Linearly Independent Set is Contained in some Basis/Finite Dimensional Case/Proof 1

Theorem

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

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

There exists a basis $B$ for $E$ such that $H \subseteq B$.

Proof

By hypothesis there is a basis $B$ of $E$ with $n$ elements.

Then $H \cup B$ is a generator for $E$.

So by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set there exists a basis $C$ of $E$ such that $H \subseteq C \subseteq H \cup B$.

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$