Sufficient Conditions for Basis of Finite Dimensional Vector Space

Theorem
Let $K$ be a division ring.

Let $n\geq0$ be a natural number.

Let $G$ be an $n$-dimensional vector space over $K$.

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


 * $(1): \quad$ $B$ is a basis of $G$.
 * $(2): \quad$ $B$ is linearly independent.
 * $(3): \quad$ $B$ is a generator for $G$.

1 implies 2 and 3
Let $B$ be a basis of $G$.

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

2 implies 1
Let $B$ be linearly independent.

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 Size of Linearly Independent Subset is at Most Size of Finite Generator.

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

3 implies 1
Let $B$ be a generator for $G$.

By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, $B$ contains a basis $B\,'$ of $G$.

But $B\,'$ has $n$ elements and hence $B\,' = B$ by Bases of Finitely Generated Vector Space have Equal Cardinality.

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