anonymous
  • anonymous
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
Mathematics
jamiebookeater
  • jamiebookeater
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Loser66
  • Loser66
The order of eigenvalues doesn't matter!! but the order of theirs eigenvectors is important!!
ganeshie8
  • ganeshie8
If \(\large \mathrm{e}\) 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
  • ganeshie8
It helps to keep in mind that there is no "the basis" for a vector space any linearly independent vectors are fine for "a basis"

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Loser66
  • Loser66
|dw:1437765054056:dw|
Loser66
  • Loser66
|dw:1437765276787:dw|
Loser66
  • Loser66
got what I mean?
anonymous
  • anonymous
so this is the same base?
ganeshie8
  • ganeshie8
what is your actual matrix that you're trying to diagonalize ?
anonymous
  • anonymous
1 3 3 -3 -5 -3 -3 3 1
anonymous
  • anonymous
everything goes perfect until i need to make bases for my found eigenvalues
Loser66
  • Loser66
Post the original problem, please, by snapshot or scanning!!
anonymous
  • anonymous
ganeshie8
  • ganeshie8
1, -2, -2 are your eigenvalues ?
anonymous
  • anonymous
and this is the characteristic equation : |dw:1437765738465:dw|
anonymous
  • anonymous
yes
ganeshie8
  • ganeshie8
and the corresponding eigenvector matrix is \[\begin{bmatrix} 1&-1&-1\\ -1&1&0\\ 1&0&1 \end{bmatrix}\] ?
anonymous
  • anonymous
yes but i got this: -1 1 1 1 -1 0 1 0 -1
ganeshie8
  • ganeshie8
doesn't matter, both your matrix and mine are same up to a scalar multiple, which is fine.
anonymous
  • anonymous
thank 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
  • ganeshie8
what do you mean by making a variable 0 ?
ganeshie8
  • ganeshie8
this 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
  • anonymous
well 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=-y-z 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
  • ganeshie8
Exactly! the nullspace of \(A-(-2)I = 0\) has dimension of \(2\), thats the reason you got 2 independent eigenvectors for the eigenvalue \(-2\)
ganeshie8
  • ganeshie8
so as you can see, repeated eigenvalues aren't bad always
anonymous
  • anonymous
well it would be even more fun if we get complex forms with a bigger multiplicity than 1 :p i guess hha
anonymous
  • anonymous
that's what proven here indeed
ganeshie8
  • ganeshie8
you can get multiple "linearly independent" eigenvectors for the same repeated eigenvalue
ganeshie8
  • ganeshie8
Yea..
anonymous
  • anonymous
well u really helped me a lot! and loser66 you too :) ! Thank you!!
ganeshie8
  • ganeshie8
np

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