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Finding an eigenvector of 3x3 matrix

  1. Feb 1, 2013 #1
    Hi,

    I'm trying to find an eigenvector of a matrix. I know that λ = 1, so my matrix (A - λI) is
    [tex][-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236][/tex]

    And from rows 2 and 3 I get these simultaneous equations

    [tex]-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0[/tex]
    [tex]0.1817t_{1}+0.1161t_{2}+0.0236t_{3}=0[/tex]

    I eliminate to find [itex]t_{2} = -4.02t_{3} [/itex] and [itex]t_{1}=-2.23t_{3}[/itex]

    Thus the eigenvector is

    t=[itex]k [-2.23, -4.02, 1][/itex]

    But using an online solver gives the eigenvector as (-0.016, 0.206, 0.978).

    Thanks for any pointers.
     
    Last edited: Feb 1, 2013
  2. jcsd
  3. Feb 1, 2013 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Before you do anything else, check to see if it is an eigenvector (At = λt ?).
    Also how do know λ = 1?
     
  4. Feb 1, 2013 #3
    Hi,

    The reason I'm assuming λ = 1 is because A is a rotation matrix (I didn't mention this in the original post), and I'm looking for the equivalent axes eigenvector.
     
  5. Feb 1, 2013 #4
    As a sanity check, I have checked the eigenvalues and λ = 1 is one of them.
     
  6. Feb 1, 2013 #5
    I realised I typo in my original matrix (matrix term 3,3 in the first post should be negative). I've correct it.

    I get a closer answer, but my first term appears to be incorrect:

    My answer: (-0.0137, 0.225, 1)
    Given answer: (-0.0088, 0.216, 1)
     
  7. Feb 1, 2013 #6

    Mark44

    Staff: Mentor

    Since your matrix is a rotation matrix, the entries are sin and cos values. I would try using more precision than the 4 decimal places you show, and see if that makes a difference in your resulting eigenvector.
     
  8. Feb 2, 2013 #7
    I've just been given the numbers as is, and told that it's a rotation matrix.

    Can you tell if my method for computing the eigenvectors is correct?

    My approach is

    λ = 1, so (A - λI) is
    [tex][-0.5253, 0.8593, -0.1906; -0.8612, -0.5018, 0.1010; 0.1817, 0.1161, -0.0236][/tex]

    From rows 2 and 3:

    [tex]-0.8612t_{1}-0.5018t_{2}+0.1010t_{3}=0[/tex]

    [tex]0.1817t_{1}+0.1161t_{2}-0.0236t_{3}=0[/tex]

    Eliminate to find [tex]t_{2}=0.225t_{3} [/tex] and [tex]t_{1}=-0.0137t_{3}[/tex]

    Thus the eigenvector is

    t=[tex]k (-0.0137, 0.225, 1)[/tex]

    But the actual answer is given as (-0.0088, 0.216, 1).
     
  9. Feb 2, 2013 #8

    Mark44

    Staff: Mentor

    Here's your matrix in a nicer form. To see what I did, right-click the matrix to view the underlying script.
    $$A - I =\begin{bmatrix} -0.5253 & 0.8593 & -0.1906 \\ -0.8612 & -0.5018 & 0.1010 \\ 0.1817 & 0.1161 & -0.0236\end{bmatrix}$$

    To check your work, I multiplied A - I above times your eigenvector x. That multiplication should produce a zero vector, but what I got was off by a little. My result was approximately < .0099, .0088, .00003>.

    When you were row-reducing A - I, if you weren't careful with your arithmetic, you could have introduced a certain amount of imprecision in your results.


    This is easy to check. Just multiply (A - I) times <-0.0088, 0.216, 1>.
     
  10. Feb 2, 2013 #9
    Thanks Mark44. I have used more precise values, and using your technique for testing the eigenvector, I get an answer much closer to a zero vector.

    That you've confirmed my method is correct will do, especially since the discrepancies seem only due to rounding errors. I would have much preferred the question to use round numbers, but never mind :)
     
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