If the elements v1 v2 ... vn form a basis for the vector space V, then the elements must span V and be linearly independant.

Also, the number of elements must be equal to the dimension of the vector space.

So my question is this:

If the number of elements is equal to the dimension of the vector space, and they are all linearly independant, will they always span V?

I feel pretty certain that they will, but I am not completely sure.

If they don't, what are some examples where they would not span but still be linearly independant?

In an n-dimensional vector space, V, any linearly independent set of vectors with n-elements will span V. Here's a proof:

 

Suppose you have a linearly independent set {a₁,...,a_n} which does not span an vector space V (dimV = n).

 

This implies that there is at least one vector a_n+1 in V which is not a linear combination of {a₁,...,a_n}. Therefore, add a_n+1 to your linearly independent set to obtain a new linearly independent set. Repeat the process until you obtain a linearly independent set which spans V.

 

This new set is a basis. However, you have a contradiction because you new set has at least n+1 elements, but a basis can only have n elements. Therefore, there can be no vector a_n+1 in V, which implies that all vectors in V are linear combinations of {a₁,...,a_n}.

 

Therefore {a₁,...,a_n} spans V.