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

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Electrical Engineering and Computer Science 64: Dynamic Systems Spring Homework 4 Solutions Exercise 47 Given a complex square matrix A, the definition of the structured singular value function is as follows µ Δ (A) = minδ Δ {σ max (Δ) det(i ΔA) = } where Δ is some set of matrices a) If Δ = {αi : α C}, then det(i ΔA) = det(i αa) Here det(i αa) = implies that there exists an x = such that (I αa)x = Expanding the left hand side of the equation yields x = αax x = Ax Therefore is an eigenvalue of A Since σ (Δ) = α, α Therefore, µ Δ (A) = λ max (A) α max arg min{σ max (Δ) det(i ΔA) = } = α = δ Δ λmax (A) b) If Δ = {Δ C n n }, then following a similar argument as in a), there exists an x = such that (I ΔA)x = That implies that x = ΔAx x = ΔAx Δ Ax Ax σmax (A) Δ x σ max (Δ) σ max (A) Then, we show that the lower bound can be achieved Since Δ = {Δ C n n }, we can choose Δ such that σ max(a) Δ = V U where U and V are from the SVD of A, A = UΣV Note that this choice results in I ΔA = I V V = V V which is singular, as required Also from the construction of Δ, σ max (Δ) = µ Δ (A) = σ max (A) Therefore, σ max(a)

2 c) If Δ = {diag(α,, α n ) α i C} with D {diag(d,, d n ) d i > }, we first note that D exists Thus: det(i ΔD AD) = det(i D ΔAD) = det((d D ΔA)D) = det(d D ΔA)det(D) = det(d (I ΔA))det(D) = det(d )det(i ΔA)det(D) = det(i ΔA) Where the first equality follows because Δ and D are diagonal and the last equality holds because det(d ) = /det(d) Thus, µ Δ (A) = µ Δ (D AD) Now let s show the left side inequality first Since Δ Δ, Δ = {αi α C} and Δ = {diag(α,, α n )}, we have that which implies that min {σ max (Δ) det(i ΔA) = } min {σ max (Δ) det(i ΔA) = }, Δ Δ Δ Δ But from part (a), µ Δ (A) = ρ(a), so, µ Δ (A) µ Δ (A) ρ(a) µ Δ (A) Now we have to show the right side of inequality Note that with Δ 3 = {Δ C}, we have Δ Δ 3 Thus by following a similar argument as above, we have Hence, min {σ max (Δ) det(i ΔA) = } min {σ max (Δ) det(i ΔA) = } Δ Δ Δ Δ 3 µ Δ (A) = µ Δ (D AD) µ Δ3 (D AD) = σ max (D AD) Exercise 48 We are given a complex square matrix A with rank(a) = According to the SVD of A we can write A = uv where u, v are complex vectors of dimension n To simplify computations we are asked to minimize the Frobenius norm of Δ in the definition of µ Δ (A) So µ Δ (A) = minδ Δ { Δ F det(i ΔA) = } Δ is the set of diagonal matrices with complex entries, Δ = {diag(δ,, δ n ) δ i C} Introduce the column vector δ = (δ,, δ n ) T and the row vector B = (u v,, u n v n ), then the original problem can be reformulated after some algebraic manipulations as µ Δ (A) = minδ C n { δ Bδ = }

3 To see this, we use the fact that A = uv, and (from excercise 3(a)) det(i ΔA) = det(i Δuv ) = det( v Δu) = v Δu Thus det(i ΔA) = implies that v Δu = Then we have = v Δu δ u ( ) v n = v δ n un δ ( = v u v ) n u n δ n = Bδ Hence, computing µ Δ (A) reduces to a least square problem, ie, min { Δ F det(i ΔA) = } min δ st = Bδ Δ Δ We are dealing with a underdetermined system of equations and we are seeking a minimum norm solution Using the projection theorem, the optimal δ is given from δ o = B (BB ) Substituting in the expression of the structured singular value function we obtain: n µ Δ(A) = u i v i In the second part of this exercise we define Δ to be the set of diagonal matrices with real entries, Δ = {diag(δ,, δ n ) δ i R} The idea remains the same, we just have to ( alter the ) constraint Re(B) equation, namely Bδ = +j Equivalently one can write Dδ = d where D = and d = Im(B) ( ) Again the optimal δ is obtained by use of the projection theorem and δ o = D (DD T ) d Substituting in the expression of the structured singular value function we obtain: i= µ Δ (A) = d T (DD T ) d Exercise 5 Suppose that A C m n is perturbed by the matrix E C m n Show that σ max (A + E) σ max (A) σ max (E) 3

4 Also find an E that achieves the upper bound Note that Also, A = A + E E A = A + E E A + E + E A A + E E (A + E) = A + E A + E A + E A + E A E Thus, putting the two inequalities above together, we get that A + E A E Note that the norm can be any matrix norm, thus the above inequality holds for the -induced norms which gives us, A matrix E that achieves the upper bound is σ max (A + E) σ max (A) σ max (E) σ E = U V = A, σ r where U and V form the SVD of A Here, A + E =, thus σ max (A + E) =, and is achieved + σ max (A) = σ max (E) Suppose that A has less than full column rank, ie, the rank(a) < n, but A + E has full column rank Show that σ min (A + E) σ max (E) Since A does not have full column rank, there exists x = such that (A + E)x Ex Ax = (A+E)x = Ex (A+E)x = Ex = E = σ (E) x x max But, σ min (A + E) 4 (A + E)x, x

5 as shown in chapter 4 (please refer to the proof in the lecture notes!) Thus σ min (A + E) σ max (E) Finally, a matrix E that results in A +E having full column rank and that achieves the upper bound is σ r+ E = U σ r+ V, for σ A = U σ r V Note that A has rank r < n, but that A + E has rank n, A + E = U σ σ r σ r+ σ r+ V It is easy to see that σ min (A + E) = σ r+, and that σ max (E) = σ r+ The result in part, and some extensions to it, give rise to the following procedure (which is widely used in practice) for estimating the rank of an unknown matrix A from a known matrix A + E, where E is known as well Essentially, the SVD of A + E is computed, and the rank of A is then estimated to be the number of singular values of A + E that are larger than E 5

6 Exercise 5 Using SVD, A can be decomposed as σ A = U σ k V, where U and V are unitary matrices and k r + Following the given procedure, let s select the first r+ columns of V : {v, v,, v r+ } Since V is unitary, those v i s are orthonormal and hence independent Note that {v, v,, v r+, v n } span R n, and if rank(e) = r, then exactly r of the vectors, {v, v,, v r+, v n }, span R(E ) = N (E) The remaining vectors span N (E) So, given any r + linearly independent vectors in R n, at least one must be in the nullspace of E That is there exists coefficients c i for i =,, r +, not all zero, such that E(c v + c v + c r+ v r+ ) = These coefficients can be normalized to obtain a nonzero vector z, z =, given by r+ α ( ) z = α i v i = v v r+ i= α r + and such that Ez = Thus, σ α σ α v ( (A E)z = Az = UΣ ) r+ α i v i = U σ v i= r+ By taking -norm of both sides of the above equation, r+α r+ σ α σ α σ α σ α (A E)z = U σ r+α r+ = σ r+ α r+ ( since U is a unitary matrix) ( r ) ( ) + r+ = σ i αi σr+ αi () i= But, from our construction of z, α α r+ z = ( v v r+ ) = = α i = α i= r+ α i= r+ 6 ()

7 Thus, equation() becomes (A E)z σ r+ Finally, (A E)z A E for all z such that z = Hence A E σ r+ To show that the lower bound can be achieved, choose σ E = U V σr E has rank r, and A E = σ r+ A E = U σ r+ V σ k Exercise 6 The model is linear one needs to note that the integration operator is a linear operator Formally one writes S(αu + βu )(t) = ( ) e t s (αu (s) + βu (s))ds = α e (t s) u(s) + β = α(su )(t) + β(su )(t) e (t s) u (s) It is non-causal since future inputs are needed in order to determine the current value of y Formally one writes ( ) (t s) (P Su)(t) = (P SP u)(t) + P e u(s)ds T T T T It is not memoryless since the current output depends on the integration of past inputs It is also time varying since (t T s) (Sσ T u)(t) = (σt Su)(t) + e u(s)ds one can argue that if the only valid input signals are those where u(t) = if t < then the system is time invariant T T 7

8 Exercise 64(i) linear, time varying, causal, not memoryless (ii) nonlinear (affine, tranlated linear) time varying, causal, not memoryless (iii) nonlinear, time invariant, causal, memoryless (iv) linear, time varying, causal, not memoryless (i),(ii) can be called time invariant under the additional requirement that u(t) = for t < 8

9 MIT OpenCourseWare 64J / 6338J Dynamic Systems and Control Spring For information about citing these materials or our Terms of Use, visit: