Corollary

Let $K$ be a commutative field and $M$ a vector space over $K$. For every element $z \in \Lambda(M)$, there exists a smallest vector subspace $N$ of $M$ such that $z \in \Lambda(N)$ and $N$ is finite-dimensional.

The proof is deduced from that of § 5, no. 2, Corollary to Proposition 4 replacing $T$ by $\Lambda$ throughout.

$N$ is called the vector subspace of $M$ associated with the element $z$ of $\Lambda(M)$.