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Finding a Basis for a set of vectors

  1. Jun 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Let H be the set of all vectors of the form [itex](a-3b, b-a, a, b)[/itex] where [itex]a[/itex] and [itex]b[/itex] are arbitrary real scalars. Show that H is a subspace of [itex]ℝ^4[/itex] and find a basis for it.

    Right, I've shown it's a proper subspace, just need help with finding a basis. Is [itex]{a-3b, b}[/itex] a suitable basis for this?

    I achieved this but putting the vectors into matrix form

    1 -3
    -1 1
    1 0
    0 1

    Reducing them to end up with

    1 -3
    0 0
    0 0
    0 1

    Is this a proper way to solve this question? Thanks in advanced.
     
  2. jcsd
  3. Jun 13, 2013 #2
    I believe so. That basis spans a two dimensional substance, because a-3b, and b-a are both linear combination of the remaining two vectors: a, b. By reducing you show that the two were linear combinations, so they went to zero, and you're left with the two basis vectors. Consider one last modification, though:

    ##\left[
    \begin{array}{cc}
    1 & 0 \\
    0 & 1 \\
    0 & 0 \\
    0 & 0
    \end{array}
    \right]##

    The space is just span(a,b).
     
  4. Jun 13, 2013 #3
    Great, thanks for that. Makes sense
     
  5. Jun 13, 2013 #4

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    What, exactly is your answer? Once you have row reduced, the only 4 dimensional vectors we can get from that matrix are (1, 0, 0, 0) and (-3, 0, 0, 1). That clearly cannot be a basis for this space because the second and third components would always be 0 which is not the case here.

    And what do you mean by "Is a−3b,b a suitable basis for this?". a and b, and so a- 3b, are numbers, not vectors and so cannot be a "basis".

    Here is how I would do it. H is the set of all vectors of the form (a- 3b, b- a, a, b).

    (a- 3b, b- a, a, b)= (a, -a, a, 0)+ (-3b, b, 0, b)= a(1, -1, 1, 0)+ b(-3, 1, 0, 1).

    And now a basis, and the dimension, are obvious.
     
    Last edited: Jun 13, 2013
  6. Jun 13, 2013 #5
    It's been so long since I've had linear algebra, that is exactly how it's done. I got caught up with the linear dependence. But, HallsofIvy is exactly correct

    $$
    a \left(\begin{array}{c}1 \\ -1 \\ 1 \\ 0\end{array}\right) + b\left(\begin{array}-3 \\ 1 \\ 0 \\ 1\end{array}\right) $$gives you the basis vectors. While the subsspace is still dim = 2, it's formed by span##\{\left(\begin{array}{c}1 \\ -1 \\ 1 \\ 0\end{array}\right),\left(\begin{array}-3 \\ 1 \\ 0 \\ 1\end{array}\right)\}##.

    Sorry about the dodgy reply.
     
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