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anonymous
 one year ago
when you compute eigenvalues, the part that you need to form a base . Does the order matter? because i mostly get contradictionary results like
a(1,0,1) while in the textbook it says a(1,01) because i extracted an other variable. Does this make a difference? because the next step is combining those bases to do a diagonalization
anonymous
 one year ago
when you compute eigenvalues, the part that you need to form a base . Does the order matter? because i mostly get contradictionary results like a(1,0,1) while in the textbook it says a(1,01) because i extracted an other variable. Does this make a difference? because the next step is combining those bases to do a diagonalization

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Loser66
 one year ago
Best ResponseYou've already chosen the best response.3The order of eigenvalues doesn't matter!! but the order of theirs eigenvectors is important!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4If \(\large \mathrm{e}\) is an eigenvector for eigenvalue \(\lambda\), then any scalar multiple of it, \(\large a \mathrm{e}\) is also an eigenvector of the same eigenvalue \(\lambda\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4It helps to keep in mind that there is no "the basis" for a vector space any linearly independent vectors are fine for "a basis"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so this is the same base?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4what is your actual matrix that you're trying to diagonalize ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01 3 3 3 5 3 3 3 1

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0everything goes perfect until i need to make bases for my found eigenvalues

Loser66
 one year ago
Best ResponseYou've already chosen the best response.3Post the original problem, please, by snapshot or scanning!!

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.41, 2, 2 are your eigenvalues ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and this is the characteristic equation : dw:1437765738465:dw

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4and the corresponding eigenvector matrix is \[\begin{bmatrix} 1&1&1\\ 1&1&0\\ 1&0&1 \end{bmatrix}\] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes but i got this: 1 1 1 1 1 0 1 0 1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4doesn't matter, both your matrix and mine are same up to a scalar multiple, which is fine.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you! and the multiplicity is 2 which implies that you can make a variable 0 ? because it's a linear combination of the other ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4what do you mean by making a variable 0 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4this is a nice matrix because the geometric multiplicity for eigenvalue 2 is also 2 : you got two linearly independent eigenvectors for the eigenvalue 2!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well i find it a bit confusing . so your eigenvalue is 2 so you get after all one equation out of the matrix x+y+z=0 which gives x=yz and then they say : we make two bases , one where z is zero and one where y is zero which gives us (1,1,0) and (1,0,1)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4Exactly! the nullspace of \(A(2)I = 0\) has dimension of \(2\), thats the reason you got 2 independent eigenvectors for the eigenvalue \(2\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4so as you can see, repeated eigenvalues aren't bad always

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well it would be even more fun if we get complex forms with a bigger multiplicity than 1 :p i guess hha

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that's what proven here indeed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.4you can get multiple "linearly independent" eigenvectors for the same repeated eigenvalue

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well u really helped me a lot! and loser66 you too :) ! Thank you!!
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