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Finding eigenvalues

  1. Oct 27, 2009 #1
    [tex]
    \left(
    \begin{array}{ccc}
    3 - \lambda & 1 & -1 \\
    -4 & 2 - \lambda & 2 \\
    -2 & 2 & 2 - \lambda
    \end{array}
    \right)[/tex]

    [tex](3 - \lambda) \left|
    \begin{array}{cc} 2 - \lambda & 2 \\
    2 & 2 - \lambda
    \end{array}
    \right| + 4 \left|
    \begin{array}{cc} 1 & -1 \\
    2 & 2 - \lambda
    \end{array}
    \right| - 2 \left|
    \begin{array}{cc} 1 & -1 \\
    2 - \lambda & 2
    \end{array}
    \right|[/tex]

    [tex]det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0[/tex]

    [tex]-3x^3 + 7x^2 -14x + 8 = 0[/tex]

    ???
     
  2. jcsd
  3. Oct 27, 2009 #2

    Pengwuino

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    Gold Member

    You can factor out a [tex]\lambda - 4[/tex] to determine one eigenvalue.
     
  4. Oct 27, 2009 #3
    Can you elaborate? I don't know how to work out cubic polynomials.
     
  5. Oct 27, 2009 #4

    lanedance

    User Avatar
    Homework Helper

    pengwuino means that 4 is a solution to the charcteristic equation, so you can re-write it as

    [tex]-3x^3 + 7x^2 -14x + 8 = (\lambda -4)(ax^2 +bx + c) = 0[/tex]

    multiply out & equate co-efficients
     
  6. Oct 27, 2009 #5
    Yes, but I want to know how he got that 4.
     
  7. Oct 27, 2009 #6

    lanedance

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    Homework Helper

    test that putting 4 into your equation gives 0
     
  8. Oct 27, 2009 #7
    Never mind. Somehow I solved it by not opening too many brackets.

    [tex](3-x)((2-x)(2-x)-4)+4((2-x)+2)-2(4-x)[/tex]
    [tex](3-x)(4-4x+x^2-4)+4(4-x)-2(4-x)[/tex]
    [tex](3-x)(-4x+x^2)+(4-2)(4-x)[/tex]
    [tex](3-x)(-x)(4-x)+2(4-x)[/tex]
    [tex](3-x)(2-x)(4-x)[/tex]
     
  9. Oct 27, 2009 #8

    lanedance

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    Homework Helper

    in general I don't think its that easy, though you could probably attempt polynomial long division, one trick is say you have the following polynomial and p is a factor

    [tex] ax^3 + bx^2 + cx + d = (x+p)(qx^2 + rx + s) [/tex]

    multplying out gives
    [tex] ax^3 + bx^2 + cx + d = (x+p)(qx^2 + rx + s) = qx^3 + (pq +r)x^2 (pr+s)x + ps [/tex]

    note by equating coefficients
    [tex] p = \frac{d}{s} [/tex]

    so if there are integer factors and everything stays nice, p must divide d - giving options here of 8,4,2,1,-1,-2,-4,-8. You could test these reasonably quickly to see if any work

    in your simplification step though you could have done the following, though not exactly obvious
    [tex]det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0[/tex]

    [tex]
    = (3 - \lambda)[(\lambda^2-4 \lambda ] - 4[(\lambda-4)] + 2[\lambda-4] = 0[/tex]

    [tex]
    = (3 - \lambda)[(\lambda-4 )\lambda ] - 4[(\lambda-4)] + 2[\lambda-4] = 0[/tex]
     
  10. Oct 27, 2009 #9

    lanedance

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    Homework Helper

    the next step you take doesn't look right
    it should be
    [tex](3-x)(-x)(4-x)+2(4-x)[/tex]
    [tex]=(x^2-3x)(4-x)+2(4-x)[/tex]
    [tex]=(x^2-3x+2)(4-x)[/tex]

    note if your initial multiplication was correct, then all the factors should solve the equation...

    though I'm not so convinced about you original polynomial now - i don't see how you get a -3x^3 term
     
  11. Oct 27, 2009 #10

    lanedance

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    Homework Helper

    assuming your determinant is correct, then expanding this as a check
    [tex]det(A - \lambda I) = (3 - \lambda)[(2 - \lambda)(2 - \lambda) - 4] + 4[(2 - \lambda) + 2] - 2[4 - \lambda] = 0[/tex]

    [tex]= (3 - \lambda)(\lambda^2 -4 \lambda) + 4(4 - \lambda) - 2(4 - \lambda) [/tex]

    [tex]=(-\lambda^3+7\lambda^2 -12 \lambda) + 2(4 - \lambda) [/tex]

    [tex]=-\lambda^3+7 \lambda^2 -14 \lambda +8 = 0[/tex]

    checking 4 is a solution
    [tex]=-(4^3)+7.(4^2) -10.(4) +8 = 4.(-4^2 +7.4-10+2) = 4(-16+28-8) = 4.0 = 0 [/tex]
     
  12. Oct 27, 2009 #11
    Thanks! I'm horridly careless at these meticulous calculations.
     
  13. Oct 27, 2009 #12

    Mark44

    Staff: Mentor

    He probably used the rational root theorem. If p/q is a root of the polynomial anxn + an-1xn-1 + ... + a1x + a0 = 0, then p must be a divisor of a0 and q must be a divisor of an.

    For the polynomial equation in question, [itex]-3x^3 + 7x^2 -14x + 8 = 0[/itex], any rational root p/q must be such that p divides 8, and q divides -3. Possible choices for p are 1, -1, 2, -2, 4, -4, 8, and -8. Possible choices for q are 1, -1, 3, -3. Possible values for p/q are any of the 32 possibilities.
     
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