# Cardinality of Linearly Independent Set is No Greater than Dimension

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## 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): $\S 27$: 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 $\text{A}.7$