# Dimension of Proper Subspace is Less Than its Superspace

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## Theorem

Let $G$ be a vector space whose dimension is $n$.

Let $H$ be a subspace of $G$.

Then $H$ is finite dimensional, and $\map \dim H \le \map \dim G$.

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If $H$ is a proper subspace of $G$, then $\map \dim H < \map \dim G$.

## Proof

Let $H$ be a subspace of $G$.

Every linearly independent subset of the vector space $H$ is a linearly independent subset of the vector space $G$.

Therefore, it has no more than $n$ elements by Size of Linearly Independent Subset is at Most Size of Finite Generator.

So the set of all natural numbers $k$ such that $H$ has a linearly independent subset of $k$ vectors has a largest element $m$, and $m \le n$.

Now, let $B$ be a linearly independent subset of $H$ having $m$ vectors.

If the subspace generated by $B$ were not $H$, then $H$ would contain a linearly independent subset of $m + 1$ vectors.

This follows by Linearly Independent Subset also Independent in Generated Subspace.

This would contradict the definition of $m$.

Hence $B$ is a generator for $H$ and is thus a basis for $H$.

Thus $H$ is finite dimensional and $\map \dim H \le \map \dim G$.

Now, if $\map \dim H = \map \dim G$, then a basis of $H$ is a basis of $G$ by Sufficient Conditions for Basis of Finite Dimensional Vector Space, and therefore $H = G$.

$\blacksquare$

## Sources

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- 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.13$