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Finding a basis for a subspace

  1. Jan 19, 2008 #1
    [SOLVED] Finding a basis for a subspace

    1. The problem statement, all variables and given/known data
    We have a subspace U in R^3 defined by:

    U = {(x_1 , x_2 ; x_3) | x_1 + 2*x_2 + x_3 = 0 }.

    Find a basis for U.

    3. The attempt at a solution

    We have the following homogeneous system:

    (1 2 1 | 0).

    From this I find the solution to be written as a*(2,1,0)^T + b(-1,0,1)^T, where a and b are arbitrary constants. From this I find three linearly independant vectors.

    Am I correct?
  2. jcsd
  3. Jan 19, 2008 #2
    (2,1,0) isn't a solution (but it's close). Also, if your subspace contains 3 linearly independent vectors then it's all of R^3. This is obviously not the case right?
  4. Jan 19, 2008 #3
    The solution was (-2,1,0).

    Yeah, I need a condition too so it is just the subspace where x_1 + 2*x_2 + x_3 = 0 - but how do I include that?
    Last edited: Jan 19, 2008
  5. Jan 19, 2008 #4
    Well, in the answer you gave I see two vectors (one with the correction you mentioned) that appear to be linearly independent. They must be a basis of the solution space, right?
    Last edited: Jan 19, 2008
  6. Jan 19, 2008 #5


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    You solved the homogeneous system and found the solution a*(-2,1,0)+b*(-1,0,1). That means (-2,1,0) and (-1,0,1) are in the subspace and they span it. They are also independent. 1) Anything in their span already automatically solves x1+2x2+x3=0, since that relation is linear. 2) What do you mean 'three' independent vectors??
  7. Jan 19, 2008 #6
    They must be basis for U. I only have two linearly independant vectors so far (not three as I wrote originally, sorry) - I need three linearly independant vectors that satisfy x_1 + 2*x_2 + x_3 = 0, right? How do I find the third?
  8. Jan 19, 2008 #7
    Doesn't U have to be spanned by three (3) vectors, since it is a subspace of R^3? If not, why?
  9. Jan 19, 2008 #8
    You can't find a third unless the solution space is all of R^3. You have two independent vectors that solve the equation so they must span the solution space.
  10. Jan 19, 2008 #9
    Ok, thanks to both of you.
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