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Homework Help: Vectors forming a Subspace

  1. Dec 9, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine whether or not the set of vectors:

    [tex]\left\{\bar{x}=t
    \left(
    \begin{array}{cc}1\\2\\1\end{array}
    \right)

    +s\left(
    \begin{array}{cc}1\\1\\1\end{array}
    \right)

    +\left(
    \begin{array}{cc}1\\0\\2\end{array}
    \right),-\infty < t,s < \infty
    \right\}[/tex]

    is a subspace.


    I really have no idea how to go about this. Do I just check if they are linear combinations of each other?
     
  2. jcsd
  3. Dec 9, 2009 #2

    Dick

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    What properties does a set of vectors have to have to be a subspace? Look it up and start checking the properties.
     
  4. Dec 9, 2009 #3
    1) If A belongs to W, and B belongs to W, then A+B belongs to W
    2) If A belongs to W and r is in the space Rn, then rA belongs to W

    I don't understand how this applies to the problem. How can apply property 1 for example if I don't know if A or B belongs to W to begin with?
     
  5. Dec 9, 2009 #4

    Dick

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    For the second property, you mean "r is in R", I think. That means if W is a vector space then 0*A=(0,0,0) is in the space. That's usually listed as one of the conditions. I would first try checking if (0,0,0) is in W. Is there any choice of s and t that will produce the vector (0,0,0)?
     
  6. Dec 9, 2009 #5
    No, because when you row-reduce the matrix all the columns are linearly independent, which means they can only equal zero if their coefficients are zero. Since vector three doesn't have a coefficient, they can never equal zero. Therefore, they are not a subspace.

    Is my reasoning correct?
     
  7. Dec 9, 2009 #6

    Dick

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    Sure. If the zero vector isn't in your space, then it's not a subspace.
     
  8. Dec 9, 2009 #7
    But that is how I would go about solving it on a test? Just row reduce the vectors and determine whether or not a zero vector is possible?
     
  9. Dec 9, 2009 #8

    Dick

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    That's not the only thing. But it's a good starting point. Then you also have to prove the set has closure under addition, if a and b are in W then a+b is in W, and under scalar multiplication, a is in W then r*a is in W. Just what you said. But the zero vector is a good starting point. If zero isn't in the set, you can forget the other two.
     
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