Connection of linear equation to span.

  1. Let E = span{v1, v2} be the linear subspace of R3 spanned by the vectors v1 = (0,1,-2) and v2 = (1, 1, 1). Find numbers a, b, c so that
    E = {(x, y, z) of R3 : ax + by + cz = 0}

    I tried doing this question, but I am totally lost. I know that any vector in E can be represented as a linear combination of v1 and v2, but then how do I interpret (x, y, z) in terms of v1 and v2? @.@
    If someone could please give me an idea of how I should be going about with the solution, I'd really appreciate it! I dont need a full solution, just an idea of how to do the question.
    and please help! T.T
  2. jcsd
  3. x, y, and z are just elements of a vector. So in the case of [tex]v_1=(0,1,-2)[/tex], x, y, and z equal 0, 1, and -2 respectively.

    Furthermore, your next step is to find the span of [tex]\{v_1, v_2\}[/tex].
  4. So do I take (x, y, z) as the respective components of v1 and v2 and then get 2 equations and find a, b and c? o_O
  5. You won't need to worry about (x, y, z) until you find the span of (v_1, v_2).
  6. But I dont know how to find the span of v1 and v2 o_O
    I know that if I take v1 and v2 and put them in the form of a linear combination:

    av1 + bv2 = (m, n) where m and n are arbitrary,
    I can solve for a and b in terms of m and n.

    Is that what you call solving for the span? o_O
  7. Well we know that E is the span [tex]\{v_1, v_2\}[/tex], and we also know that the span of a set of vectors is the linear combination of those vectors, so we can say that:

    [tex]E = a_1v_1 + a_2v_2=a_1(0, 1, -2) + a_2(1, 1, 1) = (0, a_1, -2a_1)+(a_2, a_2, a_2)[/tex]
    Last edited: Nov 8, 2009
  8. Right, I did try using that but ended up with 5 variables. Since we dont know what the values of a1 and a2 are, how can we use the idea to compute a, b and c?

  9. WOAH! THANKS A TON MAN!!! xD I finally know how to do the question!
    Thanks again!
  10. (note: I switched a,b,c, to [tex]b_1,b_2,b_3[/tex].)

    E is defined by two things, [tex]E=\{(x, y, z) | b_1x + b_2y + b_3z = 0\}[/tex], as well as, [tex]E=(a_2, a_1 + a_2, a_2-2a_1)[/tex], so inserting the values of the linear combination into the constraint of E we have:

    [tex]b_1a_2 + b_2(a_1 + a_2) + b_3(a_2-2a_1) = 0[/tex]

    Rewriting it we have:

    [tex]a_2(b_1+b_2+b_3) +a_1(b_2-2b_3)= 0[/tex]

    Analyzing this, we see that in order for it to be true both of the following statements must hold:

    [tex]b_1 + b_2 + b_3 = 0[/tex] and [tex]b_2 - 2b_3=0[/tex]

    Now all you need to do is find a particular set of numbers [tex]\{b_1, b_2, b_3\}[/tex] that satisfy the above two equations.

    One such example could be [tex]\{b_1, b_2, b_3\}=\{-3, 2, 1\}[/tex]
    Last edited: Nov 8, 2009
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