Question regarding ability to write a vector in a linearly dependent set as a linear combination of other vectors A theorem states: If $(v_1,...,v_m)$ is linearly dependent in $V$ and $v_1 \neq 0$, then there exists $j \in \{ 2,...,m\}$ such that the following hold: (a) $v_j \in \operatorname{span}(v_1,...,v_{j−1})$; (b) if the jth term is removed from $(v_1,...,v_m)$, the span of the remaining list equals $\operatorname{span}(v_1,...,v_m)$.
My confusion with this comes from why the order of the index here is seemingly important. I feel like I am really missing something obvious here, maybe having trouble with the notation being used can someone please explain to me why the definition talks about finding an index $j$ and being able to write the vector with this index as a linear combination of the preceding vectors? 
I think the root of my confusion comes from the idea that there is a vector in the set that is a linear combination of the preceding vectors. Why are all the vectors that succeed that particular vector unimportant here, what if a vector in a set is a linear combination of vectors listed after that particular vector? With subscripts greater than that particular $j$?
 A: If a vector is linearly dependent that means that you have a vector $V_i$ that you can write as the linear combination of other vectors. 
If a vector is linearly independent then you cannot write the vector $V_i$ as the linear combination of other vectors. $V_1, V_2, ... V_k$. Such that $i$ is between 1 and $k$. In other words the only way we can express the vector zero is if all scalars $\alpha_1 = \alpha_2 = ... = \alpha_k = 0$. 
Furthermore if you have the vector zero expressed as any $V_i$ then your system will be linearly dependent (I believe this is what you meant). Therefore if you have a system that is linearly dependent and you know your vector $V_1$ is not the vector zero then you surely will have a vector with any arbitrary index (in your case $j$) that will be zero. 
A: I can understand your confusion! Read more carefully part b)!
What does part a) say? It says that, if you start looking at your linearly dependent vectors in a certain prefixed order, at a  certain point, you will find a vector $v_j$ that is in in the span of the previous ones. 
Part b) says that if you remove that particular vector $v_j$ (and only that!) the span of the remaining list (excluding only $v_j$) does not change. The remaining vectors are still ‘important’, and also the following ones!
The ordering issue is not really an issue, and maybe with an example you will easily understand why. This example below is not proof but just a way to show you that the theorem actually makes sense :)
**Example 1 **
Consider a simple 4 dimensional space (you can assume $\mathbb{R}^4$ with the standard basis $\{e_1, e_2,e_3,e_4\}$.
Assume your set is $\{3e_1, e_2, e_3+e_1, e_1+e_2+e_3, e_4, e_1+e_2+e_4\}$. You already know that in a 4 dimensional space you can not have more than 4 independent vectors.
Indeed, if you look at those vectors above ‘in order’, you will find a $j$ such that your theorem is satisfied! The choice of such a $j$ is not unique. For example $v_4=e_1+e_2+e_3$ is in the span of the preceding vectors and removing it from the list will make no harm at all. But also $v_6= e_1+e_2+e_4$ is in the span of the previous ones and removing it makes no harm at all (from the point of view of preserving the span). 
The theorem basically tells you that you can remove from your list some of your vectors that are span of others :) I would suggest you maybe to read the proof, and then you will maybe be more comfortable with those concepts.
P.S. if not familiar with the basis concept yet, consider $e_1=(1,0,0,0), e_2 =(0,1,0,0), ...$
