Linearly Independent Set is Basis iff of Same Cardinality as Dimension
Jump to navigation
Jump to search
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 $67 \ \text{(v)}$