12.2. SOME APPLICATIONS OF EIGENVALUES AND EIGENVECTORS 273

Example 12.2.8 Suppose a dynamical system is of the form(x(n+1)y(n+1)

)=

(1.5 −0.51.0 0

)(x(n)y(n)

)

Find solutions to the dynamical system for given initial conditions.

In this case, the eigenvalues of the matrix are 1, and .5. The matrix is of the form(1 11 2

)(1 00 .5

)(2 −1−1 1

)

and so given an initial condition (x0

y0

)the solution to the dynamical system is(

x(n)y(n)

)=

(1 11 2

)(1 00 .5

)n(2 −1−1 1

)(x0

y0

)

=

(1 11 2

)(1 00 (.5)n

)(2 −1−1 1

)(x0

y0

)

=

(y0 ((.5)

n−1)− x0 ((.5)n−2)

y0 (2(.5)n−1)− x0 (2(.5)

n−2)

)

In the limit as n→ ∞, you get (2x0− y0

2x0− y0

)Thus for large n, (

x(n)y(n)

)≈

(2x0− y0

2x0− y0

)Letting the initial condition be (

2010

)one can graph these solutions for various values of n. Here are the solutions for values of nbetween 1 and 5(

25.020.0

)(27.525.0

)(28.7527.5

)(29.37528.75

)(29.68829.375

)

12.2. SOME APPLICATIONS OF EIGENVALUES AND EIGENVECTORS 273Example 12.2.8 Suppose a dynamical system is of the formx(n+1) \ f 1.5 —0.5 x(n)y(n+1) 10 0600 y(n)Find solutions to the dynamical system for given initial conditions.In this case, the eigenvalues of the matrix are 1, and .5. The matrix is of the formCia)to JU]and so given an initial condition_ ( yo ((.5)"= 1) = 20 ((.In the limit as n — co, you getThus for large n,Letting the initial condition be(1one can graph these solutions for various values of n. Here are the solutions for values of nbetween | and 525.0 27.5 28.75 29.375 29.68820.0 25.0 27.5 28.75 29.375