For any subset S of a vector space (in this case the "parent space" appears to be R^{3}), span(S) will be a subspace...but S can be "too big" to be a basis (since R^{3}has dimension 3, any set with more than 3 elements is too big to be a basis of ANY subspace).

Since S has 2 linearly independent elements: (0,2,4) and (-1,3,6), dim(span(S)) = 2 or 3. The notation:

{(x,|x|,2|x|):x is a real number} obscures the fact that essentially this part of S is just two "rays" (half-lines terminating at the origin): one containing (1,1,2) and one containing (-1,1,2). So:

span(S) = span({(1,1,2),(-1,1,2),(0,2,4),(-1,3,6)}).

Since this set has cardinality 4, at least one of the elements HAS to be linearly dependent on the other 3. In fact, it is easy to see that:

(0,2,4) = (1,1,2) + (-1,1,2), so we can eliminate (0,2,4) without affecting span(S):

span(S) = {(1,1,2),(-1,1,2),(-1,3,6)}. Is THIS set linearly independent? Suppose that:

a(1,1,2) + b(-1,1,2) + c(-1,3,6) = 0, that is:

a - b - c = 0

a + b + 3c = 0

2a + 2b + 6c = 0 <--same as equation 2, just doubled.

We can add equations 1&2 to get:

2a + 2c = 0, that is: a = -c. Substituting in equations 1 or 2 gives us:

b + 2c = 0, or b = -2c. If we choose c = 1 (for example) this gives us the solution a = -1, b = -2, c = 1, and we can verify:

(-1)(1,1,2) + (-2)(-1,1,2) + (1)(-1,3,6) = (-1,-1,-2) + (2,-2,-4) + (-1,3,6) = (-1+2-1,-1-2+3,-2-4+6) = (0,0,0), so we have linear dependence, in fact:

(-1,3,6) = (1,1,2) + 2(-1,1,2).

Hence span(S) = span{(1,1,2),(-1,1,2)}. I leave it to you to prove that THIS set is linearly independent, and therefore forms a basis for span(S).

(Warning: other methods can lead to a different basis for span(S). Bases, are not, in general, unique...only their SIZE is invariant).