Can a Set of Four Vectors in ℝ³ Span the Space?

[NewtonianAlch]

Homework Statement

Consider a set of vectors:

S = {v$_{1}$, v$_{2}$, v$_{3}$, v$_{4}$$\subset$ ℝ$^{3}$

a) Can S be a spanning set for ℝ$^{3}$? Give reasons for your answer.
b) Will all such sets S be spanning sets? Give a reason for your answer.

The Attempt at a Solution

a) Yes, because a linear combination of these vectors can form any given vector in ℝ$^{3}$.

b) Yes, don't really know a reason besides something similar to the one above, I can't see why not.

I'm not really sure of these answers, can anyone confirm this, or given any insight to understand this better?

Cheers

 

[HallsofIvy]

What about, {(1, 1, 0), (1, 2, 0), (3, 1, 0)}?

 

[NewtonianAlch]

Well I guess since the 3rd elements are zero, you can't form a vector in ℝ$^{3}$ which has a non-zero 3rd element.

So part b is definitely a no. However part a "can" be since they haven't explicitly defined the vectors, so it's possible given that zero/non-zero condition, is that the only reason?

 

[HallsofIvy]

In general, a set with fewer than n vectors cannot span a vector space of dimenson n but a set with n or more vectors may. A set with more than n vectors cannot be independent but a set with n or fewer may. Only with sets with exactly n vectors is it possible to both span and be independent (a basis).