## Mimi_x3 one year ago something is wrong

1. Mimi_x3

|dw:1439098716867:dw|

2. Mimi_x3

I'm trying to find the eigenvector

3. Mimi_x3

so 0v1 + 1v2 = 0 v2 = 0 v1 = 0

4. Mimi_x3

https://www.wolframalpha.com/input/?i=eigenvectors+ {{2%2C1}%2C{0%2C2}} BUT LOOK HERE

5. UsukiDoll

did you compute the eigenvalues first?

6. UsukiDoll

|dw:1439100292165:dw|

7. UsukiDoll

then using $\det(A- \lambda I) = 0$ |dw:1439100348042:dw|

8. UsukiDoll

|dw:1439100402741:dw|

9. UsukiDoll

$(-\lambda)(-\lambda) -v_2(0) = 0$ $\lambda^2-0=0$ $\lambda^2 = 0$ $\lambda = \sqrt{0}$ $\lambda = 0$ both of our eigenvalues so be 0 $\lambda_1 = 0, \lambda_2=0$

10. UsukiDoll

we will have 2 eigenvectors... and one of them all 0's.

11. UsukiDoll

we should have at least a row of zeros after doing row operations to find eigenvectors so when $\lambda = 0$ |dw:1439100836674:dw| I drew the matrix in augmented format

12. UsukiDoll

we have a row of zeros, so we're on the right track. We have a single equation which is v_2 = 0

13. UsukiDoll

eigenvectors are in the form of $[v_1,v_2...v_n]$ except it's written vertically. |dw:1439101413091:dw|

14. Mimi_x3

wait

15. Mimi_x3

my eigen values is 2 from |dw:1439101590357:dw|

16. UsukiDoll

huh? what was the original question?

17. Mimi_x3
18. Mimi_x3

I found the eigen value => 2

19. Mimi_x3

but the eigen vector does not seem to be right

20. UsukiDoll

ok ... so we have to find eigenvalues and eigenvectors of 2 1 0 2 is this the right 2 x 2 matrix?

21. Mimi_x3

yeah

22. UsukiDoll

ok gonna take a shortcut |dw:1439101734333:dw| ah! we have a repeated eigenvalue $\lambda = 2$

23. Mimi_x3

yeah so one eigen value

24. UsukiDoll

I remember this.... if I remembered correctly, there's two eigenvector... one all 0's and one we just have to place $\lambda = 2$ back inside the matrix, do some row operations, and see if we can obtain a row of zeros.

25. Mimi_x3

there's only one eigen vector..

26. Mimi_x3

so i get this at the end |dw:1439101932129:dw|

27. UsukiDoll

there could be 1 repeat eigenvalue 1 eigenvector ( the phase portrait is an improper node) or 1 repeat eigenvalue 2 eigenvectors (the phase portrait is a proper node) . we need to do a bit of matrix multiplication for this |dw:1439102328212:dw|

28. UsukiDoll

so if we have v_2 =0 v_1 must have been set at 0 (I'm starting to feel weak in real life... gotta eat dinner)

29. Mimi_x3

ok |dw:1439102632011:dw|

30. Mimi_x3

0v1 + v2 = 0 0v1 + 0v2 = 0

31. UsukiDoll

@Michele_Laino

32. Michele_Laino

what is the starting matrix?

33. UsukiDoll
34. Michele_Laino

or better: if the starting matrix is: $M = \left( {\begin{array}{*{20}{c}} 0&0 \\ 0&1 \end{array}} \right)$ then the eqiuation to eigenvalues is: $\left\{ {M - \lambda \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)} \right\}v = 0$ being v a non null vector

35. UsukiDoll

?! nugh. we know there is a repeated eigenvalue which is 2 however there could be 1 eigenvector or 2 eigenvectors. It's the eigenvectors that I got stuck on. We do know that v_2 = 0... but what is v_1?

36. Michele_Laino

ok! if the starting matrix is: $N = \left( {\begin{array}{*{20}{c}} 2&1 \\ 0&2 \end{array}} \right)$ then we have to solve this vector equation: $\left\{ {N - \lambda \left( {\begin{array}{*{20}{c}} 1&0 \\ 0&1 \end{array}} \right)} \right\}v = 0$ where, as usually, v can not be the zero vector

37. UsukiDoll

isn't v is the right eigenvector and w the left eigenvector?

38. Michele_Laino

so, developing that formula, we get: $\left( {\begin{array}{*{20}{c}} {2 - \lambda }&1 \\ 0&{2 - \lambda } \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{v_1}} \\ {{v_2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$

39. Michele_Laino

we have to request this condition: $\det \left( {\begin{array}{*{20}{c}} {2 - \lambda }&1 \\ 0&{2 - \lambda } \end{array}} \right) = 0$

40. Michele_Laino

namely: ${\left( {2 - \lambda } \right)^2} = 0$

41. UsukiDoll

yeah where there is a repeated eigenvalue and that's 2.

42. UsukiDoll

so now we plug 2 back into the matrix... we already have our row of zeros which is a good sign that we've done this correctly.. to get eigenvector

43. Michele_Laino

the value \lambda =2 is double

44. Michele_Laino

yes! we have to substitute \lambda=2 into that equation, and find the component of v, namely v_1 and v_2

45. UsukiDoll

that's what drives me nuts 0 1 l 0 0 0 l 0 0v_1+v_2=0 0_v1+0v_2=0 we only have one equation 0v_1+v_2=0 and that's just v_2 = 0

46. Michele_Laino

we got this: $\left( {\begin{array}{*{20}{c}} 0&1 \\ 0&0 \end{array}} \right)\left( {\begin{array}{*{20}{c}} {{v_1}} \\ {{v_2}} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} {{v_2}} \\ 0 \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 0 \\ 0 \end{array}} \right)$

47. UsukiDoll

yeah

48. Michele_Laino

right! v_2=0, and v_1 can be any real number, so we have a line of eigenvector

49. UsukiDoll

I did matrix multiplication way earlier got |dw:1439107968867:dw|

50. UsukiDoll

can v_1 be 1? but how?!

51. Michele_Laino

namely our eigenvectors are these ones: $\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right),\quad k \in \mathbb{R}$

52. UsukiDoll

where k belong in a set of reals.

53. Michele_Laino

yes! that's right!

54. UsukiDoll

so k belongs in a set of reals k must be a real number

55. Michele_Laino

yes! more precisely, since our matrix can be viewed as an operator between real vector spaces, then k has to be real

56. UsukiDoll

like a 1?

57. Michele_Laino

k is varying in the real line, namely k is not fixed

58. UsukiDoll

:/ stuck! k can be anything 1 2 3 4 5 6 7 8 9

59. Michele_Laino

k can be any real number

60. UsukiDoll

so it can be rational or irrational.

61. Michele_Laino

both rational and irrational

62. UsukiDoll

so what are we doing ? just leaving the other vector as |dw:1439108587318:dw| ?

63. Michele_Laino

yes!

64. UsukiDoll

ok but does it count as another eigenvector?

65. Michele_Laino

the algebraic multiplicity is 2, whereas the geometric multiplicity is 1, namely the dimension of the subspace is 1, since it is a line, so our matrix can not be diagonalized

66. Michele_Laino

more precisely we hav an infinity of eigenvectors, since we can write this: $\left( {\begin{array}{*{20}{c}} 2&1 \\ 0&2 \end{array}} \right)\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right) = 2\left( {\begin{array}{*{20}{c}} k \\ 0 \end{array}} \right),\quad k \in \mathbb{R}$

67. Michele_Laino

geometrically speaking, we have a line of eigenvectors, and that line has to pass at the origin of the cartesian system

68. phi

a few thoughts: an eigenvector is only defined up to a constant. What we care about is its direction; its magnitude can be any value (except zero) for example if we have this eigen equation $A \vec{v} = \lambda \vec{v}$ and we scale the eigenvector by a scalar c: $A (c\vec{v} ) = \lambda (c\vec{v} )$ the equation is still satisfied. a triangular matrix has its eigenvalues on the diagonal. We immediately see that we have a repeated eigenvalue of 2 According to Prof Strang see http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/lecture-21-eigenvalues-and-eigenvectors/ at the end of the lecture he points out that a version of your matrix (his has 3 where yours has 2) has *only one* eigenvector. (the [0 0 ] vector does not carry any info for us) thus if you are solving the differential equation ${\dot{x}}= Ax$ with eigenvectors, you will run into trouble. notice you can solve the 2nd equation $\frac{d \ x_2}{dt}= 2x_2$ using separation of variables to get $x_2 = c_2 e^{2t}$ substitute that into the first equation and solve "normally?" to get $x_1 = c_1 e^{2t} + c_2 t e^{2t}$