Rank of Matroid Circuit is One Less Than Cardinality
Jump to navigation
Jump to search
This article needs proofreading. Please check it for mathematical errors. If you believe there are none, please remove {{Proofread}} from the code.To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Proofread}} from the code. |
Theorem
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $C \subseteq S$ be a circuit of $M$.
Let $\rho: \powerset S \to \Z$ denote the rank function of $M$.
Then:
- $\map \rho C = \card C -1$
Proof
By definition of a circuit:
- $C$ is dependent
By matroid axiom $(\text I 1)$:
- $C \ne \O$
Let $x \in C$.
Lemma
- $C \setminus \set x$ is a maximal independent subset of $C$
We have:
\(\ds \map \rho C\) | \(=\) | \(\ds \card{C \setminus \set x}\) | Cardinality of Maximal Independent Subset Equals Rank of Set | |||||||||||
\(\ds \) | \(=\) | \(\ds \card C - \card{\set x}\) | Cardinality of Set Difference with Subset | |||||||||||
\(\ds \) | \(=\) | \(\ds \card C - 1\) | Cardinality of Singleton |
$\blacksquare$
Sources
- 1976: Dominic Welsh: Matroid Theory ... (next) Chapter $1.$ $\S 9.$ Circuits