- #1
richyw
- 180
- 0
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.