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

- anonymous

- chestercat

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- anonymous

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

- anonymous

so solving for the eigenvalues, I got lamda(1,2)= -6
Then trying to make the eigenvector got me to
0n_1=n_2

- anonymous

and i'm not sure how to proceed

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- anonymous

I got double root L =6. not -6. can you check and we work together?

- anonymous

hmm....(-6-L)^2...wouldnt L=6 give you (-6-(6))^2=(-12)^2
I go tthat first too

- anonymous

what is your characteristic equation? mine is L^2 -12L +36 = (L -6)^2 =0 ---> l =6

- anonymous

oh, sorry, you are right, my bad.

- anonymous

so, L = -6 and replace to matrix Ax -Lx

- anonymous

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?

- anonymous

OOOHH no nvm its -12L not +.....ok, so its L=-6....now what haha

- anonymous

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

- anonymous

ok, anyway, it's yours, you check and choose which one is right, then we move.

- anonymous

shoot i just got yours again....whatever, lets move on and Ill try both (I have unlimited attempts so its all good!)

- anonymous

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

- anonymous

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

- anonymous

so, i guessed
(1)
(0) is my v_1
(0)
(1) is my "generalized" eigenvector

- anonymous

to count eigenvector, you use the matrix |dw:1364753722957:dw|

- anonymous

you got 1 free variable, and the eigenvector is |dw:1364753882341:dw|

- anonymous

@niksva please, contribute , now is your turn. mine is done.

- anonymous

by the way, thanks for your medal.

- anonymous

@Hoa yeah u r right we are going to get only one eigen vector corresponding to repeated eigen value

- anonymous

sometimes, we can get 2 but not this case. (i've been taught that)

- anonymous

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?

- anonymous

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

- anonymous

@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

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!

- anonymous

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

- anonymous

:-P :-)

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