Defective eigenvalues?

1. Dec 6, 2005

µ³

if you have a differential equation of the form
x' = Ax
where A is the coefficient matrix, and you get a triple eigenvalue with a defect of 1. (meaning you get v1 and v2 as the associated eigenvector). How do you get v3 and how do you set up the solutions?
I tried finding v3 such that $(\mathbf{A}-\lambda \mathbf{I})^2 \mathbf{v_3} = \mathbf{0}$ but not exactly sure what to do after that. I got another v_2 (called it v_2') by using $(\mathbf{A}-\lambda \mathbf{I}) \mathbf{v_3} = \mathbf{v_2'}$ but again not sure how to set up solution.
$$\mathbf{x_1(t)} = \mathbf{v_1} e^{\lambda t}$$
$$\mathbf{x_2(t)} = \mathbf{v_2'} e^{\lambda t}$$a
$$\mathbf{x_3(t)} = (\mathbf{v_2'} + \mathbf{v_3}t)e^{\lambda t}$$
but it is not what is in the back of the book. I also tried using the original v_2:
$$\mathbf{x_1(t)} = \mathbf{v_1} e^{\lambda t}$$
$$\mathbf{x_2(t)} = \mathbf{v_2} e^{\lambda t}$$a
$$\mathbf{x_3(t)} = (\mathbf{v_2} + \mathbf{v_3}t)e^{\lambda t}$$
Can anyone help?

2. Dec 6, 2005

Hurkyl

Staff Emeritus
Well, it's awfully brutish, but if you think that the general solution can be written in terms of your $v_1$, $v_2$, and $v_3$, possibly with an extra factor of t on some terms, then why not just make the most general candidate possible:

$$(C_1 v_1 + C_2 t v_1 + C_3 v_2 + C_4 t v_2 + C_5 v_3 + C_6 t v_3)e^{\lambda t}$$

and plug it into the original equation, to see what relations must hold between the various coefficients?

3. Dec 8, 2005

CarlB

Maybe it would help to examine the solutions to a simpler version of the same sort of problem. For example:

$$A = \left(\begin{array}{ccc}2&0&0\\0&2&1\\0&0&2\end{array}\right)$$

The difficulty is the same as for this problem:

$$A = \left(\begin{array}{cc}2&1\\0&2\end{array}\right)$$

Carl