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How to find eigenvectors of 2x2 by gauss jordan method

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

    how to find eigenvectors by using gauss jordan

    A=[1 1; 2 2]

    2. Relevant equations
    I know how to use gauss jordan but don´t know how to use it to find eigenvectors


    3. The attempt at a solution
    First I find the eigenvalues: ((y-1)(y-2)-(1*2)=> y1=0 and y2=3

    Then I don´t know how to use the G.J method to find the eigenvectors.
     
  2. jcsd
  3. Mar 17, 2009 #2

    HallsofIvy

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    If 3 is an eigenvalue for A then there exist a non-zero vector v such that Av= 3v or (A- 3I)v= 0. Use Gauss-Jordan to find a non-zero solution to that equation and that solution is an eigenvector.
    Here, A- 3I is
    [tex]\left[\begin{array}{cc}-2 & 1 \\ 2 & -1\end{array}\right][/tex]
    (I haven't used the "augmented" matrix since the third column would always be 0.)
    Row reduce that. Because 3 is an eigenvalue, the last row will become all 0s and then the other rows give the eigenvectors. Here just adding the first row to the second will do that and the first row remains -2, 1. That is the same as -2x+ y= 0 or y= 2x. Taking x= 1, an eigenvector is <1, 2>. See what happens: What is
    [tex]\left[\begin{array}{cc} 1 & 1 \\ 2 & 2\end{array}\right]\left[\begin{array}{c}1 \\ 2\end{array}\right][/tex]?

    Do the same thing with eigevalue 0.
     
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