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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

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  1. Loser66
    • one year ago
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    The order of eigenvalues doesn't matter!! but the order of theirs eigenvectors is important!!

  2. ganeshie8
    • one year ago
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    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 \(\lambda\)

  3. ganeshie8
    • one year ago
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    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"

  4. Loser66
    • one year ago
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    |dw:1437765054056:dw|

  5. Loser66
    • one year ago
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    |dw:1437765276787:dw|

  6. Loser66
    • one year ago
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    got what I mean?

  7. anonymous
    • one year ago
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    so this is the same base?

  8. ganeshie8
    • one year ago
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    what is your actual matrix that you're trying to diagonalize ?

  9. anonymous
    • one year ago
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    1 3 3 -3 -5 -3 -3 3 1

  10. anonymous
    • one year ago
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    everything goes perfect until i need to make bases for my found eigenvalues

  11. Loser66
    • one year ago
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    Post the original problem, please, by snapshot or scanning!!

  12. anonymous
    • one year ago
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  13. ganeshie8
    • one year ago
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    1, -2, -2 are your eigenvalues ?

  14. anonymous
    • one year ago
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    and this is the characteristic equation : |dw:1437765738465:dw|

  15. anonymous
    • one year ago
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    yes

  16. ganeshie8
    • one year ago
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    and the corresponding eigenvector matrix is \[\begin{bmatrix} 1&-1&-1\\ -1&1&0\\ 1&0&1 \end{bmatrix}\] ?

  17. anonymous
    • one year ago
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    yes but i got this: -1 1 1 1 -1 0 1 0 -1

  18. ganeshie8
    • one year ago
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    doesn't matter, both your matrix and mine are same up to a scalar multiple, which is fine.

  19. anonymous
    • one year ago
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    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 ?

  20. ganeshie8
    • one year ago
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    what do you mean by making a variable 0 ?

  21. ganeshie8
    • one year ago
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    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!

  22. anonymous
    • one year ago
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    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)

  23. ganeshie8
    • one year ago
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    Exactly! the nullspace of \(A-(-2)I = 0\) has dimension of \(2\), thats the reason you got 2 independent eigenvectors for the eigenvalue \(-2\)

  24. ganeshie8
    • one year ago
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    so as you can see, repeated eigenvalues aren't bad always

  25. anonymous
    • one year ago
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    well it would be even more fun if we get complex forms with a bigger multiplicity than 1 :p i guess hha

  26. anonymous
    • one year ago
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    that's what proven here indeed

  27. ganeshie8
    • one year ago
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    you can get multiple "linearly independent" eigenvectors for the same repeated eigenvalue

  28. ganeshie8
    • one year ago
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    Yea..

  29. anonymous
    • one year ago
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    well u really helped me a lot! and loser66 you too :) ! Thank you!!

  30. ganeshie8
    • one year ago
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    np

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