How to Prove the Second Part of a Multi-Part Question on Complex Eigenvalues?

rocketboy
Messages
243
Reaction score
1

Homework Statement



This is the second part of a multi-part question. Part (a) shows that:

x'' = Ax = \left(\stackrel{-2}{4/3}\stackrel{3/2}{-3}\right)x

Part (b): Assume x = \epsilone^{rt} and show that (A - r^{2}I)\epsilon = 0

x is the solution to the second order differential equation above, and \epsilon is an eigenvector corresponding to the eigenvalue r^{2} of A.

The Attempt at a Solution



Part (b): Alright, so given the above, I stated that \epsilon = e^{-rt}x

I then substituted everything into the left side of the equation I'm trying to prove to obtain:

(A - r^{2}I)\epsilon = \left(\stackrel{-2-r^{2}}{4/3}\stackrel{3/2}{-3-r^{2}}\right)\left(\stackrel{x_{1}}{x_{2}}\right)e^{-rt}

from here, I can see that I am not going in the right direction... any suggestions to get me moving along?

Thank-you!
-J

EDIT: for clarification, the above are 2x2 matrices... latex put the entries fairly close together. The matrix A has entries (-2 3/2) on the top and (4/3 -3) on the bottom.
 
Last edited:
Physics news on Phys.org
What is \epsilon? A vector? But since you also don't tell us what x is, what see no way to understand x= \epsilon e^{rt}.
 
HallsofIvy said:
What is \epsilon? A vector? But since you also don't tell us what x is, what see no way to understand x= \epsilon e^{rt}.

Sorry, thanks, I'll edit my above post to be more complete.

Note: for some reason my \epsilon appear to be superscripts... nowhere should this be the case.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top