# Thread: Question regards to subset, span and basis

1. ## Question regards to subset, span and basis

Hi I got stuck with this homework question so I'm here to ask you guy for some advice

if x is a real number, the absolute value of x is

absulute value (x)=x, if x is equal or more than 0
=-x, if x is less than or equal to 0

let S={(x,/x/,2/x/)/x is a real number}U{0,2,4),(-1,3,6)} and let V = span(S)

Show that V is a plane which contains the origin. Find a subset of S that is a basis of V.

So I know that V = span(s) thereby it is a zero vector, thus it will contains the origin but how would I write some mathematics equation to show that?

so V=(0,0,0)+s(0,2,4)+t(-1,3,6) since it passed though the origin, how would I find the basis for a subset of V? since V is a subset of S

[0 0 -1]
[0 2 3]
[0 4 6] <--- the matrix for V, row 2 and 3 are linearly dependent to each other (did I made the matrices right?) is it possible for me to get a basis out of that matrices?

p,s I'm an international student so my English isn't great. to me, subspace, span and basis are very confusing so please dont yell at me

2. ## Re: Question regards to subset, span and basis

For any subset S of a vector space (in this case the "parent space" appears to be R3), span(S) will be a subspace...but S can be "too big" to be a basis (since R3 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).