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Eigenvalues and Eigenlines

  1. May 7, 2014 #1
    1.
    Find the corresponding Eigenline

    A= (3 -3 2 -4)



    2. Relevant equations
    A=(a b c d)
    k2-(a+d)k+(ad-bc)=0


    3. The attempt at a solution

    k2-(3-(-4))k+(3(-4)-(-3)2)=0
    k2+k-6=0
    (k+2)(k-3)

    So k=-2 and k=3

    Eigenvector for k=3
    (3 -3 2 -4)(x y) = 3(x y)
    (3x -3y 2x -4y)= (3x 3y)

    Hence 3y=0 and 2x-7y=0

    Eigenvector for k=-2
    (3 -3 2 -4)(x y) = -2(x y)
    (3x -3y 2x -4y)=(-2x -2y)

    Hence 5x-3y=0 and 2x-2y=0

    This is where I get confused because my notes say that 3y=0 and 2x-7y=0 need to reduce to a single equation, and similarly 5x-3y=0 and 2x-2y=0 need to reduce to a single equation to give the eigenline for the given eigenvalue?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 7, 2014 #2

    pasmith

    User Avatar
    Homework Helper



    Do you mean [tex]
    A = \begin{pmatrix} 3 & - 3 \\ 2 & -4 \end{pmatrix}[/tex]
    Here is your problem: [tex](-2)^2 + (-2) - 6 = -4 \neq 0 \\
    (3)^2 + 3 - 6 = 6 \neq 0[/tex] Therefore neither -2 nor 3 is an eigenvalue of [itex]A[/itex]. Try solving [itex]k^2 + k - 6 = 0[/itex] again.
     
  4. May 7, 2014 #3
    Thanks for that! Couldn't see the wood through the trees. Got it now.

    K^2+k-6=0
    (k-2)(k+3)=0
    So k=2 and k=-3

    So eventual both equations reduce to x=y/2

    Can't believe I missed that.
     
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