Mth 8 - Lecture 3 Dyln Zwick Fll 3 In our lst lecture we delt with solutions to the system: x = Ax where A is n n n mtrix with n distinct eigenvlues. As promised, tody we will del with the question of wht hppens if we hve less thn n distinct eigenvlues, which is wht hppens if ny of the roots of the chrcteristic polynomil re repeted. This lecture corresponds with section 5.4 of the textbook, nd the ssigned problems from tht section re: Section 5.4 -, 8, 5, 5, 33 The Cse of n Order Root Let s strt with the cse n n order root. So, our eigenvlue eqution hs repeted root, λ, of multiplicity. There re two wys this cn go. The first possibility is tht we hve two distinct (linerly independent eigenvectors ssocited with the eigenvlue λ. In this cse, ll is good, nd we just use these two eigenvectors to crete two distinct solutions. Admittedly, not one of Sherlock Holmes s more populr mysteries.
Exmple - Find generl solution to the system: x = 9 4 6 6 4 3 x Solution - The chrcteristic eqution of the mtrix A is: A λi = (5 λ(3 λ. So, A hs the distinct eigenvlue λ = 5 nd the repeted eigenvlue λ = 3 of multiplicity. For the eigenvlue λ = 5 the eigenvector eqution is: (A 5Iv = 4 4 6 6 6 4 b c = which hs s n eigenvector v = Now, s for the eigenvlue λ = 3 we hve the eigenvector eqution: 6 4 6 4 6 4 b c = For this eigenvector eqution we hve two linerly independent eigenvectors:
v = nd v 3 = 3 So, we hve complete set of linerly independent eigenvectors, nd ssocited solution: x(t = c v e 5t + c v e 3t + c 3 v 3 e 3t Well, tht ws no problem. Unfortuntely, s we will see momentrily, it isn t lwys the cse tht we cn find two linerly independent eigenvectors for the sme eigenvlue. Exmple - Clculte the eigenvlues nd eigenvectors for the mtrix: A = ( 3 3 7 Solution - We hve chrcteristic eqution (λ 4 =, nd so we hve root of order t λ = 4. The corresponding eigenvector eqution is: (A 4I = ( 3 3 3 3 ( b = (. We get one eigenvector: v = ( nd tht s it! In this sitution we cll this eigenvlue defective, nd the defect of this eigenvlue is the difference beween the multiplicity of the root nd the 3
number of linerly independent eigenvectors. In the exmple bove the defect is of order. Now, how does this relte to systems of differentil equtions, nd how do we del with it if defective eigenvlue shows up? Well, suppose we hve the sme mtrix A s bove, nd we re given the differentil eqution: x = Ax. For this differentil eqution we know how to find one solution: ( e 4t but for complete set of solutions we ll need nother linerly independent solution. How do we get this second solution? Bsed on experience we my think tht, if v e λt is solution, then we cn get second solution of the form v te λt. Let s try out this solution: v e λt + λv te λt = Av te λt. Unfortuntely for this to be true it must be true when t =, which would imply tht v =, which wouldn t be linerly independent solution. So, this pproch doesn t work. 3 But wit, there s hope. We don t hve to bndon this method entirely. 4 Suppose insted we modify it slightly nd replce v t with v t+v, where v is the one eigenvector tht we were ble to find. Well, if we plug this into our differentil eqution we get the reltions: v e λt + λv te λt + λv e λt = Av te λt + Av e λt. No, run nd hide is not n cceptble nswer. 3 Nuts! 4 Hoory! 4
If we equte the e λt terms nd the te λt terms we get the two equlities: (A λiv = nd (A λiv = v. Let s think bout this. Wht this mens is tht, if we hve defective eigenvlue with defect, we cn find two linerly independent solutions by simply finding solution to the eqution (A λi v = such tht (A λiv. If we cn find such vector v, then we cn construct two linerly independent solutions: x (t = v e λt nd x (t = (v t + v e λt where v is the non-zero vector given by (A λiv. For our prticulr sitution we hve: (A 4I = ( So, ny non-zero vector could potentilly work s our vector v. If we try 5
( v = then we get: ( 3 3 3 3 ( = ( 3 3 So, our linerly independent solutions re: x (t = v e 4t = nd x (t = (v t + v e 4t = ( 3 3 e 4t, ( 3t + 3t e 4t. with corresponding generl solution: x(t = c x (t + c x (t. We note tht our eigenvector v is not our originl eigenvector, but is multiple of it. Tht s fine. The Generl Cse The vector v bove is n exmple of something clled generlized eigenvector. If λ is n eigenvlue of the mtrix A, then rnk r generlized eigenvector ssocited with λ is vector v such tht: (A λi r v = but (A λi r v. 6
We note tht rnk generlized eigenvector is just our stndrd eigenvector, where we tret mtrix rised to the power s the identity mtrix. We define length k chin of generlized eigenvectors bsed on the eigenvector v s set {v,...,v k } of k generlized eigenvectors such tht (A λiv k = v k (A λiv k = v k. (A λiv = v. Now, fundmentl theorem from liner lgebr sttes 5 tht every n n mtrix A hs n linerly independent generlized eigenvectors. These n generlized eigenvectors my be rrnged in chins, with the sum of the lengths of the chins ssocited with given eigenvlue λ equl to the multiplicity of λ (.k.. the order of the root in the chrcteristic eqution. However, the structure of these chins cn be quite complicted. The ide behind how we build our solutions is tht we clculte the defect d of our eigenvlue λ nd we figure out solution to the eqution: (A λi d+ u =. We then successively multiply by the mtrix (A λi until the zero vector is obtined. The sequence gives us chin of generlized eigenvectors, nd from these we build up solutions s follows (ssuming there re k generlized eigenvectors in the chin: x (t = v e λt x (t = (v t + v e λt ( x 3 (t = v t + v t + v 3 e λt 5 This is mthspek for A theorem tht s true but we re not going to prove. So just trust me. 7
. t x k (t = (v k (k + + v t k! + v k t + v k e λt We then mlgmte ll these chins of generlized eigenvectors, nd these gives us our complete set of linerly independent solutions. 6 This lwys works. We note tht in ll the exmples we re doing we re ssuming ll our eigenvlues re rel, but tht ssumption isn t necessry. This method works just fine if we hve complex eigenvlues, s long s we llow for complex eigenvectors s well. Exmple - Find the generl solution of the system: x = 5 3 7 x. Solution - The chrcteristic eqution of the coefficient system is: A λi = λ 5 3 λ 7 λ = (λ + 3 nd so the mtrix A hs the eigenvlue λ = with multiplicity 3. The corresponding eigenvector eqution is: 5 7 b c = The only eigenvectors tht work here re of the form: 6 In prctice we ll only be deling with smller (x, 3x3, mybe 4x4 systems, so things won t get too wful. 8
nd so the eigenvlue λ = hs defect. In order to figure out the generlized eigenvectors, we need to clculte (A λi nd (A λi 3 : (A λi = (A λi 3 = 3 3 3 So, wht we wnt to do is find vector v such tht (A λi 3 v = (not hrd to do but lso such tht (A λi v (lso not hrd to do, but perhps not trivil. Let s try the simplest vector we cn think of: v = If we test this out we get: (A λi v = (A λi 3 v = (obviously 3 3 = 3 So, ll is good here. We sy v = v 3, nd use it to get the other vectors in our chin: 9
5 7 5 7 5 = = 5 = v = v Using these three generlized eigenvectors we recover our three linerly independent solutions: x (t = x 3 (t = x (t = e t te t + t e t + 5 5 e t te t + e t nd so our generl solution will be: x(t = c x (t + c x (t + c 3 x 3 where the constnts c, c nd c 3 re determined by initil conditions. Notes on Homework Problems Problem 5.4. is system with repeted eigenvlue. Shouldn t be too hrd. As in seciton 5., the problem lso sks you to use computer system or grphing clcultor to construct direction field nd typicl solution curves for the system. A sketch is fine.
Problems 5.4.8 nd 5.4.6 del with lrger systems thn problem 5.4., but the pproch is the sme. They re lrger, so they re bit hrder, but on the upside you don t hve to drw ny direction fields! For problem 5.4.5 you ve got to find some chins. You should tke some powers of the coefficient mtrix. You ll find it s nilpotent, nd tht should help you lot in generting these chins! Problem 5.4.33 investigtes wht you do when you ve got defective complex root. Plese note tht there s typo in the textbook! The eqution v = [ 9 i ] T should be v = [ i ] T.