If the set $X$ is linearly independent in and does not span $V=\Bbb{R}^n$ then there is a $v\in V$ s.t. $\{v\}\cup X$ is linearly independent

Question

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

Answer 1

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.

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Indeed, this could be stated as a general lemma:

Lemma: If $V$ is a vector space and $X\subset V$, then, for any $v\in V\setminus X$, $X\cup\{v\}$ is linearly independent if and only if $X$ is linearly independent and $v\notin\operatorname{Span}(X).$

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 $\operatorname{Span}(\emptyset)=\{0\}.$

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The question has two conditions:

1. $X$ is linearly independent
2. $X$ does not span $\mathbb R^n$

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.

Answer 2

Your attempt is basically good, but there's no need to go that route.

Assume $X=\{v_1,\dots,v_r\}$ is linearly independent and let $v\in\mathbb{R}^n$ such that $v$ does not belong to the span of $X$.

Let $\alpha v+\alpha_1v_1+\dots+\alpha_rv_r=0$.

Suppose $\alpha\ne0$: then you get a contradiction (how?).

Therefore $\alpha=0$ and from here it is easy to conclude.

Answer 3

The set $X$ does not generate $\Bbb{R}^n\implies\exists v\in\Bbb{R}^n\backslash X$ such that $v\notin \text{Span}(X)$.
I will show that $X\cup\{v\}$ is linearly independent.
Let us assume on contrary that, $X\cup\{v\}$ is linearly dependent. Then $\exists u_1, u_2, ..., u_k\in X\cup\{v\}$ scalers $a_1, a_2, ..., a_k\text{(not all zero)}\in\Bbb{R}$ such that $a_1 u_1+ a_2u_2+...+a_k u_k=0$
If $u_i\ne v\forall i\in \{1,2,..,k\}$, then it would imply $X$ is linearly dependent (because in that case $\{u_1, u_2, ...,u_k\}\subseteq X$).
So at least one of $u_i$'s must be $v$ (say $u_j)$.
We get $a_j v=a_1 u_1+...+a_{j-1} u_{j-1}+a_{j+1} u_{j+1}+...+a_k u_k$.
Not that $a_j\ne 0$(if not, suppose $a_j=0\implies a_1 u_1+...+a_{j-1} u_{j-1}+a_{j+1} u_{j+1}+...+a_k u_k=0\implies X$ is linearly dependent, contradiction).
So, $a_j^{-1}$ exists in $\Bbb{R}$, then $v=(a_j^{-1} a_1) u_1+...+(a_j^{-1} a_{j-1}) u_{j-1}+(a_j^{-1} a_{j+1})u_{j+1}+...+(a_j^{-1} a_k) u_k\implies v\in\text{Span}(X)$, contradiction.
So, we must have $X\cup\{v\}$ is linearly independent.