Generalized eigenvectors and differential equations

  • Thread starter drawar
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  • #1
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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!
 

Answers and Replies

  • #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##.
 
  • #3
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That's brilliant, thanks!
 

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