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anonymous

  • one year ago

I am getting the incorrect answer for determining the eigenvectors of a matrix. I will post my work in the next post.

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  1. amistre64
    • one year ago
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    eugene vectors are such that: Ax = Ix, or succinctly; (A-I)x = 0 and im sposing we start with the characteristic equation of A-I ... if memory serves

  2. amistre64
    • one year ago
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    curious on what your matrix is tho :)

  3. anonymous
    • one year ago
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    The matrix in question is \[\left[\begin{matrix}1 & 2 \\ 2 & 3\end{matrix}\right]\] It's easy to work out the eigenvalues with the formula \[\left| A - \lambda I \right| = 0\] They are \[2+\sqrt{5}\] and \[2-\sqrt{5}\] However, when I try to determine the eigenvectors, I get the wrong answer. Please note that the eigenvectors are unit vectors, so we can use \[v _{1}^{2} + v_{2}^{2} = 1\] The correct eigenvectors are \[\left[\begin{matrix}.5257 & .8507 \\ -.8506 & .5257\end{matrix}\right]\] Attached is my work which gives me the incorrect answer. I would like to understand what I'm doing wrong.

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  4. beginnersmind
    • one year ago
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    Where do the values \(2+\sqrt{5} \text{ and } 2-\sqrt{5}\) come from?

  5. beginnersmind
    • one year ago
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    |dw:1442189681242:dw| Sorry, too lazy to use LaTex. Solving the last line should give the value of lambda. Is this what you did?

  6. anonymous
    • one year ago
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    Yes, that is what I did, but my question is not about lambda. My question is about the eigenvectors (values of V1 and V2).

  7. beginnersmind
    • one year ago
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    Ok apparently that does give \(\lambda = 2 \pm \sqrt{5} \) I'll go through the math in a second but I have a comment first. \[v _{1}^{2} + v_{2}^{2} = 1\] You should almost never use this to express v_1 as \(\sqrt{1-v_2^2{}} \). Instead you note that if v is an eigenvector then so is cv for any nonzero scalar c. So there is an eigenvector for which the first coordinate is 1.

  8. beginnersmind
    • one year ago
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    So you solve for the eigenvector by setting one of the coordinates to 1, then normalize at the end.

  9. beginnersmind
    • one year ago
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    So anyway, here's what I did. Assuming v2 = 1, v1 = 2/(1+sqrt(5)) Normalizing (v1,v2) gives (0.5257, 0.85065)

  10. beginnersmind
    • one year ago
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    This is for \(\lambda = 1+ \sqrt{5} \). I don't really want to go through the calculation in detail, because it's kind of easy, but a pain to write out. The only trick you need to remember is that you can chose one component freely, because all the eigenvectors for a given eigenvalue lie on a line (except when that component is 0, of course).

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