19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg4}$_$\key{Final}$_Build…


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As the previous activity illustrates, computing Atξ\A^t \, \vXi is relatively simple, when the vector is an eigenvector.

Once you've seen an example like this, it's not hard to see that, in general, if ξ\vXi is an eigenvector of the matrix A\A, with eigenvalue λ\lambda, then:
Atξ  =  λtξ \A^{\bct{t}} \, \vXi \; = \; \lambda^{\bct{t}} \, \vXi
Returning to the context of our dynamical system where:
x(t+1)=Ax(t)x(t)  =  Atx(0) \vx(t+1) = \A \,\vx(t) \quad \Longleftrightarrow \quad \vx(\bct{t}) \;=\; \A^{\bct{t}} \, \bcth{\vx(0)}

we can see that these problems would be easy to solve when x(0)\bcth{\vx(0)} is an eigenvector of the system.

\bct{\underline{\hspace{6cm}}}


For example, consider the dynamical system given by:
x(t+1)=[17301018]x(t). \vx(t+1) = \sm{17}{-30}{10}{-18} \vx(t).
Find x(1)\vx(1), x(2)\vx(2) and x(3)\vx(3) if x(0)=[21]\vx(0) = \colvec{2}{1} is an eigenvector of the system with eigenvalue λ=2\lambda = 2.

Only enter integer values for the following:
  • x(1)=\quad \vx(1) =
    x(0)=\vx(0) =
  • x(2)=\quad \vx(2) =
    x(0)=\vx(0) =
  • x(3)=\quad \vx(3) =
    x(0)=\vx(0) =



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