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Linear Dynamical Systems

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Wize University Linear Algebra Textbook > Eigenvalues and Eigenvectors (Spectral Theory)

Linear Dynamical Systems

3 Activities

Practice: Recurrence Relations

Suppose a sequence is defined rescursively as:
xn+2=4xn,x0=1,  x1=4x_{n+2} = 4x_n, \quad x_0 =1,\ \ x_1 =4
Compute x20x_{20} by converting the recurrence to a linear dynamical system.
More Linear Dynamical Systems Questions:
Linear Dynamical Systems
Consider the linear dynamical system defined by A=[1211],v⃗0=[11]A=\left[\begin{array}{rr}1&2\\1&1\end{array}\right],\vec v_0=\left[\begin{array}{r}1\\1\end{array}\right]
Give a general formula for v⃗n\vec v_n , as well as a simpler formula for v⃗n\vec v_n that is valid when n is large.
Linear Dynamical Systems
Example: Linear Dynamical Systems
Consider the linear dynamical system given by
xn+1=2xn+ynyn+1=xn+2yn\begin{array}{l}x_{n+1}=2x_n+y_n\\y_{n+1}=x_n+2y_n\end{array}
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg12}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{12}$_
The following system of equations define a dynamic system between the variables x(t)x(t) and y(t)y(t):
x(t+1)=5x(t)+6y(t)y(t+1)=−2x(t)−2y(t) \begin{array}{ll} x(t+1) & = \quad 5x(t) + 6 y(t) \\[15pt] % y(t+1) & = \quad -2x(t) -2 y(t) \end{array}
If x⃗(0)=[10]\vx(0) = \colvec{1}{0}, find the general solution x⃗(t)\vx(t) for the state of the system at any time tt.
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg4}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{4}$_
As the previous activity illustrates, computing A‾t ξ⃗\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 ξ⃗\vXi is an eigenvector of the matrix A‾\A, with eigenvalue λ\lambda, then:
A‾t ξ⃗  =  λt ξ⃗ \A^{\bct{t}} \, \vXi \; = \; \lambda^{\bct{t}} \, \vXi
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg3}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{3}$_
Although the recursive relationship x⃗(t+1)=A‾  x⃗(t)\vx(t+1) = \A \; \vx(t) is not complicated, computing the vector x⃗(t)\vx(t) at a general time tt isn't very practical, as it would require the computation of a general power on a matrix, i.e.:
x⃗(t)=A‾t  x⃗(0) \vx(t) = \A^t \; \vx(0)
And, in general, A‾t x⃗(0)\A^t \, \vx(0) is not easy to compute (especially for large n×nn\times n matrices!). However, if the vector x⃗ (0)\vx\,(0) were an eigenvector of A‾\A, then the computation would be relatively simple.
19.4F_Final_Builder_Ch_12.6_Dynamical_Systems_$\tkcth{eg2}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{2}$_
Given the dynamic system defined by x⃗(t+1)=A‾  x⃗(t)\vx(t+1) = \A \;\vx(t), the initial condition x⃗(0)=[1−1]\vx(0) = \colvec{1}{-1} and the matrix
A‾=[1−111], \A = \sm{1}{-1}{1}{1},
find x⃗(3)\vx(3).
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg11}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{11}$_
The number of students who enrol in Math 133 and Linear Algebra 2 in a given semester is a function of the number of first year students currently enrolled in Math 133 f(t)f(t) and the number of second year students who already took Math 133 s(t)s(t), according to the following simplified model:
90%90\% of first year students who took Math 133 this semester will take Linear Algebra 2 next semester (the rest will have to repeat Math 133);
10%10\% of second year students who took Linear Algebra 2 will have to take it again;
19.4F_Final_Builder_Ch_12.7_Dynamical_Systems_$\tkcth{eg10}$_$\key{Final}$_Builder_$\tkcth{12.7.}\tkcf{10}$_
The number of students who enrol in Math 133 and Linear Algebra 2 in a given semester is a function of the number of first year students currently enrolled in Math 133 f(t)f(t) and the number of second year students who already took Math 133 s(t)s(t), according to the following simplified model:
90%90\% of first year students who took Math 133 this semester will take Linear Algebra 2 next semester (the rest will have to repeat Math 133);
10%10\% of second year students who took Linear Algebra 2 will have to take it again;