A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

Mimi_x3

  • one year ago

something is wrong

  • This Question is Closed
  1. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    |dw:1439098716867:dw|

  2. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I'm trying to find the eigenvector

  3. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  5. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    did you compute the eigenvalues first?

  6. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1439100292165:dw|

  7. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  8. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    |dw:1439100402741:dw|

  9. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  10. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  11. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  14. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    wait

  15. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  16. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    huh? what was the original question?

  17. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6

  18. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    I found the eigen value => 2

  19. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but the eigen vector does not seem to be right

  20. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah

  22. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  23. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    yeah so one eigen value

  24. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    there's only one eigen vector..

  26. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  27. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    ok |dw:1439102632011:dw|

  30. Mimi_x3
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  31. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    @Michele_Laino

  32. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    what is the starting matrix?

  33. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    https://gyazo.com/3f46ed9dd85e81fc5b637b27dcfbe1b6

  34. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ?! 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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  38. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  41. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  42. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    the value \lambda =2 is double

  44. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  45. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    yeah

  48. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  49. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  50. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can v_1 be 1? but how?!

  51. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  52. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    where k belong in a set of reals.

  53. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes! that's right!

  54. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  55. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  56. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    like a 1?

  57. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  58. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  59. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    k can be any real number

  60. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    so it can be rational or irrational.

  61. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    both rational and irrational

  62. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  63. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    yes!

  64. UsukiDoll
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    ok but does it count as another eigenvector?

  65. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  68. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 \[ x_1 = c_1 e^{2t} + c_2 t e^{2t} \]

  69. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.