1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook