Linearly Independent Set is Basis iff of Same Cardinality as Dimension

Theorem

Let $E$ be a vector space of $n$ dimensions.

Let $H$ be a linearly independent subset of $E$.

$H$ is a basis for $E$ if and only if it contains exactly $n$ elements.

Proof

By hypothesis, let $H$ be a linearly independent subset of $E$

Necessary Condition

Let $H$ be a basis for $E$.

By definition of dimension of vector space, a basis for $E$ contains exactly $n$ elements.

By Bases of Finitely Generated Vector Space have Equal Cardinality, it follows that $H$ also contains exactly $n$ elements.

$\Box$

Sufficient Condition

Let $H$ contain exactly $n$ elements.

By Sufficient Conditions for Basis of Finite Dimensional Vector Space $H$ is itself a basis for $E$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$
• 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 34$. Dimension: Theorem $67 \ \text{(v)}$