CAAM 335 Matrix Analysis Solutions to Homework 8 Problem (5+5+5=5 points The partial fraction expansion of the resolvent for the matrix B = is given by (si B = s } {{ } =P + s + } {{ } =P + (s (5 points Write the P j s into a vector outer-product form: P j = v j w T j for j =,,3. (5 points Give 3 eigen-pairs, (λ j,x j, of B with right eigen-vectors x j C 3. (5 points Give 3 eigen-pairs, (λ j,y j, of B with left eigen-vectors y j C 3. The eigenvalues are, and. P = = P = P 3 = = = (, v = (, v = (, v 3 = The v j s are right eigenvectors (so are any nonzero multiples of them. λ =, x = ; λ =, x =. } {{ } =P 3, w =, w =, w 3 = ; λ 3 =, x 3 =
The w j s are left eigenvectors (so are any nonzero multiples of them. λ =, y = Problem (5+=5 points The solution of the dynamical system is given by x(t = exp(btx. ; λ =, y = ; λ 3 =, y 3 = x (t = Bx(t, t >, x( = x ( (a Let B R 3 3 be a matrix with three distinct eigenvalues λ,λ,λ 3. The partial fraction expansion of the resolvent is Since B = λ P + λ P + λ 3 P 3, we have R(s = (si B = s λ P + s λ P + s λ 3 P 3. tb = tλ P +tλ P +tλ 3 P 3. Apply the matrix exponential formula to exp(tb and express x(t = exp(btx in terms of t, x, P,P,P 3 and the eigenvalues. (b Let B be the matrix in problem and x = (,, T. Determine lim t x(t. From the notes, exp(b =e λ P + e λ P + e λ 3 P 3. thus exp(btx = (e tλ P + e tλ P + e tλ 3 P 3 x. exp(btx = e t = + e t + e t As t, the second term goes to zero and thus lim x(t = t (,,T. + e t
3 Problem 3 (+5+5= points a Compute by hand (show your steps the eigenvalues and eigenvectors of the matrix 4 A = 4. b What are the algebraic and geometric multiplicities of the eigenvalue(s? c Give Jordan blocks of this matrix. To determine the eigenvalues we compute the characteristic polynomial of A, λ + 4 p(λ = det(λi A = det 4 λ. λ Computing the determinant with respect to the last row or column gives, Thus A has one eigenvalue, λ =. p(λ = (λ 4λ + 4(λ = (λ 3. To determine the geometric multiplicity we determine the dimension of the nullspace of (A I. N (A I = N 4 = t + s : t,s R Thus λ = has algebraic multiplicity of 3 and a geometric multiplicity of. Based on the geometric multiplicity as computed above, the Jordan blocks of the matrix are as follows ( J = ( J = Problem Set 6. from the MIT Notes (p. 93 Online Problems, 3, 4, 8, 6, 7, 9, 9. 5 points each.. Find the eigenvalues and the eigenvectors of these two matrices: [ ] [ ] 4 4 A = and A + I = 3 4 A + I has the same eigenvectors as A. Its eigenvalues are increased by. ( λ 4 p A (λ = det λ 3 = λ 4λ 5 = (λ 5(λ + λ =,5
4 ( 4 (A + I = 4 x ( 4 4 (A 5I = x For the eigenvectors we see that ( λ 4 p A+I (λ = det λ 4 = λ 6λ = λ(λ 6 λ =,6 ((A + I I = (A + I ((A + I 6I = (A 5I, which were handled above. Thus A and A + I have the same eigenvectors. 3. Compute the eigenvalues and the eigenvectors A and A. Check the trace: [ ] [ ] A = and A = A has the same eigenvectors as A. When A has eigenvalues λ and λ, its inverse has eigenvalues, λ and λ. ( λ p A (λ = det λ = λ λ = (λ + (λ λ =, ( (A + I = x ( (A I = x The trace of A is + = + = λ + λ. ( λ + p A (λ = det λ For the eigenvectors we see that = λ + λ = (λ + (λ λ =, ( (A + I = x ( {}} { (A + I = ( x. The trace of A is + = + = λ + λ. Thus A and A have the same eigenvectors and the eigenvalues of A are the reciprocals of the eigenvalues of A.
5 4. Compute the eigenvalues and the eigenvectors A and A. Check the trace: [ ] [ ] 3 A = and A 7 3 = 6 A has the same eigenvectors as A. When A has eigenvalues λ and λ, A has eigenvalues λ and λ. In this example, why is λ + λ = 3? ( λ + 3 p A (λ = det = λ + λ 6 = (λ + 3(λ λ = 3, λ ( 3 (A + 3I = 3 x ( {}} { 3 ( 3 3 (A I = x ( λ 7 3 p A (λ = det λ 6 For the eigenvectors we see that = λ 3λ + 36 = (λ 9(λ 4 λ = 9,4 ( (A 3 9I = 3 x 3 ( (A 3 3 4I = x. Thus A and A have the same eigenvectors and the eigenvalues of A are the eigenvalues of A squared. Since the trace of the matrix is the sum of the diagonal entries as well as the sum of the eigenvalues, the trace of A is 7 + 6 = λ + λ. 8. (a If you know that x is an eigenvector, the way to find λ is to compute Ax, the ratio between each entry of Ax and the corresponding entry of x should be constant and equal to λ. One could also compute x Ax/(x x, note x by definition. (b If you know that λ is an eigenvalue, the way to find x is to determine the null space of A λi. Any nonzero vector in the null-space of A λi is a suitable eigenvector. 6. The determinant of A equals the product λ λ λ n. Start with the polynomial det(a λi separated into is n factors (always possible. Then set λ = : det(a λi = (λ λ(λ λ...(λ n λ so deta = λ λ λ n Check this rule in Example where the Markov matrix has λ = and. Example [.8.3..7 ] λ = λ = det(a = (.8(.7 (.3(. = = λ λ
6 7. The sum of the diagonal entries (the trace equals the sum of the eigenvalues [ ] a b A = has det(a λi = λ (a + dλ + ad bc =. c d The quadratic formula gives the eigenvalues λ = (a + d + (a + d 4(ad bc/ and λ = (a + d (a + d 4(ad bc/. Their sum is a + d. If A has λ = 3 and λ = 4 then det(a λi = (λ 3(λ 4 = λ 7λ +. 9. A 3 by 3 B is known to have eigenvalues,,. This information is enough to find three of these (give the answers where possible: the rank of B is [using, say, the expansion B = x y + x 3y 3 ] the determinant of B T B is [det(b T B = det(b = ( = ] the eigenvalues of B T B cannot be determined Take for example B = α B T B = α + for which B has the prescribed eigenvalues and B T B has eigenvalues,α + and with α being arbitrary. the eigenvalues of (B + I are,/ and /5 [by the formula /(λ + ]. 9. (Review Find the eigenvalues of A, B, and C: 3 A = 4 5, B = 6 3, C = Since A is upper triangular the eigenvalues are simply, 4, and 6. The characteristic polynomial of B is, det(λi B = λ(λ λ ( ( 3(λ = (λ (λ 3. hence the eigenvalues of B are, 3, and 3. 3 Let e T = [,, ]. Clearly C = ee T is a rank- and symmetric, and Ce = e(e T e = 6e. Hence the eigenvalues of C are 6 and where has a multiplicity of.