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zonazoo

  • 2 years ago

Finding eigenvectors: Can some just show me the steps to find the eigenvector of the following matrix, with (lambda)=2. Matrix is given below.

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  1. zonazoo
    • 2 years ago
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    \[\left[\begin{matrix}-1 & 2 & -1\\ 3 & 0 & 1\\ -3 & -2 & -3\end{matrix}\right]\]

  2. brinethery
    • 2 years ago
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    I hated linear algebra and don't remember anything, but I will suggest Paul's online notes for these (just Google is). Also, if you can get your hands on a pdf copy of 'Elementary Linear Algebra' by Larson and Falvo.

  3. zzr0ck3r
    • 2 years ago
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    first do A - bI where b is some constant variable and I is the identity matrix

  4. zzr0ck3r
    • 2 years ago
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    so [-1-b,2,-1;3,-b,1;-3,-2,-3-b]

  5. zzr0ck3r
    • 2 years ago
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    understand? b is your lamda

  6. zonazoo
    • 2 years ago
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    yes, but I thought you used A-bI=0, in order to find all the lambdas... and then use Ax=bx, in order to find the eigenvectors.

  7. zzr0ck3r
    • 2 years ago
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    o sorry you labda is given

  8. zzr0ck3r
    • 2 years ago
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    you have your lambda its 2

  9. zonazoo
    • 2 years ago
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    and I have all three eigenvalues, but for some reason when i am calculating for the eigenvector, i am not getting the right answer

  10. zzr0ck3r
    • 2 years ago
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    so solve for the null space of that equation

  11. zzr0ck3r
    • 2 years ago
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    one sec

  12. zonazoo
    • 2 years ago
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    and if i can see how to do it just for lambda = 2 , i can figure out the other two eigenvectors.

  13. TuringTest
    • 2 years ago
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    if I remember this right, you just plug in each lambda and solve (A-b_1*I)=0 (A-b_2*I)=0

  14. zzr0ck3r
    • 2 years ago
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  15. zonazoo
    • 2 years ago
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    ahhh, got it, okay... i see where i went wrong, thank you for the help

  16. zzr0ck3r
    • 2 years ago
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    np

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