1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

LA: Finding a basis

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data

    Let S = [x y z w] [itex]\in[/itex] [tex]R^4[/tex] , 2x-y+2z+w=0 and 3x-z-w=0

    Find a basis for S.


    2. Relevant equations


    3. The attempt at a solution

    I started by putting the system into reduced row form:

    [2 -1 2 1]
    [3 0 -1 -1]

    [2 -1 2 1]
    [0 3 -8 -5]

    [6 0 -2 -2]
    [0 3 -8 -5]

    [1 0 -1/3 -1/3]
    [0 1 -8/3 -5/3]

    Now have:

    x - 1/3z - 1/3w = 0
    y - 8/3z - 5/3w = 0

    Letting z = s, and w = t, we get:

    x = 1/3s + 1/3t
    y = 8/3s + 5/3t
    z = s
    w = t

    And this gives:

    s[1/3 8/3 1 0] and t[1/3 5/3 0 1]

    Where the basis vectors are:

    [1/3 8/3 1 0] and [1/3 5/3 0 1]

    and are linearly independent.


    Did I do this correctly? I'm really struggling with these concepts and I feel like I'm missing something. Thanks in advance.
     
  2. jcsd
  3. Jul 7, 2011 #2

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    I didn't check all your arithmetic, but assuming your numbers are OK, yes. That is exactly how to work the problem.
     
  4. Jul 8, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Personally, I would prefer this:
    2x-y+2z+w=0 and 3x-z-w=0

    From the second equation, w= z- 3x
    Putting that into the first equation, 2x- y+ 2z+ z- 3x= -x- y+ 3z= 0 so y= 3z- x.
    Having solved for y and w in terms of x and z,
    [x, y, z, w]= [x, 3z- x, z, z- 3x]= [x, -x. 0, -3x]+ [0, 3z, z, z]= x[1, -1, 0, -3]+ z[0, 3, 1, 1]
    so a basis is {[1, -1, 0, -3], [0, 3, 1, 1]} which is essentially what you have, multiplied by 3.
     
  5. Jul 8, 2011 #4
    Thanks for the replies. Although, I'm not sure I understand your basis HallfofIvy. I can't seem to find how it's related to the one I ended up with. I understand how you got there with the substitution, but is it just another basis for S?
     
  6. Jul 9, 2011 #5

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    It is just your method using basic algebra to solve the equations rather than "row-reduction".

    Every member of this subspace is of the form <x, y, z, w> and we must have
    2x-y+2z+w=0 and 3x-z-w=0

    I just solved for y and w in terms of x and z and then used x and z as multipliers where you solved for x and y and used z and w as multipliers.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: LA: Finding a basis
  1. Finding a basis (Replies: 10)

  2. Finding basis (Replies: 1)

  3. Finding basis (Replies: 9)

  4. Finding a Basis (Replies: 20)

Loading...