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Eigenvalues and eigenvectors

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    Given the linear transformation l : R 2 → R 2 defined below, find characteristic equation, real eigenvalues and corresponding eigenvectors. a) l(x, y) = (x + 5y, 2x + 4y)

    2. Relevant equations
    characteristic equation = det (A-λI) = 0

    3. The attempt at a solution
    l(x, y) = (x + 5y, 2x + 4y)

    A =
    [ 1 + 5 ]
    [ 2 + 4 ]

    det (A-λI) = 0 =
    [ 1-λ + 5 ]
    [ 2 + 4-λ ]

    determinant works out to be

    D =
    2 - 5λ - 6) = 0

    so eigenvalues are -1 and 6.

    D =
    [-1 0]
    [ 0 6]

    to get eigenvectors we multiply the original matrix A by D

    [1 5 ] [ -1 0]
    [ 2 4] [ 0 -1]

    =
    [ -1 -5 ]
    [ -2 -4 ]

    This is where I'm stuck

    =
    [ -1 -5 ] * [x] = 0
    [ -2 -4 ] [y]

    -x-5y = 0
    -2x-4y = 0

    I can't see how x in the first equation can = x in the second equation. The same with both y's. How can I find the eigenvector here? I'm going back through my steps now to see if I made an error somewhere. I'm having the same trouble finding eigenvectors with the second eigenvalue of 6 as well.
     
    Last edited: Dec 9, 2015
  2. jcsd
  3. Dec 9, 2015 #2

    Mark44

    Staff: Mentor

    No.
    To get the eigenvector for ##\lambda = -1## solve this matrix equation:
    ##\begin{bmatrix} 1- (-1) & 5 \\
    2 & 4 - \lambda\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0\end{bmatrix}##
     
    Last edited: Dec 9, 2015
  4. Dec 9, 2015 #3
    Sorry, how did you get that matrix? Should the 6 be a 5?
     
  5. Dec 9, 2015 #4
    I got an eigenvector for λ=−1 of (-5,2)
     
  6. Dec 9, 2015 #5
    eigenvector for λ=6 of (1,1)
     
  7. Dec 9, 2015 #6

    Mark44

    Staff: Mentor

    Yes, I must have hit the wrong key. I have fixed it in my earlier post.

    Yes.

    Yes.

    You can check these yourself. If ##\lambda## is an eigenvalue associated with an eigenvector of x, it must be true that Ax = λx.
     
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