## Carissa15 one year ago I have a few questions about eigenvalues and eigenvectors in matrices.

1. Carissa15

Matrix b} $\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]$

2. Carissa15

Matrix a)$\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]$

3. jim_thompson5910

how far did you get with the eigenvalues of $\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]$

4. Loser66

What is the question?

5. Carissa15

I really haven't done anything with eigenvectors or eigenvalues. Very new to me. Seems though if I

6. jim_thompson5910

are you familiar with the idea that you need to find the determinant of the matrix $$\LARGE A - \lambda I$$ to get the eigenvalues?

7. Carissa15

$\det( \lambda \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]-\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right])=0$

8. jim_thompson5910

it should be the other way around A - (lambda)*I not (lambda)*I - A

9. jim_thompson5910

so you need to solve $\Large \det( \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]-\lambda\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right])=0$

10. Loser66

@jim_thompson5910 Yes, it is. I had 2 profs who used different ways to find out the eigenvalues. One used lambda -A and other used A - lambda. Both worked well.

11. jim_thompson5910

oh I did not know that

12. jim_thompson5910

well however you solve the equation, what values of lambda do you get?

13. Carissa15

just working through now

14. Carissa15

I get $\lambda^2-4\lambda-45=0$ not sure where to go from there

15. jim_thompson5910

use the quadratic formula to solve for lambda

16. jim_thompson5910

you can also factor the left side if you want

17. Carissa15

is the quadratic formula $\lambda^2-\lambda-4=0$?

18. jim_thompson5910

if you factored, you'll find that $\Large \lambda^2-4\lambda-45=0$ turns into $\Large (\lambda-9)(\lambda+5)=0$

19. jim_thompson5910

so you agree that $\Large \lambda = 9 \text{ or } \lambda = -5$ or no?

20. Carissa15

yes

21. Carissa15

so these are the eigenvalues?

22. jim_thompson5910

correct

23. Carissa15

I am not sure how to now find eigenectors using these

24. jim_thompson5910

$\Large A\vec{v} = \lambda \vec{v}$ $\Large A\vec{v} - \lambda \vec{v}=0$ $\Large A\vec{v} - \lambda I\vec{v}=0$ $\Large (A - \lambda I)\vec{v}=0$ That last equation leads us to this matrix equation $\Large \begin{bmatrix}2-\lambda & 7\\7&2-\lambda\end{bmatrix} \vec{v} = \begin{bmatrix}0\\0\end{bmatrix}$ which leads to this augmented matrix $\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]$

25. jim_thompson5910

If lambda = 9, then $\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]$ $\Large \left[\begin{array}{cc|c}2-9 & 7 & 0 \\7 & 2-9 & 0\end{array}\right]$ $\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7 & -7 & 0\end{array}\right]$ row reduce the augmented matrix until it's in rref form

26. Carissa15

So we would end up with R1+R2 and all values zero?

27. jim_thompson5910

you do compute R1+R2, yes but we only need to do that for one row R1+R2 = [0 0] replace R2 with that result $\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7&-7 & 0\end{array}\right]$ $\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\0&0 & 0\end{array}\right]\begin{array} \ \\ R1+R2 \rightarrow R2\end{array}$

28. Carissa15

Oh yes, of course. not both oops.

29. jim_thompson5910

the top row -7 7 0 means -7x+7y = 0 solve for y to get 7y = 7x y = x so the solution set is the column vector $$\LARGE \left[\begin{array}{c}x \\x\end{array}\right]$$ this describes all of the eigenvectors for the eigenvalue lambda = 9

30. jim_thompson5910

a lot of books may say let x = 1, so a particular eigenvector for lambda = 9 is $\LARGE \left[\begin{array}{c}1 \\1\end{array}\right]$ all other eigenvectors for lambda = 9 are simply scalar multiples of that column vector

31. Carissa15

Would this always be the case? Or do you sometimes get actual values instead of x? And Do I then follow the same steps for lambda -5?

32. jim_thompson5910

yes I've had teachers who would assign values to x. 1 is the smallest possible to allow for other multiples to be generated

33. jim_thompson5910

and yes you follow the same steps for lambda = -5

34. Carissa15

ok great. Thank you. I will work my way through them and see how I go. :)

35. jim_thompson5910

ok tell me what you get

36. Carissa15

for lambda -5 i got$\left[\begin{matrix}7 & 7 \\ 7 & 7\end{matrix}\right]\$ with Identity matrix

37. Carissa15

then did R1-R2 gives new R2 of all zeros.

38. Carissa15

I then got 7x-7y=0, which then ends up with the same result as lambda 9

39. jim_thompson5910

I'm getting the same, so you have 7 7 0 up top over 0 0 0 the top row implies that 7x + 7y = 0

40. jim_thompson5910

7x-7y=0 is incorrect

41. Carissa15

it is always x+y=0 ?

42. jim_thompson5910

I don't know what you mean by "always", but 7x + 7y = 0 does reduce to x+y = 0

43. jim_thompson5910

then x+y = 0 turns into y = -x when you solve for y

44. jim_thompson5910

so $\Large \left[\begin{array}{cc|c}x \\ y\end{array}\right]=\left[\begin{array}{cc|c}x \\ -x\end{array}\right]$

45. jim_thompson5910

if we let x = 1, then $$\Large \left[\begin{array}{cc|c}1 \\ -1\end{array}\right]$$, and all of its scalar multiples, represent the other set of eigenvectors.

46. Carissa15

oh ok.

47. jim_thompson5910

I recommend computing $\Large A*\vec{v} = \lambda*\vec{v}$ $\Large \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]*\left[\begin{matrix}1 \\ 1\end{matrix}\right] = 9*\left[\begin{matrix}1 \\ 1\end{matrix}\right]$ to verify the first eigenvalue + eigenvector (lambda = 9)

48. jim_thompson5910

same for lambda = -5

49. Carissa15

I have taken the same steps for matrix b) but get stuck as I cannot factor to find eigenvalues. Please help.

50. Carissa15

$\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]$ I end up with $\lambda^2-2\lambda-1=0$ but get stuck trying to factor to find eigenvalues

51. Carissa15

eigenvalues : $$\lambda = 1\pm \sqrt{2}$$ eigenvectors : $\begin{bmatrix} 3-(1+\sqrt{2})&-2\\1&-1-(1+\sqrt{2})\end{bmatrix} \begin{pmatrix} x_1\\x_2\end{pmatrix}= 0$

52. Loser66

|dw:1443535039866:dw|

53. Loser66

then they are same, just pick one row to calculate $$(2-\sqrt 2 )x_1= 2 x_2$$ so if $$x_1 = 2 , then~~x_2 = 2-\sqrt2$$ that is $$\left[\begin{matrix} 2\\2-\sqrt 2\end{matrix}\right]$$ is your eigenvector w.r.t $$1+\sqrt2$$

54. Loser66

gtg. good luck

55. Carissa15

Thank you everyone. Solved :)