Can someone help me with complex eigenvectors and phase portraits?

Question

anonymous · Answer

$$x=\left[\begin{matrix}3 & -13 \ 5 & 1\end{matrix}\right] x(0)=\left(\begin{matrix}3 \ -10\end{matrix}\right)$$
I used det(A-Ir) to get $$r=2\pm8i$$ I was just wondering since this a complex eigenvector does it matter which value you choose to solve? The answer key just used 2+8i was just wondering why you won't need both r solutions

anonymous · Answer

http://tutorial.math.lamar.edu/Classes/DE/ComplexEigenvalues.aspx Example 3

anonymous · Answer

As well I was wondering how do you know which n to pick, they decided to solve for n2

mr.math · Answer

You can pick either of the two eigenvalue and then find its associated eigenvector. You will get the second eigenvector by replacing the complex number in the vector by its conjugate.

anonymous · Answer

conjugate as in isin(_t)+cos(_t) right?

anonymous · Answer

Are you suppose to get the same answer? I'm not getting it could you possibly explain it a bit more?

anonymous · Answer

What I did was I took r=2+8i then got $$\left[\begin{matrix}3-(2+8i) & -13 \ 5 & 1-(2+8i)\end{matrix}\right]\left(\begin{matrix}\eta1 \ \eta2\end{matrix}\right)$$

anonymous · Answer

since the example said to use the second equation 5n1+(1-(2+8i)n2=0 I tried the other equation to see if it works but I don't get the same values in the end?