A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
When finding eigenvectors, what do you do when you reach a dead end like this? (example shown in comment)
anonymous
 3 years ago
When finding eigenvectors, what do you do when you reach a dead end like this? (example shown in comment)

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(6 1) (0 6) Lets say thats a matrix haha.....

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0and i'm not sure how to proceed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I got double root L =6. not 6. can you check and we work together?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm....(6L)^2...wouldnt L=6 give you (6(6))^2=(12)^2 I go tthat first too

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh, sorry, you are right, my bad.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so, L = 6 and replace to matrix Ax Lx

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0wait 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0OOOHH no nvm its 12L not +.....ok, so its L=6....now what haha

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hahaaha... 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, anyway, it's yours, you check and choose which one is right, then we move.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0shoot i just got yours again....whatever, lets move on and Ill try both (I have unlimited attempts so its all good!)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hi 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0alright 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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0to count eigenvector, you use the matrix dw:1364753722957:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you got 1 free variable, and the eigenvector is dw:1364753882341:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@niksva please, contribute , now is your turn. mine is done.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by the way, thanks for your medal.

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sometimes, we can get 2 but not this case. (i've been taught that)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok, 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?

anonymous
 3 years 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok 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!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0good. @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
Ask your own question
Sign UpFind more explanations on OpenStudy
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
 Engagement 19 Mad Hatter
 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.