## anonymous one year ago what is the eigenvector of this matrix. The eigenvalue of 3 has already been plugged in

1. anonymous

|dw:1433715559882:dw|

2. anonymous

i've simplified it to 2x-y+z=0 how do i make that into an eigenvector?

3. anonymous

@iambatman

4. freckles

I'm having trouble seeing how got an egivenvalue of 3

5. anonymous

Okay, so let's check and see if 3 works then

6. anonymous

the original matrix was |dw:1433716009320:dw|

7. anonymous

I found that one eigenvalue was 3 and the other two were complex numbers

8. anonymous

I'm just trying to find the eigenvector for 3 right now

9. anonymous

I plugged 3 into the matrix to get the matrix that i originally posted

10. anonymous

I just realized the other two eigenvalues aren't complex btw

11. freckles

for this other matrix you have I see how you got 3 now

12. freckles

for the first matrix all the egienvalues were 0

13. anonymous

sorry i wasn't clear. The first matrix is the second matrix with the eigenvalue 3 plugged in

14. anonymous

|dw:1433716555159:dw|

15. anonymous

which produced

16. anonymous

|dw:1433716643997:dw|

17. anonymous

after i simplified it a bit

18. freckles

ok so lambda=3 I'm going to do A-Lambda*I=0 $\left[\begin{matrix}7 & -2 & 2 \\ 8 &-1 & 4 \\ -4 & 2 & 1\end{matrix}\right]- \left[\begin{matrix}3 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 3\end{matrix}\right] \\ =\left[\begin{matrix}4 & -2 & 2 \\ 8 & -4 & 4 \\ -4 & 2 & -2\end{matrix}\right] =\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$ $\left[\begin{matrix}2 & -1 & 1 \\ 2 & -1 & 1 \\ -2 & 1 & -1 \end{matrix}\right]=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$ ok I'm cool with that

19. freckles

oops forgot to put my one column vector in

20. anonymous

after simplifying i got 2x-y+z=0

21. anonymous

the second and third row should cancel

22. freckles

$(A- \lambda I)x=0 \\ \left[\begin{matrix}2 & -1 & 1 \\ 2 & -1 & 1 \\-2 & 1 & -1 \end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)$ so we have only need to look at $2z-y+z=0$ as you said

23. freckles

err 2x-y+z=0

24. freckles

$2x-y+z=0 \\ \text{ one solution is } (1,1,-1) \\ \text{ now also say we replace } x=t \text{ then } 2t=y-z \\ \text{ so we could replace } y=3t \text{ and } z=t \\ \text{ so we have } \left(\begin{matrix}1 \\ 1 \\ -1 \end{matrix}\right)+t \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right)$ I think I remember something like this

25. freckles

that 2 should be 3

26. anonymous

so can i use ( 1 , 1 , -1) as the eigenvector?

27. freckles

as one egienvector

28. freckles

3 is repeated twice

29. freckles

ok so we have to find a second egienvector

30. freckles

for lambda=3

31. freckles

we can use the (1,1,-1) as the first

32. anonymous

ok I get it now, thanks a lot!

33. freckles

$\left[\begin{matrix}4 & -2 & 2 \\ 8 & -4 & 4 \\ -4 & 2 & -2\end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) =\left(\begin{matrix}1 \\ 1 \\ -1\end{matrix}\right)$ to find the second one you will have to solve this for (x,y,z)

34. freckles

notice I replace that zero vector with the other egienvector we found for lambda=3

35. anonymous

I think i just need one eigenvector, i need it to solve for a system of differential equations problem

36. anonymous

Thanks for the help!

37. freckles

http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iv-first-order-systems/matrix-methods-eigenvalues-and-normal-modes/MIT18_03SCF11_s33_8text.pdf could be wrong I think you will need to other egienvector

38. freckles
39. freckles

sorry the first link might not work

40. anonymous

Was 3 a repeated eigenvalue?

41. freckles

yah yah!

42. freckles

I think he is going to need that other egienvector

43. anonymous

Then I would think youd definitely need another eigenvector. Well, not technically an eigenvector, but a generalized one. If I recall, we'd need this: $(A-3I)\eta = \xi$ Where $$\xi$$ is the first eigenvector. I have no idea how to type up matrices on here though, lol.