Bases of Vector Space have Equal Cardinality

Theorem
Let $R$ be a division ring.

Let $V$ be a vector space over $R$.

Let $X$ and $Y$ be bases of $V$.

Then $X$ and $Y$ are equinumerous.

Proof
We will first prove that there is an injection from $X$ to $Y$.

Let $x \in X$.

By Expression of Vector as Linear Combination from Basis is Unique: General Result, there is a unique finite subset $C_x$ of $R \times Y$ such that:


 * $\displaystyle x = \sum_{\tuple {r, v} \mathop \in C_x} r \cdot v$ and
 * $\forall \tuple {r, v} \in C_x: r \ne 0_R$

Define $\Phi: X \to \powerset Y$ by:


 * $\map \Phi x := \Img {C_x}$

We next show that $\sequence {\map \Phi x}_{x \mathop \in X}$ satisfies the marriage condition.

That is, for every finite subset $F$ of $X$:


 * $\displaystyle \card F \le \card {\bigcup \map \Phi F}$

Since $X$ is a basis, it is linearly independent.

By Subset of Linearly Independent Set is Linearly Independent, $F$ is also linearly independent.

By the definition of $\Phi$:
 * $\displaystyle F \subseteq \operatorname {span} \bigcup \map \Phi F$.

By Finite Union of Finite Sets is Finite, $\displaystyle \bigcup \map \Phi F$ is finite.

Thus by Size of Linearly Independent Subset is at Most Size of Finite Generator:


 * $\displaystyle \size F \le \size {\bigcup \map \Phi F}$

By Hall's Marriage Theorem, there is an injection from $X$ into $Y$.

Precisely the same argument with $X$ and $Y$ interchanged shows that there is an injection from $Y$ into $X$ as well.

Thus by the Cantor-Bernstein-Schröder Theorem, $X$ and $Y$ are equinumerous.

Also known as
Some authors refer to this as the dimension theorem for vector spaces, but it should not be confused with e.g. the Rank Plus Nullity Theorem.

Also see

 * Bases of Finitely Generated Vector Space have Equal Cardinality