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Eigenvalues and eigenvectors of a matrix

  1. Aug 23, 2011 #1
    Hello i have this matrix [itex]\in Z [/itex] mod [itex] 7[/itex],
    M = \begin{pmatrix} 0&6\\ 5&0 \end{pmatrix}
    always modulo [itex]7[/itex] in [itex]Z[/itex].
    I found characteristic polynomial [itex]x^2+5[/itex].
    Eigenvalues are [itex]\lambda = 3, \lambda' = 4[/itex]
    Eigenvectors related to [itex]\lambda = 3 [/itex] are the non-zero solution of the system:
    [itex]4x +6y = 0,[/itex]
    [itex]5x+4y = 0 [/itex]
    I get:
    [itex]4x = y,[/itex]
    I don't know if it is correct, but how can i find the eigenvectors?
  2. jcsd
  3. Aug 23, 2011 #2
    Yes, y=4x is correct. Now you just need to find a non-zero vector (0,0) that satisfies this. So, we can take y=1 for example. In that case (x,y)=(4,1) satisfies the equation. So (4,1) is an eigenvector.
    We could also take y=2, then y=(1,2) was an eigenvector. etc.
  4. Aug 24, 2011 #3
    now i have a problem to solve this:
    in [itex]Z[/itex] mod [itex]11[/itex]

    M = \begin{pmatrix} 0&2\\ 7&0 \end{pmatrix}
    The characteristic polinomyal is : [itex]x^2+8[/itex],
    eigenvalues [itex]\lambda=5, \lambda'=6[/itex],
    eigenvectors related to [itex]\lambda=5[/itex] are the non-zero solution of the system:
    I don't know how to solve this system because i took a look at the solution of the exercise and it is:
    [itex]V={(y, 8y)}[/itex]
    I don't know how to get this solution.
    Oh i get now i'm wrong to write the first equation.:redface:
    Last edited: Aug 24, 2011
  5. Aug 24, 2011 #4


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    Science Advisor

    You have this backwards. If y= 4x, then the eigenvector is (1, 4), not (4, 1).
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