All Finite Bases for a Vector Space are the same Size

Last updated Nov 1, 2022

Let $V$ be a Finite-Dimensional Vector Space and let distinct $\mathbf{a}{1}, \dots, \mathbf{a}{n} \in V$ for some $n \in \mathbb{N}$ be a Vector Space Basis for $V$. Then, if $S \subset V$ is a Vector Space Basis for $V$, $|S| = n$.

Since $\mathbf{a}{1}, \dots, \mathbf{a}{n}$ is a Vector Space Basis for $V$, their Subspace Span is $V$. Since $S$ is another Vector Space Basis for $V$, $S$ is Linearly Independent. Because Spanning Set Size bound Linearly Independent Set Size, $|S| \leq n$. But by the same argument $|S| \geq n$. Therefore $|S| = n$. $\blacksquare$

• Natural Numbers