Sufficient Conditions for Basis of Finite Dimensional Vector Space
Theorem
Let $K$ be a division ring.
Let $n \ge 0$ be a natural number.
Let $E$ be an $n$-dimensional vector space over $K$.
Let $B \subseteq E$ be a subset such that $\card B = n$.
The following statements are equivalent:
- $(1): \quad$ $B$ is a basis of $E$.
- $(2): \quad$ $B$ is linearly independent.
- $(3): \quad$ $B$ is a generator for $E$.
Proof
1 implies 2 and 3
Let $B$ be a basis of $E$.
Then conditions $(2)$ and $(3)$ follow directly by the definition of basis.
$\Box$
2 implies 1
Let $B$ be linearly independent.
Suppose $B$ does not generate $E$.
Then, because $\card B = n$, by Linearly Independent Subset also Independent in Generated Subspace there would be a linearly independent subset of $n + 1$ vectors of $E$.
But this would contradict Size of Linearly Independent Subset is at Most Size of Finite Generator.
Thus condition $(2)$ implies $(1)$.
$\Box$
3 implies 1
Let $B$ be a generator for $E$.
By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, $B$ contains a basis $B'$ of $E$.
But $B'$ has $n$ elements and hence $B' = B$ by Bases of Finitely Generated Vector Space have Equal Cardinality.
Thus condition $(3)$ implies $(1)$.
$\blacksquare$
Examples
Example: $2$ Dimensions
Let $V$ be a vector space of $2$ dimensions.
Let $\mathbf u, \mathbf v \in V$ be vectors
Let $\mathbf u$ and $\mathbf v$ be such that neither $\mathbf u$ nor $\mathbf v$ is a scalar multiple of the other..
Let $\mathbf w \in V$.
Then there exist scalars such that:
- $\mathbf w = a \mathbf u + b \mathbf v$
Sources
- 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.12$
- 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.7$
- For a video presentation of the contents of this page, visit the Khan Academy.