Cardinality of Linearly Independent Set is No Greater than Dimension

Theorem

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

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

Then:

$H$ has at most $n$ elements.

Proof

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

By definition of dimension of vector space, $E$ has a basis with exactly $n$ elements.

By Sufficient Conditions for Basis of Finite Dimensional Vector Space, $B$ is a generator for $E$.

Then by Size of Linearly Independent Subset is at Most Size of Finite Generator, $H$ has at most $n$ elements.

$\blacksquare$

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 $67 \ \text{(ii)}$
• 1996: John F. Humphreys: A Course in Group Theory ... (previous) ... (next): $\text{A}.2$: Linear algebra and determinants: Theorem $\text{A}.7$