MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science : Dynamic Systems Spring 2011

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Joel Gibbs
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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering an Computer Science 624: Dynamic Systems Spring 20 Homework Solutions Exercise a Given square matrices A an A 4, we know that A is square as well: A A 2 A = 0 A 4 I 0 A A = 2 0 A 4 0 I Note that I 0 et = etieta 4 = eta 4, 0 A 4 which can be verifie by recursively computing the principal minors Also, by the elementary operations of rows, we have A A 2 A 0 et = = et = eta 0 I 0 I Finally note that when A an B are square, we have that etab = etaetb Thus we have eta = eta eta 4 b Assume A an A 4 exist Then A A 2 B B 2 I 0 AA = = 0 A 4 B 3 B 4 0 I, which yiels four matrix equations: A B + A 2 B 3 = I, 2 A B 2 + A 2 B 4 = 0, 3 A 4 B 3 = 0, 4 A 4 B 4 = I From Eqn 4, B4 = A 4, with which Eqn 2 yiels B 2 = A A 2 A 4 Also, from Eqn 3 B 3 = 0, with which from Eqn B = A Therefore, A = A 4 A A 2 A 0 A 4

2 Exercise 2 a 0 I A A 2 A3 A4 = I 0 A 3 A 4 A A 2 b Let us fin B = B B 2 B 3 B 4 such that BA = A A 2 0 A A A 4 3 A 2 The above equation implies four equations for submatrices B A + B 2 A 3 = A, 2 B A 2 + B 2 A 4 = A 2, 3 B 3 A + B 4 A 3 = 0, 4 B 3 A 2 + B 4 A 4 = A A 3 A 4 A 2 First two equations yiel B = I an B 2 = 0 Express B 3 from the thir equation as B 3 = B 4 A 3 A an plug it into the fourth After gathering the terms we get B4 A 4 A 3 A A 2 = A 4 A 3 A A 2, which turns into ientity if we set B 4 = I Therefore I 0 B = A 3 A I c Using linear operations on rows we see that et B = Then, eta = etbeta = et BA = et A et A 4 A 3 A A 2 Note that A4 A 3 A A 2 oes not have to be invertible for the proof Exercise 3 We have to prove that eti AB = eti BA Proof: Since I an I BA are square, I 0 eti BA = et B I BA I A I A = et B I 0 I I A I A = et et, B I 0 I yet, from Exercise, we have I A et = etieti = 0 I Thus, I A eti BA = et B I 2

3 Now, I A I AB 0 et = et = eti AB B I B I Therefore eti BA = eti AB Note that I BA is a q q matrix while I AB is a p p matrix Thus, when one wants to compute the eterminant of I AB or I BA, s/he can compare p an q to pick the prouct AB or BA with the smaller size b We have to show that I AB A = AI BA Proof: Assume that I BA an I AB exist Then, This completes the proof Exercise 6 of limits, ie At + ΔtBt + Δt AtBt AtBt = lim Δt 0 Δt We substitute first orer Taylor series expansions At + Δt = At + Δt At + oδt Bt + Δt = Bt + Δt Bt + oδt to obtain [ ] AtBt = AtBt + Δt AtBt + ΔtAt Bt + hot AtBt Δt Here hot stans for the terms [ hot = At + Δt A = A I = AI BAI BA = A ABAI BA = I ABAI BA I AB A = AI BA a The safest way to fin the element-wise erivative is by its efinition in terms ] [ At oδt + oδt ] Bt + Δt Bt + oδt 2, a matrix quantity, where lim Δt 0 hot/δt = 0 verify Reucing the expression an taking the limit, we obtain [AtBt] = AtBt + At Bt b For this part we write the ientity A tat = I Taking the erivative on both sies, we have [ A tat ] = A tat + A t At = 0 3

4 Rearranging an multiplying on the right by A t, we obtain A t = A t AtA t Exercise 8 Let X = {g x = α 2 M 0 + α x + α 2 x + + α M x α i C} a We have to show that the set B = {, x,, x M } is a basis for X Proof : First, let s show that elements in B are linearly inepenent It is clear that each element in B can not be written as a linear combination of each other More formally, c + c x + + c M x M = 0 i c i = 0 Thus, elements of B are linearly inepenent 2 Then, let s show that elements in B span the space X Every polynomial of orer less than or equal to M looks like M px = α i x i for some set of αi s Therefore, {, x,, x M } span X x b T : X X an T gx = gx i=0 Show that T is linear Proof: T ag x + bg 2 x = ag x + bg 2 x x = a g + b g 2 x x = at g + bt g 2 Thus, T is linear 2 gx = α α x + α 2 x + + α M M x, so Thus it can be written as follows: 2 M x M T gx = α + 2α x + + Mα α 0 α α 2α α 2 3α 3 0 α 3 = M Mα M α M 0 4

5 The big matrix, M, is a matrix representation of T with respect to basis B The column vector in the left is a representation of gx with respect to B The column vector in the right is T g with respect to basis B 3 Since the matrix M is upper triangular with zeros along iagonal in fact M is Hessenberg, the eigenvalues are all 0; λ i = 0 i =,, M + 4 One eigenvector of M for λ = 0 must satisfy MV = λ V = 0 0 V = 0 is one eigenvector Since λ i s are not istinct, the eigenvectors are not necessarily inepenent Thus in orer to computer the M others, ones uses the generalize eigenvector formula 5

6 MIT OpenCourseWare 624J / 6338J Dynamic Systems an Control Spring 20 For information about citing these materials or our Terms of Use, visit:

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