# 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