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chambek Group Title

When finding eigenvectors, what do you do when you reach a dead end like this? (example shown in comment)

  • one year ago
  • one year ago

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  1. chambek Group Title
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    (-6 1) (0 -6) Lets say thats a matrix haha.....

    • one year ago
  2. chambek Group Title
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    so solving for the eigenvalues, I got lamda(1,2)= -6 Then trying to make the eigenvector got me to 0n_1=n_2

    • one year ago
  3. chambek Group Title
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    and i'm not sure how to proceed

    • one year ago
  4. Hoa Group Title
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    I got double root L =6. not -6. can you check and we work together?

    • one year ago
  5. chambek Group Title
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    hmm....(-6-L)^2...wouldnt L=6 give you (-6-(6))^2=(-12)^2 I go tthat first too

    • one year ago
  6. Hoa Group Title
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    what is your characteristic equation? mine is L^2 -12L +36 = (L -6)^2 =0 ---> l =6

    • one year ago
  7. Hoa Group Title
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    oh, sorry, you are right, my bad.

    • one year ago
  8. Hoa Group Title
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    so, L = -6 and replace to matrix Ax -Lx

    • one year ago
  9. chambek Group Title
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    wait no, I did what you said and now im torn.... If you find the DET you get (-6-L)(-6-L), which is (-6-L)^2.....(obviously haha) but if you multiply them together you get (L^2+12L+36), which could be turned into (L+6)^2, thus making you right......HHHOOOWW?

    • one year ago
  10. chambek Group Title
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    OOOHH no nvm its -12L not +.....ok, so its L=-6....now what haha

    • one year ago
  11. Hoa Group Title
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    hahaaha... so weird . I am confused many times, when my discrete pro asks me using L-A while linear prof said that I must come up with A -L

    • one year ago
  12. Hoa Group Title
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    ok, anyway, it's yours, you check and choose which one is right, then we move.

    • one year ago
  13. chambek Group Title
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    shoot i just got yours again....whatever, lets move on and Ill try both (I have unlimited attempts so its all good!)

    • one year ago
  14. Hoa Group Title
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    hi guy!! your brother must go somewhere to eat something, I cannot stay here to witness you switch many times like that. I'll be back in 15 minutes. do it carefully

    • one year ago
  15. chambek Group Title
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    alright enjoy your meal, I got my final answer, and it's still L=-6. But I really just need to know how to go about turning that into and eigenvector and a generalized eigenvector. thanks for your help so far

    • one year ago
  16. chambek Group Title
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    so, i guessed (1) (0) is my v_1 (0) (1) is my "generalized" eigenvector

    • one year ago
  17. Hoa Group Title
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    to count eigenvector, you use the matrix |dw:1364753722957:dw|

    • one year ago
  18. Hoa Group Title
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    you got 1 free variable, and the eigenvector is |dw:1364753882341:dw|

    • one year ago
  19. Hoa Group Title
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    @niksva please, contribute , now is your turn. mine is done.

    • one year ago
  20. Hoa Group Title
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    by the way, thanks for your medal.

    • one year ago
  21. niksva Group Title
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    @Hoa yeah u r right we are going to get only one eigen vector corresponding to repeated eigen value

    • one year ago
  22. Hoa Group Title
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    sometimes, we can get 2 but not this case. (i've been taught that)

    • one year ago
  23. chambek Group Title
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    ok, so you pretty much do the same thing as if the "free variable" was any other, and create the most simple eigenvector from that....makes sense, but then how does the "generalized" eigenvector of (0) (1) come to be?

    • one year ago
  24. niksva Group Title
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    yeah we should always get same number of eigen values and eigen vectors but in this case it is not possible as after solving the equation we are getting x = 0 and y = any imaginary number or u can say any constant value

    • one year ago
  25. Hoa Group Title
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    @chambek to each eigenvalue, you have a "family" of eigenvector, yours is (0,1) is simplest one, so to generalize, just time it with a scalar . I set t so you have |dw:1364754299747:dw|

    • one year ago
  26. chambek Group Title
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    ok I get that now Thank You both for the help....now on to part 2! Finding the most generalized real valued solution using t as my independent variable! think i might have this on my own, ill make a new question if i need help, thanks!

    • one year ago
  27. Hoa Group Title
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    good. @niksva you will be disappointed when being my fan, sooner or later, you will click on unfan button. do it now is better, save time for future

    • one year ago
  28. niksva Group Title
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    :-P :-)

    • one year ago
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