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Carissa15

  • one year ago

I have a few questions about eigenvalues and eigenvectors in matrices.

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  1. Carissa15
    • one year ago
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    Matrix b} \[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\]

  2. Carissa15
    • one year ago
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    Matrix a)\[\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]\]

  3. jim_thompson5910
    • one year ago
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    how far did you get with the eigenvalues of \[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\]

  4. Loser66
    • one year ago
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    What is the question?

  5. Carissa15
    • one year ago
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    I really haven't done anything with eigenvectors or eigenvalues. Very new to me. Seems though if I

  6. jim_thompson5910
    • one year ago
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    are you familiar with the idea that you need to find the determinant of the matrix \(\LARGE A - \lambda I\) to get the eigenvalues?

  7. Carissa15
    • one year ago
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    \[\det( \lambda \left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right]-\left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right])=0\]

  8. jim_thompson5910
    • one year ago
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    it should be the other way around A - (lambda)*I not (lambda)*I - A

  9. jim_thompson5910
    • one year ago
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    so you need to solve \[\Large \det( \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]-\lambda\left[\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right])=0\]

  10. Loser66
    • one year ago
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    @jim_thompson5910 Yes, it is. I had 2 profs who used different ways to find out the eigenvalues. One used lambda -A and other used A - lambda. Both worked well.

  11. jim_thompson5910
    • one year ago
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    oh I did not know that

  12. jim_thompson5910
    • one year ago
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    well however you solve the equation, what values of lambda do you get?

  13. Carissa15
    • one year ago
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    just working through now

  14. Carissa15
    • one year ago
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    I get \[\lambda^2-4\lambda-45=0\] not sure where to go from there

  15. jim_thompson5910
    • one year ago
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    use the quadratic formula to solve for lambda

  16. jim_thompson5910
    • one year ago
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    you can also factor the left side if you want

  17. Carissa15
    • one year ago
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    is the quadratic formula \[\lambda^2-\lambda-4=0\]?

  18. jim_thompson5910
    • one year ago
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    if you factored, you'll find that \[\Large \lambda^2-4\lambda-45=0\] turns into \[\Large (\lambda-9)(\lambda+5)=0\]

  19. jim_thompson5910
    • one year ago
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    so you agree that \[\Large \lambda = 9 \text{ or } \lambda = -5\] or no?

  20. Carissa15
    • one year ago
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    yes

  21. Carissa15
    • one year ago
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    so these are the eigenvalues?

  22. jim_thompson5910
    • one year ago
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    correct

  23. Carissa15
    • one year ago
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    I am not sure how to now find eigenectors using these

  24. jim_thompson5910
    • one year ago
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    \[\Large A\vec{v} = \lambda \vec{v}\] \[\Large A\vec{v} - \lambda \vec{v}=0\] \[\Large A\vec{v} - \lambda I\vec{v}=0\] \[\Large (A - \lambda I)\vec{v}=0\] That last equation leads us to this matrix equation \[\Large \begin{bmatrix}2-\lambda & 7\\7&2-\lambda\end{bmatrix} \vec{v} = \begin{bmatrix}0\\0\end{bmatrix}\] which leads to this augmented matrix \[\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]\]

  25. jim_thompson5910
    • one year ago
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    If lambda = 9, then \[\Large \left[\begin{array}{cc|c}2-\lambda & 7 & 0 \\7 & 2-\lambda & 0\end{array}\right]\] \[\Large \left[\begin{array}{cc|c}2-9 & 7 & 0 \\7 & 2-9 & 0\end{array}\right]\] \[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7 & -7 & 0\end{array}\right]\] row reduce the augmented matrix until it's in rref form

  26. Carissa15
    • one year ago
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    So we would end up with R1+R2 and all values zero?

  27. jim_thompson5910
    • one year ago
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    you do compute R1+R2, yes but we only need to do that for one row R1+R2 = [0 0] replace R2 with that result \[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\7&-7 & 0\end{array}\right]\] \[\Large \left[\begin{array}{cc|c}-7 & 7 & 0 \\0&0 & 0\end{array}\right]\begin{array} \ \\ R1+R2 \rightarrow R2\end{array}\]

  28. Carissa15
    • one year ago
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    Oh yes, of course. not both oops.

  29. jim_thompson5910
    • one year ago
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    the top row -7 7 0 means -7x+7y = 0 solve for y to get 7y = 7x y = x so the solution set is the column vector \(\LARGE \left[\begin{array}{c}x \\x\end{array}\right]\) this describes all of the eigenvectors for the eigenvalue lambda = 9

  30. jim_thompson5910
    • one year ago
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    a lot of books may say let x = 1, so a particular eigenvector for lambda = 9 is \[\LARGE \left[\begin{array}{c}1 \\1\end{array}\right]\] all other eigenvectors for lambda = 9 are simply scalar multiples of that column vector

  31. Carissa15
    • one year ago
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    Would this always be the case? Or do you sometimes get actual values instead of x? And Do I then follow the same steps for lambda -5?

  32. jim_thompson5910
    • one year ago
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    yes I've had teachers who would assign values to x. 1 is the smallest possible to allow for other multiples to be generated

  33. jim_thompson5910
    • one year ago
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    and yes you follow the same steps for lambda = -5

  34. Carissa15
    • one year ago
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    ok great. Thank you. I will work my way through them and see how I go. :)

  35. jim_thompson5910
    • one year ago
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    ok tell me what you get

  36. Carissa15
    • one year ago
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    for lambda -5 i got\[\left[\begin{matrix}7 & 7 \\ 7 & 7\end{matrix}\right]\\] with Identity matrix

  37. Carissa15
    • one year ago
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    then did R1-R2 gives new R2 of all zeros.

  38. Carissa15
    • one year ago
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    I then got 7x-7y=0, which then ends up with the same result as lambda 9

  39. jim_thompson5910
    • one year ago
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    I'm getting the same, so you have 7 7 0 up top over 0 0 0 the top row implies that 7x + 7y = 0

  40. jim_thompson5910
    • one year ago
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    7x-7y=0 is incorrect

  41. Carissa15
    • one year ago
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    it is always x+y=0 ?

  42. jim_thompson5910
    • one year ago
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    I don't know what you mean by "always", but 7x + 7y = 0 does reduce to x+y = 0

  43. jim_thompson5910
    • one year ago
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    then x+y = 0 turns into y = -x when you solve for y

  44. jim_thompson5910
    • one year ago
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    so \[\Large \left[\begin{array}{cc|c}x \\ y\end{array}\right]=\left[\begin{array}{cc|c}x \\ -x\end{array}\right]\]

  45. jim_thompson5910
    • one year ago
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    if we let x = 1, then \(\Large \left[\begin{array}{cc|c}1 \\ -1\end{array}\right]\), and all of its scalar multiples, represent the other set of eigenvectors.

  46. Carissa15
    • one year ago
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    oh ok.

  47. jim_thompson5910
    • one year ago
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    I recommend computing \[\Large A*\vec{v} = \lambda*\vec{v}\] \[\Large \left[\begin{matrix}2 & 7 \\ 7 & 2\end{matrix}\right]*\left[\begin{matrix}1 \\ 1\end{matrix}\right] = 9*\left[\begin{matrix}1 \\ 1\end{matrix}\right] \] to verify the first eigenvalue + eigenvector (lambda = 9)

  48. jim_thompson5910
    • one year ago
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    same for lambda = -5

  49. Carissa15
    • one year ago
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    I have taken the same steps for matrix b) but get stuck as I cannot factor to find eigenvalues. Please help.

  50. Carissa15
    • one year ago
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    \[\left[\begin{matrix}3 & -2 \\ 1 & -1\end{matrix}\right]\] I end up with \[\lambda^2-2\lambda-1=0\] but get stuck trying to factor to find eigenvalues

  51. Carissa15
    • one year ago
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    eigenvalues : \(\lambda = 1\pm \sqrt{2}\) eigenvectors : \[\begin{bmatrix} 3-(1+\sqrt{2})&-2\\1&-1-(1+\sqrt{2})\end{bmatrix} \begin{pmatrix} x_1\\x_2\end{pmatrix}= 0\]

  52. Loser66
    • one year ago
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    |dw:1443535039866:dw|

  53. Loser66
    • one year ago
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    then they are same, just pick one row to calculate \((2-\sqrt 2 )x_1= 2 x_2\) so if \(x_1 = 2 , then~~x_2 = 2-\sqrt2\) that is \(\left[\begin{matrix} 2\\2-\sqrt 2\end{matrix}\right]\) is your eigenvector w.r.t \(1+\sqrt2\)

  54. Loser66
    • one year ago
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    gtg. good luck

  55. Carissa15
    • one year ago
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    Thank you everyone. Solved :)

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