Γ-Extension of Binary Matroids

Abstract

We extend the notion of a point-addition operation from graphs to binary matroids. This operation can be expressed in terms of element-addition operation and splitting operation. We consider a special case of this construction and study its properties. We call the resulting matroid of this special case a Γ-extension of the given matroid. We characterize circuits and bases of the resulting matroids and explore the effect of this operation on the connectivity of the matroids.

1. Introduction

Slater [1] defined few operations for graphs which preserve connectedness of graphs. One such operation is a point-addition (vertex-addition) operation. This operation is defined in the following way. Let 𝐺 be a graph and 𝑉(𝐺) be the set of vertices of 𝐺. Let 𝐻 be the graph obtained from 𝐺 by adding a new vertex 𝑣 adjacent to vertices 𝑣₁,𝑣₂,…,𝑣_𝑛 of 𝐺. The graph 𝐻 is said to be obtained from 𝐺 by point-addition operation. Letting 𝑋={𝑣₁,𝑣₂,…,𝑣_𝑛}, for convenience, we denote the graph 𝐻 by 𝐺^𝑋. Thus, 𝑉(𝐺^𝑋)=𝑉(𝐺)∪{𝑣} and 𝐸(𝐺^𝑋)=𝐸(𝐺)∪{𝑣𝑣₁,𝑣𝑣₂,…,𝑣𝑣_𝑛}.

Point-addition operation has several applications in graph theory. For example, Slater classified 𝑛-connected graphs using point-addition operation along with some other operations [2].

If |𝑉(𝐺)|=𝑛, then the new vertex 𝑣 can be joined to at most 𝑛 vertices of the graph. That means, we can add at most 𝑛 edges in the original graph.

Definition 1.1. Let 𝑀=𝑀[𝐴] be a binary matroid of rank 𝑟 on a set 𝑆. Let 𝐴^ be the matrix obtained from 𝐴 by the following way.(1)Adjoin 𝑘 columns to 𝐴(𝑘≤𝑟) with labels say 𝛾₁,𝛾₂,…,𝛾_𝑘. Let the resulting matrix be denoted by 𝐴^.(2)Adjoin a new row to 𝐴^ with entries zero except in the columns corresponding to 𝛾₁,𝛾₂,…,𝛾_𝑘, where it takes the value 1.

Let 𝑀^ be the vector matroid of the matrix 𝐴^. We say that 𝑀^ is obtained from 𝑀 by the point-addition operation. We call the matroid 𝑀^ point-addition matroid or Γ-extension of 𝑀. Let us denote by Γ, the set of columns 𝛾₁,𝛾₂,…,𝛾_𝑘 which are adjoined to 𝐴 in the first step. That is, Γ={𝛾₁,𝛾₂,…,𝛾_𝑘}. Then, second step consists of splitting the matroid 𝑀[𝐴^] with respect to the set Γ (see [3, 4]).

In fact, the matroid 𝑀^ is obtained by elements addition and generalized splitting operation [5]. As an immediate consequence of the definition, we have the following result.

Let 𝑣_𝑖 and 𝑣_𝑗 be two vertices of 𝐺. Then, the addition of an edge 𝑣_𝑖𝑣_𝑗, results in the smallest supergraph of 𝐺 containing edge 𝑣_𝑖𝑣_𝑗.

Proposition 1.2. Let 𝑀=𝑀(𝐺) be a cycle matroid of rank 𝑟. Let 𝐺^ be the graph obtained from 𝐺 by adding adjacent edges 𝛾₁,𝛾₂,…,𝛾_𝑛(𝑛≤𝑟) to 𝐺. Let Γ={𝛾₁,𝛾₂,…,𝛾_𝑛}. Then, the point-addition matroid 𝑀^ is graphic and 𝑀^=𝑀(𝐺^Γ).

Proof. Let 𝐴 be representation matrix of 𝑀 over 𝐺𝐹(2). Let the matrix 𝐴^ be obtained from 𝐴 by adding column vectors say, 𝛾₁,𝛾₂,…,𝛾_𝑛. Suppose that 𝐴^ is obtained from 𝐴^ by adding a new row where entries are zero, except in the columns corresponding to 𝛾₁,𝛾₂,…,𝛾_𝑛, where it takes the value 1. Thus 𝑀^=𝑀[𝐴^] is a binary matroid with ground set 𝐸(𝐺)∪Γ. Since 𝛾₁,𝛾₂,…,𝛾_𝑛 are adjacent edges in 𝐺^, the splitting of 𝑀(𝐺^) with respect to Γ is graphic (see [5]), and we have 𝑀(𝐺^)_Γ=𝑀(𝐺^Γ), where 𝐺^Γ is the graph obtained from 𝐺 by splitting operation with respect to Γ. It follows that 𝑀^=[𝑀(𝐺)]^=𝑀(𝐺^Γ).

We assume that the reader is familiar with elementary notions in matroid theory, including minors, binary, and connectivity. For an excellent introduction to the subject, read Oxley [6].

2. Γ-Extension of a Binary Matroid

If a matroid 𝑀 is obtained from a matroid 𝑁 by adding a nonempty subset 𝑇 of 𝑆(𝑁), then 𝑁 is called an extension of 𝑀. In particular, if |𝑇|=1, then 𝑁 is a single-element extension of 𝑀 (see [6]). Another term, that is sometimes used instead of single-element extension, is addition (see [7]).

Now we consider a special case of the operation that is introduced in the first section.

Definition 2.1. Let 𝑀=𝑀[𝐴] be a binary matroid of rank 𝑟 on a set 𝑆, and let 𝐴=[𝐼_𝑟∣𝐽] be the standard representation of 𝑀 over 𝐺𝐹(2). Let 𝐵 be a base of 𝑀, and let 𝑋={𝑒_𝑖1,𝑒_𝑖2,…,𝑒_𝑖𝑚} be a subset of 𝐵. We obtain the matrix 𝐴^𝑋 by the following way.(1)Obtain a matrix 𝐴₁ from 𝐴 by adjoining 𝑚(𝑚≤𝑟) columns say 𝛾_𝑖1,𝛾_𝑖2,…,𝛾_𝑖𝑚 to 𝐴, parallel to 𝑒_𝑖1,𝑒_𝑖2,…,𝑒_𝑖𝑚, respectively.(2)Split the matrix 𝐴₁ with respect to the set Γ, where Γ={𝛾_𝑖1,𝛾_𝑖2,…,𝛾_𝑖𝑚}. Denote the resulting matrix by 𝐴^𝑋.

Let 𝑀^𝑋 be the vector matroid of the matrix 𝐴^𝑋. We say that 𝑀^𝑋 is Γ-extension of 𝑀. Note that 𝑀^𝑋 is a binary matroid with ground set 𝑆∪Γ, where 𝑆∩Γ=𝜙, and |𝑋|=|Γ|. The transition from 𝑀 to 𝑀^𝑋 is called Γ-extension operation on 𝑀. In particular, if |𝑋|=ℓ, it is called ℓ-Γ-extension operation, and, for |𝑋|=1, we call it single-Γ-extension operation.

The next example illustrates this construction for the dual of Fano matroid.

Example 2.2. Let 𝑀=𝐹^∗7 be the dual of the Fano matroid 𝐹₇, and let 𝑆={1,2,3,4,5,6,7} be the ground set of 𝑀. The matrix 𝐴 that represents 𝑀 over 𝐺𝐹(2) is given by. 𝐴=⎡⎢⎢⎣11000201003001040001510116110170111⎤⎥⎥⎦.(2.1)

Consider the set 𝑋={1,3,4} contained in the base of 𝑀. Then, the corresponding matrix 𝐴^𝑋 is given by 1234567𝛾₁𝛾₃𝛾₄𝐴^𝑋=⎡⎢⎢⎢⎣10000010000010000010101101101001110100010010100011⎤⎥⎥⎥⎦.(2.2)

The vector matroid of 𝐴^𝑋 is the matroid (𝐹^∗7)^𝑋.

Corollary 2.3. Let 𝑀=𝑀[𝐴] be a binary matroid on 𝑆. Let 𝑋 be a subset of a base of 𝑀, and 𝑀^𝑋 be the Γ-extension of 𝑀 on the set 𝑆∩Γ. Then, 𝑀^𝑋⧵Γ=𝑀, that is, 𝑀^𝑋 is an extension of 𝑀.

Corollary 2.4. Let 𝑟 and 𝑟^ be the rank functions of the matroids 𝑀 and 𝑀^𝑋, respectively. Then 𝑟^(𝑀^𝑋)=𝑟(𝑀)+1.

With the help of Lemma 2.5, we characterize the circuits of the matroid 𝑀^𝑋.

Lemma 2.5. (1) Every circuit of 𝑀 is a circuit of 𝑀^𝑋.
(2) Every circuit of 𝑀^𝑋 contains at least one element of 𝑆.
(3) Every circuit of 𝑀^𝑋 contains even number of elements of Γ.

The proof follows from the construction of the matrix 𝐴^𝑋.

Remark 2.6. Let 𝑀^𝑋 be a single-Γ-extension of 𝑀 (i.e., |𝑋|=1). Then every circuit of 𝑀^𝑋 is a circuit of 𝑀 and vice versa. In fact, the added element 𝛾 is a coloop in the resulting matroid.

Theorem 2.7. Let 𝑀 be a binary matroid on 𝑆 with representation matrix 𝐴=[𝐼_𝑟∣𝐸] over 𝐺𝐹(2) and 𝑋 be a subset of a base of 𝑀. Then, a subset 𝑍 of 𝑆∪Γ is a circuit of 𝑀^𝑋 if and only if one of the following conditions hold:(1)𝑍={𝑒_𝑖,𝑒_𝑗,𝛾_𝑖,𝛾_𝑗}, where 𝑖≠𝑗,𝛾_𝑖,𝛾_𝑗∈Γ and 𝑒_𝑖,𝑒_𝑗∈𝑋 for 1≤𝑖,𝑗≤𝑟,(2)𝑍=𝐽∪𝐷, where 𝐽⊆Γ, |𝐽| is an even integer and 𝜙≠𝐷⊆𝑆 is such that 𝐷∪𝑋_𝐽 is a circuit in 𝑀, where 𝑋_𝐽={𝑒_𝑖∈𝑋∣𝛾_𝑖∈𝐽}.

Proof. If 𝑍={𝑒_𝑖,𝑒_𝑗,𝛾_𝑖,𝛾_𝑗}, then, by Definition 2.1 of 𝐴^𝑋, 𝑍 is a circuit of 𝑀^𝑋. Now, let 𝑍=𝐽∪𝐷 be as stated in (2). If |𝐽|=0, then 𝐽=𝜙,𝑋_𝐽=𝜙, and 𝑍=𝐷 is a circuit of 𝑀. Suppose that 𝐽⊆Γ,𝐽≠𝜙, and 𝑋_𝐽∪𝐷 is a circuit of 𝑀. Then clearly, 𝑍=𝐷∪𝐽 is a circuit of 𝑀^𝑋.
Conversely, let 𝑍⊆𝑆∪Γ be a circuit of 𝑀^𝑋, we have two cases:(I)𝑍∩Γ=𝜙. Then 𝐽=𝜙,𝑋_𝐽=𝜙. Thus, 𝑍=𝐷 is a circuit in 𝑀 and the condition (2) in the result holds.(II)Let 𝑍∩Γ≠𝜙, and suppose that 𝑍∩Γ=𝐽. We have two subcases:(i)𝑍∩𝑋≠𝜙. Then, 𝑍∩𝑋={𝑒_𝑖,𝑒_𝑗}=𝐷=𝑋_𝐽 and 𝐽={𝛾_𝑖,𝛾_𝑗}. Thus, 𝑍={𝑒_𝑖,𝑒_𝑗,𝛾_𝑖,𝛾_𝑗} and condition (1) in the result holds.(ii)𝑍∩𝑋=𝜙. Take 𝑍∩𝑆=𝐷. Then 𝐷∪𝑋_𝐽 is a circuit of 𝑀 and 𝐷∪𝐽 is a circuit of 𝑀^𝑋. Thus, 𝑍=𝐷∪𝐽, and the condition (2) in the result holds.

We characterize the independent sets of 𝑀^𝑋 in terms of independent sets of 𝑀. Firstly, we have the following lemma.

Lemma 2.8. (1) Every independent set of 𝑀 is independent in 𝑀^𝑋.
(2) Every subset of Γ is independent in 𝑀^𝑋.

The proof is straightforward.

Remark 2.9. Let 𝑀^𝑋 be a single-Γ-extension of 𝑀. Then, every independent set in 𝑀^𝑋 is also independent in 𝑀 and vice versa.

Theorem 2.10. Let 𝑀 be a binary matroid on 𝑆 and 𝑀^𝑋 be the Γ-extension matroid of 𝑀 with respect to 𝑋. Let ℐ be a collection of independent sets of 𝑀. Then, a subset 𝐼 of 𝑆∪Γ is an independent set of 𝑀^𝑋 if and only if one of the following conditions hold:(1)𝐼=𝐼₁∪{𝛾}, where 𝛾∈Γ and 𝐼₁∈ℐ.(2)𝐼=𝐼₁∪𝐽, where 𝐽⊆Γ, 𝐼₁∈ℐ and 𝐼₁∪𝑋_𝐽 contains no circuit of 𝑀.

Proof. If 𝐼₁∈ℐ, then clearly 𝐼₁∪{𝛾} for 𝛾∈Γ is an independent set in 𝑀^𝑋. Now, suppose that 𝐼₁∪𝑋_𝐽 contains no circuit of 𝑀. On the contrary, suppose that 𝐼₁∪𝐽 contains a circuit say 𝐶^ of 𝑀^𝑋. Then 𝐶^⧵𝐽⊆𝐼₁ and (𝐶^⧵𝐽)∪𝑋_𝐽⊆𝐼₁∪𝑋_𝐽. But (𝐶^⧵𝐽)∪𝑋_𝐽 is a circuit of 𝑀 and is contained in 𝐼₁∪𝑋_𝐽, a contradiction.
Conversely, let 𝐼 be an independent set in 𝑀^𝑋 and 𝐼⊆𝑆∪Γ. We have two cases.(I)Let 𝐼∩Γ=𝜙. Then 𝐼⊆𝑆 and 𝐼 is independent in 𝑀.(II)Let 𝐼∩Γ≠𝜙, and let 𝐼∩Γ=𝐽. Then 𝐽⊆Γ and 𝐽⊆𝐼.
We prove that 𝐼−𝐽⊆𝑆 is an independent set in 𝑀. On the contrary, suppose that 𝐼−𝐽 contains a circuit of 𝑀, say 𝐶, then 𝐶⊆𝐼−𝐽 and 𝐶⊆𝐼 gives a contradiction.
Letting 𝐼−𝐽=𝐼₁, we have 𝐼=𝐼₁∪𝐽. We claim that 𝐼₁∪𝑋_𝐽 does not contain any circuit of 𝑀. If 𝐼₁∪𝑋_𝐽 contains a circuit of 𝑀, say 𝐶^, then 𝐶⧵𝑋_𝐽⊆𝐼₁. Further, (𝐶⧵𝑋_𝐽)∪𝐽⊆𝐼₁∪𝐽, and thus (𝐶⧵𝑋_𝐽)∪𝐽⊆𝐼 leads to a contradiction as (𝐶⧵𝑋_𝐽)∪𝐽 is a circuit of 𝑀^𝑋. This completes the proof of the theorem.

Corollary 2.11. Let ℐ and ℐ^𝑋 denote the collection of independent sets of 𝑀 and 𝑀^𝑋, respectively. If |𝑋|=1, then ℐ^𝑋={𝐼∪{𝛾}∣𝐼∈ℐ,𝛾∈Γ}.

Corollary 2.12. A subset 𝐼 of 𝑆 is independent in 𝑀 if and only if 𝐼∪{𝛾} for 𝛾∈Γ is independent in 𝑀^𝑋.

Corollary 2.13. Let 𝑟 and 𝑟^ be the rank functions of 𝑀 and 𝑀^𝑋, respectively. Then 𝑟^(𝑍∪{𝛾})=𝑟(𝑍)+1 for 𝑍⊆𝑆.

In the next theorem, we characterize the bases of the matroid 𝑀^𝑋 in terms of the bases of 𝑀.

Lemma 2.14. Let 𝑍⊆𝑆, then 𝑍 is an independent set in 𝑀^𝑋 if and only if 𝑍 is an independent set in 𝑀.

The proof is straightforward.

Corollary 2.15. Let 𝑍 be a subset of 𝑆. Then 𝑟^(𝑍)=𝑟(𝑍) where 𝑟^ and 𝑟 are rank functions of 𝑀^𝑋 and 𝑀, respectively.

Theorem 2.16. A subset 𝐵^ of 𝑆∪Γ is a base for 𝑀^𝑋 if and only if 𝐵^=(𝐵⧵𝐷)∪𝐽, where 𝜙≠𝐽⊆Γ,𝐷⊂𝐵,|𝐷|=|𝐽|−1 and (𝐵⧵𝐷)∪𝑋_𝐽 contains no circuit of 𝑀.

Proof. Let 𝐵 be a base for 𝑀. Then 𝐵 is an independent set in 𝑀, and so 𝐵⧵𝐷 is independent in 𝑀. Let 𝐵⧵𝐷=𝐼₁. Then, by Theorem 2.10, 𝐼₁∪𝐽, where 𝐽⊆Γ,|𝐷|=|𝐽|−1 and (𝐵⧵𝐷)∪𝑋_𝐽 contains no circuit of 𝑀 and hence is independent in 𝑀^𝑋. Moreover, 𝑟^((𝐵⧵𝐷)∪𝐽)=|(𝐵⧵𝐷)∪𝐽|=|𝐵|−|𝐷|+|𝐽|=|𝐵|+1=𝑟^(𝑀^𝑋).
We conclude that 𝐵^=(𝐵⧵𝐷)∪𝐽 is a base for 𝑀^𝑋.
Conversely, let 𝐵^⊆𝑆∪Γ be a base for 𝑀^𝑋. Firstly, we show that 𝐵^∩Γ≠𝜙. On the contrary, suppose that 𝐵^∩Γ=𝜙. Then 𝐵^⊆𝑆 and is independent in 𝑀^𝑋. So by Lemma 2.14, 𝐵^ is independent in 𝑀. Also by Corollary 2.15, 𝑟^𝐵^=𝑟(𝐵)=||𝐵^||=𝑟(𝑀)+1.(2.3) This shows that 𝐵^⊆𝑆 and 𝑟(𝐵^)>𝑟(𝑀); a contradiction.
Now, let 𝐵′∩Γ=𝐽. Then 𝐵^−𝐽⊆𝑆 is independent in 𝑀^𝑋 as well as in 𝑀. It can be extended to form the base of 𝑀. Let 𝐷⊆𝑆−(𝐵^−𝐽) be such that (𝐵^−𝐽)∪𝐷=𝐵 is a base for 𝑀. Then 𝐵′=(𝐵⧵𝐷)∪𝐽. We claim that |𝐷|=|𝐽|−1. Now, ||𝐵||=||𝐵^−𝐽∪𝐷||=||𝐵^||−||𝐽||+||𝐷||=𝑟(𝑀)+1−||𝐽||+||𝐷||.(2.4) Since |𝐵|=𝑟(𝑀), we conclude that 1−|𝐽|+|𝐷|=0, that is, |𝐷|=|𝐽|−1. Finally, we show that (𝐵⧵𝐷)∪𝑋_𝐽 contains no circuit of 𝑀. On the contrary, suppose that (𝐵⧵𝐷)∪𝑋_𝐽 contains a circuit, say 𝐶 of 𝑀. Then 𝐶⧵𝑋_𝐽⊆𝐵⧵𝐷 and (𝐶⧵𝑋_𝐽)∪𝐽⊆(𝐵⧵𝐷)∪𝐽. Thus, (𝐶⧵𝑋_𝐽)∪𝐽⊆𝐵^ leads to a contradiction. This completes the proof of theorem.

Corollary 2.17. Every base of 𝑀^𝑋 contains at least one element of Γ.

3. Connectivity of 𝑀^𝑋

Let 𝑀 be a binary matroid on a set 𝑆 and 𝐴 be the representation matrix of 𝑀 over 𝐺𝐹(2). If 𝑀 is bridgeless, then Γ-extension of 𝑀 with respect to a singleton subset 𝑋 of 𝐵 yields a disconnected matroid.

Lemma 3.1. Let 𝑒_𝑖∈𝑆 be a coloop in a matroid 𝑀 and 𝑒_𝑖∉𝑋. Then 𝑒_𝑖 is a coloop in 𝑀^𝑋.

The proof is straightforward.

Corollary 3.2. Suppose that no element of 𝑆−𝑋 is a coloop of 𝑀. Then 𝑀^𝑋 has no coloops.

Theorem 3.3. Let |𝑋|≥2. If 𝑀 is connected matroid, then, so is 𝑀^𝑋.

Proof. Assume that 𝑀 is connected. We show that for every pair of elements 𝑢,𝑣∈𝑆∪Γ there is a circuit of 𝑀^𝑋 containing 𝑢 and 𝑣. We have three cases.(1)Let 𝑢,𝑣∈𝑆. By hypothesis, 𝑀 is connected. So there is a circuit of 𝑀 say 𝐶, containing 𝑢 and 𝑣. Since 𝐶 is a circuit in 𝑀^𝑋, we are through.(2)Let 𝑢,𝑣∈Γ and let 𝑢=𝛾_𝑖 and 𝑣=𝛾_𝑗. Then the 4-circuit {𝑒_𝑖,𝑒_𝑗,𝛾_𝑖,𝛾_𝑗} in 𝑀^𝑋 contains 𝑢 and 𝑣.(3)Let 𝑢∈𝑆 and 𝑣∈Γ. By assumption |𝑋|≥2. So there is an element say 𝑤∈Γ. Let 𝐶₁ be a circuit of 𝑀^𝑋 containing 𝑣 and 𝑤. By Lemma 2.5, 𝐶₁ contains at least one element of 𝑆, say 𝑒. Now 𝑢,𝑒∈𝑆, and 𝑀 is connected, so there is a circuit of 𝑀, say 𝐶₂ which contains 𝑢 and 𝑒. Thus 𝐶₁∩𝐶₂≠𝜙 and 𝑢∈𝐶₁,𝑣∈𝐶₂. Then there is a circuit in 𝑀^𝑋, say 𝐶₃, such that 𝐶₃⊆𝐶₁∪𝐶₂ and 𝑢,𝑣∈𝐶₃. This completes the proof of the theorem.

Remark 3.4. Converse of the above theorem is not true.

Theorem 3.5. Let 𝑀^𝑋 be a 3-Γ-extension matroid of 𝑀 and Γ={𝛾_𝑖,𝛾_𝑗,𝛾_𝑘}. If 𝑀 is a 3-connected matroid on 𝑆, then 𝑀^𝑋 is 3-connected.

Proof. On the contrary, suppose that 𝑀^𝑋 is not 3-connected, then 𝑀^𝑋 has a 1-separated or 2-separated partition. Let (𝐴,𝐵) be a 2-separated partition of 𝑆∪Γ. That means, min{|𝐴|,|𝐵|}≥2 and 𝑟^(𝐴)+𝑟^(𝐵)−𝑟^𝑀^𝑋≤1.(∗) We consider three cases.(i)Let 𝐴=Γ and 𝐵=𝑆. By Lemma 2.8, 𝐴 is independent in 𝑀^𝑋, so 𝑟^(𝐴)=||𝐴||=||Γ||=3.(3.1) Also, by Lemma 2.14, 𝑟^(𝐵)=𝑟^(𝑆)=𝑟(𝑆)=𝑟(𝑀).(3.2) Thus, 𝑟^(𝐴)+𝑟^(𝐵)−𝑟^𝑀^𝑋=3+𝑟(𝑀)−𝑟(𝑀)−1=2.(3.3) This is a contradiction to (*).(ii)Let 𝐴={𝛾_𝑖,𝛾_𝑗} and 𝐵=𝑆∪{𝛾_𝑘}.By Lemma 2.14, 𝑟^(𝐵)=𝑟(𝑀)+1. So 𝑟^(𝐴)+𝑟^(𝐵)−𝑟^𝑀^𝑋=2+𝑟(𝑀)+1−𝑟(𝑀)−1=2(3.4)gives a contradiction to (*).(iii)Let 𝐴=𝑆₁∪{𝛾_𝑖} and 𝐵=𝑆₂∪{𝛾_𝑗,𝛾_𝑘}, where 𝑆₁,𝑆₂⊆𝑆, and 𝑆₁≠𝜙,𝑆₂≠𝜙. Then min{|𝑆₁|,|𝑆₂|}≥1 and 𝑟(𝑆₁)+𝑟(𝑆₂)−𝑟(𝑀)≤𝑟^(𝐴)−1+𝑟^(𝐵)−1−𝑟^(𝑀^𝑋)+1≤0. Moreover, 𝑟𝑆₁+𝑟𝑆₂≥𝑟(𝑀).(3.5) Thus, 𝑟(𝑆₁)+𝑟(𝑆₂)−𝑟(𝑀)=0, and we conclude that (𝑆₁,𝑆₂) is a 1-separated partition for 𝑀.This is a contradiction to the fact that 𝑀 is 3-connected. By the same argument, we can show that 𝑀^𝑋 does not have 1-separated partition.

In the last theorem, the condition that |𝑋|=3 is necessary. Consider the following example.

Example 3.6. 𝑀(𝐾₄) is a 3-connected matroid. Let a representation matrix of 𝑀(𝐾₄) be 𝐴=110020103001411051016011.(3.6) Let 𝑋⊆𝐵 and 𝑋={1,2},Γ={𝛾_𝑖,𝛾₂}. Then 123456𝛾₁𝛾₂𝐴^𝑋=⎡⎢⎢⎣10000100001011001010011010010101⎤⎥⎥⎦.(3.7) By row operations on 𝐴^𝑋, we can show that 𝐴^𝑋=⎡⎢⎢⎢⎣𝐼₄||||||1100101001101101⎤⎥⎥⎥⎦.(3.8) By [], 𝑀^𝑋=𝑀[𝐴^𝑋] is not 3-connected.

If |𝑋|=1, then 𝑀^𝑋 has a coloop, and it is not 3-connected.

In general, we state the following result whose proof is immediate.

Corollary 3.7. Let 𝑀 be a 𝑛-connected binary matroid and |𝑋|<𝑛. Then 𝑀^𝑋 is not 𝑛-connected.