Are vectors linearly independent iff they form a basis? I understand how to show that if a set of vectors form a basis, they must necessarily be linearly independent, but is the converse true, and how would you show it?
 A: As an extension to the comments, any subset of linearly independent vectors is itself linearly independent, however if you remove a vector from any linearly independent set, it will no longer be in the span of that set. So to get a set of linearly independent vectors that is not a basis, just remove a vector from any basis for the specific vector space.
If however you have $n$ linearly independent vectors and your vector space is $n$-dimensional, then they form a basis.
Proof: Suppose there are $n$ linearly independent vectors and the vector space is $n$ dimensional. If you have another vector $r$, then the $n$ linearly independent vectors and $r$ are linearly dependent, so we can write $r$ as a linear combination of the original $n$ vectors i.e. they form a basis.
To your question in the comments:suppose that the dimension of our vector space is $n$. Then there exists basis vectors $\{v_1,...,v_n\}$ (else the dimension of the vector space would not be $n$). Consider $v_{n+1}$ which is different to all the vectors $v_1,...v_n$ in the basis. Then $v_{n+1} = a_1 v_1+...+a_n v_n$ since $v_1,...v_n$ form a basis (i.e. we can write any other vector in the vector space  as a linear combination of these vectors). This means exactly that $v_{n+1}$ is linearly dependent to $v_1,...v_n$. 
Now interchange $v_{n+1}$ with $r$ and "$n$ linearly independent vectors" with $\{v_1,...,v_n\}$. 
A: By definition, a subset forms a basis if it is linearly independent AND spans the space. If you are working with finite dimensional vector spaces, then span is very easy to describe, namely, it is just all linear combinations of the vectors. To read more, including the general definition, you can start here: https://en.wikipedia.org/wiki/Linear_span
You can alternatively define it, in the finite dimensional case, as 

A subset such that every vector can uniquely be written as a linear combination. 

Say the vector space is $V$ and the subset is $S = \{v_1, \dots, v_n\}$
Suppose the former definition. Then $\forall v \in V$, $v$ is expressible as some linear combination, although not necessarily unique. Let's suppose it had two such linear combinations, $v = \sum c_i v_i = \sum d_i v_i $ Now subtract the two to get $\sum (c_i - d_i ) v_i = 0$. By linear independence then, $c_i - d_i = 0$. 
Conversely, suppose every vector can uniquely be written as a linear combination. Then span is satisfied so it remains to check independence. Suppose $\sum c_i v_i = 0 \in V$. Note that the key here is that $0$ is an element of the vector space, namely the zero vector. By assumption, zero has a unique expression as a linear combination in terms of $S$, namely $0 = \sum 0 \cdot v_i$. Thus by unicity, $c_i = 0$ $\forall i$, thus satisfying independence. 
