Question

I need help with getting Eigenvectors of using complex eigenvalues.

Answer 1

So I got the eigenvalues at they are complex numbers, \[x= 1 \pm 3i\] I need help getting the eigvenvectors

Answer 2

Hmm I had to brush up from Pauls Notes :) lol

Answer 3

ok lol

Answer 4

Your eigenvalues look correct, yay

Answer 5

So then we have some particular vector which satisfies:\[\large\rm (A-\lambda I)\vec x=\vec 0\]

Answer 6

Or vectors* I suppose

Answer 7

\[\large\rm \left(\begin{matrix}3-3i & 3 \\ -6 & 3-3i\end{matrix}\right)\left(\begin{matrix}x_1 \\ x_2\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

Answer 8

yes. This is what I got \[\left[\begin{matrix}3+3i & -3 \\ 6 & -3+3i\end{matrix}\right]\] Now I need help specifically with row reducing

Answer 9

Did I make a boo boo on mine? Hmm I better check

Answer 10

Ok let's use yours then,\[\large \left(\begin{matrix}3+3i & -3 \\ 6 & -3+3i\end{matrix}\right)\left(\begin{matrix}x_1 \\ x_2\end{matrix}\right)=\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]

Answer 11

If we do the multiplication,
we can see that the row times the column gives us:\[\large (3+3i)x_1-3x_2=0\]

Answer 12

yes that correct

Answer 13

Solving for x2 gives us something like this,\[\large\rm x_2=(1+i)x_1\]ya?

\[\left(\begin{matrix}x_1 \\ x_2\end{matrix}\right)=\left(\begin{matrix}x_1 \\ (1+i)x_1\end{matrix}\right)\]
So maybe here is a nice vector we could use,\[\large \left(\begin{matrix}1 \\ 1+i\end{matrix}\right)\]

Answer 14

one sec let me do that on my paper to check if I got the same

Answer 15

ah ok yea makes sense to me

Answer 16

so then we can write it like this to \[x _{1}\left(\begin{matrix}1 \\ 1+i\end{matrix}\right)\]

Answer 17

Where x1 is just some scalar value?
Yah that seems like a good idea :)

hmm

Answer 18

And then for the other vector, you could work it out,
but I think we're just supposed to get this,\[\large \left(\begin{matrix}1 \\ 1-i\end{matrix}\right)\]right..?

Answer 19

yea ill take the same steps and see if I get that

Answer 20

Thank you for your help

Answer 21

np! ୧ʕ•̀ᴥ•́ʔ୨

Does that finish it for you?
Like... we had a lambda squared, so we should only expect to get 2 vectors right?
I dunno, I'm so rusty on this stuff :P

Answer 22

Yup it was correct, Thanks :)