How Do You Solve Defective Eigenvalues in Differential Equations?

[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.
my answer was
\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?
Thanks in advance.

 

[Hurkyl]

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?

 

[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