18.06 Problem Set 7 Solution Due Wednesday, 15 April 2009 at 4 pm in Total: 150 points.

Transcription

1 8.06 Problem Set 7 Solution Due Wednesday, 5 April 2009 at 4 pm in Total: 50 points. ( ) 2 Problem : Diagonalize A = and compute SΛ 2 k S to prove this formula for A k : A k = ( ) 3 k + 3 k 2 3 k 3 k. + Solution (5 points) Step : eigenvalues. det(a λi) = λ 2 trace(a) + det(a) = λ 2 4λ + 3 = 0. The solutions are λ =, λ 2 = 3. Step 2: solve for eigenvectors. ( ) ( ) λ =, A λi =, v = λ 2 = 3, A λi = ( ), v 2 = ). ( ) Step 3: let S = be the matrix whose columns are eigenvectors. Let ( ) 0 Λ = be the matrix for eigenvalues. Then we have 0 3 ( ( ) ( ) k ( ) 0 A k = SΛ k S = 0 3 ( ) ( ) ( ) 0 = 0 3 k = ( ) 3 k + 3 k k 3 k. + Problem 2: Consider the sequence of numbers f 0, f, f 2,..., defined by the recurrence relation f n+2 = 2f n+ + 2f n, starting with f 0 = f = (giving,, 4, 0, 28, 76, 208,...). (a) As we did for the Fibonacci numbers in class (and in the book), express this process as repeated multiplication of a vector u k = (f k+, f k ) T by a matrix A: u k+ = A u k, and thus u k = A k u 0. What is A?

2 (b) Find the eigenvalues of A, and thus explain that the ratio f k+ /f k tends towards as k. Check this by computing f k+ /f k for the first few terms in the sequence. (c) Give an explicit formula for f k (it can involve powers of numbers, but not powers of matrices) by expanding f 0 in the basis of the eigenvectors of A. (d) If we apply the recurrence relation in reverse, we use the formula: f n = f n+2 /2 f n+ (just solving the previous recurrence formula for f n ). Show that you get the same reverse formula if you just compute A. (e) What does f k /f k+ tend towards as k (i.e. after we apply the formula in reverse many times)? (Very little calculation required!) Solution (30 points = ) (a) The recurrence relation gives f k+2 = 2f k+ + 2f k f k+ = f k+. Another way to write this is ( ) ( ) ( ) fk fk+ u k+ = = = A u 0 k, A = f k+ f k ( ) (b) Solving det(a λi) = λ 2 2λ 2 = 0 gives λ = and λ 2 = Since λ > λ 2, f k+ /f k tends towards λ 2.732, as k. Check first few terms:,, 4, 0, 28, 76, 208, 568,... / = 4/ = 4 0/4 = /0 = / / /

3 REMARK: Another phenomenon one should notice is that, in the sequence, f k+ /f k > λ when k is odd and f k+ /f k < λ when k is even. This is because the second eigenvalue λ 2 is negative. We will see in the explicit formula of f k below. (c) Find the eigenvectors. λ = + ( ) ( ) 3 2 3, A λi = + 3, v = 3 λ 2 = ( ) ( ) , A λi = + 3, v 2 =. 3 Then, we expand u 0 as follows. ( ) u 0 = = ( ) ( ) 3 = v + 2 v 2. Hence, we have In particular, u k = A k u 0 = 2 λk v + 2 λk 2v 2. f k = 2[ ( + 3) k + ( 3) k]. REMARK: From the explicit formula for f k, we see that f k > ( + 3) k if 2 k is even and f k < ( + 3) k if k is odd. This explains the earlier remark that 2 f k+ /f k < + 3 if k is even and f k+ /f k > + 3 if k is odd. (d) It is easy to compute A = ( ) ( ) =. 2 2 /2 Applying the recurrence relation in reverse gives f n+ = f n+ f n = f n+2 /2 f n+. That is u n = ( fn+ f n ) = ( 0 /2 ) ( ) fn+2 = A u n+. f n+ (e) Applying the process in reverse is dominated by the biggest-magnitude eigenvalue of A. The eigenvalues of A are just the reciprocals of the eigenvalues of 3

4 A, so its biggest-magnitude eigenvalue is the reciprocal of the smallest-magnitude eigenvalue of A, i.e. /λ 2. Hence, f k f k+ λ k 2 λ k+ 2 = λ 2 = Problem 3: Suppose that A = SΛS. Take determinants to prove that det A is the product of the eigenvalues of A. (This quick proof only works when A is.) Solution (5 points) Since the determinant is multiplicative, we have det(a) = det(sλs ) = det(s) det(λ) det(s) = det(λ) = λ λ n, where λ,..., λ n are eigenvalues of A. The proof only works when A is diagonalizable. REMARK: Note that the determinant is always the product of the eigenvalues, even for non-diagonalizable matrices. However, the proof for the non-diagonalizable case is a bit trickier. Problem 4: In this problem, you will show that the trace of a matrix (the sum of the diagonal entries) is equal to the sum of the eigenvalues, by first showing that AB and BA have the same trace for any matrices A and B. Follow the following steps: (a) The explicit formula for the entries c ij of C = AB is c ij = k a ikb kj (where a ik and b kj are the entries of A and B, respectively). The trace of C is i c ii. Write down the explicit formula for the entries d ij of the product D = BA. By plugging these matrix-multiply formulas into the formulas for the trace of C = AB and D = BA, and comparing them, prove that AB and BA have the same trace. (b) A = SΛS, assuming A is. Combining this factorization with the fact you proved in (a), show that the trace of A is the same as the trace of Λ, which is sum of the eigenvalues. 4

5 Solution (0 points) (a) An explicit formula for entries d ij of D is d ij = k b ika kj, just switching the role of a and b in the expression of c ij. So, trace(c) = i trace(d) = i c ii = i d ii = i a ik b ki k b ki a ik. They are the same because we can change the indices i and k in the summation. (b) For this to work, we have to assume that A is a diagonalizable n n matrix. Using the identity above, we have (by viewing SΛ as one matrix) trace(a) = trace(sλs ) = trace(s SΛ) = trace(λ) = λ + + λ n, where λ,..., λ n are eigenvalues of A. REMARK: This again works even if A does not have a full set of independent eigenvectors. One may use generalized eigenvectors to do the same trick. We again will omit the details. k Problem 5: Suppose A 2 = A. (This does not mean A = I, since A might not be invertible; it might be a projection onto a subspace, for example.) (a) Explain why any eigenvector with λ = 0 is in the space of A, and vice versa (any nonzero vector in that space is an eigenvector with λ = 0). (b) Explain why any eigenvector with λ = is in the space of A, and vice versa (any nonzero vector in that space is an eigenvector with λ = ). (Hint: first explain why each column of A is an eigenvector.) (c) Conclude from the dimensions of these subspaces that any such A must have a full set of independent eigenvectors and hence be diagonalizable. Solution (5 points = 5+5+5) (a) Any eigenvector with λ = 0 is in the nullspace of A. This is because v N(A) if and only if A v = 0 = 0 v. 5

6 (b) Any eigenvector with λ = is in the column space of A. For each column x of A, A 2 = A implies that A x = x. Hence, each column vector is an eigenvector with λ =, so is any vector in the column space. Conversely, if v is an eigenvector with λ =, that means that Av = v, which implies that v is in C(A). REMARK: Note that for a more general matrix with nonzero eigenvalues, it is still true that any eigenvector for a nonzero eigenvalue is in the column space of A, since A(v/λ) = v. However, the span of these eigenvectors may only be a subspace of the column space (if A is not diagonalizable). (c) Say that A is an n n matrix. Note that the dimension of the column space is the rank of A, whereas the dimension of the nullspace of A is n rank(a). The dimensions add up to n. Hence, the eigenspace for λ = 0 and the eigenspace for λ 0 span the whole space R n. Therefore, A much have a full set of independent eigenvectors and hence diagonalizable. Problem 6: A genderless alien society survives by cloning/budding. Every year, 5% of young aliens become old, 3% of old aliens become dead, and % of the old aliens and 2% of the dead aliens are cloned into new young aliens. The population can be described by a Markov process: young old dead year k+ young = A old dead year k (a) Give the Markov matrix A, and compute (without Matlab) the steady-state young/old/dead population fractions. (b) In Matlab, enter your matrix A and a random starting vector x = rand(3,); x = x / sum(x) (normalized to sum to ). Now, compute the population for the first 00 years, and plot it versus time, by the following Matlab code: p = []; for k = 0:99 p = [ p, A^k * x ]; end plot([0:99], p ) legend( young, old, dead ) xlabel( year ); ylabel( population fraction ); 6

7 Check that the final population p(:,end) is close to your predicted steady state. (c) In Matlab, compute A to a large power A 000 (in Matlab: A^000). Explain why you get what you do, in light of your answer to (a). Solution (5 points = 5+5+5) (a) The Markov matrix A is given by Computing the steady-state is equivalent to find the eigenvector for λ =. We do a Gaussian elimination A I = A solution is given by v = ( 8, 2, 5 3)T. If we normalize the vector so that the sum of the coordinates is, we get v = (8, 0, 33 5)T (0.2424, , ) T. In other words, the steady-state young/old/dead population ratio is 8 : 0 : 5. (b) The code and the result is as follows. >> A = [0.95, 0.0, 0.02; 0.05, 0.96, 0; 0, 0.03, 0.98] A = >> x = rand(3, ); x = x / sum(x) x =

8 >> p = []; >> for k = 0:99 p = [p, A^k * x ]; end >> plot([0:99], p ) >> legend( young, old, dead ) >> xlabel( year ); ylabel( polulation fraction ); >> p(:, end) ans = young old dead polulation fraction year (c) >> A^000 ans =

9 For a large power like A 000, we should expect any initial vector to converge to the steady state. That means that the column space of A 000 should just be the steadystate eigenvector, which means that each column of A should be approximately this eigenvector (normalized so that each column sums to ). Problem 7: If A is both a symmetric matrix and a Markov matrix, why is its steady-state eigenvector (,,..., ) T? Solution (5 points) A very important property of a Markov matrix is that [ ]A = [ ]. Taking transpose, we have A T [ ] T = [ ] T. But A = A T is symmetric. Hence, [..., ] T is an eigenvector with λ =. It is a steady-state eigenvector. Problem 8: Find the λ s and x s so that u = e λt x is a solution of ( ) d u 2 3 dt = u. () 0 Make a linear combination of these solutions to solve this equation with the initial condition u(0) = (5, 2) T. Solution (5 points) Step : find the eigenvalues and the eigenvectors of the matrix A = Solving det(a λi) = λ 2 λ 2 = 0 gives λ = 2 and λ 2 =. ( ) ( ) 0 3 λ = 2, A λi =, v 0 3 = 0 ( ) ( ) 3 3 λ 2 =, A λi =, v =. ( ) Step 2: we have ( ) e u = e λt 2t v = 0 ( ) e and u 2 = e λ2t t v 2 = e t to be the solution of (). 9

10 Step 3: solve the initial value problem for u = c u + c 2 u 2 ; this requires to solve ( ) ( ) ( ) c 5 =. 0 2 We then have c = 3, c 2 = 2. Hence, ( ) 3e u = c u + c 2 u 2 = 2t + 2e t 2e t. c 2 Problem 9: Explain how to write an equation α d2 y dt 2 equation M d u = A u. dt Solution (5 points) ( dy ) Write u = dt. Then d u y ( d2y = dt 2 dt dy dt + β dy dt ). The equation gives that α d2 y dy = β dt2 dt γy, dy dt = dy dt. This translates to say ( ) α 0 d u 0 dt = ( ) β γ. 0 + γy = 0 as a vector Remark to the graders, it is okay for the students to assume that α 0 and write the final result as follows. ( ) d u β dt = γ α α. 0 Problem 0: A matrix A is antisymmetric, or skew symmetric, which means that A T = A. Prove that the matrix Q = e At is orthogonal: transpose the series for Q = e At to show that you get the series for e At, and thus Q T Q = I. Therefore, if u(t) = e At u(0) is any solution to the system d u = A u, then we know dt that u(t) / u(0) =. 0

11 Solution (5 points=0+5) Write out the series that defines Q = e At and transpose. ( Q T = (e At ) T (At) n ) T (A T t) n = = n! n! n=0 n=0 ( At) n = n! n=0 = e At. Hence Q T Q = e At e At = I. This implies that Q is orthogonal. Since orthogonal matrices preserve norms, we must have e At u(0) = u(0). Hence, u(t) / u(0) =. REMARK: it is not in general true that e Bt e At = e (B+A)t for any square matrices A and B. In fact, this is only true if AB = BA. But this is certainly true for B = A as it is here. More simply, e At is the matrix that propagates the solution forward in time by t, while e At propagates the solution backwards in time by t, so the two matrices must be inverses. Problem ( : ) If A 2 = A, show from the infinite series that e At = I + (e t )A. For A =, this gives e 0 0 At =. Solution (0 points) Since A 2 = A, we have A k = A for any k > 0. e At = A n t n = I + n! n=0 ( ) When A =, we have 0 0 n= At n n! e At = I + (e t )A = t n ) = I + A( = I + A(e t ). n! n= ( ) e t e t 0 Problem 2: Assume A is diagonalizable with real eigenvalues. What condition on the eigenvalues of A ensures that the solutions of d u = A u will not blow up dt for t? In comparison, what condition on the eigenvalues of A ensures that solutions of the linear recurrence relation u k+ = A u k will not blow up for k?

12 Solution (0 points = 5+5) For ODE problem, if the solutions do not blow up as t, the eigenvalues λ of A has to have real part less than or equal to 0, i.e. Re(λ) 0. This is because the solution looks like e λt v, where v is an eigenvector with eigenvalue λ. For the linear recurrence problem, if u k does not blow up, the eigenvalues λ of A has to have absolute value less than or equal to, that is λ. This is because the main term in u k looks like λ k v, where v is an eigenvector with eigenvalue λ. 2