Math 19b: Linear Algebra with Probability Oliver Knill, Spring 2011Lecture 27: Discrete dynamical systemsEigenvectors and EigenvaluesMarkets, population evolutions or ingredients in a chemical reaction are often nonlinear. A lineardescription often can give a good approximation and solve the system explicitly. Eigenvectors andeigenvalues provide us with the key to do so.A nonzero vector v is called an eigenvector of a n × n matrix A if Av = λv forsome number λ. The later is called an eigenvalue of A.We first look at real eigenvalues but also consider complex eigenvalues.1 A vector v is an eigenvector to the eigenvalue 0 if and only if ~v is in the kernel of A becauseA~v = 0~v means tha t ~v is in the kernel.2 A rota t ion A in three dimensional space has an eigenvalue 1, with eigenvector spanning theaxes of rotation. This vector satisfies A~v = ~v.3 Every standard basis vector ~wiis an eigenvector if A is a diagonal matrix. For example,"2 00 3#"01#= 3"01#.4 For an orthogonal projection P onto a space V , every vector in V is an eigenvector to theeigenvalue 1 and every vector p erpendicular to V is an eigenvector to the eigenvalue 0.5 For a reflection R at a space V , every vector v in V is an eigenvector with eigenvalue 1.Every vector perpendicular to v is an eigenvector to the eigenvalue −1.Discrete dynamical systemsWhen applying a linear map x 7→ Ax again and again, we obtain a discretedynamical system. We want to understand what happens with the orbitx1= Ax, x2= AAx = A2x, x3= AAAx = A3x, ... and find a closed formula forAnx6 The one-dimensional discrete dynamical system x 7→ ax or xn+1= axnhas the solutionxn= anx0. The value 1.0 320· 1000 = 1806.11 for example is t he balance on a bank accountwhich had 1000 dollars 20 years ago if the interest rate was a constant 3 percent.7 Look at the recursionun+1= un− un−1with u0= 0, u1= 1. Because"1 −11 0#"unun−1#="un+1un#we have a discrete dynamicalsystem. Lets compute some orbits: A ="1 −11 0#A2="0 −11 −1#A3="−1 00 −1#.We see that A6is the identity. Every initial vector is mapped after 6 iterations back to itsoriginal starting point.8 A ="1 20 1#. ~v ="11#. A~v ="31#, A2~v ="51#. A3~v ="71#. A4~v ="91#etc.Do you see a pattern?The following example shows why eigenvalues and eigenvectors are so important:9 If ~v is an eigenvector with eigenvalue λ, then A~v = λ~v, A2~v = A(A~v)) = Aλ~v = λA~v = λ2~vand more generally An~v = λn~v.For an eigenvector, we have a closed form solution for An~v. It is λn~v.10 The recursionxn+1= xn+ xn−1with x0= 0 and x1= 1 produces the Fibonacci sequence(1, 1, 2, 3, 5, 8, 13, 21, ...)This can be computed with a discrete dynamical system because"xn+1xn#="1 11 0#"xnxn−1#.Can we find a formula for the n’th term?In the third section of Liber abbaci, published in1202, the mathematician Fibonacci, with real nameLeonardo di Pisa (1170-1250) writes: ”A certainman put a pair of rabbits in a place surroundedon all sides by a wall. How many pairs of rabbitscan be produced from that pair in a year if itis supposed that every month each pair begetsa new pair which from the second month on be-comes productive?”Markov caseWe will discuss the following situatio n a bit more in detail:An n ×n matrix is called a Markov matrix if all entries are nonnegative and eachcolumn adds up to 1.11 Customers using Apple IOS and Google Android are represented by a vector"AG#.Each cycle 1/3 of IOS users switch to Android and 2/3 stays. Also lets assume t hat 1/2of the Android OS users switch to IOS and 1/2 stay. The ma t r ix A ="2/3 1/21/3 1/2#is aMarkov matrix. What customer ra t io do we have in the limit? The matrix A has aneigenvector (3/ 5, 2/5) which belongs to the eigenvalue 1.A~v = ~vmeans that 60 to 40 percent is the final stable distribution.The following fact motivates to find good methods to compute eigenva lues and eigenvectors.If A~v1= λ1~v1, A~v2= λ2~v2and ~v = c1~v1+ c2~v2, we have closed form solutionAn~v = c1λn1~v1+ c2λn2~v2.Lets try this in the Fibonacci case. We will see next time how we find the eigenvalues andeigenvectors:12 Lets try to find a number φ such that"1 11 0#"φ1#= φ"φ1#This leads to the quadra t ic equation φ + 1 = φ2which has the solutions φ+= (1 +√5)/2and φ−= (1 −√5)/2. The number φ+is one of the most famous and symmetric numbersin mathematics called the golden ratio. We have found our eigenvalues and eigenvectors.Now find c1, c2such that"10#= c1"φ+1#+ c2"φ−1#We see c1= −c2= 1/√5. We can write"xn+1xn#= An"10#=1√5φn+"φ+1#−1√5φn−"φ−1#and can read off xn= (φn+− φn−)/√5.Homework due April 13, 20111Compare Problem 52 in Chapter 7.1 ofBretscher’s Book. The glucose and excess hor-mone concentration in your blood are modeledby a vector ~v ="gh#. Between meals theconcentration changes to ~v → A~v, whereA ="0.978 −0.0060.004 0.992#.Check that"−12#and"3−1#are eigenvec-tors of A. Find the eigenvalues.2Compare Problem 54 in Chapter 7.1 ofBretscher’s Book. The dynamical systemvn+1= AvnwithA ="0 21 1#models the growth of a lilac bush. The vector~v ="na#models the number of new branchesand the number of old branches. Verify that"11#and"2−1#are eigenvectors of A. Findthe eigenvalues a nd find the close form solu-tion starting with ~v = [2, 3]T.3Compare problem 50 in Chapter 7.1 ofBretscher’s Book. Two interacting popula-tions of hares and foxes can be modeledby the discrete dynamical system vn+1= AvnwithA ="4 −21 1#Find a closed form solutions in the followingthree cases: a) ~v0="h0f0#="100100#.b) ~v0="h0f0#="200100#.c)