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A clarification in linear algebra

  1. Jan 11, 2014 #1
    1. The problem statement, all variables and given/known data

    This is a general question in linear algebra

    Determine whether the following subset of R^3 is a subspace

    The elements go vertically but I can't show them in this way and will show them horizontally however.

    ( s-2t, s, t+s ) / s, t ε R ....

    2. Relevant equations

    The book attempt to show closure under addition and multiplication

    3. The attempt at a solution

    Is it possible to use another argument. I claim that these three elements can be divided into two vectors which are also subsets of R-3 ... They are the span of (1,1,1)(-2,0,1).. Since they are lin. independent, they are vector spaces???
     
  2. jcsd
  3. Jan 11, 2014 #2

    tiny-tim

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    hi ehabmozart! :wink:
    you're saying that anything of the form ( s-2t, s, t+s ) is s(1,1,1) + t(-2,0,1) and vice versa

    so yes, if you've already proved the relevant theorems (which if you're still on page 1 of the chapter, you haven't!), that does immediately prove they are a vector subspace :smile:
     
  4. Jan 11, 2014 #3

    PeroK

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    Hi, you might want to think about it this way:

    Suppose you have the set:

    [itex](a_1s + b_1t, a_2s + b_2t, a_3s + b_3t): s, t \in R[/itex]

    Where the a's and b's are any constants. Is this set a subspace? Try showing closure under addition and scalar multiplication.

    Then, you might want to consider a mapping, L, from [itex]R^2 to R^3[/itex]

    [itex]L(s, t) = (a_1s + b_1t, a_2s + b_2t, a_3s + b_3t)[/itex]

    Then, you might want to consider what sort of mapping this is and how L could be represented.

    Then, you might look at your original problem in new light.

    Interesting? Or, not?
     
  5. Jan 11, 2014 #4
    Excuse me, what are the relevant theorems you are talking about?
     
  6. Jan 11, 2014 #5
    I am sorry but I don't your point here... Please clarify!
     
  7. Jan 11, 2014 #6

    PeroK

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    I assume you've done matrices? If so, you might notice that the original set can be seen as the action of the matrix:

    [itex]\begin{pmatrix}
    1&-2\\
    1&0\\
    1&1\\
    \end{pmatrix}[/itex]

    On vectors (s, t). And, those columns should look familiar!
     
    Last edited: Jan 11, 2014
  8. Jan 11, 2014 #7

    tiny-tim

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    direct sum of two linear spaces is a linear space?
     
  9. Jan 11, 2014 #8

    HallsofIvy

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    Yes. What your are saying is the set of all vectors of the form (s- 2t, s, t+s)= s(1, 1, 1)+ t(-2, 0, 1) is the span of vectors (1, 1, 1) and (-2, 0, 1) and so is a vector space.

    (The span of any set of vectors is a vector space. The fact that those two vectors are independent tells you that this is a two dimensional subspace of [itex]R^3[/itex] and that {(1, 1, 1) (-2, 0, 1)} is a basis.)
     
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