Dimension of a Vector Space upper bounds Linearly Independent Set sizee

Last updated Nov 1, 2022

Let $V$ be a Vector Space with dimension $\dim V = n$ for some $n \in \bar{\mathbb{N}}$. Then if $S \subset V$ s.t. $|S| > n$, $S$ is Linearly Dependent.

Suppose $\dim V = n$ for $n \in \mathbb{N}$. Then there exists a Vector Space Basis $\mathbf{a}{1}, \dots, \mathbf{a}{n} \in V$. Since a Vector Space Basis is a spanning set, because Spanning Set Size bound Linearly Independent Set Size, any Linearly Independent Set must have size less than or equal to $n$. But $|S| > n$, so it cannot be Linearly Independent. Therefore $S$ is Linearly Dependent. $\checkmark$

This is vacuously true for $\dim V = \infty$ since there does not exist $S \subset V$ such that $|S| > \infty$ (every Set is either infinite or finite). $\checkmark$ $\blacksquare$

• Set Cardinality