Basis of Vector Space is Linearly Independent and a Generator

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.

$\Box$

• 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.

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

$\Box$

• 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 $(1)$.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 27$: Theorem $27.12$
• For a video presentation of the contents of this page, visit the Khan Academy.