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 $(1)$.