# ch9_6 - 9.6 Matrix Exponential Repeated Eigenvalues x′ =...

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Unformatted text preview: 9.6: Matrix Exponential, Repeated Eigenvalues x′ = Ax, A:n×n (1) Def.: If x1(t), . . . , xn(t) is a fundamental set of solutions (F.S.S.) of (1), then X (t) = [x1(t), . . . , xn (t)] (n × n) is called a fundamental matrix (F.M.) for (1). General solution: (c = [c1, . . . , cn]T ) Ex.: A = −4 2 −3 1 Eigenvalues and eigenvectors: λ1 = −1 ↔ v1 = [2, 3]T λ2 = −2 ↔ v2 = [1, 1]T F.S.S.: 2 1 x1 (t) = e−t , x2(t) = e−2t 3 1 F.M.: If we set X (t ) = 2e−t e−2t 3e−t e−2t x(t) = c1x1(t) + . . . + cnxn(t) = X (t)c Thm.: If X (t) is a F.M. for (1) and C is a constant nonsingular matrix, then X (t)C is also a F.M. Proof: Each column of X (t)C is a linear combination of the columns of X (t) and so is a solution of (1), and X (0)C is nonsingular. y1 (t) = 2x2 (t), y2(t) = 3x2 (t), y1(t), y2 (t) are also F.S.S. with F.M. Y (t ) = = 3e−2t 3e−2t 4e−t 6e−t 02 30 1 2e−t e−2t 3e−t e−2t Matrix Exponential Consider IVP: −4 2 Ex.: A = −3 1 x′ = Ax, x(0) = x0 (2) 2e−t e−2t Solution of IVP: If X (t) is a F.M.: X (t) = 3e−t e−2t F.M., the general solution is 21 X (0) = , (X (0))−1 = 31 x(t) = X (t)c Match c to IC: eAt = X (t)(X (0))−1 −1 1 3 −2 x(0) = X (0)c = x0 ⇒ c = (X (0))−1 x0 ⇒ x(t) = X (t)(X (0))−1 x0 Def.: Given a F.M. X (t), then eAt = X (t)(X (0))−1 is the matrix exponential of At. Thm.: The solution of (2) is def IVP: = = 2e−t e−2t 3e−t e−2t 3e−2t − 2e−t 3e−2t − 3e−t −1 1 3 −2 2e−t − 2e−2t 3e−t − 2e−2t x′ = −4 2 −3 1 x, x(0) = 2 1 2 1 Solution: x(t) = = 3e−2t − 2e−t 2e−t − 2e−2t 3e−2t − 3e−t 3e−t − 2e−2t 4e−2t − 2e−t 4e−2t − 3e−t 2 x(t) = e x0 At Properties of the Matrix Exponential • Exponential series (A0 = I ): eAt = ∞ • If AB = BA, then e(A+B )t = eAteBt = eBt eAt Note: If AB = BA, then in general e(A+B )t = eAteBt = eBteAt (At)m/m! m=0 Convergence for any matrix A • d eAt = AeAt = eAtA dt • If D = [dij ] is a diagonal matrix (dij = 0 for i = j ), then eDt is a diagonal matrix with entries ediit. Ex.: exp a0 0b t= eat 0 0 ebt • eAt is nonsingular, and (eAt)−1 = e−At • If V is nonsingular, then (V AV −1 )t = V eAtV −1 e • Special case (dii = r): e(rI )t = ert I • If v is an eigenvector for an eigenvalue λ, then eAtv = eλt v 3 Matrices with only one eigenvalue Thm.: If A has only one eigenvalue λ, then there is an integer k, 0 < k ≤ n, such that (A − λI )k = 0 Use this to compute eAt as follows. Write A = λI + (A − λI ). Then eAt = e(λI )t+(A−λI )t = e(λI )t e(A−λI )t λt (A−λI )t =ee = eλt k −1 = ⇒ Ex.: A = 1 4 : T = −2, D = 1 −1 −3 p(λ) = λ2 − T λ + D = λ2 + 2λ + 1 = (λ + 1)2 ⇒ only one eigenvalue λ = −1 A+I (A + I )2 = = 2 4 −1 −2 2 4 −1 −2 00 00 2 4 −1 −2 (k = 2) (A − λI )j (tj /j !) ⇒ j =0 e(A+I )t = I + (A + I )t 1 + 2t 4t = −t 1 − 2t eAt = e−t e(A+I )t 1 + 2t 4t = e−t −t 1 − 2t 4 ⇒ only k terms of exponential series required Generalized Eigenvectors and Associated Solutions If A has repeated eigenvalues, n linearly independent eigenvectors may not exist → need generalized eigenvectors Def.: Let λ be eigenvalue of A. (a) The algebraic multiplicity, m, of λ is the multiplicity of λ as root of the characteristic polynomial (CN Sec. 9.5). (b) The geometric multiplicity, mg , of λ is dim null(A − λI ). Need: m linearly independent solutions of x′ = Ax associated with λ. • If mg = m ⇒ m linearly independent eigenvector solutions. • What if mg < m? Def.: Any nonzero vector v in null((A − λI )k ) is a generalized eigenvector for λ. Solution associated with v: (A − λI )k v = 0 ⇒ eAtv = eλt k −1 (tj /j !)(A − λI )j v j =0 Thm.: Let v1, . . . , vm be a basis of null(A − λI )k . Then the Thm.: If λ is an eigenvalue with algebraic multiplicity m, then there is an integer k, 0 < k ≤ m, such that dim null((A − λI )k ) = m dim null((A − λI )k−1) < m xi(t) = eλt k −1 (tj /j !)(A−λI )j vi, j =0 1 ≤ i ≤ m, are m linearly independent solutions of x′ = Ax. 5 2d Systems: (Sec. 9.2) ab T = a+d A= cd D = ad − bc Assume T 2 − 4D = 0 ⇒ p(λ) = (λ − λ1)2, λ1 = T /2 (a) If A = λ1I ⇒ mg = 2 ⇒ x(t) = eλ1 tx(0) (any vector is eigenvector) (b) If A = λ1I ⇒ mg = 1: • Compute eigenvector v • Pick vector w that is not a multiple of v ⇒ (A − λ1I )w = av for some a = 0 (any w ∈ R2 is generalized eigenvector) • ⇒ F.S.S.: x1(t) = eλ1 tv x2(t) = eλ1 t(w + avt) Ex.: A = 1 4 : T = −2, D = 1 −1 −3 2 4 −1 −2 ⇒ T 2 − 4D = 0 ⇒ eigenvalue λ = −1 A+I = ⇒ eigenvector v = [−2, 1]T Choose w = [1, 0]T (simple form) ⇒ (A + I )w = ⇒ F.S.S.: 2 4 −1 −2 1 2 = = −v 0 −1 x1(t) = e−tv = e−t[−2, 1]T x2(t) = e−t(w − vt) = e−t([1, 0]T − t[−2, 1]T ) = e−t[1 + 2t, −t]T Other Method: Compute (c.f. p.4) eAt = e−t (I +(A+I )t) = e−t 1 + 2t 4t −t 1 − 2t Columns of eAt are also F.S.S. 6 −1 2 1 0 Ex.: A = 0 −1 −1 −3 −3 p(λ) = −1 − λ 1 A + 2I → eigenvector v2 = [1, 0, −1]T −1 −3 − λ = (−1 − λ)[(1 + λ)(3 + λ) + 1] ⇒ eigenvector solution: x2(t) = e−2t [1, 0, −1]T = −(λ + 1)(λ2 + 4λ + 4) = −(λ + 1)(λ + 2)2 For x3 (t) use generalized eigenvector v3 that is not an eigenvector. ⇒ eigenvalues λ1 = −1, m1 = 1 T λ2 = −2, m2 = 2 2): u1 = [1, 0, 0] Basis of null((A+2I ) u2 = [0, 0, 1]T Compute A − λ1 I = A + I : Note: v2 = u1 − u2 0 2 1 1 0 1/2 = (A + 2I )u1 = (A + 2I )u2 0 0 → 0 1 1/2 A +I = 0 −1 −3 −2 00 0 v3 can be any vector c1 u1 + c2u2 s.t. (A +2I )(c1u1 + c2 u2) = (c1 + c2 )v2 = 0 Set x3 = −2 ⇒ eigenvector v1 = [1, 1, −2]T Choose v3 = u2 = [0, 0, 1]T (text: u1) Since m1 = 1 ⇒ x3 (t) = e−2t (I v3 + t(A + 2I )v3 ) ⇒ one (eigenvector) solution: = e−2t (v3 + tv2 ) x1 (t) = e−t [1, 1, −2]T = e−2t [t, 0, 1 − t]T 7 = (−1 − λ) −1 − λ 2 1 0 −1 − λ 0 −1 −3 −3 − λ m2 = 2 → check A − λ2I , (A − λ2 I )2 : 1 2 1 101 1 0 → 0 1 0 A+2I = 0 −1 −3 −1 000 0 10 010 2 1 0 → 0 0 0 (A+2I ) = 0 0 −2 0 000 6 6 −3 2 2 0 −4 −4 Ex.: A = 8 7 −4 4 1 0 −1 −2 Matlab → p(λ) = ((λ + 1)2 + 1)2 ⇒ single complex pair of eigenvalues λ1 = −1 + i, λ2 = λ1 (m = 2). 2. Check B 2 = (A − λ1 I )2 : 2 − 14i 3 − 12i −2 + 6i −4i −2 + 6i −4i 0 8i B 2= 8 − 16i 6 − 14i −6 + 6i −8i −2 − 2i −1 1 + 2i −2 + 2i Matlab → basis for null(B 2): u1 = [2, 0, 4, −1 + i]T = v1 u2 = [−3 − i, 4, 0, −2 + 2i]T 1. Check B ≡ A − λ1I = A − (−1+ i)I : ⇒ u2 is generalized eigenvector that 7−i 6 −3 2 is not an eigenvector. 2 0 −4 −3 − i Pick v3 = u2 = [−3 − i, 4, 0, −2 + 2i]T B= 8 7 −3 − i 4 Need: B v3 = [−2, 0, −4, 1 − i]T = −v2 1 0 −1 −1 − i Complex solution associated with v3: Matlab → basis for null(B ): z2(t) = e(−1+i)t (I v3 + tB v3 ) T v1 = [2, 0, 4, −1 + i] = e(−1+i)t (v3 − tv2 ) ⇒ Complex eigenvector solution: = e(−1+i)t [−3 − i − 2t, 4, −4t, −2 + 2i + (1 − i)t]T z1(t) = e(−1+i)t [2, 0, 4, −1 + i]T 3. Take real and imaginary parts of z1(t) and z2(t) to obtain F.S.S: x1 (t) = Re z1(t) = e−t[2 cos t, 0, 4 cos t, − cos t − sin t]T x2 (t) = Im z1(t) = e−t[2 sin t, 0, 4 sin t, cos t − sin t]T x3 (t) = Re z2(t) = e−t[sin t − (3 + 2t) cos t, 4 cos t, −4t cos t, (t − 2)(cos t + sin t)]T x4 (t) = Im z2(t) = e−t[− cos t − (3 + 2t) sin t, −4 sin t, −4t sin t, (t − 2)(sin t − cos t)]T 8 7 5 −3 2 1 0 0 0 Ex.: A = 12 10 −5 4 −4 −4 2 −1 Matlab → p(λ) = (λ + 1)(λ − 1)3 ⇒ eigenvalues λ1 = −1, m1 = 1 λ2 = 1, m2 = 3 Find eigenvector for λ1 : 8 5 −3 2 2 0 0 0 A+I = 12 10 −4 4 −4 −4 20 Matlab → basis of null(B ): v2 = [1, 0, 2, 0]T v3 = [1, −2, 0, 2]T Associated eigenvector solutions: x2 (t) = et [1, 0, 2, 0]T x3 (t) = et [1, −2, 0, 2]T To ﬁnd 4th solution check B 2 : −8 −8 4 −4 0 0 0 0 B2 = −16 −16 8 −8 8 8 −4 4 Matlab → eigenvector (basis vector for null(A + I )): v1 = [1, 0, 2, −1]T Associated eigenvector solution: x1 (t) = e−t [1, 0, 2, −1]T For λ2 = 1 → check powers of A − I : 6 5 −3 2 0 0 0 0 B ≡A−I = 12 10 −6 4 −4 −4 2 −2 ⇒ RREF (B 2) has only one nonzero row [1, 1, −1/2, 1/2]. Construct basis of null(B 2 ) by setting x2, x3 = 0, x4 = 2 → u1 = [−1, 0, 0, 2]T x2, x4 = 0, x3 = 2 → u2 = [1, 0, 2, 0]T x3, x4 = 0, x2 = 1 → u3 = [1, −1, 0, 0]T Check which are not eigenvectors: B u1 = −2v2 , B u2 = 0, B u3 = v2 ⇒ Can choose v4 = u1 (simple). Associated solution: x4(t) = et(I v4 + tB v4) = et (u1 − 2tv2 ) = et[−1 − 2t, 0, −4t, 2]T 9 Advanced Theory: Chains of Generalized Eigenvectors Thm.: Let λ be an eigenvalue of a n × n-matrix A with • algebraic multiplicity m • geometric multiplicity mg Let B = A − λI and k be s.t. dim null(B k ) = m dim null(B k−1) < m There are mg chains of vectors ( (i v1i), . . . , vri ), Computation of chains: Assume i − 1 chains have been computed. Let q be the largest integer for which there is a vector v in null(B q ) s.t. B q−1 v = 0, and v and all previously computed chain-vectors are linearly independent. Set ri = q . Then the ith chain is computed as ( vrii) = v ( (i) vj i) = B vj +1 for ri > j ≥ 1 i i Note: B v1 = 0 ⇒ v1 is eigenvector. 1 ≤ i ≤ mg Solutions of x′ = Ax: j −1 s.t. r1 + r2 + · · · + rmg = m, (i ) B vj +1 = (i ) B v1 = ( vj i), 1 ≤ j < ri x(i) (t) j x(i) (t) 1 =e = λt ( [vj i) + l=1 ( (tl /l!)vl i) ] if j > 1 ( eλtv1i) 0 (i ) and all the vectors vj basis of null(B k ). are a Single chain: If k = m ⇒ only one chain v1, . . . , vm, vj = B vj +1 (j < m) 10 3 −3 −6 5 2 5 −4 −3 Ex.: A = 2 −6 −4 7 −3 0 5 −2 Matlab → p(λ) = (λ + 1)3 (λ − 2) ⇒ eigenvalues λ1 = −1, m1 = 3 λ4 = 2, m4 = 1 Eigenvector for λ4 : v4 = [−1, 1, 1, 2]T → x4(t) = e2t [−1, 1, 1, 2]T λ1 = −1: Set B = A + I . Matlab → dim null(B 2 ) = 2 ⇒ k = 3 ⇒ 1 chain Matlab’s null → basis for null(B 3 ): u1 = [1, 0, 0, 0]T u2 = [0, 0, 1, 0]T u3 = [0, 1, 0, 1]T Find B 2 u1 = 0 ⇒ chain can be generated by u1 (simple form). Set v3 = u1 = [1, 0, 0, 0]T v2 = B v3 = [4, −3, 2, −3]T v1 = B v2 = [−2, 1, −2, 1]T Solutions associated with 3-chain: x1 (t) = e−t v1 = e−t [−2, 1, −2, 1]T e−t(v2 + tv1) e−t[4 + t, −3, 2, −3]T e−t(v3 + tv2 + t2v1 /2) e−t[1 + 4t − t2, −3t + t2/2, 2t − t2, −3t + t2/2]T Ex.: In example on p.8: mg = 2, k = 2, m = 3 ⇒ 2 chains: • u1, −2v2 is 2-chain • v3 is 1-chain x2(t) = = x3(t) = = Note: Approach via chains is “useful” if m >> mg , especially if mg = 1 and k = m >> 1. Summary: 1. For any matrix A, a F.S.S x1(t), . . . , xn(t) for x′ = Ax can be computed using eigenvalues and (generalized) eigenvectors. 2. X (t) = [x1(t), . . . , xn(t)] is a F.M: x(t) = X (t)c is gen. sol. 3. M.E.: eAt = X (t)(X (0))−1 x(t) = eAtx(0) 11 Matlab Tools Eigenvalues & Eigenvectors Matrix Exponential • eig(A): vector of eigenvalues of A • [V,D]=eig(A) → outputs V: matrix of eigenvectors (columns) D: diagonal matrix of eigenvalues Symbolic Computation: >> A=[1 1;-1 1];[V,D]=eig(sym(A)) V= D= [ 1, 1] [ 1+i, 0] [ i, -i] [ 0, 1-i] >> A=sym([-2 1 -1;1 -3 0;3 -5 0]); >> [V,D]=eig(A) V= D= [ 1, -2] [ -2, 0, 0] [ 1, -1] [ 0, -2, 0] [ 1, 1] [ 0, 0, -1] Numerical Computation: >> A=[-2 1 -1;1 -3 0;3 -5 0]; >> [V,D]=eig(A) V= 0.5774 0.5774 -0.8165 0.5774 0.5774 -0.4082 0.5774 0.5774 0.4082 D= -2.0000 0 0 0 -2.0000 0 0 0 -1.0000 Note: no generalized eigenvectors • expm(A): matrix exponential of A Symbolic Computation: >> A=[1 1;-1 1];syms t;expm(sym(A)*t) ans = [ exp(t)*cos(t), exp(t)*sin(t)] [ -exp(t)*sin(t), exp(t)*cos(t)] Note: t must be declared symbolically >> A=sym([-2 1 -1;1 -3 0;3 -5 0]);syms t; >> expm(A*t);ans(:,1) ans = [ 3*exp(-2*t)-2*exp(-t)+2*t*exp(-2*t)] [ 2*t*exp(-2*t)-exp(-t)+exp(-2*t)] [ 2*t*exp(-2*t)+exp(-t)-exp(-2*t)] In this example only ﬁrst column of eAt is displayed Numerical Computation: >> A=[-2 1 -1;1 -3 0;3 -5 0];t=2;expm(A*t) ans = -0.1425 0.3582 -0.1974 -0.0438 0.1425 -0.0804 0.1903 -0.3439 0.1720 Use loop to compute solution array: >> t=linspace(0,1,20);x0=[1;0;0];x=[ ]; >> for n=1:20;x=[x expm(A*t(n))*x0];end First entry of solution can be plotted via >> plot(t,x(1,:)) 12 Worked Out Examples from Exercises Ex. 9.2.31: Find general solution of y′ = Ay for A = T = 4, D = 4 ⇒ T 2 = 4D ⇒ single eigenvalue λ = 2 A − 2I = Pick w = ⇒ F.S.S.: 1 0 1 −1 1 −1 ⇒ eigenvector v = 1 −1 1 −1 1 0 1 1 = 1 1 =v 3 −1 1 1 ⇒ (A − 2I )w = y1 (t) = e2t v = e2t [1, 1]T y2 (t) = e2t (w + tv) = e2t([1, 0]T + t[1, 1]T ) = e2t[1 + t, t]T 1 1+t 1 t General solution: y(t) = c1 y1(t) + c2 y2(t) = Y (t)c with F.M. Y (t) = e2t Ex. 9.2.37: Find solution of system of Ex. 31 with IC y(0) = [2, −1]T 11 c1 2 1st method: Match c to IC: Y (0)c = = 10 c2 −1 ⇒ c1 c2 = 11 10 −1 2 −1 = 0 1 1 −1 −1 3 2 −1 = e2t 2 −1 = −1 3 ⇒ y(t) = e2t 1 1+t 1 t 2 + 3t 3t − 1 2 + 3t 3t − 1 2nd method: y(t) = eAt y(0) = e2t (I + (A − 2I )t)y(0) = e2t 1+t −t t 1−t = e2t 13 Ex. 9.6.1: Compute eA for A = A2 = −2 −4 1 2 −2 −4 1 2 −2 −4 1 2 = 00 00 using the exponential series ⇒ Am = 0 if m > 1 ⇒ eA = I + A = Ex. 9.6.6: Let A = using A2 = 0 −1 1 0 −1 −4 1 3 cos t − sin t sin t cos t Compute: . Show that eAt = . cos t = 1 − t2 /2! + t4/4! − t6/6! + · · · sin t = t − t3 /3! + t5/5! − t7/7! + · · · 0 −1 1 0 0 −1 1 0 = −I, A3 = A(−I ) = −A, A4 = −A2 = −(−I ) = I ⇒ A4m = I, A4m+1 = A, A4m+2 = −I, A4m+3 = −A for m = 0, 1, 2, . . . ⇒ e At t4m t4m+1 t4m+2 t4m+3 = (At) /n! = (I +A −I −A ) (4m)! (4m + 1)! (4m + 2)! (4m + 3)! n=0 m=0 n ∞ ∞ = I (1 − t2 /2! + t4/4! − t6/6! + · · ·) + A(t − t3/3! + t5 /5! − t7 /7! + · · ·) = I cos t + A sin t = cos t − sin t sin t cos t 14 Ex. 9.6.11: Compute eAt by diagonalizing A for A = −2 6 0 −1 A upper triangular ⇒ eigenvalues are diagonal entries: λ1 = −2, λ2 = −1 06 −1 6 A − (−2)I = ⇒ v1 = [1, 0]T . A − (−1)I = ⇒ v2 = [6, 1]T 01 00 16 1 −6 Set V = [v1, v1 ] = ⇒ V −1 = . Verify V −1AV is diagonal: 01 0 1 V −1 AV = 1 −6 0 1 −1 −2 6 0 −1 At 16 −2 12 = 01 0 −1 Dt −1 16 −2 0 = ≡D 01 0 −1 e−2t 0 = 0 e−t e−2t 0 1 −6 0 1 6e−t − 6e−2t e−t ⇒ A = V DV ⇒e = Ve V = e−2t 0 = 16 01 1 −6 0 1 6e−t e−t Ex. 9.6.14: Compute eAt for A = C.P.: p(λ) = −2 − λ 1 −1 −λ −2 1 −1 0 = (−2 − λ)(−λ) + 1 = λ2 + 2λ + 1 = (λ + 1)2 −1 1 −1 1 . Compute: 1−t t −t 1+t 15 ⇒ eigenvalue λ = −1. Set B = A + I = B2 = −1 1 −1 1 −1 1 −1 1 = 00 00 ⇒ eAt = e−t(I +Bt) = e−t Ex. 9.6.18: Compute eAt −1 0 0 for A = −1 1 −1 −2 4 −3 = −(λ + 1) 1−λ −1 4 −3 − λ C.P.: p(λ) = −1 − λ 0 0 −1 1−λ −1 −2 4 −3 − λ = −(λ + 1)[(λ − 1)(λ + 3) + 4] = −(λ + 1)(λ2 + 2λ + 1) = −(λ + 1)3 ⇒ single eigenvalue λ = −1. Set B = A + I 00 0 00 0 00 0 000 2 B = −1 2 −1 ⇒ B = −1 2 −1 −1 2 −1 = 0 0 0 −2 4 −2 −2 4 −2 −2 4 −2 000 B 2 = 0 (k = 2) ⇒ eAt Note: That B 2 = 0 follows also directly from dim null(B ) = 2 1 0 0 −t = e−t(I + Bt) = e−t −t 1 + 2t −2t 4t 1 − 2t 16 Ex. 9.6.22: Compute eAt 1 −1 2 0 1 0 0 0 for A = 0 0 1 0 0 −1 2 1 = (1 − λ) 1−λ 0 0 0 1−λ 0 −1 2 1−λ C.P.: p(λ) = 1−λ −1 2 0 0 1−λ 0 0 0 0 1−λ 0 0 −1 2 1−λ = (1 − λ)4 ⇒ single eigenvalue λ = 1. 0 −1 0 0 ⇒ B2 = 0 0 0 −1 ⇒ eAt 1 −t 2t 0 1 0 = et(I + Bt) = et 0 01 0 −t 2t 0 −1 2 0 0 0 0 0 Set B = A − I = 0 0 0 0 0 −1 2 0 20 0 −1 2 0 00 0 0 0 0 0 0 0 0 0 = 0 0 00 000 20 0 −1 2 0 00 0 0 0 1 0 0 0 0 0 0 0 0 17 −2 1 −1 0 by hand Ex. 9.6.26: Do the 6 tasks below for A = 1 −3 3 −5 0 1. Find eigenvalues: p(λ) = −2 − λ 1 −1 1 −3 − λ 0 3 −5 −λ 1 −1 −5 −λ + (−1)2+2(−3 − λ) −2 − λ −1 3 −λ = (−1)2+11 = −(−λ − 5) − (λ + 3)[(λ + 2)λ + 3] = λ + 5 − (λ + 3)(λ2 + 2λ + 3) = λ + 5 − (λ3 + 4λ2 + 9λ + 9) = −(λ3 + 5λ2 + 8λ + 4) = −(λ + 1)(λ + 2)2 ⇒ eigenvalues λ = −1 and λ = −2 2. Find algebraic (m) and geometric (mg ) multiplicities for each eigenvalue: λ = −1 → m = 1; λ = −2 → m = 2 (from p(λ)) Find geometric multiplicities: −1 1 −1 −1 1 −1 10 0 → 0 −1 −1 → 0 1 λ = −1: A + I = 1 −2 3 −5 1 0 −2 −2 00 ⇒ mg = 1 0 1 −1 1 −1 0 1 0 → 0 1 −1 → 0 λ = −2: A + 2I ≡ B = 1 −1 3 −5 2 0 −2 2 0 ⇒ mg = 1 0 −1 1 −1 0 0 2 1 0 18 ⇒ dim null(B 2 ) = 2 = m. Since dim null(B ) = 1 ⇒ k = 2 4. For each eigenvalue, ﬁnd m linearly independent generalized eigenvectors. For λ = −1: From RREF (A + I ) ⇒ eigenvector v1 = [−2, −1, 1]T For λ = −2: From RREF (B ) ⇒ eigenvector v2 = [1, 1, 1]T ; (B = A + 2I ) m = 2 → need solution of B 2 v = 0 that is not a multiple of v2 Use RREF (B 2 ): set y2 = 0, y3 = 1 ⇒ y1 = −1 ⇒ v3 ≡ [−1, 0, 1]T is in null(B 2) Since v2 , v3 are linearly independent, they are linearly independent generalized eigenvectors for λ = −2. 5. Verify linear independence of all generalized eigenvectors from 4. −2 1 −1 0. Compute determinant of V : Set V = [v1, v2 , v3] = −1 1 11 1 det(V ) = (−1)2+1(−1) 1 −1 1 1 + (−1)2+21 −2 −1 1 1 =2−1=1=0 19 Ex. 9.6.26 continued 1 3. For each eigenvalue ﬁnd smallest k s.t. dim null((A − λI )k ) = m For λ = −1: k = 1 (since m = mg = 1) For λ = −2: 0 1 −1 0 1 −1 −2 4 −2 1 −2 1 2 0 1 −1 0 = −1 2 −1 → 0 0 0 B = 1 −1 3 −5 2 3 −5 2 1 −2 1 0 00 ⇒ v1 , v2, v3 are linearly independent. Ex. 9.6.26 continued 2 6. Find fundamental set of solutions for y′ = Ay λ = −1, v1 → eigenvector solution: y1 (t) = e−t [−2, −1, 1]T λ = −2, v2 → eigenvector solution: y2 (t) = e−2t [1, 1, 1]T λ = −2, v3 → generalized eigenvector solution y3 (t) = e−2t(v3 + tB v3 ) Compute 0 1 −1 −1 −1 0 0 = −1 = −v2 B v3 = 1 −1 3 −5 2 1 −1 ⇒ y3(t) = e−2t (v3 − tv2) = e−2t ([−1, 0, 1]T − t[1, 1, 1]T ) = e−2t [−1 − t, −t, 1 − t]T and y1 (t), y2 (t), y3(t) are F.S.S. for y′ = Ay. 20 2 0 0 0 0 −14 −2 −7 11 −9 −3 7 −6 −9 −3 Ex. 9.6.33: 6 tasks as in Ex. 26 for A= −9 17 −12 −19 −5 −29 −7 −13 23 −16 19 5 9 −15 12 1. Find eigenvalues: >> A=[2 0 0 0 0 1;-14 -2 -7 11 -9 -8;-9 -3 -3 7 -6 -4;... ⇒ -19 -5 -9 17 -12 -9;-29 -7 -13 23 -16 -15;19 5 9 -15 12 11]; λ = 1, 2 >> A=sym(A);factor(poly(A)) both with ans = m=3 (x-1)^3*(x-2)^3 2. Find algebraic (m) and geometric (mg ) multiplicities: From p(λ): m = 3 for λ = 1 and λ = 2. Find mg : >> B1=A-eye(6);null(B1)’ ans = [ -1, 0, -2, -1, 1, 1] >> B2=A-2*eye(6);null(B2)’ ans = [ 1, 7, -6, 0, 0, 0] [ 0, 3, -3, 0, 1, 0] [ 0, -6, 5, 1, 0, 0] ⇒ mg = 1 for λ = 1 mg = 3 = m for λ = 2 1 −8 −4 −9 −15 11 3. For each eigenvalue ﬁnd smallest k s.t. dim null((A − λI )k ) = m For λ = 2 we are done: from 2. ⇒ k = 1. Check λ = 1: >> null(B1^3)’ >> null(B1^2)’ ans = ans = [ -1, 2, 1, 0, 0, 0] ⇒ k=3 [ 1, -2, 0, -1, -3, 1] [ -1, 3/2, 0, 0, 1, 0] [ 0, 1, 1, 1, 1, -1] [ -2, 5/2, 0, -1, 0, 1] 21 Ex. 9.6.33 continued 1 4. For each eigenvalue, ﬁnd m linearly independent generalized eigenvectors. (a) For λ = 2 we are done: mg = m = 3 ⇒ every generalized eigenvector is in null(A − 2I ) and so is an eigenvector. Assign variables to the eigenvectors in Matlab and denote them by v1 , v2, v3 : >> null(B2);v1=ans(:,1);v2=ans(:,2);v3=ans(:,3); >> [v1 v2 v3]’ ans = [ 0, 3, -3, 0, 1, 0] [ 1, 7, -6, 0, 0, 0] [ 0, -6, 5, 1, 0, 0] v1 = [0, 3, −3, 0, 1, 0]T v2 = [1, 7, −6, 0, 0, 0]T v3 = [0, −6, 5, 1, 0, 0]T (b) For λ = 1 we need a basis of null((A − I )3). In view of task 6 it makes sense to determine a chain of 3 generalized eigenvectors. Let’s check the chains for each of the basis vectors: >> null(B1^3);u1=ans(:,1);u2=ans(:,2);u3=ans(:,3); >> [u1 B1*u1 B1^2*u1]’,[u2 B1*u2 B1^2*u2]’,[u3 B1*u3 B1^2*u3]’ ans = [-1,2, 1, 0,0,0] [-1,1,-1, 0,2,0] [-1,0,-2,-1,1,1] ans = [ -1,3/2, 0, 0, 1, 0] [ -1,1/2,-3/2,-1/2,3/2,1/2] [-1/2, 0, -1,-1/2,1/2,1/2] ans = [ -2,5/2, 0, -1, 0, 1] [ -1,3/2,-1/2, 1/2,5/2,-1/2] [-3/2, 0, -3,-3/2,3/2, 3/2] All three basis vectors generate full chains (this needs not to be the case in general). Let’s choose the simplest chain which is the ﬁrst. Assign names to chain vectors in Matlab; denote them v6, v5 , v4 : 22 Ex. 9.6.33 continued 2 >> v6=u1;v5=B1*v6;v4=B1*v5; >> [v6 v5 v4]’ ans = [ -1, 2, 1, 0, 0, 0] [ -1, 1, -1, 0, 2, 0] [ -1, 0, -2, -1, 1, 1] v6 v5 v4 0 = = = = [−1, 2, 1, 0, 0, 0]T (A − I )v6 = [−1, 1, −1, 0, 2, 0]T (A − I )v5 = [−1, 0, −2, −1, 1, 1]T (A − I )v4 (v4 is eigenvector) 5. Verify linear independence of all generalized eigenvectors from 4. Since det([v1, v2 , v3 , v4, v5 , v6]) = −1 = 0, >> det([v1 v2 v3 v4 v5 v6]) the six generalized eigenvectors are linearly ans = independent. -1 6. Grand Finale: Find fundamental set of solutions for y′ = Ay Vectors v1, v2 , v3 are linearly independent eigenvectors for λ = 2 ⇒ three linearly independent eigenvector solutions: y1 (t) = e2t v1 = e2t[0, 3, −3, 0, 1, 0]T y2 (t) = e2t v2 = e2t[1, 7, −6, 0, 0, 0]T y3 (t) = e2t v3 = e2t[0, −6, 5, 1, 0, 0]T Solutions associated with chain v4 , v5, v6 : y4 (t) = etv4 = et [−1, 0, −2, −1, 1, 1]T y5 (t) = et(v5 + tv4) = et[−1 − t, 1, −1 − 2t, −t, 2 + t, t]T y6 (t) = et(v6 + tv5 + t2 v4/2) = et[−1 − t − t2 /2, 2 + t, 1 − t − t2, −t2/2, 2t + t2/2, t2/2]T 23 Note: Ex. 35-45 require same tasks as Ex. 26-33, except task 5. ...
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## This note was uploaded on 04/11/2011 for the course MECHANICAL MAT 219 taught by Professor Benjaminwalter during the Fall '10 term at Middle East Technical University.

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