Results 1 to 5 of 5

Math Help - 3x3 eigenvectors. Clueless.

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    2

    Angry 3x3 eigenvectors. Clueless.

    I admit this isn't too urgent, so apologies, but I'm posting here because I have exams quite soon and my inability to do this particular problem is getting a bit frustrating, not to mention worrying.

    Anyway, I understand how to work out eigenvectors for 2x2 matrices well enough, I think, but have absolutely no idea how to go about it for 3x3 matrices. I've looked around online but I either don't understand the notation used or it's just not dumbed down enough for me, I suppose...

    Specifically, anyway, the problem is as follows:

    Given the 3x3 matrix:

    ( 2 -4 0 )
    ( -4 2 0 )
    ( 0 0 4 )

    I have to work out the eigenvalues and corresponding eigenvectors. I've come up with the eigenvalues 4, 6 and -2, and used Mathematica to check these which are apparently right. But where do I go from here to get the eigenvectors?

    I only need help with one of them if someone could just explain the method to me - in especially simple terms note, since I'm apparently incapable of understanding anything more complicated.

    Starting with the eigenvalue lamda = 4, for example, and calculating (A - lamda I)X (where A is the original matrix, I is the identity) I've come up with these equations:

    -2x_11 - 4x_21 = 0
    -4x_11 - 2x_21 = 0


    I'm using _ to indicate subscript by the way, not sure if there's a clearer way to do this or not. But anyway these equations seem to contradict each other, so I'm not even sure I've started out properly. I assume the bottom value of the resulting column vector would be 0, but that's as far as I can get.

    Any help?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    A= \begin{pmatrix} 2 & -4 & 0 \\ -4 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix}

    The eigenvalues are found the same way as for a 2x2 matrix :

    \chi (\lambda)=det(A-\lambda I_3)

    Hence \lambda, eigenvalue, verify :

    \begin{vmatrix} 2-\lambda & -4 & 0 \\ -4 & 2-\lambda & 0 \\ 0 & 0 & 4-\lambda \end{vmatrix} = 0

    Supposing your eigenvalues are correct, now let's look for the eigenvectors.

    The property of an eigenvalue is that :

    AX=\lambda X

    So you have to solve, for each eigenvalue :

    \begin{pmatrix} 2 & -4 & 0 \\ -4 & 2 & 0 \\ 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \begin{pmatrix} \lambda x_1 \\ \lambda x_2 \\ \lambda x_3 \end{pmatrix}

    Is this more clear ?
    I can try showing you how to do for 4, eg
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    So we have :

    2x_1-4x_2=4x_1
    -4x_1+2x_2=4x_2
    4x_3=4x_3

    x_1=-2x_2
    -4x_1-2x_2=0
    x_3=x_3

    -> 8x_2-2x_2=0

    -> x_2=0

    -> x_1=0

    So an eigenvector to the eigenvalue 4 will be \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}


    For 6, we have :
    \begin{bmatrix} 2x_1-4x_2=6x_1 \\ -4x_1+2x_2=6x_2 \\ 4x_3=6x_3 \rightarrow x_3=0 \end{bmatrix}

    And solve it as to find a generator for x1 and x2...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Apr 2008
    Posts
    2
    Ahh, that helps a lot. So the eigenvectors are (0, 0, 1), (-1, 1, 0) and (1, 1, 0) I'm pretty sure. My mistake was trying to solve (A - lamda I)X=0 instead of AX = lamda X. Thanks for your help, it's much appreciated!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    You're welcome
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Absolutely clueless...
    Posted in the Algebra Forum
    Replies: 13
    Last Post: January 20th 2011, 07:12 AM
  2. Optimizations... Clueless
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 9th 2010, 06:04 PM
  3. Clueless on this proof, please help!!!
    Posted in the Discrete Math Forum
    Replies: 4
    Last Post: June 21st 2010, 12:02 PM
  4. Clueless
    Posted in the Algebra Forum
    Replies: 3
    Last Post: November 24th 2009, 09:20 AM
  5. integration problem..plz help, I'm clueless...
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 10th 2008, 10:18 AM

Search Tags


/mathhelpforum @mathhelpforum