Eigenvalues and eigenvectors.

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1 Eigenvalues and eigenvectors. Example. A population of a certain animal can be broken into three classes: eggs, juveniles, and adults. Eggs take one year to become juveniles, and juveniles take one year to become adults. Each year, each adult produces an average of 5 eggs. Only 40% of the eggs survive to become juveniles, 50% of the juveniles survive to adulthood, and 80% of adults survive to the next year. Starting with a population of 20 adults, find the long term growth rate of the population. Solution The Leslie Matrix modeling this situation is.4 0 0, and the initial population is Let s see what the population looks like after 25, 26, and 27 years. > (L %^%25) %*% x [1,] [2,] [3,] > sum((l %^%25) %*% x) [1] > (L %^%25) %*% x/sum((l %^%25) %*% x) [1,] [2,] [3,] > (L %^%26) %*% x [1,] [2,] [3,] > sum((l %^%26) %*% x) [1]

2 > (L %^%26) %*% x/sum((l %^%26) %*% x) [1,] [2,] [3,] > (L %^%27) %*% x [1,] [2,] [3,] > sum((l %^%27) %*% x) [1] > (L %^%27) %*% x/sum((l %^%27) %*% x) [1,] [2,] [3,] We ll summarize our findings below. Year Population Total Population Population structure eggs juvs adults eggs juvs adults eggs juvs adults % eggs 19% juvs 17% adults % eggs 19% juvs 17% adults % eggs 19% juvs 17% adults Growth of population since prior year About 35% About 35% It appears that the population is growing at a rate of about 35% per year, and the population seems to have a stable age distribution where about 64% of the population are eggs, 19% are juveniles, and 17% are adults. It would be nice, however, to be able to do all of this a bit more efficiently.

3 Solution 2. If the population reaches a stable age structure, and each group age group in the population is growing by a factor of each year, we ll have In addition, the entries of vector should add to 1 to give the percentage of the population in each age group. In Mathematics, we call an eigenvalue of the matrix L, and we call the vector an eigenvector corresponding to the eigenvalue. As you d guess, R easily does these things, although to get the vector x to look exactly the way we want, we have to work just a little. In R, the eigen command will do these things. > eigen(l) [1] i i i [,2] [,3] [1,] i i i [2,] i i i [3,] i i i In general, an n by n matrix has n eigenvalues, and it is true (but by no means obvious!) that every Leslie matrix has a dominant (or largest) real eigenvalue, which tells the long term rate at which the population is changing. In this case, it s 1.35 which is telling us that in the long run the population will grow by a factor of 1.35 (or 35%) each year. Under, column 1 is an eigenvector for 1.35, column 2 is an eigenvector for , and column 3 is an eigenvector for To find the stable age distribution, we want the eigenvector corresponding to the dominant real eigenvalue. However, we want to modify that eigenvector so its entries sum to 1. Here s where your R knowledge is put to the test. The command > x = eigen(l) will lift off the eigenvector in question. We look inside the object eigen(l), extract the information in the part of that data frame named vectors (), and since that gives the three eigenvectors as a 3 by 3 matrix, we lift out column 1 () of that. > x [1] i i i Finally, the easiest way to make x sum to 1 is to take each entry in x and divide by the sum of x s entries. > x/sum(x) [1] i i i

4 Not surprisingly, this vector tells us that in the long run, about 64% of the population will be in age class 1 (eggs), 19% will be in age class 2 (juveniles), and 17% will be in age class 3 (adults). Example. Take the previous example with some modifications. Only 10% of eggs survive, only 20% of juveniles survive to adulthood, and each adult produces an average of only 2 eggs. Here the Leslie matrix is > L = matrix(c(0,0,2,.1,0,0,0,.2,.8), nrow=3, ncol=3, byrow=true) > eigen(l) [1] i i i [,2] [,3] [1,] i i i [2,] i i i [3,] i i i Here the dominant eigenvalue is 0.85, indicating that in the long run, the next year s population will be only 85% of what it was the year before. Thus, this population will eventually go extinct. Example. See the example on the previous handout which discussed the Moran model. 1 2/ /9 2/9 0 The matrix of that process was 0 2/9 5/ /9 1 and the four states corresponded to 0, 1, 2, and 3 type A individuals in the population. Let find the eigenvalues and corresponding eigenvectors of this matrix.

5 > N = matrix(c(1,0,0,0,2/9,5/9,2/9,0,0,2/9,5/9,2/9,0,0,0,1), nrow=4, ncol=4) > eigen(n) [1] [,2] [,3] [,4] [1,] [2,] [3,] [4,] Notice that 1 shows up twice as an eigenvalue, or has multiplicity 2 as you d say in college algebra (in fact 1 will always be an eigenvalue of any matrix of a Markov chain), and there are two different eigenvectors corresponding to this eigenvalue and are telling you that there are two possible long term states to this system. Either we end up in state 1 (where the A s die out) forever or we end up in state 4 (where the B s die out) forever. Note from the earlier example, the probability of eventually reaching either of these so called absorbing states depends heavily on which state we started in. Example. A certain disease has 4 possible states. The disease is either in remission, stage 1, stage 2, or stage 3. During any given year: If the disease is currently in remission, there is a 5% chance it will advance to stage 1 during the next year; otherwise it stays in remission. If it s currently in stage 1, there is a 40% chance it will go into remission, a 50% chance it will stay stage 1, and a 10% chance it will advance to stage 2. If it s currently in stage 2, there s a 30% chance it goes into remission, a 30% chance it stays stage 2, and a 40% chance it goes to state 3. Stage 3 is when the condition is terminal. Once the disease is at this stage, it stays that way until the death of the diseased. Discuss how this disease runs its course. Solution First, note that is the matrix of this Markov chain. And, from here on, I ll stop writing down the R commands needed to produce matrices.

6 > eigen(d) [1] [,2] [,3] [,4] [1,] [2,] [3,] [4,] This is not horribly enlightening, as this tells us that eventually (if they live long enough!) the patient will end up in the terminal state. This is why when dealing with terminal conditions, one usually speaks of 5 year or 10 year prognoses. > D %^% 5 [,2] [,3] [,4] [1,] [2,] [3,] [4,] > D %^% 10 [,2] [,3] [,4] [1,] [2,] [3,] [4,] For instance, if the patient is currently in remission, there is about an 89% probability they will also be in remission 5 years from now. If the patient is currently in remission, there s about a 3.5% probability they will be terminal in 10 years, and if they are currently stage 2, there is a 58% probability they will be terminal in ten years. Now, the following is impractical (for a human patient anyway!), but it illustrates something useful. > round(d %^% 5000, 3) [,2] [,3] [,4] [1,] [2,] [3,] [4,] If you let this process do its thing for 5000 years, we basically end up in the terminal state with probability 1, regardless of the starting state.

7 Example. There are three geographic areas in which a certain mammal lives, and these critters migrate. In any given year, 15% of the animals in area A migrate to area B, 5% migrate to C, and the rest stay put. 5% of the animals in B migrate to A, 5% migrate to C, and the rest stay put. 10% of the animals in C migrate to A, 20% migrate to B, and the rest stay put. What will be the long term state of such a system (of course, conveniently ignoring births, deaths and a hundred other things!)? Solution. First set up the transition matrix; we ll call it D. > D = matrix(c(.8,.05,.1,.15,.9,.2,.05,.05,.7), nrow=3, ncol=3, byrow=true) > D [,2] [,3] [1,] [2,] [3,] > eigen(d) [1] Not surprisingly, the dominant eigenvalue is 1, which tells us that the population is neither growing or shrinking (there s no births or deaths in this model!). [,2] [,3] [1,] e [2,] e [3,] e As before, the eigenvector for the eigenvalue 1 (scaled appropriately!) should give the long term state of the system. > x = eigen(d) > x/sum(x) [1] This tells us that eventually, about 23% of the population will end up in area A, 63% will reside in B, and the other 14% will be in C. The calculation > D %^% 100 [,2] [,3] [1,] [2,] [3,] shows that if you let this process run for 100 years, the population will have the same distribution (or at least very close to it) regardless of the initial distribution of the population.

Leslie matrices and Markov chains.

Leslie matrices and Markov chains. Example. Suppose a certain species of insect can be divided into 2 classes, eggs and adults. 10% of eggs survive for 1 week to become adults, each adult yields an average