Higher-dimensional linear systems (ordinary differential equations)

$$X'=\left[\begin{matrix} 0 & 1 & 0 \ -1 & 0 &0 \0 & 0 & -1\end{matrix}\right]X$$

Question

Higher-dimensional linear systems (ordinary differential equations)

$$X'=\left[\begin{matrix} 0 & 1 & 0 \ -1 & 0 &0 \0 & 0 & -1\end{matrix}\right]X$$

richyw · Answer

So clearly I have eigenvalues 1, and $\pm i$.  Now what do I do?

richyw · Answer

my book jumps straight to the solution satisfying the initial condition $X_0=(x_0,y_0,z_0)$$$Y(t)=x_0\left[\begin{matrix} \cos t \ -\sin t \0 \end{matrix}\right]+y_0\left[\begin{matrix} \sin t \ \cos t \0 \end{matrix}\right]+z_0\left[\begin{matrix} 0 \ 0\1 \end{matrix}\right]$$

no explanation (and what is X(t)?)

richyw · Answer

also never purchase hirsch's textbook. It's terrible and lacking so much content that they use like thin cardstock for the pages to make it thick enough! (don't even think about trying to lay this book open while you take notes!)

anonymous · Answer

0+-i = rotation .. 
by de-moivre conversion of imaginary to argand

richyw · Answer

sorry what are you saying?

richyw · Answer

oops and I meant -1 for the real eignevalue

richyw · Answer

nevermind can see that I missread the last term. now I can clearly see what is going on. thanks!