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Finding eigenvectors.

  1. Oct 10, 2009 #1
    1. The problem statement, all variables and given/known data
    I have matrix
    A = 4 2
    0 1

    Whose eigenvalues I found to be 4 & 1

    I need to find the eigenvectors for the same matrix
    2. Relevant equations
    (A-lambdaI)V=0


    3. The attempt at a solution
    Lambda = 4 gives
    0 2 x V1 = 0
    0 -3 V2

    0v1 + 2v2 = 0
    0v1 - 3v2 = 0

    Eigenvector = 0 , 0
    How does that give me any choice? Either theres some fundamental rule im missing or my maple skills need a workout as according to maple the vector is 1, 0
    which would give 1+0 = 0 ? No?

    Lambda = 1 gives

    3 2 x V1 = 0
    0 0 V2

    3v1 + 2v2 = 0
    0v1 + 0 v2 = 0

    eigenvector = -2, 3
    Ok I get that this works, why couldnt I say that the eigenvector is 2, -3?
    wouldnt this give 3(2) + 2(-3) = 0
    Whats the difference?

    Clarifications greatly appreciated.

    P.s: Sorry for the painful format, couldnt figure out how to draw matrices with latex.
     
    Last edited: Oct 10, 2009
  2. jcsd
  3. Oct 10, 2009 #2
    How did you find those eigenvalues?
     
  4. Oct 10, 2009 #3
    Using det(a - lambdaI) = 0
     
  5. Oct 10, 2009 #4
    [tex]\lambda =4 \Rightarrow V_2=0[/tex] so [tex]V_1[/tex] can take any value, including 1.

    [tex]\lambda =1 \Rightarrow 3V_1=-2V_2[/tex]. Therefore the eigenvector you suggested is just as valid as the one you found using Maple.
     
    Last edited: Oct 10, 2009
  6. Oct 11, 2009 #5

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    The eigenvalues of a trianguar (or diagonal) matrix are just the numbers on the main diagonal.

    No, you only have 2v2= 0 and -3v2= 0 which are satisfied as long as v2= 0. v1 can be anything. An eigenvector for eigenvalue 4 is any vector of the form <x, 0>.

    The definition of "eigenvalue" is that there exist a non-zero vector such that [itex]Av= \lambda v[/itex]
    Here you must have
    [tex]\begin{bmatrix}4 & 2 \\ 0 & 1\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}[/tex]
    [tex]= \begin{bmatrix}4x+ 2y \\ y\end{bmatrix}= \begin{bmatrix}4x \\ 4y\end{bmatrix}[/tex]
    which gives you the two equations 4x+ 2y= 4x and 4y= 4y. The first equation reduces to 2y= 0 which gives y= 0 and the second equation is automatically satisfied. Again, x can be anything.
     
    Last edited: Oct 12, 2009
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