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Math Help - Question regards to subset, span and basis

  1. #1
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    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
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  2. #2
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    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).
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