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 $S \cup \set b$ is linearly independent.

Proof

Suppose that:

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

where $\sequence {x_n}$ is a sequence of distinct vectors of $S$.

If $\lambda \ne 0$, then $\ds b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of $b$.

Hence $\lambda = 0$, and so:

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

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

$\blacksquare$

Sources

• 1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.11$

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