Linear dependent subset? Need help understanding a nuance of a definition Let $V$ be a vector space over a field $F$, and let $S \subset V$.
$S$ is linearly dependent if there exist a finite number of distinct vectors $x_1,x_2,...,x_n \in S$ and scalars $a_1,a_2,...,a_n \in F$(not all zero) such that
$\sum\limits_{i=1}^n a_ix_i = 0$
I don't fully comprehend it.
Say, $0 \in S$. Then $\forall a \in F \ \ 0 = a0$.The vector is zero, but the scalar $a$ needn't be. Or does it means "not all zero" for both vectors and scalars?
Now, whatever the answer to the first issue is, say $x_1,...,x_n$ are linearly dependent, that is, $\exists a_1,...,a_n$ such that
$\sum\limits_{i=1}^n a_ix_i = 0$
Does this mean that $\{x_1,...,x_n\} \cup A$ is linearly dependent $\forall \ \ A \subset V$? 
That being said, does this mean that a union of linearly dependent subset with ANY other subset of $V$ is linearly dependent, that is, we don't care about the properties of an other subset?
 A: To your first question, your definition of linear dependence is correct, and what you have done is proved that any set containing the zero vector is indeed a linearly dependent set. To your second question, yes. (And it looks like you already know how to prove it! :) )
A: Not all zero is only for scalars. 
Look at it in the following way: Let us first understand what it means to say that two distinct vectors $x_1$ and $x_2$ are linearly dependent. It simply means that one of them is a scalar multiple of the other, i.e., $x_1=\lambda x_2$. In this case we don't have any restriction on $\lambda$. $\lambda can be zero as well.$ Let us now see what this yields: It says that there exist scalars &\lambda, -1& such that $\lambda x_1-x_2=0$. Notice that not both scalars are zero($-1 \neq 0$). Here, we began with a given set of two distinct vectors.
If we have a given set of $n$ distinct vectors, say $x_1,x_2,...,x_n$, then they are said to be dependent if at least one of them can be written as some linear combination of the others. This  implies that there exists scalars $\lambda_1,...,\lambda_{n-1}$ such that at least one of the vectors (say $x_n$ wlog) is given by $x_n = \lambda_1 x_1+...+\lambda_{n-1}x_{n-1}$. Again in this definition we don't need $\lambda_k,1\leq k \leq n-1$ to be not all zero! All of them can be zero and yet the vectors $x_1,x_2,...,x_n$ will be linearly dependent. But then how do you translate this definition into the standard one? Well in this case you have $ \lambda_1 x_1+...+\lambda_{n-1}x_{n-1}-x_n=0$ where $-1\neq 0.$ Can you see now that why they say that "a set of $n$ distinct vectors are linearly dependent if there exist a set of $n$ scalars not all zero such that $ \lambda_1 x_1+...+\lambda_{n-1}x_{n-1}+\lambda_nx_n=0$?"
Now suppose there is a set $S\subseteq V$. How to define the notion of a dependent set? $S$ can be infinite! Here is a simple way. A set $S$ will be called dependent if you are able to find a finite subset of $S$ (consisting of distinct vectors) which are linearly dependent in the above sense. This answers your last question as well. Because if you already have a linearly dependent set then its union with any other set will still have those finite set of linearly dependent distinct vectors and hence will be dependent.
A: To address the motivation for the concept (i.e., why do we care about linear dependence?), it's basically a matter of redundancy. In a linearly independent set, no vector can be expressed as a combination of the others, so there is no redundancy; each one is "important", so to speak.
Imagine an Etch A Sketch: there are two knobs, one which moves the pen in the direction $\mathbf x_1 = [1\ 0]$ (or back), and the other in the direction $\mathbf x_2 = [0\ 1]$. The two knobs are independent in the sense that there is no amount of turning the first knob that will be equivalent to turning the second. To put it another way, the only way to remain fixed while turning the knobs is not to turn them at all. This is exactly the statement "the only solution to $a_1\mathbf x_1 + a_2\mathbf x_2 = 0$ is $a_1=a_2=0$."
Now suppose you add a third knob that moves in the direction $\mathbf x_3 = [1\ 1]$, i.e., along a diagonal. Putting aside the question of dexterity, this third knob is unnecessary because you can simply operate the first two at an equal speed to achieve the same result. Expressing this as an equation, we have $\mathbf x_3 = \mathbf x_1 + \mathbf x_2$, which we can rearrange to agree with the form of the equation in the definition of linear dependence:
$$1\mathbf x_1 + 1\mathbf x_2 + (-1)\mathbf x_3 = 0.$$
This is saying that by turning the first two knobs at a rate of $1$ unit while turning the third knob at a rate of $-1$ (i.e., turning it backwards at the same rate), we will keep the pen fixed.
In fact, any new knob that keeps the pen in the plane will be redundant, since the direction $[a\ b]$ can be attained with the first two knobs alone by going $a$ units horizontally and $b$ units vertically. To add a third knob while keeping them all linearly independent, you would have to move into the third dimension.
Finally, in this metaphor, the concept of a "zero vector" on the Etch A Sketch would be a knob that moves the pen in the direction $[0\ 0]$, which is to say that it does nothing at all except frustrate children.
