I couldn't fit the whole question in the title, so here is what I was asked:
Suppose $S = \{ v_1, v_2\}$ is a linearly independent set in a finite dimensional vector space $V$. Show that if $[S] \ne V$, then there is a vector $v_3 \in V$ such that $\{v_1,v_2,v_3\}$ is linearly independent.
I tried visualizing this in the $\mathbb{R}^3$ vector space, and it makes sense because the span of $S$ would be the vectors that are on a plane in the 3-dimensional space, and if you introduce a new vector so that the three vectors are now linearly independent, then the vector must be outside of that plane. But I'm struggling to formalize this into words.
I tried proving this statement by the contrapositive: If $\forall v_3\in V$ $\{v_1,v_2,v_3\}$ is linearly dependent, then $[S]=V$. This is how I did it so far:
Suppose that for all vectors $v_3\in V$, the set $\{v_1,v_2,v_3\}$ is linearly dependent. This means that there is a nontrivial solution $\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}$ to the equation
\begin{align*}
\alpha_1v_1+\alpha_2v_2+\alpha_3v_3=\Theta
\end{align*}
However, we know that $S$ is linearly independent, so
\begin{align*}
c_1v_1+c_2v_2=\Theta\end{align*}
where $c_1=c_2=0$. Therefore, we have that $\alpha_1=\alpha_2=0$. But since there must be a nontrivial solution to the equation, we know $|\alpha_1|+|\alpha_2|+|\alpha_3|\neq0\neq0+0+|\alpha_3|$. Therefore,
\begin{align*}
0v_1+0v_2+a_3v_3=\Theta\\
\Theta+\Theta+a_3v_3=\Theta\\
a_3v_3=\Theta
\end{align*}
Since $a_3\neq0$, $v_3$ must be $\Theta$.
Once I got up to this point, I started feeling like I made an incorrect assumption somewhere or that I'm going in the wrong direction. Is it true, given my initial assumption, that $v_3=\Theta$? If not, how can I show that $[S]=V$? Or maybe my entire approach is incorrect. I'd appreciate any pointers.
Best Answer
Using the contrapositive approach: if $v_3\in V$ is arbitrary and $v_1,v_2,v_3$ are linearly dependent, then as you note there are scalars $\alpha_1,\alpha_2,\alpha_3\in\mathbb{R}$ not all zero with $$\alpha_1v_1+\alpha_2v_2+\alpha_3v_3=0$$ If $\alpha_3=0$, then this would imply that $$\alpha_1v_1+\alpha_2v_2=0$$ where $\alpha_1,\alpha_2$ are not both zero, contradicting the assumption that $v_1,v_2$ are linearly independent. So we must have $\alpha_3\ne0$, and therefore $$v_3=-\frac{\alpha_1}{\alpha_3}v_1-\frac{\alpha_2}{\alpha_3}v_2$$ which shows that $v_3$ is in the span of $v_1,v_2$. Since $v_3$ was arbitrary, this means $v_1,v_2$ span $V$.