Question

Can anyone help me solve for the eigenvectors of complex numbers to obtain a vector? for the following matrix ( below) i had obtained lamda = 3+2i and 3-2i

Answer 1

\[\left[\begin{matrix}9 & -20 \\ 2 & -3\end{matrix}\right]\]

Answer 2

So I guess the first step is to \[\left| A-rI \right|\] to get the eigenvalues

Answer 3

Or have you already done that?

Answer 4

yep, my eigenvalues are 3+ 2i and 3-2i

Answer 5

Okay, and we know that for complex values, our solutions are
e^(at)cos(bt)(vec-c) -e^(at)sin(bt)(vec-d) and e^(at)sin(bt)(vec-c) -e^(at)cos(bt)(vec-d)

Answer 6

hollddd up say what?

Answer 7

im solving this in simple r.r.e.f to obtain a vector haha

Answer 8

Oh woops. I was going to get the solution for the diffy q.

Answer 9

haha no its my bad, i didnt make it clear enough . let me edit it

Answer 10

Okay,
When r=3+2i
then rI = \[\left[\begin{matrix}6-2i & -20 \\ 2 & 2i\end{matrix}\right]\]
next we can take
z=\[z=\left(\begin{matrix}-20 \\ 6-2i\end{matrix}\right) = \left(\begin{matrix}-20 \\ 6\end{matrix}\right) +i \left(\begin{matrix}0 \\ -2\end{matrix}\right)\]

Answer 11

And well the other vector associated with that would be the same except that 2i wouldbe positive so the i matrix would be [ 0/2]

Answer 12

wouldnt it be -6 instead of +6 in your row 1 column 2?

Answer 13

and i dont understand where you got that z

Answer 14

\[\left[\begin{matrix}6-2i & -20 \\ 2 & 2i\end{matrix}\right] \left(\begin{matrix}z1 \\ z2\end{matrix}\right) =\left(\begin{matrix}0 \\ 0\end{matrix}\right)\]
z1=-20 and z2=(6-2i)s with, s=some variable,
eigenvectors associated with r= 6 +/- 2i are z=s*col(-20, 6-2i)

Answer 15

Then you take s=1, and you get z from there.

Answer 16

interesting, but still it would be \[\left[\begin{matrix}-6+2i & -20 \\ 2 & 2i\end{matrix}\right]\]

Answer 17

Ah! yes it would. Damn negative signs.

Answer 18

lol not a probl

Answer 19

so in conclusion, the vector for [-20, -6+2i]

Answer 20

but why don't you rr it into r.e f?

Answer 21

Yes that's the correct vector.
I don't know what rr or ref means. Sorry.

Answer 22

row reduce it into row echlon form

Answer 23

I guess you could if that's what your professor wants. But when I learned it in diffy q's we did it like that.

Answer 24

I'm logging off, good luck!

Answer 25

thank you!