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Homework Help: Proof that vectors span R3

  1. Apr 2, 2010 #1
    Show that vectors v1, v2, and v3 span R3.


    I'm pretty sure I'm doing this wrong?

    a(V1) +b(V2) +c(V3) = [x,y,z]

    for (a= 0, b = 0, c = 1/3)

    [0,0,0] +[0,0,0] +[1,1,1] = [x,y,z]

    [1,1,1] = [x,y,z]

    Attached Files:

    Last edited: Apr 2, 2010
  2. jcsd
  3. Apr 2, 2010 #2


    Staff: Mentor

    Not right. In a nutshell you want to show that for an arbitrary vector <x, y, z>, there are some constants a, b, and c so that aV1 +bV2 +cV3 = <x,y,z>.

    You can do this by solving the matrix equation Ab = x for b, where the columns of matrix A are your vectors V1, V2, and V3. The vector I show as b is <a, b, c>, and the vector I show as x is <x, y, z>.
  4. Apr 2, 2010 #3
    Try showing that you can generate the standard basis of R3, {(1,0,0), (0,1,0), (0,0,1)}, using the elements v1, v2, v3. For example, what combination of these vectors will give you (0,1,0)?
  5. Apr 2, 2010 #4
    Is there any way u dumb it down just a lil more , I still feel very lost.
  6. Apr 2, 2010 #5
    Ok, I got this far


    Attached Files:

    • dfae.jpg
      File size:
      10.3 KB
  7. Apr 2, 2010 #6
    You want to show that {v1, v2, v3} = V spans R3. You already know that the vectors (1,0,0), (0,1,0), and (0,0,1) span R3. So you can try showing that V generates (0,1,0) and (0,0,1) and thus generates R3
  8. Apr 2, 2010 #7
    How do I know they span R3?
  9. Apr 2, 2010 #8


    Staff: Mentor

    The work in the attachment looks fine. If you can row-reduce your matrix to the identity matrix [1 0 0; 0 1 0; 0 0 1], that's enough to guarantee that your three vectors span R3.
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