Question

I need help with eigenvectors and eigenvalues. first order differential equations system:

X' = [0 1 2] * x
[-5 -3 -7]
[1 0 0]

is the system, with initial value problem of:

X(0) = [ 2 ]
[ -9 ]
[ 4 ]

Answer 1

I may have forgotten how to continue afterwards, but the eigenvalus of a matrix \(A\) are constants \[\lambda:\det(A-\lambda I)=0\]

Answer 2

Can you use a polynomial equation solver for this problem? I am trying to work the eigenvalues by hand, but the resulting (cubic) equation has inexact roots. Ignoring this problem, and using TuringTest's equation, you should be able to arrive at 2 complementary imaginary solutions and a real one. The general solution of the system should be of the form \[X(t) = c_1 e^{\lambda_1} \eta_1 + something\], where lambda 1 is the real eingevalue and n1its eigenvector. Ignore the something for now, can you try to solve the \(\lambda_1 \) case? Find the eigenvector, then we can work out the complex solutions.

Answer 3

PLEASE!! I don't know Turning Test equation... Also what's inexact roots?? Sorry, as you can see I really need help asap!! This is due tomorrow :( And I am pulling my second all-nighter!

Answer 4

He just posted above me. The one to find the eigenvalues. And what I meant is that most of the problems I saw during my DE course had at least one 'nice' root, like -1, 0, 1. Your eigenvalues are 'ugly', meaning that or your professor wants you to struggle with the algebra or you should use a polynomial equation solver (i.e., a computer or a pretty good calculator).

Answer 5

Yes, she doesn't like us!! She gives us 1 example and 2 exercises with no solutions!!! So I am soooooo lost!!

Answer 6

Ok then, I will try to guide you from the start. Do you how to calculate the eigenvalues? To setup the determinant, at least?

Answer 7

AWESOME! Ok,

to work out eigenvalues:

det(A-lamda I )=0 (for sake ofeasy typing lets make lamda=r

thus my 'matrix' becomes

det ( [ -r 1 2 ] ) = 0
( [-5 -3-r -7 ] )
( [ 1 0 0-r ] )

Answer 8

correct

Answer 9

ok computing
.....

Answer 10

ok I get r = -1 of multiplicity 3

Answer 11

That's weird. My result matches this: http://www.wolframalpha.com/input/?i=determinant [{-x%2C+1%2C+2}%2C+{-5%2C+-3-x%2C+-7}%2C+{1%2C+0%2C+-x}]
Maybe I mistyped something. I will redo my work.

Answer 12

that's right... which sucks
http://www.wolframalpha.com/input/?i=eigenvalues%20%5B0%20%2C1%2C%202%5D%2C%5B-5%2C-3%2C-7%5D%2C%5B1%2C0%2C0%5D&t=crmtb01
I have enough trouble with a double eigenvalue, don't think I know how to deal with a triple...

Answer 13

My lecturer is evil! Honestly!

Answer 14

Yeah, my bad. So, it's actually easier than what I thought.

Answer 15

Ok please share what to do next!!?

Answer 16

how do you deal with a tripl eigenvalue?
you need to find another vector when you have a double, so you must need at least twi extras for a triple :/

Answer 17

two*

Answer 18

sigh yip, ok, I'm going to try and find the first normal one...

Answer 19

http://tutorial.math.lamar.edu/Classes/DE/RepeatedEigenvalues.aspx

Answer 20

blegh...

Answer 21

cool will see what I can do!!Thanks you so much for helping!

Answer 22

welcome, good luck!
sorry I don't really have the time to go through it with you

Answer 23

Its fine! Have an awesome day form a very cold South Africa!

Answer 24

enjoy it
nice day from hot and humid central Mexico

Answer 25

:D

Answer 26

Yeah, this becomes messy quite quickly. If your lecturer puts something like this in your exam, do yourself a favor and laugh. Really, from scratch this should be an hour long problem, if you know what to do. On a exam, probably more. But the catch here is that there isn't a 'right' way out, you will have to try some solutions, and solve lots of systems of equations for different vectors, as showed in Paul's Note. Conceptually, I don't think it's hard, it's just really confusing and messy.

Answer 27

sigh, if you're bored you know whereI am!! Hee hee Thanks!

Answer 28

No problem. I will be around for some time, ask away if you are stuck somewhere, I will try to help you out. I almost regret not taking this part of the subject seriously :-). Good luck!

Answer 29

Thanks! Here I go....

Answer 30

DUDE!

Answer 31

O my freaken word! I just found the solution in a old textbook!

Answer 32

Haha, that's better, huh. I think that, in this case, it's better to see other people's work than doing this hairy problem on your own.

Answer 33

This kind of problem is almost as bad as integrating square root of tangent of x dx by hand :-).

Answer 34

Ha ha you don't want to see the other question from first principles!

Answer 35

Yeah, I think I will pass. I gotta go now, message me if you need anything, mate. See you, and good luck on your exam.

Answer 36

Thanks!!! xxx