left(C_3\setminus \left\{e\right\}\right)\cup \left\{a\right\}\subseteq B$ or $C_3\subseteq \left\{e\right\}$, contradictions to hypothesis in (2). Further,

This shows that $B\cup \left\{\gamma \right\}$ is a base of $M_X^e$.

(3) Let $\left(B\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ satisfies the conditions in (3). By the argument similar to one as given above, we show that it is a independent subset of $M_X^e$. Moreover,

Thus, $\left(B\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is a base of $M_X^e$.

(4) Let $B\cup \left\{z\right\}$, where $z\in S-B$, satisfies the condition given in (4). By Theorem 4.2.7, $B\cup \left\{z\right\}$ is independent in $M_X^e$ and so

Therefore $B\cup \left\{z\right\}$ is a base for $M_X^e$.

Conversely, let $B^{\prime }$ be a base for $M_X^e$. We consider the following cases:

(I) Let $a\in B^{\prime }$ and $\gamma \notin B^{\prime }$. Then $B^{\prime }-\left\{a\right\}$ is independent in $M_X^e$. We show that $B^{\prime }-\left\{a\right\}$ is also independent in $M$. On the contrary, suppose $B^{\prime }-\left\{a\right\}$ contains a circuit $C$ of $M$. We have two subcases.

• $C$ contains an even number of elements of $X$. Then $C\subseteq B^{\prime }$; a contradiction.
• $C$ contains an odd number of elements of $X$. Then $C\subseteq B^{\prime }-\left\{a\right\}$ and $C\cup \left\{a\right\}\subseteq B^{\prime }$; a contradiction, because $C\cup \left\{a\right\}$ is a circuit of $M_X^e$. Next, $$r\left(B^{\prime }-\left\{a\right\}\right)=\mid B^{\prime }-\left\{a\right\}\mid =\mid B^{\prime }\mid -1=r^{\prime }\left(M_X^e\right)-1=r\left(M\right).$$

  Therefore, $B^{\prime }-\left\{a\right\}$ is a base for $M$.

(II) Let $a\notin B^{\prime }$ and $\gamma \in B^{\prime }$. We show that $B^{\prime }-\left\{\gamma \right\}$ is a base for $M$. Firstly, we prove that $B^{\prime }-\left\{\gamma \right\}$ is independent in $M$. On the contrary, suppose $B^{\prime }-\left\{\gamma \right\}$ is dependent in $M$. Let $C$ be a circuit of $M$, contained in $B^{\prime }-\left\{\gamma \right\}$. We have two subcases:

• Let $C$ contains an even number of elements of $X$. Then $C$ is a circuit of $M_X^e$ and $C\subseteq B^{\prime }-\left\{\gamma \right\}$. But $C\subseteq B^{\prime }$, is a contradiction.
• Let $C$ contains an odd number of elements of $X$. Then $C\subseteq B^{\prime }-\left\{\gamma \right\}$ and so $C\setminus \left\{e\right\}\subseteq B^{\prime }-\left\{\gamma \right\}$. This implies that $\left(C\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}\subseteq B^{\prime }$ which is a contradiction, since $\left(C\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$ is a circuit of $M_X^e$. Secondly, $B^{\prime }-\left\{\gamma \right\}$ is maximal independent in $M$, follows from the fact that $r\left(B^{\prime }-\left\{\gamma \right\}\right)=r\left(M\right)$.

(III) Let $a,\gamma \in B^{\prime }$. We show that $\left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}$ is a base for $M$. Clearly $e\notin B^{\prime }$, for $e\in B^{\prime }$ implies that $\left\{e,a,\gamma \right\}\subseteq B^{\prime }$ and this is a contradiction.

Firstly, we show that $\left(B^{\prime }-\left\{a,\gamma \right\}\right)\cup \left\{e\right\}$ is independent in $M$. On the contrary, suppose it is dependent in $M$ and let $C$ be a circuit of $M$ contained in $\left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}$. We have two subcases:

• $C$ contains an even number of elements of $X$. Then $C$ is a circuit of $M_X^e$ and $C\subseteq \left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}$. Thus $\left(C\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}\subseteq B^{\prime }$; a contradiction because $\left(C\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is a circuit of $M_X^e$.
• $C$ contains an odd number of elements of $X$ and $C\subseteq \left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}$. Hence $\left(C\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}\subseteq B^{\prime }$. But $\left(C\setminus \left\{e\right\}\right)\cup \left\{\gamma \right\}$ is a circuit of $M_X^e$, so we get a contradiction. Further, $\left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}$ is maximal independent in $M$, since $$\begin{array}{rcll}r\left(\left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}\right)& =& \mid \left(B^{\prime }\setminus \left\{a,\gamma \right\}\right)\cup \left\{e\right\}\mid & \\ & =& \mid B^{\prime }\mid -1=r^{\prime }\left(M_X^e\right)-1=r\left(M\right).& \end{array}$$

(IV) Let $a,\gamma \notin B^{\prime }$. Then $B^{\prime }\subseteq S$ and $B^{\prime }$ is not independent in $M$ since $r^{\prime }\left(B^{\prime }\right)=r^{\prime }\left(M_X^e\right)=r\left(M\right)+1$. Thus $B^{\prime }$ is dependent in $M$. So there is a circuit $C$ of $M$ contained in $B^{\prime }$. If $C$ contains an even number of elements of $X$, then $C$ is a circuit of $M_X^e$ and we get a contradiction. So $C$ must contain an odd number of elements of $X$ and suppose $x_i$ be an element of $X$ contained in $C$. Then $B=B^{\prime }-\left\{x_i\right\}$ is a base of $M$. $\square$

4. Rank Function of $M_X^e$

Lemma 4.1. Let $M$ be a binary matroid on $S$ and $M_X^e$ be a es-splitting of $M$ with ground set $S\cup \left\{a,\gamma \right\}$. Let $r$ and $r^{\prime }$ be the rank functions of $M$ and $M_X^e$, respectively. Then for $Z\subseteq S$ the following properties hold:

(1) $r^{\prime }\left(Z\cup \left\{a\right\}\right)=r\left(Z\right)+1$

(2) $r^{\prime }\left(Z\cup \left\{a,\gamma \right\}\right)=\left\{\begin{array}{ccc}r\left(Z\right)+1\hfill & \text{if}\hfill & e\in Z\hfill \\ r\left(Z\right)+2\hfill & \text{if}\hfill & e\notin Z\hfill \end{array}\right.$

Proof. (1) Let $T$ be a base for $A$ in $M$. Then $r\left(Z\right)=\mid T\mid$. We show that $T\cup \left\{a\right\}$ is a base for $Z\cup \left\{a\right\}$ in $M_X^e$. On the contrary, suppose $T\cup \left\{a\right\}$ is dependent in $M_X^e$ and $C$ is circuit of $M_X^e$ contained in $T\cup \left\{a\right\}$. We consider the following cases:

(i) $C\in {\mathcal{C}}_0$ or ${\mathcal{C}}_1$. Then $C\subseteq T\cup \left\{a\right\}$ and hence $C\subseteq T$; a contradiction.

(ii) $C\in {\mathcal{C}}_2$. Then $C=C_1\cup \left\{a\right\}$, where $C_1$ is a circuit in $M$. Consequently $C_1\subseteq T$ gives a contradiction.

(iii) Let $C\in {\mathcal{C}}_3$. Then there is a circuit of $M$, say $C_1$ with $C_1\subseteq T$, a contradiction.

Now, we prove that $T\cup \left\{a\right\}$ is a maximal independent set in $M_X^e$. On the contrary, suppose $T\cup \left\{a\right\}\cup \left\{z\right\}$ is maximal independent in $M_X^e$, where $z\in Z-\left(T\cup \left\{a\right\}\right)$. Then $T\cup \left\{z\right\}\cup \left\{a\right\}\subseteq Z\cup \left\{a\right\}$. Thus $T\cup \left\{z\right\}\subseteq Z$, a contradiction. Now,

By the same argument as above, we can show that

(2) Let $T$ be a base for $Z$ in $M$. Then $r\left(Z\right)=\mid T\mid$. We have the following two cases:

(I) Let $e\in T$. Then we claim that $\left(T\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is a base for $Z\cup \left\{a,\gamma \right\}$ in $M_X^e$. On the contrary, suppose that it is dependent set of $M_X^e$ and contains a circuit $C$ of $M_X^e$. We have the following subcases:

(i) Let $C\in {\mathcal{C}}_0$ or ${\mathcal{C}}_1$. Then $C\subseteq \left(T\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ implies that $C\subseteq T\setminus \left\{e\right\}\subseteq T$; a contradiction.

(ii) Let $C\in {\mathcal{C}}_2$ and $C=C_1\cup \left\{a\right\}$, where $C_1$ is a circuit of $M$. Then $C_1\cup \left\{a\right\}\subseteq \left(T\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ and $C_1\subseteq T\setminus \left\{e\right\}\subseteq T$; a contradiction.

(iii) Let $C\in {\mathcal{C}}_3$ and $C=\left(C_1\setminus e\right)\cup \left\{a,\gamma \right\},e\in C_1$, where $C_1$ is a circuit in $M$. Clearly $C_1\subseteq T$ again; a contradiction. Now, we show that $\left(T\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is maximal independent in $M_X^e$. On the contrary, suppose $\left(T\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}\cup \left\{z\right\}$ for $z\in Z$ is maximal. Then $\left(T\setminus \left\{e\right\}\right)\cup \left\{z\right\}\subseteq Z\setminus \left\{e\right\}\subseteq T\cup \left\{z\right\}\subseteq Z$. Consequently, $T\cup \left\{z\right\}\subseteq Z$; a contradiction. If $e\in T$, then $\left(T\setminus \left\{e\right\}\right)\cup \left\{a,\gamma \right\}$ is a base for $Z\cup \left\{a,\gamma \right\}$ in $M_X^e$ and hence

(II) Let $e\notin T$. Then we show that $T\cup \left\{a,\gamma \right\}$ is a base for $Z\cup \left\{a,\gamma \right\}$ in $M_X^e$. We prove that $T\cup \left\{a,\gamma \right\}$ is maximal independent in $M_X^e$. On the contrary, suppose it is dependent in $M_X^e$. By the same argument as in case (I), we obtain a contradiction. Thus $T\cup \left\{a,\gamma \right\}$ is base of $Z\cup \left\{a,\gamma \right\}$. Finally,

This completes the proof. $\square$

Lemma 4.2. If $Z\subseteq S$, then

Proof. Let $Z\subseteq S$. We have the following cases:

(1) Suppose $Z$ does not contain any circuit of $M$. Then $Z$ is independent in $M$, and by Lemma 2.2, it is independent in $M_X^e$ and hence $r^{\prime }\left(Z\right)=\mid Z\mid =r\left(Z\right)$.

(2) Suppose $Z$ does not contain a circuit of $M$ containing an odd number of elements of $X$. Suppose $Z$ contains a circuit say $C$, containing an even number of elements of $X$. Then $C$ is a circuit in $M_X^e$ and $Z$ is dependent in $M$, as well as in $M_X^e$. Consequently, a base of $Z$ in $M$ is also a base of it in $M_X^e$. Thus, $r^{\prime }\left(Z\right)=r\left(Z\right)$.

(3) Let $Z$ contains a circuit of $M$, say $C$ containing an odd number of elements of $X$. For $\alpha \in C$, the set $C-\left\{\alpha \right\}$ is independent in $M\mid Z$. Now, we extend $C-\left\{\alpha \right\}$ to a base $T$ of $Z$. Let $T=\left(C-\left\{\alpha \right\}\right)\cup \left\{{\beta }_1,{\beta }_2,\dots ,{\beta }_k\right\}$, where ${\beta }_i\in Z,i=1,2,\dots ,k$, is a base of $Z$ in $M$. Then

On the other hand, $C$ is independent in $M_X^e$ and $T^{\prime }=C\cup \left\{{\beta }_1,{\beta }_2,\dots {\beta }_k\right\}$ is independent in $M_X^e$. If $T^{\prime }$ is not a base of $Z$ in $M_X^e$, then $T^{\prime \prime }=C\cup \left\{{\beta }_1,{\beta }_2,\dots ,{\beta }_k,\delta \right\}$ for some $\delta \in Z$ is independent subset of $Z$ in $M_X^e$. Now $C\cup \left\{{\beta }_1,{\beta }_2,\dots ,{\beta }_k\right\}$ is a dependent subset of $Z$ in $M$. Let $C^{\prime }$ be a circuit of $M$ contained in it. We have the following cases:

(i) $C^{\prime }$ contains an even number of elements of $X$. Then $C^{\prime }$ is a circuit of $M_X^e$ with $C^{\prime }\subseteq C\cup \left\{{\beta }_1,{\beta }_2,\dots ,{\beta }_k\right\}$ a, contradiction.

(ii) $C^{\prime }$ contains an odd number of element of $X$. Consider

where $C_i^{\prime }$ are circuits of $M$ and $C_i^{\prime }\cap C_j^{\prime }=\phi ,i\ne j$ and $i,j=1,2,\dots ,m$. If some $C_i^{\prime }$ contains an even number of elements of $X$, then it leads to a contradiction. If not, consider the circuits $C_j^{\prime },C_k^{\prime }$ from (**). Then either $C_j^{\prime }\cup C_k^{\prime }$ is a circuit or a subset of it, is a circuit of $M_X^e$ contained in $T^{\prime \prime }$. We conclude that $T^{\prime }$ must be a maximal independent subset of $Z$ in $M_X^e$. Now,

From $\left(\ast \right)$ and $\left(\ast \ast \ast \right)$, we deduce that $r^{\prime }\left(Z\right)=r\left(Z\right)+1$. This completes the proof. $\square$

Corollary 4.3. Let $M=\left(S,r\right)$ and $M_X^e=\left(S\cup \left\{a,\gamma \right\},r^{\prime }\right)$ be matroids with usual meaning. Let $J=\left\{a,\gamma \right\}$. If $Y\subseteq S\cup \left\{a,\gamma \right\}$, then

The proof follows from Lemmas 4.1 and 4.2.

Corollary 4.4.4. Let $M=\left(S,r\right)$ be a binary matroid. Let $r_1,r_2,r_3$ be the rank functions of the matroids $M_X,M_X^{\prime }$ and $M_X^e$, respectively. Then $r_3\left(Y\right)=r_2\left(Y\right)=r_1\left(Y\right)$ for $Y\subseteq S$.

Proof. It is known [3] that for a matroid $M$ on $S$ with $T\subseteq S$ and $X\subseteq S-T$,

where $r_M$ is a rank function of $M$. We have

Thus from (*), it follows that, $r_3\left(Y\right)=r_2\left(Y\right)$ and $r_2\left(Y\right)=r_1\left(Y\right)$ for $Y\subseteq S$. $\square$

References

Department of Mathematics, University of Urmia, Urmia - 57135 (Iran)

E-mail address: azanchiler@yahoo.com

Received March 28, 2007