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) 9 11 21 63 252 70 69 141 42684 A = 575 575 1149 3451 13801 (2) 3891 3891 7782 23345 93365 1024 1024 2048 6144 24572 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 = 0. 5 5 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 5.4 151
In he secions ha follow we illusrae appropriae Maple, Mahemaica, and MATLAB echniques o analyze he 4 4 marix 35 12 4 30 22 8 3 19 A = (3) 10 3 0 9 27 9 3 23 of Problem 31 in Secion 5.4 of he ex. You can use any of he oher problems here (especially Problems 23 30 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); 4 3 2 λ λ λ λ 4 + 6 4 + 1 (ha is, 4 ( λ 1) ) eigenvals(a); 1,1,1,1 eigenvecs(a); [1, 4, {[0 1 3 0], [ 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
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); 34 22 u2 : = 10 27 and u3 := evalm( B &* u2); 42 7 u3 : = 21 42 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 u3 + 42 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 ), 3 2 3 x ( ) = ( u + u + u ) e. 1 2 4 1 2 2 3 Applicaion 5.4 153
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] 2 3 4 1-4 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
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 u3 + 14 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 ), 3 2 3 x ( ) = ( u + u + u ) e. 1 2 4 1 2 2 3 Using MATLAB 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 proceed o explore is characerisic polynomial, eigenvalues, and eigenvecors. poly(a) ans = 1.0000-4.0000 6.0000-4.0000 1.0000 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 5.4 155
D = Thus MATLAB finds only he wo independen eigenvecors w1 = [1 0-1 -1]'; w2 = [0 1 3 0]'; 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 = [1 0 0 0]'; we calculae he furher generalized eigenvecors and u2 = B*u1 u2 = 34 22-10 -27 u3 = B*u2 u3 = 42 7-21 -42 156 Chaper 5
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 ), 3 2 3 x ( ) = ( u + u + u ) e. 1 2 4 1 2 2 3 Applicaion 5.4 157