Cardinality of Generator of Vector Space is not Less than Dimension

Theorem
Let $V$ be a vector space over a field $F$.

Let $\mathcal B$ be a generator for $V$ containing $m$ elements.

Then:
 * $\dim_F \left({V}\right) \le m$

where $\dim_F \left({V}\right)$ is the dimension of $V$.

Proof
Let $\mathcal B = \left\{{x_1, x_2, \ldots, x_m}\right\}$ be a generator for $G$.

Let $\left\{{y_1, y_2, \ldots, y_n}\right\}$ be a subset of $G$ such that $n > m$.

As $\mathcal B$ generates $G$, there exist $\alpha_{ij} \in F$ where $1 \le i \le m, 1 \le j \le n$ such that:
 * $\displaystyle \forall j: 1 \le j \le n: y_j = \sum_{i \mathop = 1}^m \alpha_{ij} x_i$

Now let $\beta_1, \ldots, \beta_n$ be elements of $F$.

Then:
 * $\displaystyle \sum_{j \mathop = 1}^n \beta_j y_j = \sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n \left({\alpha_{ij} \beta_j}\right) x_i$

As $n > m$ the result Homogeneous Linear Equations with More Unknowns than Equations can be applied.

That is, there exist $\beta_1, \ldots, \beta_n \in F$ which are not all zero such that:
 * $\displaystyle \forall i: 1 \le i \le m: y_j = \sum_{j \mathop = 1}^n \alpha_{ij} \beta_j = 0$

That is, such that:
 * $\displaystyle \sum_{j \mathop = 1}^n \beta_j y_j = 0$

So $\left\{{y_1, y_2, \ldots, y_n}\right\}$ is linearly dependent.

The result follows by definition of dimension.