ECEN 605 LINEAR SYSTEMS. Lecture 8 Invariant Subspaces 1/26

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1 1/26 ECEN 605 LINEAR SYSTEMS Lecture 8 Invariant Subspaces

2 Subspaces Let ẋ(t) = A x(t) + B u(t) y(t) = C x(t) (1a) (1b) denote a dynamic system where X, U and Y denote n, r and m dimensional vector spaces, and A, B, C are linear operators: A : X X, B : U X, C : X Y. (2) The structure of the system can be effectively explored and displayed in several different coordinate systems using various invariant subspaces. A subspace V X is A-invariant if AV V. If V has dimension K and {v 1, v 2,...v K } is a basis for V, one can choose any subspace 2/26

3 W X such that V W = X, a basis {w K+1,...w n } for W and form the coordinate transformation T = [v 1,..., v K, w K+1,..., w n ] (3) and set x = T z. (4) The system equation (1) is then transformed into y(t) = C n z(t) ż(t) = A n z(t) + B n u(t) (5a) (5b) where A n = T 1 AT, B n = T 1 B, C n = CT. (6) 3/26

4 Writing (6) as AT = TA n, TB n = B, C n = CT (7) it is easy to see that, the A-invariance of V implies that [ ] A1 A A n = 3. (8) 0 A 2 Thus (5) can be written as ] [ ] [ ] [ż1 A1 A = 3 z1 + ż 0 A 2 z 2 y = [ C 1 C 2 ] [ z 1 z 2 [ B1 B 2 ] u (9a) ]. (9b) If W also happens to be A-invariant then A 3 = 0 in (9a). 4/26

5 Now we construct two special A-invariant subspaces for the dynamic system (1). These are the controllable subspace and unobservable subspace. For this, let B denote the image of B: and let B := {Bu u U} (10) V u := B + AB + + A n 1 B. (11) Let KerC denote the null space of C: KerC := {x Cx = 0} (12) and V y = n 1 i=0 KerCA i. (13) 5/26

6 V u is called the controllable subspace and V y is called the unobservable subspace of the system (1). Lemma AV u V u B V u (14a) (14b) 6/26

7 proof That B V u follows from (11). To prove (14a) consider an arbitrary vector v V u. Then v = Bu 0 + ABu A n 1 Bu n 1 (15) for some u i, i = 0, 1,..., n 1 Therefore Av = ABu 0 + A 2 Bu A n Bu n 1. (16) Since A n can be expressed as a linear combination of lower powers of A, it follows that Av = Bū 0 + A 2 Bū A n Bū n 1. (17) for some ū i, i = 0, 1, 2,..., n 1. 7/26

8 proof (cont.) Therefore Av V u. (18) Since v was an arbitrary vector in V, it follows that AV u V u. (19) 8/26

9 Remark Consider the collection of subspaces V = {V AV V, B V}. (20) It is easy to show that V u is the smallest element of V, that is, is contained in every subspace in V u. 9/26

10 Lemma AV y V y V y KerC. (21a) (21b) 10/26

11 Proof. That (21b) is true follows from (13). To prove (21a), pick an arbitrary v V y. Then Now, it is easy to verify that CA i v = 0, i = 0, 1,..., n 1. (22) CA i (Av) = 0, i = 0, 1,..., n 1 (23) again using the Cayley-Hamilton Theorem. Thus Av V y and this proves (21a). 11/26

12 Remark Consider the collection of all subspaces which are A-invariant and contained in KerC: V = {V AV V, V KerC}. (24) It is easy to see that V y is the largest element of V, that is, every element VinV satisfies V V y. (25) 12/26

13 Lemma A(V u V y ) (V u V y ) A(V u + V y ) (V u + V y ). (26a) (26b) 13/26

14 Proof. If v V u V y, Av V u and Av V y and so Av V u V y. If v V u + V y, then v = v 1 + v 2 with v 1 V u and v 2 V y. Then Av = Av 1 + Av 2 V u + V y. 14/26

15 Now consider a decomposition of the state space as follows: X = V 1 V 2 V 3 V 4 (27) where and V 2, V 3, V 4 satisfy V 1 := V u V y (28) V 2 V 1 = V u (29) V 3 V 1 = V y (30) and V 4 V 3 V 2 V 1 = X (31) but are otherwise arbitrary. 15/26

16 Let dimv i = K i (32) and let T i denote an n K i matrix whose columns form a bases for V i, i = 1, 2, 3, 4. Let denote a coordinate transformation and set In this coordinate T = [T 1, T 2, T 3, T 4 ] (33) x = Tz. (34) ż = A n z + B n u y = C n z (35a) (35b) 16/26

17 and A n = T 1 AT B n = T 1 B C n = CT. (36a) (36b) (36c) 17/26

18 Theorem (Kalman Canonical Decomposition) In the coordinate system defined by (28)-(36); (A n, B n, C n ) have the following structure: A 1 A 3 A 5 A 7 A n = 0 A 2 0 A A 4 A A 6 B 1 (37) B n = B C n = [ 0 C 2 0 C 6 ]. 18/26

19 Proof. The proof follows from the facts: AV 1 V 1 A(V 1 V 2 ) V 1 V 2 A(V 1 V 3 ) V 1 V 3 (38a) (38b) (38c) and B V 1 V 2 V 1 V 3 KerC, (39a) (39b) and the relations (36). 19/26

20 Remark The structure (37) specifies only the zero blocks in the matrices. The non-zero blocks will depend on the actual subspaces V 2, V 3 and V 4 and the bases for these subspaces. 20/26

21 Example (Kalman Canonical Decomposition) Consider the system ẋ(t) = A x(t) + B u(t) y(t) = C x(t) (40) with 21/26

22 A = B = C = [ ]. (41) 22/26

23 The controllable subspace V u is generated by = V 1 0 u. (42) 0 0 The unobservalbe subspace V y = Ker [ ] = (43) /26

24 Then 0 0 V u V y = =: V 1 1. (44) 0 24/26

25 Define to satisfy and let V 2 =, V 0 3 = and V 0 4 = (45) X = V 1 V 2 V 3 V 4 (46) T = (47) /26

26 It is easy to verify that T 1 AT = A n = T 1 B = B n = CT = C n = [ ] (48) which is the Kalman Canonical Decomposition of the system. 26/26

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