Science:Math Exam Resources/Courses/MATH307/April 2006/Question 04 (b)

MATH307 April 2006

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Question 04 (b)

The following discrete dynamical system describes the yearly migration of wild horse populations among three areas R, G, and B. Let r(t), g(t), and b(t) be the sizes of the horse population in areas R, G, and B respectively at the tth year. ${\begin{aligned}x(t+1)&={\begin{bmatrix}r(t+1)\\g(t+1)\\b(t+1)\end{bmatrix}}\\&={\begin{bmatrix}r(t)/2+g(t)/3+b(t)/3\\r(t)/2+g(t)/3+b(t)/2\\g(t)/3+b(t)/6\end{bmatrix}}\\&={\begin{bmatrix}1/2&1/3&1/3\\1/2&1/3&1/2\\0&1/3&1/6\end{bmatrix}}{\begin{bmatrix}r(t)\\g(t)\\b(t)\end{bmatrix}}\\&=Ax(t)\end{aligned}}$ Where the Markov matrix A describes how the horses move among these areas from one year to the next. The 1st column indicates that each year 1/2 of the horses in area R remain in area R and 1/2 will migrate to area G. The 2nd column shows that horses in area G will be evenly distributed in the three areas one year later. The 3rd column implies that, of the horses in area B, 1/3 will migrate to area R, 1/2 will migrate to area G, and only 1/6 will remain in area B. We assume that no horses are lost and no new horses are added and that initially (i.e. at t = 0), there are a total of 350 horses all located in area B. Thus ${\vec {x}}(0)=[0\ 0\ 350]^{T}$. (b) Find a vector ${\vec {v}}$ such that $A{\vec {v}}={\vec {v}}$, and thus $A^{t}{\vec {v}}={\vec {v}}$ for all t > 1.

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. $A*v=v$ Is the vector v corresponding to the eigenvalue 1 To solve this we modify the equation $A*v=\lambda v$ ->$A*v-\lambda v=(A-\lambda I)*v=0$ In this case λ=1 so $(A-I)*v=0$. We also want this to be non-trivial ($v\not =0$). we now take the reduced row echelon form a $(A-I)$ $rref(A-I)=$ $rref\left(\left[{\begin{array}{ccc}-{\frac {1}{2}}&{\frac {1}{3}}&{\frac {1}{3}}\\{\frac {1}{2}}&-{\frac {2}{3}}&{\frac {1}{2}}\\0&-{\frac {1}{2}}&-{\frac {5}{6}}\end{array}}\right]\right)$ $=\left[{\begin{array}{ccc}1&0&-{\frac {7}{3}}\\0&1&-{\frac {5}{2}}\\0&0&0\end{array}}\right]$ This gives us an eigenvector $v=\left[{\begin{array}{c}{\frac {7}{3}}\\{\frac {5}{2}}\\1\end{array}}\right]$ For this application we want the components of the vectors to add to 1. To do this we divide each component by the sum of all the components. Once we do this we get our final vector $v=\left[{\begin{array}{c}{\frac {2}{5}}\\{\frac {3}{7}}\\{\frac {6}{35}}\end{array}}\right]$