Math 151 Introdution to Eigenvetors The motivating example we used to desrie matrixes was landsape hange and vegetation suession. We hose the simple example of Bare Soil (B), eing replaed y Grasses (G) and then these eing replaed y Shrus. Desriing a landsape at a partiular time we saw ould e done using a vetor that gave the area in km 2 (or fration of total area whih is then a numer etween and 1) as a vetor with three entries ( BGS)giving the fration of the landsape of eah speies. If these were fration of area, then the vetor of numers would sum to one. If these were area in eah suessional type, then they would sum to the total area on the landsape. We next desried how this vetor v = ( BGS)ould hange through time due to the proess of suession, and noted that this an e desried mathematially y multiplying a vetor times a matrix whih speifies over the underlying time period (whih ould e a deade) how muh of eah type hanges to eah other type. So if the matrix is.4 M =.5.1 then this says that eah deade 4% of the are soil stays are soil, 5% hanges to grass and 1% hanges to shru. Italso says that 8% of the grass area stays grass and 2% eomes shru in the ourse of a deade. If an area eomes shru, then it stays shru. If we start out with a ertain area in eah suessional stage, then we an use this matrix to projet forward in time one deade y multiplying the matrix times the vetor for the urrent stage distriution. That is, if y t gives the vetor of area in eah stage at deade t, then y t + 1 = My t This is alled a matrix equation sine it is an equation that has a matrix in it and it allows us to projet from one deade forward to the next deade. So for example, if we start out with 1 km 2 in a region that is are soil, and no grass or shru, then y = gives the initial distriution of stages at time and we an get the distriution of stages at time 1 deade using or.4 y 1 =.5.1 1 y 1 = My 1 1 1 4 = 5 1 so that after one deade this region has 4 km 2 in are soil, 5 km 2 in grass and 1 km 2 in shru.
Doing this again we an get how muh of the landsape is in eah time at deade 2 from -2-.4 = My 1 =M 2 y =.5.1 1 4 16 5 = 6 1 24 We an ontinue to multiply the matrix M times eah deades vetor to get the next deades vetor. Look at the entries in the vetor and note that they sum to 1. This is eause there is no land raeted or destroyed here - eah unit of area must remain as one of the three types B, G or S. What do you think will happen if we ontinue to find Y t for larger and larger values of t? Yes, you are orret, eveything eventually eomes shru, and so this means that the vetor y t gets loser and loser to the vetor 1 We will see later in the ourse that the mathematial way to state this is lim y t = t 1 This is a fany way of saying that eventually all the area eomes shru. Inthis ase it was pretty easy to intuit what would happen in the long-term in this landsape. This would not e as easy to determine if we modify the situation as we mentioned earlier to look at the effet of fire (fire is only one kind of disturane in this system that ould ause the system to swith ak to are soil - other forms of distuane in natural systems that ould ause similar effets are windstorms, hurrianes, and disease. So our ojetive in this setion of the ourse is to develop a mathematial way to see what happens after a long-time in a strutured system. We are using the example of suession, ut we ll also see that exatly the same methods work to determine the long-term fration os a population in eah of a set of age ranges, whih is part of the area of demography wehave already mentioned. Note that if we start out a landsape with 1 km 2 in shru, then after a deade we get.4.5.1 1 = 1 1 whih means the landsape doesn t hange at all over the deade - everything remains in the shru stage. We all the vetor 1 an eigenvetor for the matrix M eause if we start at time at this eigenvetor of distriutions of stages, we stay there forever. This is an equilirium state for the system of suession - one there we stay there forever. More than that, in this situation we an show that no matter what distriution of initial states we start at, the system eventually approahes this eigenvetor - in this
ase we say it is stale. -3- In general, we will say that v is an eigenvetor for a matrix A if thee is some onstant λ so that λ v = Av and we say that v has assoiated eigenvalue λ. inthe aove situation, the vetor is an eigenvetor with eigenvalue λ = 1. 1 If we were to hange the matrix M to inlude the effet of fire, one ase we disussed is.4 N =.5.1.1.7.5.95 If we start out with an initial distriution y,then we an see y iterating that y t = N t y For example, if we start out with 1 km 2 in are soil then using Matla as a tool to alulate and 8 y 5 = N 5 y = 9 83 8. 8 y 1 = N 1 y = 14. 7 76. 5 So that after a long time the vetor of states approahes and if we start out at this state we see that 8. 8.4 14. 7 =.5 76. 5.1 8. 8 v = 14. 7 76. 5.1.7 so that v is an eigenvetor (and its eigenvalue is one)..5.95 8. 8 14. 7 76. 5 There are a variety of methods to find eigenvetors and eigenvalues. One way to find an eigenvetor is numerially. Wean use Matla to find for any matrix P, the matrix P n,where n is a large numer (say 1), then multiply this times the initial vetor for the landsape, y,toget a numerial answer for the long-term state of the landsape. If the initial vetor y ontained the
-4- fration of the landsape in eah vegetation type, then P 1 y will e a vetor giving the longterm fration of the landsape in eah vegetation type. If the initial vetor y rather represents the numer of hetares or ares of eah type, then divide eah term in P 1 y y the sum of the omponents of this vetor to get the long-term fration in eah state (the eigenvetor is speified only up to a onstant multiple). Aseond way to get the eigenvetor is to realize that it arises when the long term struture of the system doesn t hange. This is expressed as P y = λy where λ is a onstant that represents how the vetor of vegetation types inreases or dereases through one time period. In our ase of a fixed landsape, land area is neither reated nor destroyed, so there is no hange from one time period to another and so λ = 1. This means to find the eigenvetor all we need to do is find a vetor y that satisfies P y = y. This is easy to do using simple algera for small matries, ut for larger ones the mathematis eomes more diffiult. In this ase you either use the theory of determinants, or else use Matla to find the answer. In this lass, we expet that you will e ale to ompute y hand the eigenvetors and eigenvalues only for the simplest matries - 2x2 ones. Consider the 2x2 matrix and lets look at the equations arising from where P = a d Py=λy y = y 1 so P y 1 = a d y 1 Then this holds if a y 1 + =λy 1 and y 1 + d =λ.inorder for these to oth hold, we need and y 1 = λ a y 1 = λ d y 2 So the only way these an oth hold is if y 1 = = (whih is not an interesting ase) or if λ a = λ d.this is a quadrati equation in λ: λ 2 (a+d)λ+ad = and this equation is alled the harateristi equation for this matrix. We all a + d the Trae of the matrix P ( Tr(P) ) and a d the Determinant of the matrix P ( Det( P) ). By solving this quadrati for the roots, we find the eigenvalues λ (there will e two ingeneral for a 2x2 matrix). To find the eigenvetor, weplug in one of these λ values to find the ratio of y 1 to,and this
gives usthe eigenvetor up to a onstant. -5- Note that in the ase of our suession model where the total area of the landsape doesn t hange, we have λ = 1, we must have 1 (a+d)+ad =. The landsape transition matrix an e written in this ase as P = a 1 a 1 eause the olumns of the matrix P sum to one (so a + = 1and + d = 1). So Tr(P) = a + 1 and Det(P) = a (1 ) (1 )and in this ase then y 1 = 1 a = 1 d gives the eigenvetor. Normalize this so they sum to 1. As an example, onsider the matrix P = 4 3 1 2 then Tr(P) = 6 and Det(P) = 5 so the haarteristi equation is λ 2 6 λ + 5 = so the eigenvalues are λ = 1 and λ = 5 with eigenvetor for λ = 1 and eigenvetor for λ = 5. 1 3 1 1