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Finding eigenvectors for a given matrix

  1. Nov 4, 2011 #1
    1. The problem statement, all variables and given/known data
    Finding Eigenvectors of the given matrix


    2. Relevant equations

    The matrix is
    [itex]A=\begin{pmatrix}
    5.956 & -1.284\\
    -1.284&0.435
    \end{pmatrix}[/itex]


    3. The attempt at a solution

    I have found the eigenvalues to be

    [itex]\lambda _{1}=0.624001[/itex] and [itex]\lambda _{2}=0.150994[/itex]

    and

    [itex]A-\lambda_{1} I =\begin{pmatrix}
    -0.2840 &-1.284 \\
    -1.284 &-5.8050
    \end{pmatrix}[/itex]

    [itex]A-\lambda_{2} I =\begin{pmatrix}
    5.8050 &-1.284 \\
    -1.284 &0.2840
    \end{pmatrix}[/itex]

    where A is the given matrix, I have rounded up the values slightly.

    Using wolfram online I know the eigenvectors I should get are
    [itex]E_{1}(-0.9764,0.2161)[/itex] and [itex]E_{2}(-0.2161, 0.9764)[/itex]

    However I can't seem to get them.

    Could somone please show me how to get those eigenvectors

    Thanks
     
  2. jcsd
  3. Nov 4, 2011 #2
    To get the eigenvectors of a matrix A use this equation and solve the system of equations for each eigenvector:
    [itex](A - \lambda_{\alpha}I)E_{\alpha} = 0 [/itex]​
     
  4. Nov 4, 2011 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Assuming that 0.624001 really is an eigenvalue, then there should be an infinite number of values of x and y such that
    [tex]\begin{bmatrix}5.956 & -1.284 \\ -1.284 & 0.435\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}= \begin{bmatrix} 0.624001x \\ 0.624001 y\end{bmatrix}[/tex]
    which is the same as the pair of equations
    5.956x- 1.284y= 0.624001x and -1.284x+ 0.435y= 0.624001y.
    Solve either of those for y= ax and take, say, x=1 to get an eigenvector.
     
  5. Nov 5, 2011 #4

    lurflurf

    User Avatar
    Homework Helper

    ^It is not
    The eigenvalues are approximately 6.24058 and 0.151.
    Probably a decimal shift to blame.
     
  6. Nov 5, 2011 #5
    Actually shouldn't that be det(A-λI)? Then he should find the roots of that characteristic polynomial and then proceed to find the bases for the Nullspace of those eigenvalues. Each basis should be the eigenvector for whatever eigenvalue you used.
     
  7. Nov 6, 2011 #6

    Mark44

    Staff: Mentor

    No, what jncarter wrote was correct. The equation above is a matrix equation, and 0 on the right side is a vector.

    The equation above implies the equation you show with the determinant. The idea is that if Mx = 0, where M is a square matrix and x is an n-vector, then |M| = 0.
     
    Last edited: Nov 6, 2011
  8. Nov 6, 2011 #7
    My bad, just never seen that notation before. Learn something new every day, hope I didn't confuse OP. Thanks for clarification.
     
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