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Nullspace of an eigenspace

  1. Jul 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Screenshot2012-07-28at80229PM.png
    2. Relevant equations
    3. The attempt at a solution
    I don't understand how [tex]u_1[/tex] = [tex][1 -1]^T [/tex]? By my reckoning [tex]u_1[/tex] =
    [tex]
    \frac{v_1}{\parallel v_1 \parallel}
    [/tex]
    which is
    [tex]
    \frac{-2}{\parallel -2+1 \parallel}, \frac{-2}{\parallel -2+1 \parallel}
    [/tex]
    which is
    [-2, -2] not [1, -1]
     
    Last edited: Jul 28, 2012
  2. jcsd
  3. Jul 28, 2012 #2
    Could you clear up your tex? It is hard to understand what your actual question is. Use [ tex] and [ /tex] tags for tex. (Without the space!)
     
  4. Jul 28, 2012 #3

    SammyS

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    There.

    I pretty much cleaned it up for you.
     
  5. Jul 28, 2012 #4
    I do not know what v_1 is, as it is not in the problem statement. But I can explain where u_1 comes from. The problem gives the matrix 2I-A. If you take u_1 and calculate (2I-A)u_1, then you get the 0 vector. Thus, u_1 lies in the null space of (2I - A). Since the null space of 2I - A has dimension one, that means that u_1 spans the null space of 2I - A, and so that 2Iu_1 = Au_1, meaning that u_1 is an eigenvector of A, and thus spans the eigenspace of A. Also, you do not have to normalize u_1, since the span of the normalized u_1 is the same as the span of u_1.
     
  6. Jul 29, 2012 #5
    Here is more of the question:
    Screenshot2012-07-29at35312AM.png
    if u_1 spans the null space of 2I - A, then that does not explain how u_1 is [tex][1 -1]^T[/tex]
     
    Last edited: Jul 29, 2012
  7. Jul 29, 2012 #6
    I think I understand what's going on now. Here's another example:

    Screenshot2012-07-29at50229AM.png

    It looks like after you reduce things to row echelon form, you reduce it again so that one of the values = 1. here, x -3/4y = 0. then you put one of the variables on the other side of the equation, so that u_1 becomes [tex][3/4 1 0]^T[/tex]. But I would think it should be [tex][1 3/4 0][/tex]

    I'm not sure about u_2 however.
     
  8. Jul 29, 2012 #7

    vela

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    You're looking for the solution to (2I-I)x=0, right? So solve
    $$(2I-A)\vec{x} = \begin{pmatrix} -2 & -2 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} x \\ y \end{pmatrix} = 0.$$ Forget about nullspaces and eigenspaces for the moment. You're just solving a system of equations.
     
  9. Jul 29, 2012 #8
    I sort of understand what is going on.
     
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