# repeated eigenvalue general solution problem

• Jan 23rd 2007, 12:09 AM
OkashiiKen
repeated eigenvalue general solution problem
Hello all,

I am having trouble solving this repeated eigenvalue problem.

$\displaystyle x'=\left|\begin{array}{ccc}0&1&1\\1&0&1\\1&1&0 \end{array} \right|{x}$

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

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

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

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
$\displaystyle x^{(2)}(t) = u.te^\lambda_2t + w.e^\lambda_2t$
where u and w are vectors, and u is actually the eigenvector from the matrix acquired using $\displaystyle \lambda = -1$

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

Thanks,
Nick
• Jan 23rd 2007, 03:19 AM
TD!
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).