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(-6 1) (0 -6) Lets say thats a matrix haha.....
so solving for the eigenvalues, I got lamda(1,2)= -6 Then trying to make the eigenvector got me to 0n_1=n_2
and i'm not sure how to proceed
I got double root L =6. not -6. can you check and we work together?
hmm....(-6-L)^2...wouldnt L=6 give you (-6-(6))^2=(-12)^2 I go tthat first too
what is your characteristic equation? mine is L^2 -12L +36 = (L -6)^2 =0 ---> l =6
oh, sorry, you are right, my bad.
so, L = -6 and replace to matrix Ax -Lx
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?
OOOHH no nvm its -12L not +.....ok, so its L=-6....now what haha
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
ok, anyway, it's yours, you check and choose which one is right, then we move.
shoot i just got yours again....whatever, lets move on and Ill try both (I have unlimited attempts so its all good!)
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
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
so, i guessed (1) (0) is my v_1 (0) (1) is my "generalized" eigenvector
to count eigenvector, you use the matrix |dw:1364753722957:dw|
you got 1 free variable, and the eigenvector is |dw:1364753882341:dw|
by the way, thanks for your medal.
sometimes, we can get 2 but not this case. (i've been taught that)
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?
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
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!