Prove or give a counterexample for linearly independent list of vectors. Let $V$ be a vector space over $\mathbb{F}$, and suppose that $\vec{v}_{1}, \vec{v}_{2}, \dots, \vec{v}_{m} \in V$
  are linearly independent.  Prove or disprove:  For every choice of non-zero constants $\alpha_{1}, \alpha_{2}, \dots, \alpha_{m} \in \mathbb{F}$,
  the vectors $\alpha_{1} \vec{v}_{1}, \alpha_{2} \vec{v}_{2}, \dots, \alpha_{m} \vec{v}_{m}$ are also linearly independent.
Update: the original proof I came up with was too verbose. It's not even good to look at it. Turns out that I over-thought about the problem...
 A: Consider  $\alpha_1v_1,\alpha_2v_2,\ldots,   \alpha_mv_m$ with  $\alpha_i\neq 0$ for all $1\le i\le m $. Suppose  $$\sum_{i=1}^m a_i\alpha_i v_i=0.$$ Since  $v_i$s are linearly independent, we have  $a_i\alpha_i=0$. As $\mathbb F $ is a field, we can conclude that  $a_i=0$ for all $1\le i\le m $.
A: A much simpler :
Suppose there are scalars $a_1,a_2,\dots,a_m$ such that the zero vector, $\textit 0$, can be written as
$$\textit{0}=a_1(\alpha_1v_1)+a_2(\alpha_2v_2)+\cdots+a_m(\alpha_mv_m)$$
Now, this is the same as
$$\textit{0}=(a_1\alpha_1)v_1+(a_2\alpha_2)v_2+\cdots+(a_m\alpha_m)v_m$$
and since $v_1,v_2,\dots,v_m$ are linearly independent vectors, it follows that
$a_i\alpha_i=0$ for all $i=1,2,\dots,m$. But, $\alpha_i\neq 0$, which means $a_i=0$. Hence, the vectors $\alpha_1v_1,\alpha_2v_2,\dots,\alpha_mv_m$ are also linearly independent.
A: Let $$ \lambda_1 \alpha_{1} \vec{v}_{1} + \lambda _2 \alpha_{2} \vec{v}_{2}+ \dots \lambda_m \alpha_{m} \vec{v}_{m} =0$$
Since  $$\vec{v}_{1},  \vec{v}_{2}, \dots,  \vec{v}_{m}$$
are linearly independent, we have
$$ \lambda_1 \alpha_{1} = \lambda _2 \alpha_{2}= \dots \lambda_m \alpha_{m}  =0$$
Since $\alpha s $ are not zero, we have $$ \lambda_1  = \lambda _2 = \dots \lambda_m  =0$$
That proves the linear independence of   $$ \alpha_{1} \vec{v}_{1} , \alpha_{2} \vec{v}_{2}, \dots , \alpha_{m} \vec{v}_{m} $$
