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Systematic way of extending a set to a basis

  1. Mar 16, 2014 #1
    1. The problem statement, all variables and given/known data

    I want to extend the below U set of vectors to R4.
    u1 = (0, 0, 0, -4), u2 = (0, 0, -4, 3), u3 = (3, 2, 3, -2).



    3. The attempt at a solution

    For a set of vectors to form a basis for Rn, the vectors must be LI and spans Rn(has n vectors)

    u1, u2 and u3 are linearly independent and spans R3 but not R4.
    To extend u1, u2 and u3 to R4. I require another vector, w. [itex]w\notin span{u1,u2,u3}[/itex]

    One of the solution is trial-and-error which I am not keen.


    I started with this:

    0λ1 + 0λ2 +3λ3 +λ4w1= 0
    0λ1 + 0λ2 +2λ3 +λ4w2= 0
    0λ1 -4λ2 +3λ3 + λ4w3= 0
    -4λ1 +3λ2 -2λ3 + λ4w4 = 0

    which is a brick wall.

    The other approach I had was UW = 0
    I want to find the W set of vectors that maps the coefficient matrix U into the zero-vector.
     
  2. jcsd
  3. Mar 16, 2014 #2

    vela

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    In the second approach, how are you defining U and what's stopping you from doing it?
     
  4. Mar 16, 2014 #3
    There's nothing stopping me from doing so. The calculation can easily be done but I'd appreciate if anyone could shed some light on the idea behind using the definition of UW=0 to solve for the W set of vectors.
     
  5. Mar 16, 2014 #4

    vela

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    Well, again, you need to define what U and W are first. We can't read your mind.
     
  6. Mar 16, 2014 #5

    Ray Vickson

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    There is a way that is easy in this problem, but is not really systematic and applicable to all problems without modification. Say the fourth vector you want is ##v = (v_1,v_2,v_3,v_4)##. If you want v to be orthogonal to all the u1, u2 and u3, then ##v \perp u1 \rightarrow v_4 = 0##.
    So, now ##v = (v_1,v_2,v_3,0)##. The condition ##v \perp u2 \rightarrow v_3=0##. So, now ##v = (v_1,v_2,0,0)##, and ##v \perp u3 \rightarrow 3v_1+2v_2 = 0##. Take any v_1 and v_2 that satisfy this last equation; then v is perpendicular to all the ui, so is linearly independent of them.
     
  7. Mar 16, 2014 #6
    A basis need not be orthogonal. i.e. (1,1), (1,0) is a basis for R2
     
  8. Mar 16, 2014 #7

    micromass

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    Nobody said it was. But adding vectors orthogonal to the original set is an easy solution.
     
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