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anonymous

  • one year ago

what is the eigenvector of this matrix. The eigenvalue of 3 has already been plugged in

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  1. anonymous
    • one year ago
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    |dw:1433715559882:dw|

  2. anonymous
    • one year ago
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    i've simplified it to 2x-y+z=0 how do i make that into an eigenvector?

  3. anonymous
    • one year ago
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    @iambatman

  4. freckles
    • one year ago
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    I'm having trouble seeing how got an egivenvalue of 3

  5. anonymous
    • one year ago
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    Okay, so let's check and see if 3 works then

  6. anonymous
    • one year ago
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    the original matrix was |dw:1433716009320:dw|

  7. anonymous
    • one year ago
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    I found that one eigenvalue was 3 and the other two were complex numbers

  8. anonymous
    • one year ago
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    I'm just trying to find the eigenvector for 3 right now

  9. anonymous
    • one year ago
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    I plugged 3 into the matrix to get the matrix that i originally posted

  10. anonymous
    • one year ago
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    I just realized the other two eigenvalues aren't complex btw

  11. freckles
    • one year ago
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    for this other matrix you have I see how you got 3 now

  12. freckles
    • one year ago
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    for the first matrix all the egienvalues were 0

  13. anonymous
    • one year ago
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    sorry i wasn't clear. The first matrix is the second matrix with the eigenvalue 3 plugged in

  14. anonymous
    • one year ago
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    |dw:1433716555159:dw|

  15. anonymous
    • one year ago
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    which produced

  16. anonymous
    • one year ago
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    |dw:1433716643997:dw|

  17. anonymous
    • one year ago
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    after i simplified it a bit

  18. freckles
    • one year ago
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    ok so lambda=3 I'm going to do A-Lambda*I=0 \[\left[\begin{matrix}7 & -2 & 2 \\ 8 &-1 & 4 \\ -4 & 2 & 1\end{matrix}\right]- \left[\begin{matrix}3 & 0 & 0 \\ 0 & 3 &0 \\ 0 & 0 & 3\end{matrix}\right] \\ =\left[\begin{matrix}4 & -2 & 2 \\ 8 & -4 & 4 \\ -4 & 2 & -2\end{matrix}\right] =\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] \[\left[\begin{matrix}2 & -1 & 1 \\ 2 & -1 & 1 \\ -2 & 1 & -1 \end{matrix}\right]=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] ok I'm cool with that

  19. freckles
    • one year ago
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    oops forgot to put my one column vector in

  20. anonymous
    • one year ago
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    after simplifying i got 2x-y+z=0

  21. anonymous
    • one year ago
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    the second and third row should cancel

  22. freckles
    • one year ago
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    \[(A- \lambda I)x=0 \\ \left[\begin{matrix}2 & -1 & 1 \\ 2 & -1 & 1 \\-2 & 1 & -1 \end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right)=\left(\begin{matrix}0 \\ 0 \\ 0\end{matrix}\right)\] so we have only need to look at \[2z-y+z=0\] as you said

  23. freckles
    • one year ago
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    err 2x-y+z=0

  24. freckles
    • one year ago
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    \[2x-y+z=0 \\ \text{ one solution is } (1,1,-1) \\ \text{ now also say we replace } x=t \text{ then } 2t=y-z \\ \text{ so we could replace } y=3t \text{ and } z=t \\ \text{ so we have } \left(\begin{matrix}1 \\ 1 \\ -1 \end{matrix}\right)+t \left(\begin{matrix}1 \\ 2 \\ 1\end{matrix}\right)\] I think I remember something like this

  25. freckles
    • one year ago
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    that 2 should be 3

  26. anonymous
    • one year ago
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    so can i use ( 1 , 1 , -1) as the eigenvector?

  27. freckles
    • one year ago
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    as one egienvector

  28. freckles
    • one year ago
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    3 is repeated twice

  29. freckles
    • one year ago
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    ok so we have to find a second egienvector

  30. freckles
    • one year ago
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    for lambda=3

  31. freckles
    • one year ago
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    we can use the (1,1,-1) as the first

  32. anonymous
    • one year ago
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    ok I get it now, thanks a lot!

  33. freckles
    • one year ago
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    \[\left[\begin{matrix}4 & -2 & 2 \\ 8 & -4 & 4 \\ -4 & 2 & -2\end{matrix}\right] \left(\begin{matrix}x \\ y \\ z\end{matrix}\right) =\left(\begin{matrix}1 \\ 1 \\ -1\end{matrix}\right)\] to find the second one you will have to solve this for (x,y,z)

  34. freckles
    • one year ago
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    notice I replace that zero vector with the other egienvector we found for lambda=3

  35. anonymous
    • one year ago
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    I think i just need one eigenvector, i need it to solve for a system of differential equations problem

  36. anonymous
    • one year ago
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    Thanks for the help!

  37. freckles
    • one year ago
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    http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-iv-first-order-systems/matrix-methods-eigenvalues-and-normal-modes/MIT18_03SCF11_s33_8text.pdf could be wrong I think you will need to other egienvector

  38. freckles
    • one year ago
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    sorry the first link might not work

  39. anonymous
    • one year ago
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    Was 3 a repeated eigenvalue?

  40. freckles
    • one year ago
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    yah yah!

  41. freckles
    • one year ago
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    I think he is going to need that other egienvector

  42. anonymous
    • one year ago
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    Then I would think youd definitely need another eigenvector. Well, not technically an eigenvector, but a generalized one. If I recall, we'd need this: \[(A-3I)\eta = \xi \] Where \(\xi\) is the first eigenvector. I have no idea how to type up matrices on here though, lol.

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