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Homework Help: What are the eingenvalues?

  1. Nov 8, 2008 #1
    1. 1 3 0
    3 -2 -1
    0 -1 1
    What are the eingenvalues? (they are 1, 3 and -4)




    2. 1-n 3 0
    3 -2-n -1
    0 -1 1-n




    I basically get 4n-n^3 -3 = 0
    I'm not sure what to do from there?

    Working: (1-n). (n+2).(n-1) -1) - 3 (3 (1-n))


    Thanks guys!
     
  2. jcsd
  3. Nov 8, 2008 #2

    HallsofIvy

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    Re: Eigenvalues!


    That is not at all the equation I get. Please show in detail how you arrived at that.

     
  4. Nov 8, 2008 #3

    tiny-tim

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    Hi Davio! :smile:
    For a start, you can see that (1 - n) divides everything, so leave that on the outside …

    (1 - n){(n+2)(n-1) - ? - ?} …

    carry on from there! :smile:
     
  5. Nov 8, 2008 #4

    HallsofIvy

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    Re: Eigenvalues!

    Yes, n= 1 satisfies that equation but clearly n= 3 and n= -4 do not. His real problem is he has the wrong equation.
     
  6. Nov 8, 2008 #5
    Re: Eigenvalues!

    Hi guys.
    1-n 3 0
    3 -2-n -1
    0 -1 1-n

    My working for determinant :
    + 1-n x (-2-n)(1-n) - (-1).(-1)
    + 1-n x (-2-n)(1-n) -1
    + 1-n x (n^2 +2n -n -2) -1
    + 1-n x (n^2 +2n -n -3)
    + 1-n x (n^2 +n -3)
    (n^2 +n -3) - (n^3 -n^2 +3n)
    (n^2 +n -3) - n^3 -n^2 +3n
    - (n^3 +4n) -3
    4n - n^3 -3
    The original question is to prove the eigen values are 1, 3 and ? (negative 4 from online calculator)
    Thanks in advance :S
    I'm not sure how polynonial division would help me :S






    -3 x (3 x 1-n) - 0
    -3 x (3 - 3n)
    -9 + 9n
     
  7. Nov 8, 2008 #6

    HallsofIvy

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    Re: Eigenvalues!

    Please use parentheses! You mean (1- n)[(-2-n)((1-n)- (-1)(-1)]
    No, if you are keeping "-" outside that last parenthesis, it is -(n^3+ n^2- 3n)

    Even if you had not messed up the sign, this is just the top left corner, 1-n, times its cofactor
    [tex]\left|\begin{array}{cc} -2-n & -1 \\ -1 & 1- n\end{array}\right|[/tex]
    If you are expanding on the first row, you should also have
    [tex]-3\left|\begin{array}{cc}3 & -1 \\ 0 & 1-n\end{array}\right][/tex]
    If instead you are expanding on the first column, you should have
    [tex]-3\left|\begin{array}{cc}3 & 0 \\ -1 & 1-n\end{array}\right][/tex]
    which is obviously the same thing.

    In general if you no "n= a" is a root of a cubic polynomial then you know the polynomial is equal to (x- a) times quadratic polynomial. Dividing the cubic polynomial by x-a gives the remaining quadratic polynomial which you can solve, for example, using the quadratic formula.


    Oh, here's your missing term! You should have (n^2 +n -3)-(n^3+ n^2- 3n)- 9+ 9n=
    -n^3+ 13n- 12= 0.

    Yes, 1 satisfies that: -1^3+ 13(1)- 12= -1+ 13- 12= 0. Now divide -n^3+ 13n- 12 by n- 1 to get a quadratic.
     
    Last edited by a moderator: Nov 8, 2008
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