## anonymous one year ago when you compute eigenvalues, the part that you need to form a base . Does the order matter? because i mostly get contradictionary results like a(1,0,-1) while in the textbook it says a(-1,01) because i extracted an other variable. Does this make a difference? because the next step is combining those bases to do a diagonalization

1. Loser66

The order of eigenvalues doesn't matter!! but the order of theirs eigenvectors is important!!

2. ganeshie8

If $$\large \mathrm{e}$$ is an eigenvector for eigenvalue $$\lambda$$, then any scalar multiple of it, $$\large a \mathrm{e}$$ is also an eigenvector of the same eigenvalue $$\lambda$$

3. ganeshie8

It helps to keep in mind that there is no "the basis" for a vector space any linearly independent vectors are fine for "a basis"

4. Loser66

|dw:1437765054056:dw|

5. Loser66

|dw:1437765276787:dw|

6. Loser66

got what I mean?

7. anonymous

so this is the same base?

8. ganeshie8

what is your actual matrix that you're trying to diagonalize ?

9. anonymous

1 3 3 -3 -5 -3 -3 3 1

10. anonymous

everything goes perfect until i need to make bases for my found eigenvalues

11. Loser66

Post the original problem, please, by snapshot or scanning!!

12. anonymous

13. ganeshie8

1, -2, -2 are your eigenvalues ?

14. anonymous

and this is the characteristic equation : |dw:1437765738465:dw|

15. anonymous

yes

16. ganeshie8

and the corresponding eigenvector matrix is $\begin{bmatrix} 1&-1&-1\\ -1&1&0\\ 1&0&1 \end{bmatrix}$ ?

17. anonymous

yes but i got this: -1 1 1 1 -1 0 1 0 -1

18. ganeshie8

doesn't matter, both your matrix and mine are same up to a scalar multiple, which is fine.

19. anonymous

thank you! and the multiplicity is 2 which implies that you can make a variable 0 ? because it's a linear combination of the other ?

20. ganeshie8

what do you mean by making a variable 0 ?

21. ganeshie8

this is a nice matrix because the geometric multiplicity for eigenvalue -2 is also 2 : you got two linearly independent eigenvectors for the eigenvalue -2!

22. anonymous

well i find it a bit confusing . so your eigenvalue is -2 so you get after all one equation out of the matrix x+y+z=0 which gives x=-y-z and then they say : we make two bases , one where z is zero and one where y is zero which gives us (-1,1,0) and (-1,0,1)

23. ganeshie8

Exactly! the nullspace of $$A-(-2)I = 0$$ has dimension of $$2$$, thats the reason you got 2 independent eigenvectors for the eigenvalue $$-2$$

24. ganeshie8

so as you can see, repeated eigenvalues aren't bad always

25. anonymous

well it would be even more fun if we get complex forms with a bigger multiplicity than 1 :p i guess hha

26. anonymous

that's what proven here indeed

27. ganeshie8

you can get multiple "linearly independent" eigenvectors for the same repeated eigenvalue

28. ganeshie8

Yea..

29. anonymous

well u really helped me a lot! and loser66 you too :) ! Thank you!!

30. ganeshie8

np