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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)
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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chambek
 one year ago
Best ResponseYou've already chosen the best response.1(6 1) (0 6) Lets say thats a matrix haha.....

chambek
 one year ago
Best ResponseYou've already chosen the best response.1so solving for the eigenvalues, I got lamda(1,2)= 6 Then trying to make the eigenvector got me to 0n_1=n_2

chambek
 one year ago
Best ResponseYou've already chosen the best response.1and i'm not sure how to proceed

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2I got double root L =6. not 6. can you check and we work together?

chambek
 one year ago
Best ResponseYou've already chosen the best response.1hmm....(6L)^2...wouldnt L=6 give you (6(6))^2=(12)^2 I go tthat first too

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2what is your characteristic equation? mine is L^2 12L +36 = (L 6)^2 =0 > l =6

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2oh, sorry, you are right, my bad.

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2so, L = 6 and replace to matrix Ax Lx

chambek
 one year ago
Best ResponseYou've already chosen the best response.1wait no, I did what you said and now im torn.... If you find the DET you get (6L)(6L), which is (6L)^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?

chambek
 one year ago
Best ResponseYou've already chosen the best response.1OOOHH no nvm its 12L not +.....ok, so its L=6....now what haha

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2hahaaha... so weird . I am confused many times, when my discrete pro asks me using LA while linear prof said that I must come up with A L

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2ok, anyway, it's yours, you check and choose which one is right, then we move.

chambek
 one year ago
Best ResponseYou've already chosen the best response.1shoot i just got yours again....whatever, lets move on and Ill try both (I have unlimited attempts so its all good!)

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2hi 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

chambek
 one year ago
Best ResponseYou've already chosen the best response.1alright 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

chambek
 one year ago
Best ResponseYou've already chosen the best response.1so, i guessed (1) (0) is my v_1 (0) (1) is my "generalized" eigenvector

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2to count eigenvector, you use the matrix dw:1364753722957:dw

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2you got 1 free variable, and the eigenvector is dw:1364753882341:dw

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2@niksva please, contribute , now is your turn. mine is done.

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2by the way, thanks for your medal.

niksva
 one year ago
Best ResponseYou've already chosen the best response.0@Hoa yeah u r right we are going to get only one eigen vector corresponding to repeated eigen value

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2sometimes, we can get 2 but not this case. (i've been taught that)

chambek
 one year ago
Best ResponseYou've already chosen the best response.1ok, 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?

niksva
 one year ago
Best ResponseYou've already chosen the best response.0yeah 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

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2@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

chambek
 one year ago
Best ResponseYou've already chosen the best response.1ok 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!

Hoa
 one year ago
Best ResponseYou've already chosen the best response.2good. @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
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