Show that if the set of linearly independent vectors ${X}$ does not span $\mathbb R^n$ then there exists a vector $v \in \mathbb R^n$ s.t. $\{v\} \cup X$ is linearly independent.
My shot at a proof for this by contraposition, let me know when I mess up:
let $X \cup \{v\}$ be linearly dependent $\forall v \in \mathbb R^n$, then
$$\exists \alpha_1 , \dots, \alpha_k\text{ s.t. }v=\alpha_1x_1 +\dots + \alpha_kx_k \forall v \in \mathbb R^n$$
thus by definition of span, $X$ spans $\mathbb R^n$
this is a contradiction of the original proposal, thus $\exists v \in R^n$ s.t. $v \cup X$ is linearly independent
Best Answer
It is mostly correct, but incomplete.
You don't state what your $v$ is in the proof. You need to pick $v$ to be an element of $\mathbb R^n$ which is not in the span of $X$. (Thus, you are using one condition.)
You should also state why $\exists \alpha_1,\dots,\alpha_k.$
If $X\cup\{v\}$ is linearly dependent, then $\exists \beta_0,\dots,\beta_k$ such that $$0=\beta_0 v + \beta_1 x_1+\dots+\beta_k x_k$$ with the $\beta_i$ not all zero.
But if $\beta_0=0$ then $X$ would be linearly dependent. So $\beta_0\neq 0$ and from there you can conclude that $\alpha_i=\frac{-\beta_i}{\beta_0}$ gives you:
$$v=\alpha_1 x_1+\cdots+\alpha_kx_k$$
showing $v\in \operatorname{Span}(X),$ which is a contradiction.
Indeed, this could be stated as a general lemma:
This lemma is exactly why you can find $\alpha_1,\dots,\alpha_k.$ Since $X$ is linearly independent, we have, by this lemma, if $X\cup \{v\}$ is linearly dependent, then $v\in\operatorname{Span}(X).$
Aside: This lemma assumes the definition that $\operatorname{Span}(\emptyset)=\{0\}.$
The question has two conditions:
Without $(2),$ you cannot find $v.$
Without $(1),$ you cannot prove $X\cup\{v\}$ is linearly independent.
Your proof does not mention where you are using either condition.