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chambek

  • one year ago

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

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  1. chambek
    • one year ago
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    (-6 1) (0 -6) Lets say thats a matrix haha.....

  2. chambek
    • one year ago
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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

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

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

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

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

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

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

  9. chambek
    • one year ago
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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?

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

  11. Hoa
    • one year ago
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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

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

  13. chambek
    • one year ago
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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!)

  14. Hoa
    • one year ago
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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

  15. chambek
    • one year ago
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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

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

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

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

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

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

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

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

  23. chambek
    • one year ago
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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?

  24. niksva
    • one year ago
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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

  25. Hoa
    • one year ago
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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|

  26. chambek
    • one year ago
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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!

  27. Hoa
    • one year ago
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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

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

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