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I can't seem to find an eigenvector for this 2x2 matrix

  1. Jul 20, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm doing an ODE for homework and I cant find the eigenvector for this matrix (sorry, I don't know how to make matrices on here. Consider these as one matrix.):

    [ 2 0] [v[itex]_{1}[/itex]] = [0]
    [ 1 1] [v[itex]_{2}[/itex]] = [0]

    2. Relevant equations

    The only way he has taught us to find eigenvectors seems mostly like guesswork to me.

    3. The attempt at a solution

    2v[itex]_{1}[/itex] + 0v[itex]_{2}[/itex] = 0

    v[itex]_{1}[/itex] + v[itex]_{2}[/itex] = 0

    So 2v[itex]_{1}[/itex] = 0 =====> v[itex]_{1}[/itex] = 0

    But v[itex]_{1}[/itex] + v[itex]_{2}[/itex] = 0, so v[itex]_{2}[/itex] = 0 as well. But eigenvectors cannot consist of all zeros?
     
  2. jcsd
  3. Jul 20, 2012 #2

    jedishrfu

    Staff: Mentor

    wouldnt your eqn be more like

    M * v = k * v where M is your matrix and k is some constant and v is your vector

    2*v1 + 0*v2 = k * v1

    v1 + v2 = k*v2
     
  4. Jul 20, 2012 #3
    Sorry maybe I didn't give the full background. (A-rI)v = 0

    A =

    [3 1]
    [2 2]

    one of the roots of the characteristic equation is r[itex]_{2}[/itex] = 1

    Since r[itex]_{2}[/itex] = 1, (A-I)v=0

    So

    [ 2 0]
    [ 1 1] v = 0
     
  5. Jul 20, 2012 #4
    !!!!

    Just found my mistake, I copied the matrix wrong!!!

    This is embarrasseing...
     
  6. Jul 20, 2012 #5

    jedishrfu

    Staff: Mentor

    isn't that just Av = Iv which is Av = v ?
     
  7. Jul 20, 2012 #6
    Found my mistake, I accidentally subtracted 1 from ALL the elements in the original matrix A, when I should have only subtracted 1 from a11 and a22
     
  8. Jul 20, 2012 #7

    jedishrfu

    Staff: Mentor

    okay that makes sense.
     
  9. Jul 20, 2012 #8

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Since you have found your error, expanding on jedishrfu's suggestion that you solve Av= v, that would be [tex]\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix}x \\ y\end{bmatrix}[/tex].
    which gives the equations 3x+ y= x and 2x+ 2y= y, both of which are equivalent to y= -2x. That is, any eigenvector, corresponding to eigenvalue 1, is of the form <x, y>= <x, -2x>= x<1, -2>.

    For the other eigenvalue, 4, we have [tex]\begin{bmatrix}3 & 1 \\ 2 & 2\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= 4\begin{bmatrix}x \\ y\end{bmatrix}[/tex]
    which gives the equations 3x+ y= 4x and 2x+ 2y= 4y, both of which are equivalent to y= x. That is, any eigenvector, corresponding to eigenvalue 4, is of the form <x, y>=<x, x>= x<1, 1>.

    Of course, solving [itex](A- \lambda)v= 0[/itex] is equivalent to solving [itex]Av= \lambda v[/itex] but the latter is more closely connected to the definitions of "eigenvalue" and "eigenvector".
     
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