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If the incidence vectors of are linearly independent over then
Statement
Let be a field and let be a family of subsets of . If the incidence vectors are linearly independent in , then .
Facts & Assumptions
Given: a field , a natural number , and a family of subsets of whose incidence vectors are linearly independent in .
The standard basis of has vectors and spans (The standard list with and for is an ordered basis of ; hence , and is the zero space with basis and dimension ).
Every linearly independent subset of a vector space has size at most that of any finite spanning set (If has a spanning set with elements, then every linearly independent subset of is finite with at most elements; in particular has no linearly independent subset equinumerous with ).
Distinct subsets have distinct incidence vectors (The incidence vector of a subset over a stated field).
Proof
By [F1], the space is spanned by a finite set of vectors.
The linearly independent set therefore has at most elements by [F2].
Since [F3] identifies distinct subsets with distinct incidence vectors, the family itself has at most members.
Remarks
- This is the master lemma for the direct incidence-vector bounds, including Oddtown and Fisher's inequality. Other bounds on the page use subspace counts, shifting, or polynomial-function spaces instead.
Depends on
- The incidence vector $v_A\in F^{n}$ of a subset $A\subseteq[n]$ over a stated field
- A finite family of subsets of $[n]$ and its incidence matrix over $F$
- Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i<n} \lambda_i v_i = 0_V$ forces every $\lambda_i = 0_F$, and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent
- Finite-dimensional vector space, and its dimension $\dim_F V$; infinite-dimensional means having no finite basis
- The standard list $e : n \to F^{n}$ with $e_i(i) = 1_F$ and $e_i(j) = 0_F$ for $j \ne i$ is an ordered basis of $F^{n}$; hence $\dim_F F^{n} = n$, and $F^{0}$ is the zero space with basis $\varnothing$ and dimension $0$
- If $V$ has a spanning set with $n$ elements, then every linearly independent subset of $V$ is finite with at most $n$ elements; in particular $V$ has no linearly independent subset equinumerous with $\mathbb{N}$
- If $\dim_F V = n$ and $U$ is a linear subspace of $V$, then $U$ is finite-dimensional, $\dim_F U \le n$, and $\dim_F U = n$ if and only if $U = V$
- Vector space over a field
Used by
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Sources
- L. Babai and P. Frankl, Linear Algebra Methods in Combinatorics, Chapter 1 (standard reference, not scraped)