Understanding a Basis I'm having a very difficult time understanding a basis.  I understand that it's basically just coordinates that span a subspace, but can someone help me to understand how a set can span a subspace and be a linearly independent set?
If a set spans a subspace doesn't this mean that any vector can be represented by a linear combination of the vectors in the subspace?  If that is the case, isn't this a direct contradiction to the set being linearly independent?  
What am I missing here?
 A: That the given vectors  span the (sub)space, means that every evector in it can be written as at least one linear combination in the given vectors.
That the given vectors are linearly independant, means that no vector can be written in more than one way as linear combination.
Thus for a basis, every vector in the subspace can be written as exactly one linear combination of the basis vectors.
A: 
If a set spans a subspace doesn't this mean that any vector can be represented by a linear combination of the vectors in the subspace? If that is the case, isn't this a direct contradiction to the set being linearly independent?

I am not sure exactly what your complaint is here, but it sounds like you are saying, "since every vector is a linear combination of the basis vectors, doesn't that mean $0$ is a linear combination also, contradicting linear independence?"  Well, sure, $0$ is a linear combination, but it is the following linear combination:
$$
0 b_1 + 0 b_2 + 0b_3 + \cdots + 0 b_n
$$
where $b_i$ are the basis.  So this does not contradict linear independence.
A: I am currently reading through Paul R. Halmos' "Finite dimensional linear algebra". The way a basis is defined in his book is as follows:
Given a finite dimensional vector space $V$, a (not "the") basis of $V$, say $S$, is a linearly independent set of vectors, such that every vector in $V$ is equal to some linear combination of the vectors in $S$.
Now, to answer your question, let us assume that a set of vectors, again say $S$, spans a vector space $V$. i.e, we assume that every vector in $V$ is equal to some linear combination of the vectors in $S$. Linear independence states that, given a linear combination of the vectors in $S$, if this combination is equal to the zero vector, then all the coefficients in the linear combination must be equal to zero.
Your statement is the following: $S$ is linearly independent $\implies$ that $S$ does not span $V$. Taking the contrapositive of this statement, we have that $S$ spans $V$ $\implies$ $S$ is linearly dependent. Now, we can simply provide a counter example. Say then that $V = \mathbb{R}^2$. Then, $V$ forms a vector space, and if we let $S = \{(1, 0), (0, 1)\}$, we have a linearly independent basis that spans $V$. $\square$
Note that some intricacies have been omitted (it's 1am!), such as proving that $S$ does actually span $V$, however I believe my answer to be sufficient as it stands. Do let me know if anything anything is missing. Clearly this idea extends to your question of a subspace if we set $V$ to be equal to the subspace you are speaking of (any will do).
