1. Limited time only! Sign up for a free 30min personal 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!

Finding a basis of a vector space

  1. Sep 16, 2009 #1
    1. The problem statement

    Let W = {(x, y, z, t): x + y + 2z - t = 0} be a vector space under R^4. Find a basis of W over R.

    2. The attempt at a solution

    To me I would think that the vector space itself could its own basis, but I know I'm probably way off. I also tried solving x = t - y - 2z and say that could be a basis but my confidence level on each possibility is not too high.


    Thank you ahead of time for your help.
     
  2. jcsd
  3. Sep 16, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The vector space itself spans the vector space. That doesn't mean it's a basis. In some sense a basis is a set containing the smallest number of vectors that span a vector space. Can you name some vectors in W? Writing them as (x,y,z,t), I'll give you one. (1,0,0,1) is in W. Can you give me another? Now how many do you really need so that linearly combinations of them cover all of W?
     
  4. Sep 16, 2009 #3
    From what I see, (a, 0, 0, a) where a is any real number could be an element of W. Other ones could be (0, a, 0, a), (0, -a, a, a), and (-a, 0, a, a).

    How about (0, -2a, a, 0)? Couple that with (a, 0, 0, a), wouldn't that cover all of W?
     
  5. Sep 16, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Why not just put a=1? You can make a basis of W with just three vectors.
     
  6. Sep 16, 2009 #5
    a = 1 would give us (1, 0, 0, 1) & (0, -2, 1, 0). What would the third vector be? I thought that these two alone could be enough for the basis.
     
  7. Sep 16, 2009 #6
    (1,1,1,4) is under W. Does (0, -2a, a, 0) and (a, 0, 0, a) cover that?

    I was replying to #3.. delayed a bit in replying.
     
  8. Sep 16, 2009 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    (1,-1,0,0) is also in the span.
     
  9. Sep 16, 2009 #8
    It is simple. Pick one vector, pick second, and pick third that is not a combination of the first and second.
     
  10. Sep 17, 2009 #9

    Mark44

    Staff: Mentor

    If you write your defining equation a certain way, the vectors will just about pop out at you.
    x = -y -2z +t
    y = y
    z = z
    t = t

    One vector I see here is (-1, 1, 0, 0), as Dick also pointed out. Another is (1, 0, 0, 1), as bruu pointed out. There is one more obvious one.
     
  11. Sep 17, 2009 #10

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Once you have x = t - y - 2z, let each variable on the right be 1 while the others are 0.

    For example, it t= 1, y= z= 0, x= 1- 0- 0= 1 so <1, 1, 0, 0> is such a vector. If y= 1, t= z= 0, x= 0- 1- 0= -1 so <0, -1, 1, 0> is such a vector. If z=1, t= y= 0, x= 0- 0- 2= -2 so <0, -2, 0, 1> is such a vector.
     
  12. Sep 17, 2009 #11
    Thank you to everyone for your help, I think I have it down!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook