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Generalized eigenvectors and differential equations

  1. Nov 15, 2013 #1
    Let [itex]A[/itex] be an 3x3 matrix such that [itex]A\mathbf{v_1}=\mathbf{v_1}+\mathbf{v_2}, A\mathbf{v_2}=\mathbf{v_2}+\mathbf{v_3}, A\mathbf{v_3}=\mathbf{v_3}[/itex] where [itex]\mathbf{v_3} \neq \mathbb{0}[/itex]. Let [itex]B=S^{-1}AS[/itex] where [itex]S[/itex] is another 3x3 matrix.
    (i) Find the general solution of [itex]\dot{\mathbf{x}}=B\mathbf{x}[/itex].
    (ii) Show that 1 is the only eigenvalue of [itex]B[/itex].

    It's clear that [itex]\mathbf{v_3},\mathbf{v_2}[/itex] and [itex]\mathbf{v_1}[/itex] form a chain of generalized eigenvectors associated with [itex]\lambda=1[/itex] and hence are linearly independent. From this I can find the general solution of [itex]\dot{\mathbf{x}}=A\mathbf{x}=SBS^{-1}\mathbf{x}[/itex] but how can I proceed from here to find the general solution of [itex]\dot{\mathbf{x}}=B\mathbf{x}[/itex]?
    Any help is much appreciated, thank you!
     
  2. jcsd
  3. Nov 15, 2013 #2

    AlephZero

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    You have ##A = S B S^{-1}##.

    If ##x## is a solution of ##\dot x = Bx##, let ##x = S^{-1} y##.

    Then ##\dot y = Ay##.
     
  4. Nov 16, 2013 #3
    That's brilliant, thanks!
     
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