Question about linear independence If S is a set of k independent vectors in V , then each vector in V can be expressed uniquely as a linear combination of vectors in S? 
A independent set of vectors means the linear combination equalling zero means the scalars must be 0 right? Does this mean that of a vector in set can't be a linear combination of the other vectors?
 A: 
If S is a set of k independent vectors in V , then each vector in V can be expressed uniquely as a linear combination of vectors in S?

No. Uniqueness is correct; existence is not.  For example, $\{(1,0,0), (0,1,0)\}$ in a linearly independent set and $(0,0,1)$ cannot be expressed as a linear combination of members of that set.
However, linear independence does imply that for those vectors that can be so expressed, the expression is unique.

Does this mean that of a vector in set can't be a linear combination of the other vectors?

Yes. If $v_1$ is a linear combination of $v_2,\ldots,v_n$ then
$$
v_1 = c_2 v_2 + \cdots + c_n v_n \text{ for some scalars }c_1,\ldots,c_n.
$$
Therefore
$$
v_1 - c_2 v_2 - \cdots - c_n v_n = 0,
$$
and the coefficients are not all $0$ since the coefficient of $v_1$ is $1$.
A: If the dimension of the vector space $V$ over a field $F$ is $k$ then any linearly independent set $S$ of $k$ vectors will be a basis for $V$. This is the same as saying that for all $v \in V$ the vector $v$ has a unique representation as 
$$v= \sum_{i}c_is_i$$
where $c_i \in F$ and $s_i \in S$. 
Since $0= \sum_{i}0s_i$, by uniqueness if we have a linear combination of linearly independent vectors equalling zero, then all the scalars must be zero. However, just because a set is linearly independent does not mean it spans the space $V$, for this we require that the set $S$ be linearly independent and have cardinality equal to that of the dimension of $V$.
