ECE 602 Exam 2 Solutions, 3/23/2011.

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1 NAME: ECE 62 Exam 2 Solutions, 3/23/211. You can use any books or paper notes you wish to bring. No electronic devices of any kind are allowed. You can only use materials that you bring yourself. You are not allowed to borrow any materials from other students during the exam. You have two hours to complete THREE problems. All answers must be fully explained and substantiated to receive full credit. There will not be any discussion of grades. All re-grade requests must be submitted in writing. Problem Points Score TOTAL 1

2 Problem 1 (1 points). Consider the following matrix A: (a) Find all eigenvalues of A. (b) How many linearly independent eigenvectors does A have? Find as many linearly independent eigenvectors as you can, normalize each of them to have unit norm, and for each of them choose the direction such that the first entry of every eigenvector is nonnegative. (c) Let v 1 and v 2 be the eigenvectors you found in part (b) that correspond to the largest eigenvalue and the smallest eigenvalue, respectively. Find an orthonormal basis for the space spanned by v 1 and v 2. (d) Compute A 3 3A 2 6A. Solution. (a) To find the eigenvalues, form the characteristic polynomial of A and set it to zero: λ det(λi A) = det λ λ 1 = (λ 1) 3 9(λ 1) = (λ 1)(λ 2 2λ 8) = (λ 1)(λ + 2)(λ 4) Thus, the eigenvalues are λ 1 = 4, λ 2 = 2, and λ 3 = 1. (b) To find the eigenvectors, look at the null space of λ i I A for i = 1,2,3. For λ 1 = 4, I A = det The null space of this matrix is one-dimensional and consists of all vectors of the form (a,,a) T. Therefore, the unit eigenvector corresponding to eigenvalue λ 1 = 4 and having a nonnegative first entry is v 1 = (1/ 2,,1/ 2) T. For λ 2 = 2, 2I A = det The null space of this matrix is one-dimensional and consists of all vectors of the form (a,, a) T. Therefore, the unit eigenvector corresponding to eigenvalue λ 2 = 2 and having a nonnegative first entry is v 2 = (1/ 2,, 1/ 2) T. For λ 3 = 1, I A = det

3 The null space of this matrix is one-dimensional and consists of all vectors of the form ( 2a,3a, 2a) T. Therefore, the unit eigenvector corresponding to eigenvalue λ 3 = 1 and having a nonnegative first entry is v 3 = (2/ 17, 3/ 17,2/ 17) T. There are no more eigenvalues and hence no more linearly independent eigenvectors. Answer: there are three linearly independent eigenvectors. Normalizing them to unit norm and orienting them so that their first entries are nonnegative, we have the following set of eigenvectors: v 1 = 1/ 2 1/ 2, v 2 = 1/ 2 1/ 2, v 3 = 2/ 17 3/ 17 2/ 17. (c) The largest eigenvalue is λ 1 = 4, and the smallest one is λ 2 = 2. The eigenvectors v 1 and v 2 corresponding to these eigenvalues were found in part (b). These two eigenvectors are orthonormal, hence they form an orthonormal basis for their span. (d) In part (a), the characteristic polynomial of A was found to be (λ 1)(λ 2 2λ 8) = λ 3 3λ 2 6λ + 8. Therefore, by Cayley-Hamilton theorem, we have: A 3 3A 2 6A + 8I = A 3 3A 2 6A = 8I =

4 Problem 2 (1 points). Consider the following nonlinear system for state x = (x 1,x 2 ) T : ẋ 1 = 2(x 1 x 2 ) 2 ẋ 2 = (x 1 x 2 ) 2 (a) Find the set of all equilibrium points for this system. (b) Linearize this system around the point x 1 = x 2 = 1. (c) For any initial condition δx(), write down the solution δx(t) of the linearized system at time t. Also, write down the solution x(t) = δx(t) + x e where x e is the equilibrium point in part (b). (d) For any initial condition x(), write down the solution x(t) of the original system at time t. (e) Let x(t) be the solution obtained in part (d), and let x lin (t) be the linearized solution obtained in part (c). For x() = (1 2) T, calculate the squared distance between these solutions are at time t = 1, i.e., calculate x(1) x lin (1) 2. Solution. (a) All constant solutions are obtained by setting the left-hand side of the equation to zero, which gives x 1 = x 2. Thus, the set of all equilibrium points is the set of all x such that x 1 = x 2. (b) The partial derivatives of the right-hand side of the first equation with respect to x 1 and x 2 are 4(x 1 x 2 ) and 4(x 2 x 1 ), respectively. At the equilibrium point x 1 = x 2 = 1, both of these are zero. Hence, the linearized equation is δx 1 =. Similarly, the partial derivatives of the right-hand side of the second equation are 2(x 1 x 2 ) and 2(x 2 x 1 ), and are both equal to zero at x 1 = x 2 = 1. Hence, the linearized version of the second equation is δx 2 =. Therefore, the linearized system is: δx = (c) Integrating the above equation, we obtain that the solutions of the linearized system are constants: δx(t) = δx(). Hence x(t) = δx() + x e = x(). (d) Subtracting the second equation in the original system from the first and denoting u = x 1 x 2, we get: u = u 2 Integrating this equation, we get u 1 = t + C, or u(t) = 1 t + C. 4

5 Setting t =, we get that C = 1/u(): 1 u(t) = t 1/u() = u() u()t 1 From the second equation of the original system, we have that ẋ 2 = u, and therefore x 2 (t) = u(t) + C 1, where C 1 is found from x 2 () = u() + C 1 : x 2 (t) = u(t) + x 2 () u() = u() u()t 1 + x 2() u() = x 1() x 2 () (x 1 () x 2 ())t 1 + x 2() (x 1 () x 2 ()) = x 1() x 2 () (x 1 () x 2 ())t 1 + 2x 2() x 1 () Finally, from the first equation of the original system, ẋ 1 = 2ẋ 2, hence x 1 (t) = 2x 2 (t) + x 1 () 2x 2 (): [ x 1 (t) = 2 x ] 1() x 2 () (x 1 () x 2 ())t 1 + 2x 2() x 1 () + x 1 () 2x 2 () x 1 () x 2 () = 2 (x 1 () x 2 ())t 1 + 2x 2() x 1 () (e) With the given initial condition, we have x 2 (t) = ( 1)/( t 1) = 1/(t + 1) + 3 and x 1 (t) = 2/(t+1)+3. Therefore, x 1 (1) = 2/(1+1)+3 = 2 and x 2 (1) = 1/(1+1)+3 = 2.. Hence, we have x(1) x lin (1) 2 = (2 1) 2 + (2. 2) 2 = 1.2.

6 Problem 3 (2 points). Consider the following system: ẋ(t) = Ax(t), (1) where x(t) R 3 and A = (a) Let v 1,v 2,v 3 R 3, and denote by x i (t) the solution to Eq. (1) with initial condition x() = v i, for i = 1,2,3. Find v 1,v 2,v 3 satisfying all the following properties: For any t >, the solution x i (t) is a multiple of v i. v 1 = v 2 = v 3 = 1. x 1 (1) > x 2 (1) > x 3 (1). The first entry of each v i is nonnegative. (b) Is the system given in Eq. (1) stable? Fully explain. (c) Let s = (2 1 1) T. Find e 1A s. Simplify as much as possible. You do not need to simplify terms of the form e d where d is a number. (d) If x() = v 1 +62v 2, find v T 3 x(211). Here v 1, v 2, and v 3 are the vectors you found in part (a). (e) Let r(m + 1) = Ar(m) where r(m) R 3. Is this discrete-time system stable? Fully explain. (f) For the discrete-time system in part (e), let the initial condition be Find r(). Simplify as much as you can. r() = 32 Solution. (a) The first step is to realize that v i s are eigenvectors of A. To find the eigenvalues and eigenvectors, first form the characteristic polynomial and set it to zero: λ 3 4 det(λi A) = det λ λ + 3 = (λ 3)(λ + 1)(λ + 3) 16(λ + 1) = (λ + 1)(λ )(λ + ) = (λ + 1)(λ )(λ + ) 6

7 Thus, the eigenvalues are 1 and ±. Since x i (t) = e λ it v i and v i = 1, we have x i (1) = e λ i. Hence, the condition x 1 (1) > x 2 (1) > x 3 (1) means that the eigenvector v 1 corresponds to the largest eigenvalue, ; the eigenvector v 2 corresponds to the second largest eigenvalue, 1; and the eigenvector v 3 corresponds to the smallest eigenvalue,. To find the eigenvectors, look at the null space of λ i I A for i = 1,2,3. For λ 1 =, 2 4 I A = det The null space of this matrix is one-dimensional and consists of all vectors of the form (2a,,a) T. Therefore, the unit eigenvector corresponding to eigenvalue λ 1 = and having a nonnegative first entry is v 1 = (2/,,1/ ) T. For λ 2 = 1, I A = det The null space of this matrix is one-dimensional and consists of all vectors of the form (,a,) T. Therefore, the unit eigenvector corresponding to eigenvalue λ 2 = 1 and having a nonnegative first entry can be either v 2 = (, 1,) T or v 2 = (,1,) T. For λ 3 =, 8 I A = det The null space of this matrix is one-dimensional and consists of all vectors of the form (a,, 2a) T. Therefore, the unit eigenvector corresponding to eigenvalue λ 3 = and having a nonnegative first entry is v 3 = (1/,, 2/ ) T. Another approach to finding the modes of the system would be to note that the given system consists of two decoupled subsystems: ( ẋ1 ẋ 3 ẋ 2 ) = x 2 and ( 3 4 = 4 3 )( x1 x 3 ) The solution for the first subsystem is given by x 2 (t) = e t x 2 (). This corresponds to the eigenvector v 2 found above. To find the modes of the solution for the second subsystem, we can find the eigenvalues and eigenvectors for the second system matrix. The characteristic polynomial is (λ 3)(λ + 3) 16 = λ 2 2, and hence the eigenvalues are ±. Finding the eigenvectors yields the same modes as found above. (b) This system is not stable because it has an eigenvalue whose real part is nonnegative (λ 1 = ). 7

8 (c) This question is asking for x(1) when x() = s = v 1 + v 2. We therefore have: x(1) = x 1 (1) + x 2 (1) = e 1λ 1 v 1 + e 1λ 2 v 2 2e = e 1 e (d) Since v 1 and v 2 are two real eigenvectors of A, their span is an invariant space. Therefore, if x() is in their span, so is x(211). Note, however, that v 3 is orthogonal to both v 1 and v 2, and is therefore also orthogonal to any linear combination of v 1 and v 2, including x(211). Hence, v3 T x(211) =. (e) This system is unstable because λ i 1 for all eigenvalues of A. (f) Since A has three linearly independent eigenvectors, it is diagonalizable, and A = Ddiag(λ 1,λ 2,λ 3)D 1, where D is the matrix whose columns are the eigenvectors v 1, v 2, and v 3. Note that these eigenvectors have unit norm and happen to be pairwise orthogonal. Also, D happens to be symmetric. Hence, D 1 = D T = D, and we have: r() = A r() = Ddiag(λ 1,λ 2,λ 3)Dr() 2/ 1/ = 1 1/ 2/ 2/ 1/ = 1 1/ 2/ 2/ 1/ = 1 1/ 2/ 4/ = 1 3/ 8 = 6 ( 1) ( ) ( 1) ( ) 1 / 2 1 / 2/ 1/ 1 1/ 2/ 32/ 2 32/ 32 8