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

chambek Group TitleBest ResponseYou've already chosen the best response.1
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

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

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

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

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

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

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

chambek Group TitleBest ResponseYou've already chosen the best response.1
wait 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?
 one year ago

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

Hoa Group TitleBest ResponseYou've already chosen the best response.2
hahaaha... 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
 one year ago

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

chambek Group TitleBest ResponseYou've already chosen the best response.1
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

Hoa Group TitleBest ResponseYou've already chosen the best response.2
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

chambek Group TitleBest ResponseYou've already chosen the best response.1
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

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

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

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

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

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

niksva Group TitleBest 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
 one year ago

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

chambek Group TitleBest ResponseYou've already chosen the best response.1
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

niksva Group TitleBest ResponseYou've already chosen the best response.0
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

Hoa Group TitleBest 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
 one year ago

chambek Group TitleBest ResponseYou've already chosen the best response.1
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

Hoa Group TitleBest ResponseYou've already chosen the best response.2
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
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