Linear system of differential equations with repeated eigenvalues

[richyw]

Homework Statement

[tex]X'=AX[/tex][tex]A=\left[\begin{matrix} 0 & 1 & 0 \\ -1 & 0 &0 \\0 & 0 & -1\end{matrix}\right][/tex]

Homework Equations

n/a

The Attempt at a Solution

The eigenvalues are -1, and [itex]\pm i[/itex]. I also can see that the matrix A is already in the form
[tex]A=\left[\begin{matrix} \alpha & \beta & 0 \\ -\beta & \alpha &0 \\0 & 0 & \lambda\end{matrix}\right][/tex] where [itex]\lambda_1=\lambda,\:\lambda_{2,3}=\alpha\pm i\beta[/itex] So I don't see the point really in computing the eigenvectors because this is already in canonical form isn't it? so I don't need to find T that would change the original matrix into its canonical form. So I think that the solution to X'=AX would just be Y(t). I have NO IDEA how to find Y(t) though. My book doesn't show the steps.

 

[Adrmay]

Also, I'm supposed to find the eigenvectors of X' and show that X' is diagonalizable. Would this just involve plugging in the eigenvalues to the matrix X' and solving for the eigenvectors?Any help would be greatly appreciated. Thanks!