## singular points in 3-dim space

For a linearized system I have eigenvalues $$\lambda_1, \lambda_2 = a \pm bi \;(a>0)$$ and $$\lambda_3 < 0$$,
then it should be an unstable spiral point. As $$t \to +\infty$$ the trajectory will lie in the plane which is parallel with the plane spanned by eigenvectors $$v_1,v_2$$ corresponding to $$\lambda_1, \lambda_2$$.

Right? I am just not very sure.

 PhysOrg.com science news on PhysOrg.com >> 'Whodunnit' of Irish potato famine solved>> The mammoth's lament: Study shows how cosmic impact sparked devastating climate change>> Curiosity Mars rover drills second rock target
 Recognitions: Gold Member Science Advisor Staff Emeritus Yes, that is correct.
 Thanks