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elica85

how do i find the eigenvector?

  • 2 years ago
  • 2 years ago

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  1. exraven
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    \[\left( A-\lambda I \right)x=0\]lamda is the eigenvalue, x is the eigenvector

    • 2 years ago
  2. elica85
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    i was able to find the eigenvalue \[\left[\begin{matrix}-6 & 1 \\ -2 & -3\end{matrix}\right]\] to be \[\lambda _{1}=-4, \lambda _{2}=-5\] now what's the next step?

    • 2 years ago
  3. myininaya
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    @nightshade please don't post off-topic answers. @exraven and elica85 we want x to not be the zero vector also

    • 2 years ago
  4. myininaya
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    Did you try to plug into exraven's formula to find the eigenvectors

    • 2 years ago
  5. elica85
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    i thought it's lambdaI-A, not A-lambdaI

    • 2 years ago
  6. elica85
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    idk what i'm doing, my school doesn't require us to take linear algebra to take diff eq so i'm suppose to know what to do but i don't

    • 2 years ago
  7. elica85
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    o if i just have to find x, i should be able to...

    • 2 years ago
  8. myininaya
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    \[(A-\lambda_n \cdot I)x=0\] For \[\lambda_1=-4\] |dw:1333932816786:dw|

    • 2 years ago
  9. myininaya
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    For \[\lambda_2=-5\] |dw:1333932896865:dw|

    • 2 years ago
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