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Repeated Eigenvalue of a n=3 system of differential equations

  1. May 2, 2013 #1
    1. The problem statement, all variables and given/known data

    x' = \begin{pmatrix}0&1&3\\2&-1&2\\-1&0&-2\end{pmatrix}*x





    3. The attempt at a solution
    I've found the repeated eigenvalues to be λ[itex]_{1,2,3}[/itex]=-1
    I've also found the first (and only non zero eigenvector) to be \begin{pmatrix}1&2&-1\end{pmatrix}, but I'm not entirely sure where to go from here. Everything I've found talks about 2x2 matrices. Even looking around on MIT opencoursewares site, they say that its possible but is beyond the scope of that class...I did find some reference to using Jourdan Canonical form may possibly be used, but I don't remember him talking too much about that.
     
  2. jcsd
  3. May 2, 2013 #2

    Mark44

    Staff: Mentor

    That's Jordan.. Here's a wiki article about Jordan Normal Form. The section on generalized eigenvectors that might be of help.
     
  4. May 3, 2013 #3

    Ray Vickson

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    For a matrix A having a single eigenvalue ##r## of multiplicity 3 the analytic matrix function [tex]f(A) = c_0 I + c_1 A + c_2 A^2 + c_3 A^3 + \cdots [/tex] (corresponding to the ordinary analytic function ##f(x) = c_0 + c_1 x + c_2 x^2 + \cdots ##) is of the form
    [tex] f(A) = E_1 f(r) + E_2 f'(r) + E_3 f''(r)[/tex] for some fixed matrices ##E_1, E_2, E_3## which are the same for all functions f. We can determine the ##E_1## by looking at special cases of f: for ##f(x) = x^0 = 1## we have ##f'x) = f''(x) = 0##, so ##I = A^0 = 1 E_1 + 0 E_2 + 0 E_3 = E_1##. For##f(x) = x## we have ##f'(x) = 1, f''(x) = 0## and so ##A = r E_1 + 1 E_2 ##. For ##f(x) = x^2## we have ##f'(x) = 2x, f''(x) = 2## and so ##A^2 = r^2 E_1 + 2r E_2 + 2 E_3.## Altogether, we have the three equations
    [tex] E_1 = I\\
    r E_1 + E_2 = A\\
    r^2 E_1 + 2r E_2 + 2E_3 = A^2.[/tex]
    Once we have the ##E_i## we can solve the differential equation system using the matrix exponential ##M(t) = e^{At}##. This will have the form
    [tex] M(t) = e^{rt} E_1 + t e^{rt} E_2 + t^2 e^{rt} E_3,[/tex]
    obtained from the function ##f(x) = e^{xt}.##

    This all follows by looking at the Jordan Canonical Form. Note, however, that the discussion above applies equally whether the matrix has a diagonal Jordan form, or one block of size 1 and one of size 2, or a single Jordan block of size 3; those separate cases will just correspond to situations in which some ##E_i## are zero. For example, if A were diagonalizable but with a single eigenvalue r of multiplicity 3, we would just have ##E_2 = E_3 = 0##.
     
  5. May 3, 2013 #4

    Ray Vickson

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    Science Advisor
    Homework Helper

    For a matrix A having a single eigenvalue ##r## of multiplicity 3 the analytic matrix function [tex]f(A) = c_0 I + c_1 A + c_2 A^2 + c_3 A^3 + \cdots [/tex] (corresponding to the ordinary analytic function ##f(x) = c_0 + c_1 x + c_2 x^2 + \cdots ##) is of the form
    [tex] f(A) = E_1 f(r) + E_2 f'(r) + E_3 f''(r)[/tex] for some fixed matrices ##E_1, E_2, E_3## which a re the same for all functions f. We can determine the ##E_1## by looking at special cases of f: for ##f(x) = x^0 = 1## we have ##f'x) = f''(x) = 0##, so ##I = A^0 = 1 E_1 + 0 E_2 + 0 E_3 = E_1##. For##f(x) = x## we have ##f'(x) = 1, f''(x) = 0## and so ##A = r E_1 + 1 E_2 ##. For ##f(x) = x^2## we have ##f'(x) = 2x, f''(x) = 2## and so ##A^2 = r^2 E_1 + 2r E_2 + 2 E_3.## Altogether, we have the three equations
    [tex] E_1 = I\\
    r E_1 + E_2 = A\\
    r^2 E_1 + 2r E_2 + 2E_3 = A^2.[/tex]
    Once we have the ##E_i## we can solve the differential equation system using the matrix exponential ##M(t) = e^{At}##. This will have the form
    [tex] M(t) = e^{rt} E_1 + t e^{rt} E_2 + t^2 e^{rt} E_3,[/tex]
    obtained from the function ##f(x) = e^{xt}.##
     
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