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