EE263: Introduction to Linear Dynamical Systems Review Session 9

Transcription

1 EE63: Introduction to Linear Dynamical Systems Review Session 9 SVD continued EE63 RS9 1

2 Singular Value Decomposition recall any nonzero matrix A R m n, with Rank(A) = r, has an SVD given by A = UΣV T, where U R m r, U T U = I V R n r, V T V = I Σ = diag(σ 1,...,σ r ) here, σ 1 σ σ r > 0 are the singular values (this is the economy or thin version of the SVD) EE63 RS9

3 Some basic facts about singular values recall A = max x =1 Ax = σ max(a) for any A, we have 1. σ max (A) σ max (A)σ max (), σ max (A) = max x =1 A(x) σ max(a) max x =1 x = σ max(a)σ max (). σ min (A) σ min (A)σ min (), σ min (A) = min x =1 A(x) σ min(a) min x =1 x = σ min(a)σ min () 3. σ min (A+) σ min (A) σ max (), Ax = (A+)x x (A+)x + x EE63 RS9 3

4 rearranging this inequality yields (A+)x Ax x σ min (A) x σ max () x If x 0, then (A+)x x σ min (A) σ max () since this is true for all x 0, it is true for the minimizing x: min x 0 (A+)x x = σ min (A+) σ min (A) σ max () EE63 RS9 4

5 A Pythagorean inequality for the matrix norm suppose that A R m n and R p n, show that A A +. under what conditions do we have equality? suppose that v R n with v = 1 is the maximum gain direction of hence but A = A ] v, [ A ], A ] v = [ Av v = Av + v A + EE63 RS9 5

6 since in general the maximum [ gain ] directions of A and are different from A the maximum direction of equality holds when the maximum gain direction of A and are the same, if v 1 R n with v 1 = 1 is the maximum gain direction of both A and then [ A [ A ]v 1 = [ Av1 v 1 = Av 1 + v 1 = A +, ] and since also A A + we should have A +, conversely, if A = A = A + then from above the maximum gain directions of A and should be equal EE63 RS9 6

7 [ ] A to the maximum gain direction of, and therefore, the maximum gain directions of A and should be equal themselves EE63 RS9 7

8 Pseudo-inverse the pseudo-inverse of A = UΣV T is given by A = VΣ 1 U T A full rank skinny - A = (A T A) 1 A T A full rank fat - A = A T (AA T ) 1 can be used when A not full rank take y R m, then x = A y has minimum norm, among all w that minimize Aw y w minimizes Aw y iff Aw y = 0, i.e., A T Aw = A T y so A y solves minimize w subject to A T Aw = A T y. EE63 RS9 8

9 Range and nullspace let u 1,...,u r be the columns of U, and let v 1,...,v r be the columns of V the set {u 1,...,u r } is an orthonormal basis for R(A) R(A) R(U) Az = UΣV T z = U(ΣV T z) so if x R(A) then x R(U) R(U) R(A) take x R(U), x = Uy for some y Σ diagonal, full rank, V T fat, full rank ΣV T fat and full rank so for any y there exists z such that y = ΣV T z and so x = UΣV T z = Az and x R(A) similarly the set {v 1,...,v r } is an orthonormal basis for R(A T ) = N(A) EE63 RS9 9

10 Example given two sets of independent vectors b 1,...,b r R m, and d 1,...,d n r R n, construct a matrix A R m n, with R(A) = span{b 1,...,b r }, and N(A) = span{d 1,...,d n r } find an orthonormal basis b 1,..., b r for span{b 1,...,b r } (using, say, Gram-Schmidt) let b 1,..., b r be the columns of similarly find an orthonormal basis d 1,..., d r for span{d 1,...,d n r } let d 1,..., d r be the columns of D now we can simply take A = D T clearly the choice of A is not unique EE63 RS9 10

11 Energy storage efficiency in a linear dynamical system consider the discrete-time linear dynamic system x(t+1) = Ax(t)+u(t), y(t) = Cx(t), where x(t) R n, and u(t),y(t) R initial state x(0) = 0, input sequence u(0),...,u(n 1) we are interested in the output over the next N samples, i.e., y(n),...,y(n 1), (u(t) = 0 for t N) define the input and output energies as E in = N 1 t=0 u(t), E out = N 1 t=n y(t) how would you choose the (nonzero) input sequence u(0),...,u(n 1) to maximize the ratio of output energy to input energy? EE63 RS9 11

12 Energy storage efficiency in a linear dynamical system for the discrete time system x(t+1) = Ax(t)+u(t), y(t) = Cx(t), we can write the state at time t in terms of the past inputs as x(t) = [ A... A t 1 ] u(t 1). u(0), and the output y(t) as y(t) = [ C CA... CA t 1 ] u(t 1). u(0) since the input is zero from time t = N onwards, the state is propagated forward by A at each sample, i.e., x(t+1) = Ax(t), and so for t N, we EE63 RS9 1

13 can write the output in terms of the input from t = 0 to t = N 1 as y(t) = [ CA t N CA t N+1... CA t 1 ] u(n 1). u(0) stacking up the y(t)s, t = N,...,N 1, we obtain the equation, y(n). y(n 1) = C CA... CA N 1 CA CA... CA N... CA N 1 CA N... CA N u(n 1). u(0), or Y = CU; and we are interested in maximizing Y / U, subject to U 1 this is an SVD problem - the maximum value of Y / U is equal to the largest singular value of C, and the sequence of inputs u is precisely the right singular vector corresponding to σ max (C)! EE63 RS9 13