# Linearly Independent Set is Contained in some Basis/Proof 2

## 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

Let $H = \set {\xi_1, \xi_2, \ldots, \xi r}$.

Consider the basis $B = \set {\alpha_1, \alpha_2, \ldots, \alpha_n}$ of $E$.

Consider the set $G = H \cup B = \set {\xi_1, \xi_2, \ldots, \xi r, \alpha_1, \alpha_2, \ldots, \alpha_n}$.

We have that $G$ is a generator of $E$.

As $B$ is a basis, it follows that each of $H$ is a linear combination of $B$.

Thus $G = H \cup B$ is linearly dependent.

Thus one of the elements $\alpha_i$ of $B$ is a linear combination of the preceding elements of $G$.

Eliminate this one, and do the same thing with with $G \setminus \set {\alpha_i}$.

Eventually there will exist a set which is a basis of $E$ containing all the elements of $H$.

$\blacksquare$

## Sources

- 1969: C.R.J. Clapham:
*Introduction to Abstract Algebra*... (previous) ... (next): Chapter $7$: Vector Spaces: $\S 34$. Dimension: Theorem $67 \ \text{(i)}$