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How to find the eigen vector

  1. Dec 16, 2007 #1
    1. The problem statement, all variables and given/known data

    Find the eigenvector for each of the matrices

    2. Relevant equations
    I have a 2X2 matrice. (4, 2) which are on the top and (2,1) which are on the bottom. I understand how to get the value of the eigen, but I am confused about getting the vector.


    3. The attempt at a solution

    I took the value of A-I(lama). Second I did the row reduction inorder to get the top row with the pivot of one and one-half. NOW this is where Im confused. Thank in advance!!!!
     
  2. jcsd
  3. Dec 16, 2007 #2

    malawi_glenn

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    You out in one eigenvalue (one of your lambdas) in the matrix: (A-I*lambda),

    Then you know that (A-I*lambda)*x=0, where x = (x_1,x_2) is the eigenvector, and 0 is the 2dim zero-vector. Now solve that system of equations.

    Can you show us how you do all of this so we can help you? Write matrices and vetors as you would do in matlab, or just write them in TeX
     
  4. Dec 16, 2007 #3
    my values for the lamda's were 0 and 5
    my 2x2 matrice was 4, 2 and for the bottom was 2,1

    for the first lamda=0
    Ill plug it into the formula, A-lamda*I which is (4 2/2 1) - (0 0 /0 0)= (4 2/2 1)
    Ill do the row-reduction to get ONLY the top row of 1, 1/2.

    Which Ill now multiply with x_1 and x_2.

    which will turn out to be x_1= 1/2 x_2

    so then my answer will turn out to be x_1= 1 and x_2= -2

    Is this correct? Am I on the right path to solving for eigen vectors?? Thanks again!
     
  5. Dec 16, 2007 #4
    Also, do I solve a 3x3 matrix the same way as i do with a 2x2????
     
  6. Dec 16, 2007 #5

    Defennder

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    Yeah I think a possible eigenvector for lamda=0 is (1, -2). Now do the same for lamda=5. Then you would have obtained the 2 eigenvectors, which are linearly independent. Note that since you have 2 linearly independent eigenvectors for 2x2 matrix, you have shown that it is diagonalizable.

    For 3x3 matrix, the exact same process applies, except that we should have at least 3 linearly independent eigenvectors if the matrix is to be diagonalizable.
     
  7. Dec 16, 2007 #6
    I am having a hard time with the 3x3 matrice with finding the eigen value. I know its similar to finding a the determinant. But for some reason, I do get the right answer/values just not with the right signs. How do I know which sign to use???

    I think with finding the eigen value, its a negative, then you have to take into consideration the sign of the place of the number inside the matrice. But my signs keep coming out messed up
     
  8. Dec 16, 2007 #7

    malawi_glenn

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    why cant you just show us the problem in its entire length and show in detail how you solve it, so we can say "hey, thats wrong".
     
  9. Dec 16, 2007 #8
    (3, 1, 1) second row: (1, 0, 2) third row: (1, 2, 0)

    which i then convert into to (3-lamda) , -lamda, -lamda : all of which are diagonally.

    -(3-lamda)*2x2matrice of top row: lamda, 2 second row: 2, lamda

    -1*(-(2x2 matrice of top row:1, 1 second row: 2, lamda))

    -1* 2x2 matrice of top row: 1, 1 second row: lamda, 2


    for the fiirst matrice I got: lamda-cube-3lamda^2-4lamda+12

    for the second matrice I got: -lamda-2

    for the third I got: -lamda-2

    my three eigen values are : 1, 4, -2



    alright, here goes the fun part: finding the eigen vector is not really the same as a 2x2 matrice


    for example for lamda =1


    I took the value of lamda and placed it diagonally in a matrice of all zeros. which i then subtracted from the original matrice.

    gave me= top row: 2, 1, 1 second row: 1,1,2 and the third row: 1,2,1

    Now Im getting confused here. because when i switch R_2 with R_1 and then do the row reduction with the other rows i am still left with the other rows which i then keep reducing to all zeros with ONLY one 1 , in the top row, left.
    that is just going to give me x_1 but it'll zero out x_2 and x_3


    What am I doing wrong here???????
     
  10. Dec 16, 2007 #9

    malawi_glenn

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    Sorry, for me its all a mess. And I have much to study my self for tomorrows exams. Either you write in TeX or matlab-style, or wait for someone else to help you :/
     
  11. Dec 16, 2007 #10

    HallsofIvy

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    It's not at all clear what you are doing. The matrix I think is
    [tex]\left[\begin{array}{ccc}3 & 1 & 1 \\ 1 & 0 & 2 \\1 & 2 & 0 \end{array}\right][/tex]
    and so to find the eigen values you solve the equation
    [tex]\left|\begin{array}{ccc}3-\lambda & 1 & 1 \\ 1 & -\lambda & 2 \\1 & 2 & -\lambda \end{array}\right|= 0[/tex]
    Expanding by the first row, that is
    [tex](3- \lambda)\left|\begin{array}{cc}-\lambda & 2 \\2 & -\lambda\end{array}\right|-(-\lambda- 2)+ (2+ \lambda)[/tex]
    [tex]= (3- \lambda)(\lambda^2- 4)+ \lambda+ 2+ \lambda+ 2[/tex]
    Seeing that there is a [itex]\lambda+ 2[/itex] in each term, I would factor that out before I continue:
    [tex]=(\lambda+ 2)((3- \lambda)(\lambda- 2)+ 2)=(\lambda+ 2)(-\lambda^2+ 5\lambda- 6)= 0[/tex]
    The roots of [itex]\lambda^2- 5\lambda+ 6= 0[/itex] are 3 and 2 so the three eigenvalues are -2, 2, and 3.
     
  12. Dec 16, 2007 #11
    thats what i did. sorry but i dont have matlab, which was why i wrote it out completely. anyhow, thats what i did, but that answer in the book for the eigen value was
    4, 3 -2

    OK-so how do i go from here inorder to get my eigen vector??? i understand how to get the eigen vector with a two by two but it gets confusing with a 3x3.

    thanks
     
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