Cardinality of Generator of Vector Space is not Less than Dimension/Proof 2

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Theorem

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

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


Then:

$\map {\dim_F} V \le m$

where $\map {\dim_F} V$ is the dimension of $V$.


Proof

From Generator of Vector Space Contains Basis there exists a basis $B$ of $E$ such that $B \subseteq G$.

From Cardinality of Basis of Vector Space, $\card B = n$.

The result follows.

$\blacksquare$


Sources