A basis IS a spanning set....but NOT vice-versa. A spanning set does not need to be linearly independent, which is a kind of "minimality" condition. For example, if you add another vector to a basis, it will no longer be a basis, but it will STILL be a spanning set.

A spanning set NEED NOT BE linearly independent....and a linearly independent set need not be a spanning set. A basis is BOTH. Bases are "just the right size": big enough to span the space (spanning), and small enough not to have any redundancy (linear independence). A basis is special..its size captures the DIMENSION of the space, which is an invariant property of a vector space (it never changes...even though you CAN change from one basis for a space, to a different basis).

Spanning is easy to describe: a set spans a vector space if and only if every vector in the space can be written as a linear combination of the spanning vectors. Linear independence is a bit more technical to describe: a set of vectors is linearly independent if and only if the only possible way to describe to 0-vector is the 0-linear combination (every "coordinate" in the basis is 0).

For example: the set {(1,0,1), (0,1,0) , (3,2,3)} in R^{3}is NOT linearly independent. Why? because:

(3)(1,0,1) + (2)(0,1,0) + (-1)(3,2,3) = (3,0,3) + (0,2,0) - (3,2,3) = (0,0,0) is a non-zero linear combination that sums up to the 0-vector.

Given a spanning set, it can be "trimmed down" to a basis, by eliminating those vectors which are linearly dependent on other ones (if any are). Given a linearly independent set, it can be extended to a basis (if it isn't one yet) by adding more linearly independent vectors. A basis cannot be added to, or subtracted from without "breaking" the basis property.

R^{3}has, not surprisingly, 3 dimensions (or "degrees of freedom"...in everyday language we often describe these as: left/right (often the x-axis), in/out (the y-axis) and up/down (the z-axis)). Any basis for R^{3}has to have exactly 3 elements, no more, no less.

It is pretty obvious the set {(1,0,0),(0,1,0),(0,0,1)} is a spanning set (or GENERATING set) for R^{3}: for any vector (x,y,z) can be written as the linear combination: x(1,0,0) + y(0,1,0) + z(0,0,1) = (x,0,0) + (0,y,0) + (0,0,z) = (x,y,z). It's not quite so obvious it is linearly independent:

Suppose a(1,0,0) + b(0,1,0) + c(0,0,1) = (a,b,c) = (0,0,0). Looking at the first coordinate, we see only (1,0,0) contributes to it, so we must have a = 0. Similarly, we must have b = c = 0, as well, so the only way we can sum these up to get the 0-vector is if a,b, and c are ALL 0. Thus it is linearly independent.

The following would be a good exercise for you. Which of the following are linearly independent sets, which span R^{3}, and which are bases?

1. S = {(1,0,0), (0,1,1)}

2. S = {(1,0,0), (1,0,1), (1,1,0)}

3. S = {(2,1,1), (2,-1,1), (0,2,0)}

4. S = {(1,0,1), (0,1,1), (-1,-1,0)}

5. S = {(1,1,0), (2,0,4), (3,-2,1), (0,0,5)}