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: Finding the eigenvectors for T()

  1. Dec 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Which of the following is not an eigenvector for [tex]
    T \left(
    \left[ {\begin{array}{cc}
    x \\
    y \\
    \end{array} } \right] \right) =
    \left[ {\begin{array}{cc}
    x + y \\
    x+ y \\
    \end{array} } \right]
    [/tex] ?

    A) v = [-2 -2]T
    B) v = [1 -1]T
    C) v = [1 2]T
    D) All are eigenvectors

    2. Relevant equations

    Ax = [tex]\lambda[/tex]x

    3. The attempt at a solution
    My problem is that all eigenvectors I've computed have come from 2x2 matrices. My best guess on starting is
    [tex]
    T \left(
    \left[ {\begin{array}{cc}
    -2 \\
    -2 \\
    \end{array} } \right] \right) =
    \left[ {\begin{array}{cc}
    -4 \\
    -4 \\
    \end{array} } \right]
    \left[ {\begin{array}{cc}
    -2 \\
    -2 \\
    \end{array} } \right]
    =
    [/tex]

    but this obviously doesn't work because of the size. How do I find an eigenvalue of a 2x1 matrix? Is it possible? Am I even looking at this correctly?

    Edit: I think I've figured it out. First, I constructed the standard matrix for T and got [tex]\left[ {\begin{array}{cc}
    1&1 \\
    1&1 \\
    \end{array} } \right] [/tex]
    Then I used Ax = [tex]\lambda[/tex]x with the vectors given to find the eigenvalues. Letter C didn't have an eigenvalue, so that is the answer.
    [tex]
    T \left(
    \left[ {\begin{array}{cc}
    -2 \\
    -2 \\
    \end{array} } \right] \right) =
    \left[ {\begin{array}{cc}
    1&1 \\
    1&1 \\
    \end{array} } \right]
    \left[ {\begin{array}{cc}
    -2 \\
    -2 \\
    \end{array} } \right] =
    \left[ {\begin{array}{cc}
    -4 \\
    -4 \\
    \end{array} } \right] = 2
    \left[ {\begin{array}{cc}
    -2 \\
    -2 \\
    \end{array} } \right]
    [/tex]
    [tex]
    T \left(
    \left[ {\begin{array}{cc}
    1 \\
    -1 \\
    \end{array} } \right] \right) =
    \left[ {\begin{array}{cc}
    1&1 \\
    1&1 \\
    \end{array} } \right]
    \left[ {\begin{array}{cc}
    1 \\
    -1 \\
    \end{array} } \right] =
    \left[ {\begin{array}{cc}
    0 \\
    0 \\
    \end{array} } \right] = 0
    \left[ {\begin{array}{cc}
    1 \\
    -1 \\
    \end{array} } \right]
    [/tex]
    [tex]
    T \left(
    \left[ {\begin{array}{cc}
    1 \\
    2 \\
    \end{array} } \right] \right) =
    \left[ {\begin{array}{cc}
    1&1 \\
    1&1 \\
    \end{array} } \right]
    \left[ {\begin{array}{cc}
    1 \\
    2 \\
    \end{array} } \right] =
    \left[ {\begin{array}{cc}
    3 \\
    3 \\
    \end{array} } \right] = ???
    [/tex]
     
    Last edited: Dec 17, 2009
  2. jcsd
  3. Dec 17, 2009 #2

    Mark44

    Staff: Mentor

    Your answer is correct, but you have made extra work for yourself in this problem. You didn't need to find a matrix representation for this transformation. All you needed to do was determine whether T(x) is equal to a scalar multiple of x.
    For example, T(-2, -2) = (-4, -4) = 2*(-2, -2).
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook