Hw8_sol - Analytical Topics − A)−1 = (sI 1 s−1 1 3 1...

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Unformatted text preview: Analytical Topics − A)−1 = (sI 1 s−1 1 3 1 s−1 − 0 1 s+2 1 s+2 Problem 1 as before. = 10 00 1 1 Solutions to Homework #8 - Fall 2009.1 + 1 1 s−1 3 0 s + 2 −3 1 6. Another formula for the matrix exponential. You might remember that for any complex number a ∈ C, ea = limn→∞ (1 + a/n)n . You will establish the matrix analog: for any A ∈ Rn×n , eA = lim (I + A/n)n . n→∞ To simplify things, you can assume A is diagonalizable. Hint: diagonalize. Solution: Assuming A ∈ Rn×n is diagonalizable, there exists an invertible matrix T ∈ Rn×n such that A = T diag(λ1 , . . . , λn )T −1 where λ1 , . . . , λn are the eigenvalues of A. Therefore (I + A/n)n = (T T −1 + T diag(λ1 /n, . . . , λn /n)T −1 )n = T (I + diag(λ1 /n, . . . , λn /n))T −1 n = T (I + diag(λ1 /n, . . . , λn /n))n T −1 . But (I + diag(λ1 /n, . . . , λn /n)) is diagonal and therefore its nth power is simply a diagonal matrix with diagonal entries equal to the nth power of the diagonal entries of (I + diag(λ1 /n, . . . , λn /n)). Thus (I + A/n)n = T diag((1 + λ1 /n)n , . . . , (1 + λn /n)n )T −1 and taking the limit as n → ∞ gives lim (I + A/n) n→∞ n 11 = nlim T diag((1 + λ1 /n)n , . . . , (1 + λn /n)n )T −1 →∞ = T diag( lim (1 + λ1 /n)n , . . . , lim (1 + λn /n)n )T −1 n→∞ λ1 λn = T diag(e , . . . , e )T = eA , −1 n→∞ and we are done. 7. Affine dynamical systems. A function f : Rn → Rm is called affine if it is a linear function plus a constant, i.e., of the form f (x) = Ax + b. Affine functions are more general than linear functions, which result when b = 0. We can generalize linear dynamical systems to affine dynamical systems, which have the form x = Ax + Bu + f, ˙ y = Cx + Du + g. Fortunately we don’t need a whole new theory for (or course on) affine dynamical systems; a simple shift of coordinates converts it to a linear dynamical system. Assuming A is invertible, define x = x + A−1 f and y = y − g + CA−1 f . Show that x, u, and y ˜ ˜ ˜ ˜ are the state, input, and output of a linear dynamical system. Solution: All we have to do is to show that x, u and y satisfy a linear dynamical system. First ˜ ˜ note that dx ˜ d dx = (x + A−1 f ) = (A−1 f ∈ Rn is constant) dt dt dt c J.C. Cockburn page 1 of 7 and therefore plot(t,[S(i+1,2);S(8,2);S(8,2)]) hold on; Analytical Topics plot(t,[S(i+1,2);S(8,2);S(8,2)],’o’) Solutions to Homework #8 - Fall 2009.1 plot(t_comp,[S_comp(i_comp+1,2);S_comp(14,2);S_comp(14,2)],’--’) plot(t_comp,[S_comp(i_comp+1,2);S_comp(14,2);S_comp(14,2)],’o’) Problem 2 axis([-1,8,-3,3]); grid on; xlabel(’t’); ylabel(’s22’) 2. Properties of the matrix exponential. (a) Show that eA+B = eA eB if A and B commute, i.e., AB = BA. The converse is also true, i.e., if eA+B = eA eB then A and B commute. (But it is hard to show.) (b) Carefully show that d At e dt = AeAt = eAt A. Solution: (a) We will show that if A and B commute then eA eB = eA+B . We begin by writing the expressions for eA and eB eA = I + A + A2 A3 + +··· 2! 3! B2 B3 + +··· 2! 3! Now we multiply both expressions and get eB = I + B + A2 B 2 A3 A2 B AB 2 B 3 + + + + + +··· 2! 2! 3! 2! 2! 3! A2 + 2AB + B 2 A3 + 3A2 B + 3AB 2 + B 3 + +··· = I +A+B+ 2! 3! Now we note that, if A and B commute, we are able to write things such as 6 (A + B )2 = A2 + 2AB + B 2 . So, if A and B commute we can finally write eA eB = I + A + B + AB + eA eB = I + (A + B ) + (A + B )2 (A + B )3 + + · · · = eA+B 2! 3! (b) It suffices to note that A commute with itself. Then one can write A 3 t2 deAt = A + A2 t + +··· dt 2! (At)2 + · · ·) = A(I + At + 2! (At)2 = (I + At + + · · ·)A 2! = AeAt = eAt A 3. Determinant of matrix exponential. (a) Suppose the eigenvalues of A ∈ Rn×n are λ1 , . . . , λn . Show that the eigenvalues of eA are eλ1 , . . . , eλn . You can assume that A is diagonalizable, although it is true in the general case. (b) Show that det eA = eTr A . Hint: det X is the product of the eigenvalues of X , and Tr Y is the sum of the eigenvalues of Y . Solution: (a) Suppose that A is diagonalizable with eigenvalues λ1 , . . . , λn . Therefore, the in- 2 of 7 c J.C. Cockburn page vertible matrix T exists such that dt Analytical Topics Problem 3 2! (At)2 + · · ·) = A(I + At + 2! Solutions to Homework #8 - Fall 2009.1 (At)2 = (I + At + + · · ·)A 2! = AeAt = eAt A 3. Determinant of matrix exponential. (a) Suppose the eigenvalues of A ∈ Rn×n are λ1 , . . . , λn . Show that the eigenvalues of eA are eλ1 , . . . , eλn . You can assume that A is diagonalizable, although it is true in the general case. (b) Show that det eA = eTr A . Hint: det X is the product of the eigenvalues of X , and Tr Y is the sum of the eigenvalues of Y . Solution: (a) Suppose that A is diagonalizable with eigenvalues λ1 , . . . , λn . Therefore, the invertible matrix T exists such that A = T diag(λ1 , . . . , λn )T −1 and we get eA = T ediag(λ1 ,...,λn ) T −1 = T diag(eλ , . . . , eλ )T −1 . 1 n As a result eA T = T diag(eλ1 , . . . , eλn ) which shows that the eigenvalues of eA are eλ1 , . . . , eλn . Note that this also shows that the eigenvectors of A (the columns of T ) and eA are the same. (b) The determinant of a matrix is equal to the product of its eigenvalues and therefore det eA = eλ1 eλ2 · · · eλn = eλ1 +λ2 +···+λn . But λ1 + λ2 + · · · + λn is the sum of the eigenvalues of A which is equal to Tr A. Thus det eA = eTr A . 7 c J.C. Cockburn page 3 of 7 Analytical Topics Solutions to Homework #8 - Fall 2009.1 Problem 4 4. Characteristic polynomial. Consider the characteristic polynomial X (s) = det(sI − A) of the matrix A ∈ Rn×n . (a) Show that X is monic, which means that its leading coefficient is one: X (s) = sn + · · ·. (b) Show that the sn−1 coefficient of X is given by − Tr A. (Tr X is the trace of a matrix: Tr X = n=1 Xii .) i (c) Show that the constant coefficient of X is given by det(−A). (d) Let λ1 , . . . , λn denote the eigenvalues of A, so that X (s) = sn + an−1 sn−1 + · · · + a1 s + a0 = (s − λ1 )(s − λ2 ) · · · (s − λn ). By equating coefficients show that an−1 = − n i=1 λi and a0 = n i=1 (−λi ). Solution: (a) Expand the determinant expression to get ˜ det(sI − A) = (s − a11 ) det A + other terms, ˜ where A is the A matrix without the first row and first column. The other terms are similar, except for the fact that the determinant is multiplied by a scalar. ˜ Expanding det A we will reach a similar equation, and after expanding all terms you will reach something like n det(sI − A) = i=1 (s − aii ) + other terms. The other terms contribute with polynomials whose order is less than n, and since the first term is a monic polynomial with order n it follows that det(sI − A) is also monic. (b) Let’s take a closer look at the relation n det(sI − A) = i=1 (s − aii ) + other terms. A little reasoning will show us that the other terms in fact are polynomials whose degree is less than n − 1 (provided that n > 1, and for n = 1 we have the trivial ˜ case). This is so because in the first expression of item (a) we have that A is the only matrix that has n − 1 entries with s, and the same applies to other expansions of the expression. Then it follows that the sn−1 term of X is the sn−1 term of (s − aii ). But this term is −aii , which is equal to − Tr A. (c) The constant coefficient is given by X (0). But X is simply det(sI − A). By taking s = 0 it follows that X (0) = det(−A). c J.C. Cockburn 8 page 4 of 7 Analytical Topics Solutions to Homework #8 - Fall 2009.1 (d) First we note that, if n = 1, the relations are valid for the polynomial s − λ1 . Now suppose the relations are valid for a monic polynomial P (s). Multiply P (s) by s − λi and expand as P (s)(s − λi ) = sP (s) − λi P (s). Suppose P (s) has degree n. Then sP (s) is monic with degree n + 1 and the constant coefficient is zero. The polynomial −λi P (s) has degree n, the sn coefficient is −λi and the constant coefficient is (−λj ). Since the constant coefficient of sP (s) is zero we conclude by induction that a0 = n=1 (−λi ). Since P (s) satisfies i the properties, the sn term of P (s) is λj and we conclude, again by induction, that an−1 = − n=1 λi . i 5. Spectral resolution of the identity. Suppose A ∈ Rn×n has n linearly independent eigenvectors p1 , . . . , pn , pT pi = 1, i = i Problem 5 T 1, . . . , n, with associated eigenvalues λi . Let P = [p1 · · · pn ] and Q = P −1 . Let qi be Matrix the ith row of Q. Functions T (a) Let Rk = pk qk . What is the range of Rk ? What is the rank of Rk ? Can you 1 2 describe the null space of Rk ? Let A = . −2 −3 2 (b) Show that Ri Rj = 0 for i = j . What is Ri ? a) Use the techniques developed in class to evaluate functions of matrices to find sin(A) and cos(A) in (c) Show that n closed form. Rk 2 (A) + cos2 (A) = I . (sI − A)−1 = b) Prove that sin s k =1If − λk numerically give an estimate of the c) Are the following formulas both correct ? show it. done accuracy of your results. a partial fraction expansion of (sI − A)−1 . For this reason the Note that this is Ri ’s are called the residue matrices of A. tan(A) = (cos(A))−1 sin(A) (d) Show that R1 + · · · + Rn tan(A)For this reason the −1 = I. residue matrices are said to = sin(A) (cos(A)) constitute a resolution of the identity. Solutions Find the residue matrices for (e) Part a) Since det(sI − A) = s2 + 2s + 1 the A has one eigenvalue λ = −1 of multiplicity two. 1 0 Let f (s) = sin(s) and since A is a 2 × 2 matrix define g1 s)−2α0 + α1 s. Then (= A= f above = sin( P = α Q α1 ( then b oth ways described (s)|s=−1(i.e., find −1)and 0 +and−1) calculate the R’s, and then do a partial fraction expansion of (sI − A)−1 to find the R’s). df Solution: ds = cos(−1) = α1 s=−1 Therefore (a) Note that because Rk may be complex, the linear spaces we work with assume α0 is cos(1) − Using Rk = pk q T that scalar multiplication = complex. sin(1) = −0.3012k , α1 = cos(1) = 0.5403 T R(Rk ) = { Rk x | x ∈ Cn } = { pk qk x | x ∈ Cn } Similarly for f (s) = cos(s) and g (s) ==0 + β1 s | α ∈ C} = span{p }. β { αp k k f (s)|s=−1 = cos(−1) = β0 + β1 (−1) df = − sin(−1) = β1 9 ds s=−1 c J.C. Cockburn page 5 of 7 Analytical Topics Solutions to Homework #8 - Fall 2009.1 Therefore β0 = cos(1) + sin(1) = 1.3818 β1 = sin(1) = 0.8415 Therefore, sin(A) = cos(A) = Part b) 2 cos(1) − sin(1) 2 cos(1) 0.2391 1.0806 = , −2 cos(1) −2 cos(1) − sin(1) −1.0806 −1.9221 cos(1) + 2 sin(1) 2 sin(1) 2.2232 1.6829 = −2 sin(1) cos(1) − 2 sin(1) −1.6829 −1.1426 From the results of part a) we obtain sin2 (A) = sin2 (1) − 4 cos(1) sin(1) −4 cos(1) sin(1) , 4 cos(1) sin(1) sin2 (1) + 4 cos(1) sin(1) cos2 (A) = cos2 (1) + 4 cos(1) sin(1) 4 cos(1) sin(1) 2 (1) − 4 cos(1) sin(1) −4 cos(1) sin(1) cos Therefore sin2 (A) + cos2 (A) = sin2 (1) + cos2 (1) 0 2 (1) + cos2 (1) = I 0 sin Part c) Let f (s) = tan(s) and g (s) = γ0 + γ1 s. Then f (s)|s=−1 = tan(−1) = α0 + γ1 (−1) df = sec2 (−1) = γ1 ds s=−1 Therefore γ0 = − tan(1) + sec2 (1) = 1.8681 γ1 = sec2 (1) = 3.4255 and tan(A) = = − tan(1) + 2 sec2 (1) 2 sec2 (1) 2 (1) − tan(1) − 2 sec2 (1) 2 sec 5.2936 6.8510 −6.8510 −8.4084 From part a) (cos(A))−1 = 2.2232 1.6829 −3.9140 −5.7649 = −1.6829 −1.1426 5.7649 7.6157 It can be verified that sin(A)(cos(A))−1 = (cos(A))−1 sin(A) = c J.C. Cockburn 5.2937 6.8511 −6.8511 −8.4086 page 6 of 7 Analytical Topics Solutions to Homework #8 - Fall 2009.1 The error between the numerical values of tan(A) and the above expressions is 0.0573 0.1253 × 10−3 −0.1253 −0.1934 This suggest that the approximation is of the order of 10−4 . This is a very poor approximation generally unacceptable. For better results more significant digits must be considered in the evaluation of each of the functions. Note however that since only four significant digits were considered this is not an unexpected result. Note: Actually parts b) and c) should not be solved as above since that approach does not prove that sin2 (A) + cos2 (A) = I or tan(A) = (cos(A))−1 sin(A) for any A ∈ Rn×n . I only shows the validity of the above formulas for the particular matrix in question. A more general approach uses power series expansions and their convergence properties. For example, 1 2 18 sin2 (x) = x2 − x4 + x6 − x + ··· 3 45 315 1 2 18 cos2 (x) = 1 − x2 + x4 − x6 + x + ··· 3 45 315 5 61 6 277 8 1 x+ x + ··· (cos(x))−1 = 1 + x2 + x4 + 2 24 720 8064 1 2 17 7 62 9 tan(x) = x + x3 + x5 + x+ x + ··· 3 15 315 2835 2 17 7 62 9 1 x+ x + ··· (cos(x))−1 sin(x) = x + x3 + x5 + 3 15 315 2835 Therefore, provided that the spectrum of A is included in the domain of analyticity of each of the above functions (which implies that the above series converge) it is clear that sin2 (A) + cos( A) = I tan(A) = (cos(A))−1 sin(A) = sin(A)(cos(A))−1 c J.C. Cockburn page 7 of 7 ...
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This note was uploaded on 07/28/2011 for the course EE 263 taught by Professor Boyd,s during the Summer '08 term at Stanford.

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