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Find two linearly independent eigenvectors for eigenvalue 1

  1. Apr 19, 2015 #1
    1. The problem statement, all variables and given/known data
    A linear transformation with Matrix A = ##
    \begin{pmatrix}
    5&4&2\\
    4&5&2\\
    2&2&2
    \end{pmatrix} ## has eigenvalues 1 and 10. Find two linearly independent eigenvectors corresponding to the eigenvalue 1.
    2. Relevant equations
    3. The attempt at a solution

    I know from the trace of matrix A that eigenvalue 1 has a multiplicity of 2 and that eigenvalue 10 has a multiplicity of 1. Solving for eigenvectors corresponding for ## \lambda = 1 ##, pretend it's an augmented matrix...
    ##
    \begin{pmatrix}
    4&4&2\\
    4&4&2\\
    2&2&1
    \end{pmatrix}
    R_2 ->R_2 - R_1;
    R_3 ->R_3 - \dfrac{1}{2} \cdot R_2 =
    \begin{pmatrix}
    4&4&2\\
    0&0&0\\
    0&0&0
    \end{pmatrix}
    R_1-> \dfrac{1}{2} \cdot R_1 =
    \begin{pmatrix}
    2&2&1\\
    0&0&0\\
    0&0&0
    \end{pmatrix}
    ##
    Therefore ##x_2## and ##x_3## are arbitrary. Solving for ##x_1## gives $$x_1=-x_2-\dfrac{1}{2} x_3$$ $$x_2=x_2$$ $$x_3=x_3$$
    $$v= \begin{pmatrix}
    -x_2-\dfrac{1}{2} x_3\\
    x_2\\
    x_3
    \end{pmatrix} $$
    ##v= x_2 \begin{pmatrix}
    -1 \\
    1 \\
    0
    \end{pmatrix} +
    x_3 \begin{pmatrix}
    -1 \\
    0 \\
    2
    \end{pmatrix} ## Now those are the two eigenvectors I got however the answer in my book says the two eigenvectors are ##v_1=<1,0,-2>## and ##v_2=<0,1,-2>##
     
  2. jcsd
  3. Apr 19, 2015 #2

    Mark44

    Staff: Mentor

    Your eigenvectors are correct, which I verified by multiplying them on the left by your matrix, with both multiplications resulting in 1 times the eigenvector.
    The book's first eigenvector is the -1 multiple of your second eigenvector. Their second eigenvector is not a scalar multiple of your first eigenvector, but it still is an eigenvector associated with the eigenvalue 1. Your two vectors and the book's two vectors span the same two-dim. subspace of R3, so all is good.
     
  4. Apr 19, 2015 #3
    Okay thanks!
     
  5. Apr 19, 2015 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    In the book's solution, they solved for ##x_3 = -2 x_1 -2 x_2##, so ##x_1## and ##x_2## ended up being arbitrary instead of ##x_2## and ##x_3## as in your case.
     
  6. Apr 19, 2015 #5
    Oh, so that's where they're getting that eigenvector from.
     
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