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Finding spaned subspaces

  1. Jun 15, 2010 #1
    Hi, I'm trying to solve this exercise.

    1. The problem statement, all variables and given/known data
    Given the following subsets of [tex]\mathbb(R)^3[/tex] find the subspaces generated by them:
    [tex]\{(1,0,-2),(-1,0,2),(3,0,-1),(-1,0,3)\}[/tex]

    3. The attempt at a solution
    I've tried to solve the linear dependence, so I've made the system:
    [tex]\begin{Bmatrix} a-b+3c-d=0 \\-2a+2b-c+3d=0 \end{matrix}[/tex]

    And I got: [tex]\begin{Bmatrix} a-b+8c=0 \\5c+4d=0 \end{matrix}[/tex]

    [tex]a=b-8c[/tex] and [tex]d=-5c[/tex]

    [tex](b-8c,b,c,-5c)[/tex]

    I don't know what to do next. I know that there is a linear dependence, but I don't know how to work with it, and what to do with the result I've found. Does it mean that the solution is a plane because there are only two free variables in [tex](b-8c,b,c,-5c)[/tex]? and how do I get the solution from here.

    Bye there, and thanks of course.
     
    Last edited: Jun 15, 2010
  2. jcsd
  3. Jun 15, 2010 #2

    Mark44

    Staff: Mentor

    Where did the d = -5c come from? How did you go from two equations to the same two equations plus one more?
     
  4. Jun 15, 2010 #3
    Lets see:

    [tex]\begin{bmatrix}{1}&{-1}&{3}&{-1}\\{-2}&{2}&{-1}&{3}\end{bmatrix}[/tex]
    Twice the first row plus the second on the second:
    [tex]\begin{bmatrix}{1}&{-1}&{3}&{-1}\\{0}&{0}&{5}&{1}\end{bmatrix}[/tex]

    [tex]\begin{bmatrix}{1}&{-1}&{8}&{0}\\{0}&{0}&{5}&{1}\end{bmatrix}[/tex]

    Thats what I did.

    Srry, there was something wrong in my first post. I've just corrected it.
     
  5. Jun 15, 2010 #4

    Mark44

    Staff: Mentor

    The matrix above is fine.
    This didn't do you any good. You want to use leading entries in a row to eliminate nonleading entries in the rows above and below. Multiply the 2nd row to get 0 0 1 1/5, and use the leading entry in the second row to eliminate the 3 in the row above it. That will give you a completely reduced, row-echelon matrix.
    The last step you did wasn't technically wrong - it just wasn't a big help.

    Since the set of vectors you started with had four vectors in R3, there are obviously too many to form a basis for R3. By inspection you can see that the second vector is a multiple of the first.

    It's not so easy to see (the work you are doing shows it), but the 4th vector is a linear combination of the first three. This means that the subspace spanned by the four vectors is the same as that spanned by the first and third.
     
  6. Jun 15, 2010 #5
    Thank you very much Mark.

    So, what I got is:

    [tex]\begin{bmatrix}{1}&{-1}&{0}&{-8/5}\\{0}&{0}&{1}&{1/5}\end{bmatrix}[/tex]

    How should I use this?
     
    Last edited: Jun 15, 2010
  7. Jun 16, 2010 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    I prefer your original form:
    [tex]\a(1,0,-2)+ b(-1,0,2)+ c(3,0,-1)+ d(-1,0,3)[/tex]
    Although I would NOT use that to "solve the independence" (you KNOW they are not independent because there are 4 vectors in [itex]R^3[/itex]).

    Any vector in that subspace can be written
    [tex](x, y, z)= a(1,0,-2)+ b(-1,0,2)+ c(3,0,-1)+ d(-1,0,3)[/tex]
    or
    [tex](x, y, z)= (a- b+ 3c- d, 0, -2a- 2b- c+ 3d)[/tex]

    One obvious result is that y= 0. Now, what further relationships can you find between x and z?
     
    Last edited: Jun 16, 2010
  8. Jun 16, 2010 #7
    Well, [tex]x=a-b+3c-d[/tex] and [tex]z=-2a-2b-c+3d[/tex]

    And as I said [tex]a=b-8c[/tex] and [tex]d=-5c[/tex], so...

    [tex]x=b-8c-b+3c-5c=-10c[/tex] and [tex]z=-2(b-8c)-2b-c+3(-5c)=-2b-16c-2b-c-15c=-32c-4b[/tex]
     
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