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1 9.5: Fundamental Sets of Eigenvector Solutions Homogenous system: x = Ax, A : n n Ex.: A = Characteristic Polynomial: (degree n) p(λ) = det(a λi) Def.: The multiplicity of a root λ i of p(λ) is the smallest positive integer m for which dm dλ m p(λ) λ=λi 0. Fundamental Thm. of Algebra: If the roots are counted with multiplicities, then p(λ) has exactly n roots λ 1,..., λ n, and p(λ) = ( 1) n (λ λ 1 ) (λ λ n ) Thm.: If λ 1,..., λ n are n real eigenvalues of A and v 1,...,v n are n linearly independent eigenvectors, then x j (t) = e λ jt v j, 1 j n are a fundamental set of solutions Find p(λ) = λ 3 + 3λ 2 + 4λ 12 = (λ + 2)(λ 3)(λ 2) λ 1 = 2, λ 2 = 3, λ 3 = 2. Eigenvectors: A+2I = v 1 = [ 1,0,1] T. Analogously: v 2 = [1,1,0] T, v 3 = [0,2,1] T. x 1 (t) = e 2t [ 1,0,1] T x 2 (t) = e 3t [1,1,0] T x 3 (t) = e 2t [0,2,1] T are a fundamental set of solutions. 1 1 Ex.: A = 1 1 p(λ) = (1 λ) 2 +1 = 0 λ 1,2 = 1 ±i Complex Eigenvalues 1

2 Review: Complex Numbers and Complex Exponential Complex Numbers C: set of complex numbers λ = α + iβ C, i = 1 α, β R; α = Re(λ), β = Im(λ) Complex conjugate: λ = α iβ Addition and multiplication If λ = α + iβ, µ = γ + iδ: λ + µ = (α + γ) + i(β + δ) like vector addition λµ = (α + iβ)(γ + iδ) = αγ βδ + i(αδ + βγ) Re(λ) = (λ + λ)/2 Im(λ) = (λ λ)/(2i) Absolute value: λ = α 2 + β 2 Inversion: 1 λ = λ λλ = λ λ 2 Euler Formula e iθ = cos θ + isin θ point on the unit circle: e iθ 2 = cos 2 θ + sin 2 θ = 1 e iθ = cos θ isin θ = 1/e iθ Complex exponential: e λ e λ = e α+iβ = e α e iβ = e α (cos β + isin β) = e α iβ = e α e iβ = e α (cos β isin β) = (e λ ) Exponential function: e λt = e αt( cos(βt) + isin(βt) ) e λt = e αt( cos(βt) isin(βt) ) = (e λt ) 2

3 Complex Eigenvalues and Eigenvectors Complex vectors: x,y R n z = x + iy C n Complex matrices: matrices with complex entries nullspace, linear independence, basis, dimension as in the real case (with complex vectors and scalars allowed) Def.: Let A: real n n-matrix. A complex root of det(a λi) = 0 is a complex eigenvalue. Any 0 v C n s.t. Av = λv is an eigenvector, and null(a λi) is the eigenspace for λ. Pairs: Av = λv Av = λv complex conjugate pairs of eigenvalues and eigenvectors Thm.: Let λ be a complex eigenvalue with eigenvector v = u + iw. Then v,v and u,w are linearly independent. { } 1 1 λ1 = 1 + i Ex.: A = 1 1 λ 2 = 1 i i 1 Compute A (1 + i)i = 1 i R3(1,i) 1 i R1(2,1,1) 1 i 1 i 0 0 v = [1, i] T = [1,0] T + i[0,1] T = u+iw is basis of null(a λ 1 I) v is eigenvector for λ 1 = 1 + i v = [1, i] T = u iw is eigenvector for λ 2 = λ 1 = 1 i { } det([v,v]) = 2i 0 Since, det([u,w]) = 1 0 v,v and u,w are linearly independent. 3

4 Solutions of Systems Derived from Complex Conjugate Pairs of Eigenvalues and Eigenvectors A: real n n-matrix System: x = Ax (1) Assume Av = λv with λ = α + iβ C, β 0 v = u + iw C n, v 0 Linearly independent complex solutions of (1): z(t) = e λt v, z(t) = e λt v Real and imaginary parts: z(t) = e (α+iβ)t (u + iw) = e αt (cos βt + isin βt)(u + iw) = e αt (ucos βt wsin βt) +ie αt (usin βt + wcos βt) = Rez(t) + iimz(t) Linearly independent real solutions of (1): x 1 (t) = e αt (ucos βt wsin βt) x 2 (t) = e αt (usin βt + wcos βt) For 2d Systems: (n = 2, Sec. 9.2) { a b T = a + d A = c d D = ad bc p(λ) = λ 2 Tλ + D Assume T 2 4D < 0 complex eigenvalues λ, λ: λ = (T + i α + iβ 4D T 2 )/2 Let v = u + iw 0 be in null(a λi). Then x 1 (t) = e αt (ucos βt wsin βt) x 2 (t) = e αt (usin βt + wcos βt) are already a fundamental set of solutions 4 }

5 Ex.: A = T = 2, D = 2 T 2 4D = 4 < 0 complex eigenvalues Characteristic polynomial: p(λ) = λ 2 2λ + 2 = (λ 1) (λ 1) 2 = 1 λ 1 = ±i Thus the eigenvalues are: λ = 1 + i, λ = 1 i (α = 1, β = 1) Find eigenvector for λ = 1 + i: i 1 A (1 + i)i = 1 i A λi is singular rows of A λi are linearly dependent 2nd row is complex multiple of 1st row ( i[ i, 1] = [ 1, i]) need only consider 1st row for finding basis v = [v 1, v 2 ] T of null(a λi) Equation from 1st row: iv 1 +v 2 = 0 Set v 2 = i v 1 = 1 eigenvector: v = = + i i u =, w = 0 1 Lin. indep. complex solutions: z(t) = e (1+i)t 1, z(t) = e i (1 i)t 1 i Fundamental set of real solutions: ( ) x 1 (t) = e t 1 0 cos t sin t 0 1 = e t cos t sin t ( ) x 2 (t) = e t 1 0 sin t + cos t 0 1 = e t [ sin t cos t Fundamental matrix: [ X(t) = e t cos t sin t sin t cos t ] ] 5

6 Fundamental Set of Eigenvector Solutions A: real n n-matrix System: x = Ax (1) Thm.: Assume 1. A has k real eigenvalues λ 1,..., λ k with linearly independent eigenvectors v 1,...,v k. 2. A has m complex conjugate pairs λ k+1, λ k+1,..., λ k+m, λ k+m of eigenvalues with eigenvectors v k+1,v k+1,...,v k+m,v k+m. 3. n = k + 2m and the eigenvectors of 2. are linearly independent. Then the n vector functions x i (t) = e λit v i, 1 i k x j (t) = e α jt (u j cos β j t w j sin β j t) x j+m (t) = e α jt (u j sin β j t + w j cos β j t) k + 1 j k + m are a fundamental set of solutions. Ex.: A = Use Matlab s poly and factor p(λ) = (λ + 1)[(λ + 2) 2 + 9] eigenvalues: λ 1 = 1 λ 2 = 2 + 3i, λ 3 = λ 2 eigenvectors (using Matlab s null): v 1 = [2,1,2] T v 2 = [i,2 2i,1] T = [0,2,1] T + i[1, 2,0] T Fundamental set of solutions: x 1 (t) = e t [2,1,2] T x 2 (t) = e 2t ([0,2,1] T cos3t [1, 2,0] T sin3t) = e 2t p(t) x 3 (t) = e 2t ([0,2,1] T sin3t +[1, 2,0] T cos3t) = e 2t q(t) p(t) = [ sin3t,2cos3t+2sin3t,cos3t] T q(t) = [cos3t,2sin3t 2cos3t,sin3t] T 6

7 Worked Out Examples from Exercises Ex : Find fundamental set of solutions of y = Ay for A = T = 6, D = 10 T 2 4D = 4 < 0 complex eigenvalues: p(λ) = λ 2 +6λ+10 = (λ+3) 2 +1 = 0 eigenvalues λ = 3+i, λ = 3 i [ 2 i i 2 1 A ( 3+i)I = eigenvector: v = = +i 5 2 i u = [2, 5] T, v = [1,0] T fundamental set of solutions: ( ) [ y 1 (t) = e 3t 2 1 cos t sin t = e 5 0 3t 2cos t sin t 5cos t ( ) [ y 2 (t) = e 3t 2 1 sin t + cos t = e 5 0 3t 2sin t + cos t 5sin t ] ] ] Ex : Find solution of system of Ex. 18 for IC y(0) = [1, 5] T c1 1 y(0) = c 1 + c 5 2 = = c 2 5 c1 = = c ( ) [ y(t) = e 3t 2cos t sin t 2sin t + cos t = e 5cos t 5sin t 3t cos t 3sin t 5sin t 5cos t ] 7

8 Ex : Find eigenvalues and eigenvectors of A and verify linear independence of eigenvectors A = λ , det(a λi) = 6 1 λ λ p(λ) = (2 λ) 1 λ λ = (λ 2)(λ 1)(λ + 1) eigenvalues: λ 1 = 2, λ 2 = 1, λ 3 = 1. Find eigenvectors: A 2I = v 1 = A I = v2 = A + I = det([v 1,v 2,v 3 ]) = v 1,v 2,v 3 are linearly independent v3 = = ( 1) = 1 0 8

9 = 3x Ex : Find general solution of = 5x + 6y 4z = 5x + 2y A = λ , det(a λi) = 5 6 λ λ p(λ) = ( 3 λ) 6 λ 4 2 λ = (λ + 3)[(6 λ)( λ) + 8] λ 1 = 3 A + 3I = v 1 = [1,1,1] T λ 2 = 2 A 2I = λ 3 = 4 A 4I = = (λ + 3)(λ 2 6λ + 8) = (λ + 3)(λ 2)(λ 4) x y z v2 = [0,1,1] T v 3 = [0,2,1] T General Solution: x(t) = c 1 e 3t [1,1,1] T + c 2 e 2t [0,1,1] T + c 3 e 4t [0,2,1] T 9

10 Ex : Find solution to IC x(0) = [ 2,0,2] T for system of Ex. 9 Write general solution as x(t) = X(t)c, c = [c 1, c 2, c 3 ] T, with F.M. X(t) = e 3t 0 0 e 3t e 2t 2e 4t ; IC X(0)c = c 1 c 2 = 2 0 e 3t e 2t e 4t c 3 2 [X(0),x(0)] = c = x(t) = 2e 3t [1,1,1] T + 6e 2t [0,1,1] T 2e 4t [0,2,1] T = [ 2e 3t, 2e 3t + 6e 2t 4e 4t, 2e 3t + 6e 2t 2e 4t ] T Ex : Find real and imaginary parts for y(t) = e 2it [1,1 + 2i, 3i] T Write y(t) = e 2it v, v = [1,1+2i, 3i] T = u+iw, u = [1,1,0] T, w = [0,2, 3] T y(t) = (cos2t + isin2t)(u + iw) = ucos2t + iusin2t + iwcos2t + i 2 wsin2t = (ucos2t wsin2t) + i(usin2t + wcos2t) Rey(t) = ucos2t wsin2t = [1,1,0] T cos2t [0,2, 3] T sin2t = [cos2t,cos2t 2sin2t,3sin2t] T Imy(t) = usin2t + wcos2t = [1,1,0] T sin2t + [0,2, 3] T cos2t = [sin2t,sin2t + 2cos2t, 3sin2t] T 10

11 = 4x + 8y + 8z Ex : Find general solution of = 4x + 4y + 2z = 2z A = λ , det(a λi) = 4 4 λ λ p(λ) = (2 λ) 4 λ λ = (2 λ)[( 4 λ)(4 λ) + 32] x y z = (2 λ)(λ ) λ 1 = 2, λ 2 = 4i, λ 3 = λ 2 A 2I = /2 1/ A 4iI = 1 1/2 1/ i i i R1(2,1,1+i),R2(2,3) i v 1 = v2 = i 0 1 i i 1 0 Set v 2 = u 2 + iw 2, u 2 = [1,1,0] T, w 2 = [ 1,0,0] T 11

12 General solution: x(t) = c 1 x 1 (t) + c 2 x 2 (t) + c 3 x 3 (t), where x 1 (t) = e 2t v 1 = e 2t [0, 1,1] T x 2 (t) = u 2 cos4t w 2 sin4t = [1,1,0] T cos4t [ 1,0,0] T sin4t = [cos4t + sin4t,cos4t,0] T x 3 (t) = u 2 sin4t + w 2 cos4t = [1,1,0] T sin4t + [ 1,0,0] T cos4t = [sin4t cos4t,sin4t,0] T Ex : Find solution to IC x(0) = [1,0,0] T for system of Ex cos4t + sin4t sin4t cos4t F.M.: X(t) = e 2t cos4t sin4t, X(0) = e 2t Solve X(0)c = x(0) for c: [X(0),x(0)] = c = Solution x(t) = x 3 (t) = [cos4t sin4t, sin4t,0] T 12

13 Ex : Find F.S.S. for y = Ay, A = , using a computer Use Matlab s poly, roots, null commands with rational format: >> A=[-6 2-3; ;4-2 1]; >> format rat,cpol=poly(a); >> eigvals=roots(cpol) eigvals = >> v1=null(a+3*eye(3), r ); >> v2=null(a+2*eye(3), r ); >> v3=null(a+eye(3), r ); >> [v1 ;v2 ;v3 ] ans = /2 1/ eigenvalues and eigenvectors (multiply second eigenvector by 2): λ 1 = 3 λ 2 = 2 λ 3 = 1 fundamental set of solutions: v 1 = [ 1,0,1] T v 2 = [ 1,1,2] T v 3 = [ 1, 1,1] T y 1 (t) = e 3t [ 1,0,1] T y 2 (t) = e 2t [ 1,1,2] T y 3 (t) = e t [ 1, 1,1] T 13

14 Ex : Find F.S.S. for y = Ay, A = This time use Matlab s symbolic toolbox: >> A=[ ; ; ]; >> poly(sym(a));eigvals=solve(ans) eigvals = [ -2] [ -3+2*i] [ -3-2*i], using a computer >> v1=null(sym(a)+2*eye(3)); >> v2=null(sym(a)-eigvals(2)*eye(3)); >> [v1 ;transpose(v2)] ans = [1, -2, -2] [1, -5/4-1/4*i, -3/4-1/4*i] eigenvalues and eigenvectors (multiply v 1 by 1, v 2 by 4): λ 1 = 2 λ 2 = 3 + 2i fundamental set of solutions: v 1 = [ 1,2,2] T v 2 = [ 4,5 + i,3 + i] T u 2 = [ 4,5,3] T, w 2 = [0,1,1] T y 1 (t) = e 2t [ 1,2,2] T y 2 (t) = e 3t (u 2 cos2t w 2 sin2t) = e 3t ([ 4,5,3] T cos2t [0,1,1] T sin2t) = e 3t [ 4cos2t,5cos2t sin2t,3cos2t sin2t] T y 3 (t) = e 3t (u 2 sin2t + w 2 cos2t) = e 3t ([ 4,5,3] T sin2t + [0,1,1] T cos2t) = e 3t [ 4sin2t,5sin2t + cos2t,3sin2t + cos2t] T Note: [5/( 3 i)]v 2 = [ 6 + 2i,8 i,5] T answer given in text. 14

15 Ex : Solve system of Ex. 37 for IC y(0) = [ 6,2,9] T via computer Let Y (0) = [y 1 (0),y 2 (0),y 3 (0)] = Solve Y (0)c = y(0): >> M=[ ; ; ]; >> rref(m) ans = c 1 = 2, c 2 = 3, c 3 = 1 y(t) = 2e 3t [ 1,0,1] T +3e 2t [ 1,1,2] T +e t [ 1, 1,1] T or y(t) = [ 2e 3t 3e 2t e t,3e 2t e t,2e 3t + 6e 2t + e t ] T Ex : Solve system of Ex. 41 for IC y(0) = [ 1,7,3] T via computer y(t) = 7e 2t [ 1,2,2] T Y (0) = 2 5 1,Y (0)c = y(0) >> M=[ ; ; ]; >> rref(m) ans = e 3t [ 4cos2t,5cos2t sin2t,3cos2t sin2t] T +11e 3t [ 4sin2t,5sin2t + cos2t,3sin2t + cos2t] T y 1 (t) = 7e 2t 4e 3t (2cos2t + 11sin2t) y 2 (t) = 14e 2t + e 3t (21cos2t + 53sin2t) y 3 (t) = 14e 2t + e 3t (17cos2t + 31sin2t) 15

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