Understanding of Definition of Linearly (In)dependent Standard definition. Let $v_{1},v_{2},...,v_{n}$ be vectors. If $v_{1},v_{2},...,v_{n}$ are linearly independent, then for all $\alpha _{1},\alpha _{2},...,\alpha _{n}$ such that
$\alpha _{1} x_{1}+...+\alpha _{n} x_{n}=0$
we must have $\alpha _{1}=...=\alpha _{n}=0$.

My definition  Let $v_{1},v_{2},...,v_{n}$ be vectors. If $v_{1},v_{2},...,v_{n}$ are linearly independent then there are $\alpha _{1},\alpha _{2},...,\alpha _{n}$ in $\mathbb{R}$ such that
$\alpha _{1} x_{1}+...+\alpha _{n} x_{n}=0$
implies $\alpha _{1}=...=\alpha _{n}=0$.
Is there a difference between standard definition and my definition?
 A: By your definition, $x_1 = (1,0)$ and $x_2 = (2,0)$ (and any set of vectors for that matter) are linearly independent.  In particular, we can take $\alpha_1 = \alpha_2 = 0$.  For these particular values, the statements 
$$
\alpha_1x_1 + \alpha_2 x_2 = 0\\
\alpha_1 = \alpha_2 = 0
$$
are both true.  So, for these particular values of $\alpha_i$, "$\alpha_1x_1 + \alpha_2 x_2 = 0$" implies "$\alpha_1 = \alpha_2 = 0$", since True $\implies$ True is a true implication.
A: Yes!
Just because
$α_1x_1+...+α_nx_n=0$
does not imply that
$a_i = 0 Ɐ i ∈ \mathbb R, 1≤ i ≤ n$
and we cannot say that the set is linearly independent.
For example, lets consider the following set
$S = \{ (1,3,4,2), (2,2,-4,0), (1,-3,2,-4), (-1,0,1,0) \}$
There exists a solution where
$4(1,3,4,2) -3(2,2,-4,0)+2(1,-3,2,-4) + 0(-1,0,1,0) = 0$
As such, only when we ascertain that $a_i = 0 Ɐ i ∈ \mathbb R, 1≤ i ≤ n$ then do we say that the set is linearly independent.
This is a case of $A$ imply $B$ does not mean $B$ imply $A$
I'm not sure if this answered your question.
Also it need not necessarily be $\mathbb R$ but may be $\mathsf F$ for fields regarding your scalars $a$.
