Linearly Independent Subset also Independent in Generated Subspace

Theorem

Let $G$ be a finitely generated $K$-vector space.

Let $S$ be a linearly independent subset of $G$.

Let $M$ be the subspace of $G$ generated by $S$.

If $M \ne G$, then $\forall b \in G: b \notin M$, the set $S \cup \left\{{b}\right\}$ is linearly independent.

Proof

Suppose that:

$\displaystyle \sum_{k=1}^n \lambda_k x_k + \lambda b = 0$

where $\left \langle {x_n} \right \rangle$ is a sequence of distinct vectors of $S$.

If $\displaystyle \lambda \ne 0$, then $\displaystyle b = - \lambda^{-1} \left({\sum_{k=1}^n \lambda_k x_k}\right) \in M$ which contradicts the definition of $b$.

Hence $\lambda = 0$, and so:

$\displaystyle \sum_{k=1}^n \lambda_k x_k = 0$

Therefore $\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda = 0$ as $S$ is linearly independent.

$\blacksquare$

Sources

• Seth Warner: Modern Algebra (1965): $\S 27$: Theorem $27.11$
• For a video presentation of the contents of this page, visit the Khan Academy.