Question

What does it mean to have a dominant eigenvalue and a dominant eigenvector?

Answer 1

@dan815

Answer 2

that is what is governing the change

Answer 3

I'll scan my attempt.. easier..

Answer 4

for example a matrix has 2 eigen values z1 and z2

Answer 5

the question was to find eigenvalues and determine the dominant eigenvalue and dominant eigenvector

Answer 6

can we say eigen values > 1 are dominant ?

Answer 7

I found the eigenvalues no problem.. but don't know what the dominant part of the question mean

Answer 8

yeah the biggest eigen value is usually the most dominant as u look at higher orders of the transfomation

Answer 9

and yeah positive

Answer 10

oh so ok I have eigenvalues as a and -1/2a

Answer 11

everything below negative will disappear at high powers

Answer 12

so a has to be the dominant eigenvalue then right?

Answer 13

what do we know about "a" ?

Answer 14

is it positive and > 1 ?

Answer 15

I'm scanning a XD hold on a bit

Answer 16

all eigenvalues and the eigenvector stuff ... . I know it's big.

Answer 17

but I don't know who is the dominant one... if it's supposed to be positive then there's no way that lambda = -1/2a can be a dominant value.

Answer 18

WHAT THE HECK! THERE'S ANOTHER EIGENVECTOR! D*** IT WOLFRAM!

Answer 19

That doesn't make any sense. One of them is a repeater... -a/2 is a repeaterrrrrrrr what the hxlllllllll

Answer 20

ugh and my signs are switched on one of my eigenvectors too. UGH!

Answer 21

but I still believe that the eigenvector with the 6a^2 is the highest ! :P thanks to lambda = a :P

Answer 22

@rational

Answer 23

Ok I found out what happened... I think it was due to a sign error somewhere when I was taking the second equation with the -a^2y = 3/2a^3 z (sorry typing out of my head) and wolfram got the extra eigenvector because we are dealing with a repeated eigenvalue... so if we let x = 0 and add the two equations we will only have the y and z left... so solving for y on the second equation y = 0 and again for z it will be z = 0 so eigenvector 3 = [0 0 0]

Answer 24

alright... I found the definition of the dominant eigenvalue and dominant eigenvector. It occurs when so since lambda_1 = a, lambda_2 = -(a)/2 [repeated]

by applying that definition I will have
Therefore, lambda_1 = a is the dominant eigenvalue and it's corresponding eigenvector v_1 =[6a^2 3a 1] is the dominant eigenvector. YAY!