## anonymous 4 years ago Can someone help me figure out the eigenvector?

1. anonymous

|dw:1327807423151:dw|

2. anonymous

the eigenvalue is 2 but I cant figure out the eigenvector

3. anonymous

After subtracting the two matrices this is what I got |dw:1327807555690:dw|

4. anonymous

What wld the answer be?

5. anonymous

preetha i thought u were going to chat with my older sis I hid her laptop so she wouldn't keep emailing you

6. anonymous

ok so the eigenvalue is 2. The eigenvector associated with 2 we will call X and we need: $AX=2X$ 0 0 -1 x 2x 0 -1 -4 y = 2y 0 0 1 z 2z So use matrix multiplication: -z = 2x, -y -4z = 2y, z=2z the last one implies z=0, and that implies that x=0. We have then -y=2y which means y=0 too, so the 0 vector is the eigenvector associated with 2...

7. anonymous

maybe someone can verify this is right...I need to sleep!

8. anonymous

ok I am gonna look this over

9. anonymous

Thanks :D

10. anonymous

Obviously here X = (x,y,z)^t and you do this for each eigenvalue...

11. anonymous

ummm i think u used the wrong matrice

12. anonymous

cuzz I was using a diff method so that is y i showed u that matrice

13. anonymous

ohh, you were using the method: $(A-2I)X = 0$ ?

14. anonymous

yes

15. anonymous

well use the correct matrix and just use my steps above as a blueprint, you can't go far wrong...

16. anonymous

idk okk whtvr

17. anonymous

Thanks I really appreciate ur help :DDDDDDDDDDD

18. anonymous

:D

19. anonymous

NO I may have sounded rude but seriously I realy appreciate ur help :DDD

20. mathmate

|dw:1327836820216:dw| For $$\lambda = 2$$, since the first column is all zeroes, this would be your free variable and the eigenvector is (1,0,0). Have you found the eigenvector corresponding to the other eigenvalues?