1Math 1b Practical — Eigenvalues and eigenvectors, IIFebruary 11, 2011Example 6. Let P be the matrix of the orthogonal projection onto a subspace U of Rn.If u ∈ U,thenPu = u,andifw ∈ U⊥,thenP w = 0. That is, elements of U are eigenvec-tors corresponding to eigenvalue 1, and elements of U⊥are eigenvectors corresponding toeigenvalue 0. The (geometric) multiplicity of 1 is dim(U) and the (geometric) multiplicityof 0 is n − dim(U).Let R be the matrix of the reflection through a subspace U of Rn.Ifu ∈ U,thenRu = u,andifw ∈ U⊥,thenRw = −w. That is, elements of U are eigenvectorscorresponding to eigenvalue +1, and elements of U⊥are eigenvectors corresponding toeigenvalue −1. (I may give a more careful explanation in class.)Example 7. If A and B are square matrices, the eigenvalues ofAOOBare those of A AND those of B. More about this in class, or in a later version of thishandout.∗∗∗In Example 1 of the part I of this handout, we saw that 5 and −2 were the eigenvaluesof A =1342 and thatAe1=134234=1520=5e1,Ae2=13421−1=−22= −2e2.As an indication of why this helps us understand A and its powers, first note thatAne1=5ne1and Ane2=(−2)ne2.Since e1and e2form a basis for R2, we can understand Anx by writing x as a linearcombination of e1and e2. For example, let x =(9, 5) .Wehave(checkthis)95=234+31−12and thenAn95=2An34+3An1−1=2·5n34+3·(−2)n1−1=6 · 5n+3· (−2)n8 · 5n−3 · (−2)n.Thatis,wehavederivedaformula for An(9, 5) . One can see from this, for example, thatthe ratio of its first and second coordinates tends to 3/4asn tends to infinity.Sometimes a matrix has an positive eigenvalue λ so that λ>|μ| for every othereigenvalue μ; such an eigenvalue is called the dominant eigenvalue. In this example, 5 isthe dominant eigenvalue.Story 1. Smoking. (From Nakos and Joyner.) Suppose that the probability that asmoker will continue smoking a year later is 65% whereas the probability that a nonsmokerwill continue not smoking is 85%. We introduce the matrix A below (this is an exampleof a probability matrix,weshallseelater).A =0.65 0.150.35 0.85.If we start with 90% smokers and 10% nonsmokers, the next year we will have 60% smokersbecauseA.9.1=0.65 0.150.35 0.85.9.1=.6.4,and the next year we have 45% smokers becauseA.6.4=0.65 0.150.35 0.85.6.4=.45.55.In general, if the initial distribution is 100p% smokers and 100q% nonsmokers (herep, q ≥ 0, p + q = 1), then after k years we will havesmokersnonsmks= Akpq.What happens in the “long run”? The answer is that no matter what p and q are, eventuallythere will be about 30% smokers and about 70% nonsmokers. (Note that if we start with30% smokers and 70% nonsmokers, these proportions do not change the next year.)The explanation is that the eigenvalues and corresponding eigenvectors of A are (checkthis)λ1=1, e1=.3.7,λ2= .5, e2=1−1.Ifpq= a.3.7+ b1−1,3then the formula for Ak(p, q) isAkpq= a.3.7+ b(.5)k1−1.If p + q =1,thena = 1 (check this), and so no matter what b is (or even whether p, q arenonnegative or not),limk→∞Akpq=.3.7.Story 2. Fibonnaci numbers.Write F0= F1= 1 and determine Fnfor n =2, 3,... by Fn= Fn−1+ Fn−2;soF0,F1,F2,...=1, 1, 2, 3, 5, 8, 13, 21, 34, 55,....For each n ≥ 1,FnFn+1=0111Fn−1Fnand it follows by induction thatFnFn+1= An11whereA =0111.The eigenvalues of A are the roots of0=det(A −λI)=λ2− λ − 1.These are τ and 1 −τ where τ =(1+√5)/2; corresponding eigenvectors are1τand11 − τ.We want to find out what happens starting with (1, 1) , so we must write that vector asa linear combination of the eigenvectors; check that11=τ2τ − 11τ−1 −τ2τ − 111 − τ.If we equate the first coordinates of the sides ofFnFn+1= An11=τ2τ − 1An1τ−1 − τ2τ − 1An11 − τ,4we get the formulaFn=12τ − 1τn+1− (1 − τ )n+1=1√51+√52n+1−1√51 −√52n+1≈ 0.724 (1.618)n.Story 3. The rabbit epidemic. This example is taken from Linear Functions andMatrix Algebra by Bill Jacob. A disease is introduced into the rabbit population on anisland somewhere. Initially there are 12 million rabbits, all ‘uninfected’. After n months,there will be U(n) ‘uninfected’ rabbits, S(n) rabbits who have become ‘sick’, and I(n)rabbits who have somehow become ‘immune’. So U(0) = 12 (million), S(0) = 0, I(0) = 0.The values of these numbers changes by the rule⎛⎝U(n +1)S(n +1)I(n +1)⎞⎠=⎛⎝3411211213000341⎞⎠⎛⎝U(n)S(n)I(n)⎞⎠.So we have⎛⎝U(0)S(0)I(0)⎞⎠=⎛⎝1200⎞⎠,⎛⎝U(1)S(1)I(1)⎞⎠=⎛⎝940⎞⎠,⎛⎝U(2)S(2)I(2)⎞⎠=⎛⎝85/1233⎞⎠, ...Do these numbers make sense? I don’t know. It’s just a story anyway. We are sayingthat of the uninfected rabbits, 3/4 remain uninfected and 1/3 become sick after one month.That’s more that we had. I guess some new rabbits were born in that month. Of the sickrabbits, 3/4 (recover and) become immune and 1/12 recover but without immunity. Noneremain ‘sick’. I guess some pass away. Etc.Here are the numerical results for the first seven months, expressed as rows for con-venience:(12, 0, 0) = 12. (1., 0., 0.)(9., 4., 0.)=13. (0.6923, 0.3077, 0.)(7.08333, 3., 3.)=13.0833 (0.5414, 0.2293, 0.2293)(5.8125, 2.36111, 5.25) = 13.4236 (0.4330, 0.1759, 0.3911)(4.99363, 1.9375, 7.02083) = 13.9520 (0.3579, 0.1389, 0.5032)(4.49175, 1.66454, 8.47396) = 14.6303 (0.3070, 0.1138, 0.5792)(4.21369, 1.49725, 9.72237) = 15.4333 (0.2730, 0.0970, 0.6300)(4.09524, 1.40456, 10.8453) = 16.3451 (0.2505, 0.0859, 0.6635)Here are the results for the 100th through 103rd month:(1405.82, 439.147, 4909.94) = 6754.9010 (0.2081, 0.0650, 0.7269)(1500.12, 468.606, 5239.3) = 7208.0226 (0.2081, 0.0650, 0.7269)(1600.75, 500.04, 5590.75) = 7691.5397 (0.2081, 0.0650, 0.7269)(1708.13, 533.583, 5965.78) = 8207.4913 (0.2081, 0.0650, 0.7269)5That’s a lot of rabbits. Remember that we are counting in millions. The data showsthat the proportions seem to stabilize, with roughly 21% uninfected, 6% sick, and 73%immune.The numbers U(n), S(n), and I(n) of uninfected, sick, and immune rabitts at the endof n months are the coordinates of⎛⎝U(n)S(n)I(n)⎞⎠=