Application 5.4 Defective Eigenvalues and Generalized Eigenvectors

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1 Applicaion 5.4 Defecive Eigenvalues and Generalized Eigenvecors The goal of his applicaion is he soluion of he linear sysems like where he coefficien marix is he exoic 5-by-5 marix x = Ax, (1) A = (2) ha is generaed by he MATLAB command gallery(5). Wha is so exoic abou his paricular marix? Well, ener i in your calculaor or compuer sysem of choice, and hen use appropriae commands o show ha: Firs, he characerisic equaion of A reduces o single eigenvalue λ = 0 of mulipliciy 5. 5 λ = 0, so A has he Second, here is only a single eigenvecor associaed wih his eigenvalue, which hus has defec 4. To seek a chain of generalized eigenvecors, show ha A 4 0 bu 5 5 zero marix). Hence any nonzero 5-vecor u 1 saisfies he equaion A 5 = 0 (he ( A λ I) u = A u = Calculae he vecors u 2 = Au 1, u 3 = Au 2, u 4 = Au 3, and u 5 = Au 4 in urn. You should find ha u 5 is nonzero, and is herefore (o wihin a consan muliple) he unique eigenvecor v of he marix A. Bu can his eigenvecor v you find possibly be independen of your original choice of he saring vecor u 1 0? Invesigae his quesion by repeaing he process wih several differen choices of u 1. Finally, having found a lengh 5 chain {u 5, u 4, u 3 } of generalized eigenvecors based on he (ordinary) eigenvecor u 5 associaed wih he single eigenvalue λ = 0 of he marix A, wrie five linearly independen soluions of he 5-dimensional homogeneous linear sysem x = Ax. Applicaion

2 In he secions ha follow we illusrae appropriae Maple, Mahemaica, and MATLAB echniques o analyze he 4 4 marix A = (3) of Problem 31 in Secion 5.4 of he ex. You can use any of he oher problems here (especially Problems and 32) o pracice hese echniques. Using Maple Firs we ener he marix in (3): wih(linalg): A := marix(4,4, [ 35, -12, 4, 30, 22, -8, 3, 19, -10, 3, 0, -9, -27, 9, -3, -23 ] ): Then we explore is characerisic polynomial, eigenvalues, and eigenvecors: charpoly(a,lambda); λ λ λ λ (ha is, 4 ( λ 1) ) eigenvals(a); 1,1,1,1 eigenvecs(a); [1, 4, {[ ], [ 1 0 ] } ] Thus Maple finds only he wo independen eigenvecors w1 := marix(4,1, [ 0, 1, 3, 0]): w2 := marix(4,1, [-1, 0, 1, 1]): associaed wih he mulipliciy 4 eigenvalue λ = 1, which herefore has defec 2. To explore he siuaion we se up he 4 4 ideniy marix and he marix B = A λ I: 152 Chaper 5

3 Id := diag(1,1,1,1): L = 1: B := evalm( A - L*Id): When we calculae B 2 and B 3, B2 := evalm(b &* B); B3 := evalm(b2 &* B); 2 we find ha B 0 bu λ = 1. Choosing B 3 = 0, so here should be a lengh 3 chain associaed wih u1 := marix(4,1,[1,0,0,0]); we calculae he furher generalized eigenvecors u2 := evalm( B &* u1); u2 : = and u3 := evalm( B &* u2); 42 7 u3 : = Thus we have found he lengh 3 chain {u 3 } based on he (ordinary) eigenvecor u 3. (To reconcile his resul wih Maple's eigenvecs calculaion, you can check ha u w2 = 7w 1.) Consequenly four linearly independen soluions of he sysem x = Ax are given by x () = w e, x ( ) = u e, 2 3 x ( ) = ( u + u e ), x ( ) = ( u + u + u ) e Applicaion

4 Using Mahemaica Firs we ener he marix in (3): A = { { 35, -12, 4, 30 }, { 22, -8, 3, 19 }, {-10, 3, 0, -9 }, {-27, 9, -3, -23 } }; Then we explore is characerisic polynomial, eigenvalues, and eigenvecors: CharacerisicPolynomial[A, r] r + 6 r - 4 r + r (ha is, 4 ( r 1) ) Eigenvalues[A] {1, 1, 1, 1} Eigenvecors[A] {{-3,-1,0,3}, {0,1,3,0}, {0,0,0,0}, {0,0,0,0}} Thus Mahemaica finds only he wo independen (nonzero) eigenvecors w1 = {-3,-1, 0, 3}; w2 = { 0, 1, 3, 0}; associaed wih he mulipliciy 4 eigenvalue λ = 1, which herefore has defec 2. To explore he siuaion we se up he 4 4 ideniy marix and he marix B = A λ I: Id = DiagonalMarix[1,1,1,1]; L = 1; B = A - L*Id; When we calculae B 2 and B 3, B2 = B.B B3 = B2.B 2 we find ha B 0 bu λ = 1. Choosing B 3 = 0, so here should be a lengh 3 chain associaed wih u1 = {{1},{0},{0},{0}} 154 Chaper 5

5 we calculae u2 = B.u1 {{34}, {22}, {-10}, {-27}} u3 = B.u2 {{42}, {7}, {-21}, {-42}} Thus we have found he lengh 3 chain {u 3 } based on he (ordinary) eigenvecor u 3. (To reconcile his resul wih Mahemaica's Eigenvecors calculaion, you can check ha u w1 = 7w 2.) Consequenly four linearly independen soluions of he sysem x = Ax are given by x () = w e, x ( ) = u e, 2 3 x ( ) = ( u + u e ), x ( ) = ( u + u + u ) e Using MATLAB Firs we ener he marix in (3): A = [ ]; Then we proceed o explore is characerisic polynomial, eigenvalues, and eigenvecors. poly(a) ans = These are he coefficiens of he characerisic polynomial, which hence is Then 4 ( λ 1). [V, D] = eigensys(a) V = [ 1, 0] [ 0, 1] [-1, 3] [-1, 0] Applicaion

6 D = Thus MATLAB finds only he wo independen eigenvecors w1 = [ ]'; w2 = [ ]'; associaed wih he single mulipliciy 4 eigenvalue λ = 1, which herefore has defec 2. To explore he siuaion we se up he 4 4 ideniy marix and he marix B = A λ I: Id = eye(4); B = A - L*Id; When we calculae B 2 and B 3, B2 = B*B B3 = B2*B 2 3 We find ha B 0 bu B = 0, so here should be a lengh 3 chain associaed wih he eigenvalue λ = 1. Choosing he firs generalized eigenvecor u1 = [ ]'; we calculae he furher generalized eigenvecors and u2 = B*u1 u2 = u3 = B*u2 u3 = Chaper 5

7 Thus we have found he lengh 3 chain {u 3 } based on he (ordinary) eigenvecor u 3. (To reconcile his resul wih MATLAB's eigensys calculaion, you can check ha u3 42 w1 = 7w 2.) Consequenly four linearly independen soluions of he sysem x = Ax are given by x () = w e, x ( ) = u e, 2 3 x ( ) = ( u + u e ), x ( ) = ( u + u + u ) e Applicaion

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