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Subspaces R^n

  1. Jun 18, 2005 #1
    Q: Determine whether U is a subspace of R^3.

    U = {[0 s t]^T | s and t in R}

    A: Yes. U = span {[0 1 0]^T, [0 0 1]}

    Can someone explain to me how the heck they come up with that answer? Seems so random.
     
  2. jcsd
  3. Jun 18, 2005 #2
    Let's write column vectors as row vectors, shall we.

    U = { (0, s, t); s and t in R } = { (0, s, 0) + (0, 0, t); s and t in R } = { s(0, 1, 0) + t(0, 0, 1); s and t in R }.

    See now how one might notice that (0, 1, 0) and (0, 0, 1) form a basis for U?

    Of course, this is not necessary to show that U is a subspace of R^3, one can directly use the definition of subspace instead.
     
  4. Jun 18, 2005 #3

    matt grime

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    not random at all - they are just noting that given two (or any set of) vectors, they span a subspace and the things they span in this case is exactly the set U. it is easy to check the axioms if you need to.
     
  5. Jun 18, 2005 #4
    So basically we got to chose arbitrary (sp?) choose a "s" and "t", so that it satisfies the three main conditions?
     
  6. Jun 18, 2005 #5
    Actually, how would you do this question?

    U = {[r 0 s]^T | r^2 + s^2 = 0, r and s in R}

    Determine if U is a subspace of R^3?
     
  7. Jun 18, 2005 #6

    HallsofIvy

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    What are the possible (real) values for r and s so that s2+ y2= 0? What are the possible vectors in this vector space?
     
  8. Jun 18, 2005 #7
    r = 0 and s = 0 correct? Possible vectors? That's what I cannot figure out. Are there many different answers for this?
     
  9. Jun 18, 2005 #8

    Hurkyl

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    Well, there's one element in {[r 0 s]^T | r^2 + s^2 = 0, r and s in R} for each r,s pair satisfying r^2 + s^2 = 0, right?
     
  10. Jun 19, 2005 #9
    So from what you are saying then yes, there is an element in {[r 0 s]^T | r^2 + s^2 = 0, r and s in R} for each r, s pair satisfying r^2 + s^2 = 0 where r and s = 0?
     
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