Results 1 to 2 of 2

Math Help - repeated eigenvalue general solution problem

  1. #1
    Newbie
    Joined
    Dec 2006
    Posts
    9

    repeated eigenvalue general solution problem

    Hello all,

    I am having trouble solving this repeated eigenvalue problem.

    <br />
x'=\left|\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0 \end{array} \right|{x}

    solving for the eigenvalues i get
    <br />
\left|\begin{array}{ccc}-\lambda&1&1\\1&-\lambda&1\\1&1&-\lambda \end{array} \right|=-(\lambda+1)^2(\lambda-2)=0
    which means
    \lambda_1 = -1, \lambda_2 = 2

    solving for the non-repeated solution using the eigenvalue
    \lambda_1 = 2
    <br />
\left|\begin{array}{ccc}-2&1&1\\1&-2&1\\1&1&-2 \end{array} \right|<br />
    which gives me the eigenvector and solution
    <br />
x_1(t)=\left|\begin{array}{ccc}1\\1\\1\end{array} \right| e^{2t}<br />

    however when I try to solve for the repeated value
    \lambda = -1
    i get the matrix
    <br />
\left|\begin{array}{ccc}1&1&1\\1&1&1\\1&1&1 \end{array} \right|<br />

    I dont know how to find a LI eigenvector from this matrix. I have the solution from the back of my textbook, but even after just arbitrarily taking the same eigenvector that they chose, I cannot come up with the same solution. I believe the repeated solution should be of the form
     x^{(2)}(t) = u.te^\lambda_2t + w.e^\lambda_2t
    where u and w are vectors, and u is actually the eigenvector from the matrix acquired using \lambda = -1

    I am stuck here, any help would be greatly appreciated.

    Thanks,
    Nick
    Follow Math Help Forum on Facebook and Google+

  2. #2
    TD!
    TD! is offline
    Senior Member
    Joined
    Jan 2006
    From
    Brussels, Belgium
    Posts
    405
    Thanks
    3
    The system comes down to the equation x+y+z = 0, from which you can get two linearly independent vectors, e.g. (1,0,-1) and (1,-1,0).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Another general solution problem
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: December 4th 2010, 08:25 AM
  2. general solution to trig problem
    Posted in the Trigonometry Forum
    Replies: 10
    Last Post: May 18th 2009, 03:28 PM
  3. i have problem in trigo general solution again
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: March 23rd 2008, 05:41 AM
  4. problem in general solution
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: March 19th 2008, 11:22 AM
  5. 5 trigo problem [general solution]
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: February 4th 2008, 06:48 AM

Search Tags


/mathhelpforum @mathhelpforum