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Find the eigenvalues of a matrix

  1. Jun 16, 2007 #1
    1. The problem statement, all variables and given/known data
    i'm trying to find the eigenvalues of a matrix and i have the solution but i don't understand how it gets from the step 1 to step 2??? could someone please explain.

    let # = lambda

    Step 1: (1-#)[(2-#)(-1-#)+1]+[3(-1-#)+2]+4[3-2(2-#)] = 0
    Step 2: (1-#)(#+2)(#-3) = 0





    2. Relevant equations



    3. The attempt at a solution
    i jst kept expanding the other half of the equation with my final result bein
    useless

    (1-#)(#^2 -#-1)+(5#-5) = 0
     
  2. jcsd
  3. Jun 16, 2007 #2

    Dick

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    Science Advisor
    Homework Helper

    If you expand the first equation completely you get -L^3+2*L^2+5*L-6=0 (L=lambda). Now you just want to factor this polynomial. Probably the easiest way to do this is to look for zeros of the polynomial. Eg L=1 is a zero so L-1 is a root (we also use the clue that if p/q is a rational root the p divides the constant polynomial coefficient and q divides the highest power coefficient). Once you found one root divide it out and try to factor what's left. It's pretty standard stuff.
     
  4. Jun 29, 2007 #3

    daniel_i_l

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    Gold Member

    If the coefficient of the variable with the highest power is one:
    [tex] x^n +a_{n-1} x^{n-1} + ... + a_1 x + a_0 [/tex]
    then if the polynomial has a integer root then the last coefficient (a_0) is a multiple of the root. For example, given the polynomial:
    [tex] x^2 + 5x + 6 [/tex]
    the possible integer roots are +/-1 , +/-2, +/-3 and +/-6
    a quick check shows that 2 and 3 are roots so the polynomial is equal to
    [tex] (x-3)(x-2) [/tex]
    This is especially useful for finding e-values from the e-polynomial since in most problems the e-values are integers and the coefficient of x^n is always 1.
     
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