Matroid Base Substitution From Fundamental Circuit
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Theorem
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $B$ be a base of $M$.
Let $y \in S \setminus B$.
Let $\map C {y,B}$ denote the fundamental circuit of $y$ in $B$.
Let $x \in B$.
Then:
- $\paren{B \setminus \set x} \cup \set y$ is a base of $M$ if and only if $x \in \map C {y,B}$
That is, $y$ can be substituted for $x$ in $B$ to form another base of $M$ if and only if $x \in \map C {y,B}$.
Proof
Necessary Condition
Let $\paren{B \setminus \set x} \cup \set y$ be a base of $M$.
By definition of the fundamental circuit:
- $\map C {y,B} \subseteq B \cup \set y$
and
- $\map C {y,B}$ is dependent
Aiming for a contradiction, suppose $x \notin \map C {y,B}$.
Then:
- $\map C {y,B} \subseteq \paren{B \setminus \set x} \cup \set y$
By matroid axiom $( \text I 2)$:
- $\map C {y,B}$ is independent
This contradicts the fact that:
- $\map C {y,B}$ is dependent
It follows that:
- $x \in \map C {y,B}$
$\Box$
Sufficient Condition
We prove the contrapositive statement.
Let $\paren{B \setminus \set x} \cup \set y$ not be a base of $M$.
We have:
\(\ds \card{\paren{B \setminus \set x} \cup \set y}\) | \(=\) | \(\ds \card{B \setminus \set x} + \card{\set y}\) | Corollary of Cardinality of Set Union | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren{\card B - \card{\set x} } + \card{\set y}\) | Cardinality of Set Difference | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren{\card B - 1 } + 1\) | Cardinality of Singleton | |||||||||||
\(\ds \) | \(=\) | \(\ds \card B\) |
From contrapositive statement of Independent Subset is Base if Cardinality Equals Cardinality of Base:
- $\paren{B \setminus \set x} \cup \set y$ is a dependent subset of $M$
From Dependent Subset Contains a Circuit there exists a circuit $C'$:
- $C' \subseteq \paren{B \setminus \set x} \cup \set y \subseteq B \cup \set y$
From Matroid Base Union External Element has Fundamental Circuit:
- $C' = \map C {y,B}$
Hence:
- $x \notin \map C {y, B}$
From Rule of Transposition we have:
- $x \in \map C {y, B} \implies \paren{B \setminus \set x} \cup \set y$ is a base of $M$
$\blacksquare$
Sources
- 1976: Dominic Welsh: Matroid Theory ... (previous) ... (next) Chapter $1.$ $\S 9.$ Circuits, Theorem $1$