Sufficient Conditions for Basis of Finite Dimensional Vector Space

Theorem
Let $$G$$ be an $n$-dimensional vector space.

Let $$B \subseteq G$$ such that $$\left|{B}\right| = n$$.

Then the following statements are equivalent:
 * $$(1) \quad B$$ is a basis of $$G$$.
 * $$(2) \quad B$$ is linearly independent.
 * $$(3) \quad B$$ is a generator for $$G$$.

Proof

 * Suppose $$B$$ is a basis of $$G$$, i.e. that condition $$(1)$$ holds.

Then conditions $$(2)$$ and $$(3)$$ follow directly by definition basis.


 * Suppose $$B$$ is linearly independent, i.e. that condition $$(2)$$ holds.

Suppose $$B$$ does not generate $$G$$.

Then, because $$\left|{B}\right| = n$$, by Linearly Independent Subset also Independent in Generated Subspace there would be a linearly independent subset of $$n + 1$$ vectors of $$G$$.

But this would contradict Linearly Independent Subset of Finitely Generated Vector Space is Finite.

Thus condition $$(2)$$ implies $$(1)$$.


 * Suppose $$B$$ is a generator for $$G$$, i.e. that condition $$(3)$$ holds.

By Linearly Independent Subset of Basis of Vector Space, $$B$$ contains a basis $$B'$$ of $$G$$.

But $$B'$$ has $$n$$ elements and hence $$B' = B$$ by Bases of Finitely Generated Vector Space.

Thus condition $$(3)$$ implies $$(1)$$.