Dimension of Proper Subspace is Less Than its Superspace

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It has been suggested that this page be renamed.
In particular: Proper subspace of finite space. If the supperspace is infinite dimensional, then it may $H \subsetneq G$ but $\map \dim G = \map \dim H = \infty$.
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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 $H = G$.

$\blacksquare$


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