Another eigenvector problem

1. Nov 26, 2006

Mindscrape

I need to solve the differential equation
$$\mathbf{x'} = \left( \begin{array}{ccc} 3 & 0 & -1\\ 0 & -3 & -1\\ 0 & 2 & -1 \end{array} \right) \mathbf{x}$$

solving for the eigenvalues by taking the determinate and using the "basketweave" yields

$$(3 - \lambda)(-3-\lambda)(-1-\lambda) + 2(3-\lambda) = 0$$

and further simplification shows that

$$-\lambda^3 - \lambda^2 +7\lambda +15 = 0$$

guessing roots, I found that 3 is one and divided the polynomial by the root

$$(\lambda - 3)(-\lambda^2 - 4\lambda -5)=0$$

so the eigenvalues are (solving for both eqns in the brackets)
$$\lambda = 3$$
$$\lambda = 2 + i$$
$$\lambda = 2 - i$$

The thing is that when I put in the real eigenvalue, it doesn't seem to have an eigenvector. Is this right? I think it would be because I don't have any idea how to get the solution for a mixture of real and complex eigenvalues. The solutions I know how to solve are when the eigenvectors are of the form v1,v2=p±iq. Though I imagine that if the real eigenvalue made an eigenvector I could just take the general solution of that and add it to the general solution of the complex eigenvectors.

Basically what I want to know is, what do I do know?

2. Nov 26, 2006

qbert

there's some miscalculation there.
eigenvalues are 3, (-2 + i), (-2 -i)

you can just look at the matrix and
pick the eigenvector that gives 3 as
the eigenvalue. (1, 0, 0).