Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set

Theorem
Let $K$ be a field.

Let $X$ be a vector space over $K$.

Let $L$ be a linearly independent subset of $X$ such that:


 * $U = \map \span L \ne X$

Let:


 * $x \in X \setminus U$

Then $L \cup \set x$ is linearly independent.

Proof
Take $x_1, \ldots, x_n \in L$ and take $\alpha_1, \ldots, \alpha_n, \alpha_{n + 1} \in K$ such that:


 * $\ds \alpha_{n + 1} x + \sum_{k \mathop = 1}^n \alpha_i x_i = 0$

If $\alpha_{n + 1} = 0$, then we have:


 * $\ds \sum_{k \mathop = 1}^n \alpha_i x_i = 0$

and so $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$ from the linearly independence of $U$.

Suppose that $\alpha_{n + 1} \ne 0$, then we would have:


 * $\ds x = -\alpha^{-1}_{n + 1} \sum_{k \mathop = 1}^n \alpha_i x_i$

This would mean that $x \in U$, (from Linear Span is Linear Subspace) contradicting that $x \in X \setminus U$.

So we have $\alpha_{n + 1} = 0$, and hence $\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$.

So $L \cup \set x$ is linearly independent.