Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set
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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.
$\blacksquare$