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Proving A subspace

  1. Sep 26, 2012 #1
    1. Let V={X [itex]\in[/itex] R2 : (1 2) X= (0)}
    ..........................(3 4) ..... (0)
    Show that V is a subspace of R^2 with the usual operations. What is the dimension of V

    2. Relevant Equations

    3. I am really kind of lost, the statement seems to make no sense. X is in R but it also = the matrix [1 2, 3 4] but also equals the matrix [0,0]?
    Is v not a subspace of R^2 because it equals the zero vector? Thanks for the help.
     
    Last edited by a moderator: Sep 26, 2012
  2. jcsd
  3. Sep 26, 2012 #2

    Mark44

    Staff: Mentor

    No, X is a vector in R2.
    No, this is a 2 x 2 matrix, so it can't possibly be equal to a vector in R2.
    No again.

    Here is your matrix:
    $$\begin{bmatrix}1 & 2 \\ 3 & 4\end{bmatrix} $$
    Let's call it A

    Here is your vector x
    $$\begin{bmatrix}x_1 \\ x_2\end{bmatrix} $$

    V is all of the vectors in R2 such that Ax = 0, where 0 is the zero vector in R2.
     
  4. Sep 26, 2012 #3
    Isn't the only solution to that equation [0,0]? Hence, it is not a subspace because the only solution is the zero vector? Thanks again
     
  5. Sep 26, 2012 #4

    Mark44

    Staff: Mentor

    Maybe. Why do you think so?
    A set with only the zero vector in it is a subspace of whatever space it's in.
     
  6. Sep 26, 2012 #5
    So essentially the dimension is then 1 because the basis is the zero vector? Is that possible and I think so because reducing the matrix [1 2, 3 4] gives the identity which set to 0 means that the only solution is at x=[0,0]
     
  7. Sep 26, 2012 #6

    Mark44

    Staff: Mentor

    You're mostly on track here. Since the basis for V is just the zero vector, the dimension of V is zero, not one. Your matrix A maps every nonzero vector in R to some other nonzero vector.
     
  8. Sep 26, 2012 #7
    Thanks! Okay, so how would I go about proving V is a subspace of R^2 though? Like in terms of addition and multiplication being closed? Do I just pretend that two vectors in V are [a,b] and [c,d]??
     
  9. Sep 26, 2012 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    The only element of V is [0,0]. Doesn't that make it pretty easy to show it's a subspace? It's a trivial subspace.
     
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