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Eigenvectors, eigenvalues and matrices

  1. Aug 2, 2007 #1
    I have:

    [tex]x' = \left(\begin{array}{cc}2&-5\\1&-2\end{array}\right) x[/tex]

    I found that the eigenvalues are [tex]r_1 = i[/tex] and [tex]r_2 = - i[/tex].
    Also, I calculated the eigenvectors to be

    [tex]\xi_1 = \left(\begin{array}{c}2 + i\\1\end{array}\right)[/tex]

    [tex]\xi_2 = \left(\begin{array}{c}2 - i\\1\end{array}\right)[/tex]

    The answer, however, is

    [tex]\xi_1 = \left(\begin{array}{c}5\\2 - i\end{array}\right)[/tex]

    [tex]\xi_2 = \left(\begin{array}{c}5\\2 + i\end{array}\right)[/tex]

    I don't understand what I did wrong.
    This is what I did for the eigenvector corresponding to [tex]r_1 = i[/tex]:

    [tex]\left(\begin{array}{cc}2 - i&-5\\1&-2 - i\end{array}\right)\left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right) = 0[/tex]

    By multiplying row 2 and subtracting it from row 1, I got:

    [tex]\left(\begin{array}{cc}0&0\\1&-2 - i\end{array}\right) \left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right) = 0[/tex]

    So now I have:

    [tex]\xi_1 + (-2 - i)\xi_2 = 0[/tex]

    [tex]\xi_1 = (2 + i)\xi_2[/tex]

    so:

    [tex]\xi = \left(\begin{array}{c}\xi_1\\\xi_2\end{array}\right) = \left(\begin{array}{c}\(2 + i)\xi_2\\\xi_2\end{array}\right)[/tex]

    Now I let [tex]\xi_2 = 1[/tex]:

    [tex]\xi = \left(\begin{array}{c}2 + i\\1\end{array}\right)[/tex]

    I think I did everything right, but I get the wrong answer. Am I missing something?
     
  2. jcsd
  3. Aug 2, 2007 #2
    If v is an eigenvector, then a*v (with scalar a) is an eigenvector too (by definition and matrices being linear maps). Multiply your solutions with 2-i and 2+i, respectively.
     
  4. Aug 2, 2007 #3
    Oh now I see what they did. But my solution is also correct, isn't it?
     
  5. Aug 2, 2007 #4

    HallsofIvy

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    Staff Emeritus
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    Yes. The set of all eigenvectors corresponding to a single eigenvalue forms a vector space. There are, necessarily, an infinite number of eigenvectors for any eigenvalue.
     
  6. Aug 2, 2007 #5
    Thanks for clearing that up!
     
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