I have this ODE system: $$X'=AX$$
$$A= \begin{pmatrix} 1 & -1 & 1 \\ 0 & 2 & 0 \\ -2 & 1 & -1 \\ \end{pmatrix} $$
I have to find all the solutions $X$.
I have founded the eigenvalues of this matrix by using the Laplace method and the Ruffini theorem to scompose the resulting third order equation.
So, the eigenvalues for $A$ are: $$\lambda_{1}=2$$ $$\lambda_{2,3}=\pm i$$
So, I'm expecting something like that for the solutions: $$c_1e^{2t}\mathbf u+c_2e^{0}[\mathbf v\cos(t)-\mathbf w\sin(t)]$$
Now I have founded the eigenvector $\mathbf u$ for $\lambda_1$. Noting that $rg(A-2I)=2$ I just consider, for example, two lines: $$x-y+z=0$$ $$-2x+y-3z=0$$
So I have
$$y=-\frac{5}{2}x$$ $$x=2$$ $$z=-3$$
So, $$\mathbf u=\begin{pmatrix} 2\\ -5\\ -3 \\ \end{pmatrix} $$
I have more troubles with $\mathbf v$ and $\mathbf w$
$$\det(A-iI)=\det \begin{pmatrix} 1-i & -1 & 1 \\ 0 & 2-i & 0 \\ -2 & 1 & -1-i \\ \end{pmatrix} $$ Which is a 3x3 matrix with $rg=3$. So I have to consider all the lines and solve the corresponding system to find the result. Moreover, since the eigenvalues are complex and coniugated, I think that it's enough this matrix to find the solution.
What do you think about that? Any hint to simplify the finding of $\mathbf v$ and $\mathbf w$? Thank you very much.
Letting $(A-2I)v_1=0$, you can easily obtain $$ v_1=(-2,5,3)^T. $$ Letting $(A-iI)x=0$, you can easily obtain $$ x=(-1-i,0,2)^T=v_2+iv_3, v_2=(-1,0,-2),v_3=(-1,0,0). $$ So the general solution is $$ X=c_1e^{2t}v_1+c_2(v_2\cos t-v_3\sin t)+c_3(v_3\cos t+v_2\sin t)$$