Lösungshinweise zur 1. Übung in Kontrolltheorie II

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1 Universität Würzburg Institut für Mathematik Dr G Dirr, G Ciaramella Sommersemester 2015 Würzburg, den Lösungshinweise zur 1 Übung in Kontrolltheorie II Aufgabe 21 (a) Sei X ein metrischer Raum Zeigen Sie, dass endliche Durchschnitte und kartesische Produkte offener und dichter Teilmengen wieder offen und dicht sind (b) Zeigen Sie, dass für m n die Menge aller reellen Matrizen vom Rang m offen und dicht in der Menge aller reellen m n-matrizen ist (c) Zeigen Sie, dass für m = p die Produkte R n (A, B)O n (A, C) generisch invertierbar sind Genauer gesagt, zeigen Sie, dass die Menge aller reellen Tripel (A, B, C), für die das Produkt R n (A, B)O n (A, C) invertierbar ist, eine offene und dichte Teilmenge von R n n R n m R m n bilden (a): Finite intersections: Let O 1,, O n open and dense subsets of X Then it is a standard result from topology that the intersection O := n O k is again open Therefore, we only have to prove (by induction) the intersection O := n O k is again dense in X Assume that the assertion holds for some n N and consider n + 1 open and dense subsets O 1,, O n+1 Moreover, let x be an arbitrary point in X and choose any open non-empty ball B r (x) around x Since O n+1 is open and dense there exists non-empty ball B r (x ) such tha B r (x ) O n+1 B r (x) Now, by induction hypothesis O := n O k is dense in X Hence, O B r (x ) and thus B r (x) ( n+1 O ) k, ie n+1 O k is dense in X Finite Cartesian products: (b): A standard result from linear algebra provides the following equivalence: A onto A one-to-one Hence, for A R m n with m n one has the equivalences: rk A = m AA one-to-one det(aa ) 0 Therefore, one can characterize the set S m of all real m n-matrizen with rank m as follows: S m = {A R m n det(aa ) 0} Hence it suffices to prove that N := {A R m n det(aa ) = 0} is closed and nowhere dense To this end, we consider the map A P (A) := det(aa ) Clearly, the map P is a polynomial in the entries of A and, moreover, one has P (A 0 ) = 1 for A 0 := ( I m 0 ) Hence, P is not the zero-polynomial and thus the assertion follows from standard result on reell polynomials (cf lecture notes) (c): A similar approach as in (b) suggests to consider the polynomial map (A, B, C) P (A, B, C) := det ( R n (A, B)O n (A, C) )

2 Certainly, the set S inv of all real triples (A, B, C) sucht that R n (A, B)O n (A, C) is invertible is given by S inv = {A R m n P (A, B, C) 0} Thus, we have to show that N := {A R m n P (A, B, C) = 0} is closed and nowhere dense This is guaranteed once we can show that P does not represent the zero-polynomial For proving P 0 consider the following triple (A 0, B 0, C 0 ) given by: A 0 := 0, B 0 := 0 0, and C 0 := Then a straightforward computation shows R n (A 0, B 0 ) := and and, finally, O n (A 0, C 0 ) := R n (A 0, B 0 )O n (A 0, C 0 ) := Hence, one has P (A 0, B 0, C 0 ) 0 and thus P 0

3 Aufgabe 22 Sei S(t) t N eine beliebige Markow-Folge Zeigen Sie die folgende Aussage: Falls S(t) t N eine endlich dimensionale kanonische Realisierung besitzt, so sind alle kanonischen Realisierung von S(t) t N endlich dimensional und besitzen die gleiche Dimension Proof: Let (A, B, C) and (A, B, C ) two canonical realizations of S(t) t N and assume that (A, B, C) is finite dimensional and (A, B, C ) finite dimensional From Lemma II8, we know O S (A, C)R t (A, B) = H s,t(s) = O S (A, C )R t (A, B ) ( ) for all s, t N and, in particular, O(A, C)R(A, B) = H(S) = O(A, C )R(A, B ) ( ) Since (A, B, C) is assumed to be finite dimensional we conclude from ( ) the following estimate rk H s,t(s) min{rk O S (A, C), rk R t (A, B)} = dim X =: n Hence, from Lemma II7 it follows rk H(S) n On the other hand, we know that (A, B, C ) is controllable and observable and, therefore, R(A, B ) is onto and O(A, C ) is one-to-one Hence, ( ) implies contrariwise that rk H(S) = Consequently, two canonical realizations of S(t) t N have to be either both finite or infite dimensional Now, the remaining statement follows immediately from Theorem III7

4 Aufgabe 23 Zeigen Sie, dass die Konstruktion im Beweis von Satz II4 eine kanonische Realisierung von S(t) t N liefert Proof:

5 Aufgabe 24 Zwei lineare Kontrollsysteme (A, B, C) und A, B, C mit Zustandsraum X, Kontrollraum U und Ausgangsraum Y heißen Feedback-äquivalent, wenn es invertierbare lineare Abbildungen R : U U, S : Y Y und T : X X sowie eine weitere lineare Abbildung F : Y U gibt, so dass die folgenden Gleichungen erfüllt sind: A = T (A + BF C)T 1, B = T BR, und C = SCT 1 Zeigen Sie, dass Feedback-Äquivalenz ist eine Äquivalenzrelation auf der Menge aller linearen Systeme (A, B, C) L(X) L(U, X) L(X, Y ) ist Proof:

6 Aufgabe 25 Sei (A, b) K n n K n kontrollierbar und sei p ein normiertes Polynom von Grad n Zeigen Sie Ackermanns Formel mit f := e n R(A, b) 1 p(a) det(λi A bf) = p(λ) For proofing of Ackermanns Formula, we need the following auxiliary result: Lemma: For all x, y K n one has det(i xy ) = 1 y x Proof of the Lemma: Without loss of generality, we can assume x 0 Thus there exists an invertible n n-matrix T with T x = e 1 It follows det(i xy ) = det(i T xy T 1 ) = det(i e 1 y T 1 ) = 1 y T 1 e 1 = 1 y T 1 T x = 1 y x Proof of Ackermann s Formula: First, we investigate how state space equivalence acts on Ackermann s FormulaTo this end, consider the pair (A := T AT 1, b := T b) and define f := e n R(A, b ) 1 P (A ) Then, a straightforward computation shows R(A, b ) = T R(A, b) and f = e n R(A, b) 1 T 1 P (T AT 1 ) = ft 1 Hence, we obtain det(λi A b f ) = det(λi T AT 1 T bft 1 ) = det(λi A bf) = p(λ) Therefore, (A, b, f) satisfies Ackermann s Formula if and only if (A, b, f ) does Consequently, we can assume without loss of generality that (A, b) is in controllability normal form, ie 0 0 α α1 A := 0 0, b := α 0 n 1 with χ A (λ) := det(λi A) = λ n n 1 k=0 α kλ k Thus, f simplifies to f = e n R(A, b) 1 p(a) = e n p(a) since one has R(A, b) = I Now, an application of the above Lemma yields det(λi A bf) = det ( (λi A)(I (λi A) 1 bf) ) = det(λi A) det(i (λi A) 1 bf) = det(λi A)(1 f(λi A) 1 b) = det(λi A)(1 + e n p(a)(λi A) 1 e 1 ) = det(λi A)(1 + e n (λi A) 1 p(a)e 1 ) = det(λi A) + e n (λi A) p(a)e 1 )

7 where (λi A) denotes the adjunct matrix (cofactor matrix) of (λi A) Finally, plugging in p(λ) = λ n + n 1 k=0 β kλ k in the above formula yields det(λi A bf) = det(λi A) + e n (λi A) p(a)e 1 ) β 0 + α 0 n 1 = λ n α k λ k + ( 1 λ λ n 1) β 1 + α 1 = p(λ) k=0 β n 1 + α n 1 and thus the proof is complete Remark: See Helmke s notes for an alternative proof

8 Aufgabe 26 Sei K ein beliebiger nicht endlicher Körper und sei A zyklisch und (A, B) K n n K n m kontrollierbar (a) Zeigen Sie, dass es dann ein u K m gibt so dass (A, Bu) kontrollierbar ist (b) Zeigen Sie, dass die Aussage für nicht zyklisches A falsch ist [Hinweis: (i) A heißt zyklisch, wenn es einen Vektor x K n gibt, so dass (ii) Sie dürfen obda K = C voraussetzen] span{x, Ax, A 2 x, } = K n (a): For proving part (a) the following characterization of a cyclic matrix is quite useful 1 Lemma 1: (a) A matrix A K n n is cyclic if and only if the characteristic polynomial χ A and the minimal polynomial µ A coincide This is equivalent to the fact that there exists only one Jordan block to each distinct value of A if K is algebraically closed (b) Let A K n n be cyclic and let µ A (λ) = ( p 1 (λ) ) e1 (p r (λ) ) e r denote the minimal polynomial of A with prime factors p k (λ) and exponents e k N Moreover, let E k := ker ( p k (A) ) e k, E k := ker ( p k (A) ) e k 1, and πk the associated projection onto E k, see ( ) Then x K n is a cyclic vector if and only if π k (x) =: x k E k \ E k for all k = 1,, r Remark: Note that, if K is algebraically closed, then E k = ker(λ k I A) e k generalized eigenspaces of A associated with the eigenvalue λ k are precisely the Proof of the Lemma 1: Fisrt, we prove the statement that if A K n n is cyclic then the minimal and characteristic polynomial have to coincide, assume χ A µ A Since µ A is always a factor of χ A one concludes that the degree of µ A is less that the degree of χ A which equals n Now it is trivial to see that for x K n and hence A cannot be cyclic dim span{x, Ax,, A n 1 x} deg µ A < n Next, we sketch the proof of the converse statement that χ A = µ A implies cyclicity of A K n n Let µ A (λ) = ( p 1 (λ) ) e1 (p r (λ) ) e r be the (unique) decomposition of the minimal polynomial into coprime factors g k (λ) := ( p k (λ) ) e k (Note that the polynomial ring K[λ] is a Euclidean ring and thus allows a unique prime factorization) Then, there exist polynomials f k (λ) such that f 1 (λ) g k (λ) + + f r (λ) k 1 g k (λ) 1 (This is the so-called Bezout identity) Now, define π l := f l (A) r g k (A) Then, the above k l identity implies π 1 (x) + π r (x) = x ( ) 1 The students are allowed to apply to this result without any proof/reference k r

9 In particular, for x E l we obatin π l (x) = x Moreover, one has g l (A) π l = g l (A)f l (A) g k (A) = f l (A)g l (A) Hence imπ l = E l := ker g l (A) for l = 1,, r and thus ( ) yields k l g k (A) = f l (A)µ A (A) 0 k l E E r = K n ( ) Now, it is easy to see that ( ) is actually a direct sum Assume x x r = 0 with x k E k for all k = 1,, r Then it follows for l = 1,, r Hence, we have shown 0 = π l (x x r) = x l E 1 E r = K n ( ) Finally, we show that π l is actually a projection, ie π l π l = π l for l = 1,, r To this end, consider π 1 π 1 = π 1 (I f 2 (A) = π 1 f 1 (A) g k (A) + + f r (A) k 2 ( g k (A) f 2 (A) k 1 k r ) g k (A) g k (A) + + f r (A) k 2 = π 1 ˆf 2 (A)µ A (A) ˆf r (A)µ A (A) = π 1 A similar compuation shows π l π l = π l for l = 1,, r Now, we show that A acts cyclic on the A-invarinat subspace E k, ie k r ) g k (A) span{x k, Ax k,, A d k 1 x k } = E k for all x k E k \ E k with d k := deg ( p k (λ) ) e k First, note that E k E k otherwise one would get a contradiction to the fact that ( p k (λ) ) e k is the minimal polynomial of A Ek Now, assume that V := span{x k, Ax k,, A d k 1 x k } E k Since V is A-invarinat by construction we can consider the minimal polynomial of A V Clearly, µ A V has to be a factor of ( p k (λ) ) e k If µ A V = ( p k (λ) ) l with l < e k we would obtain a contradiction to x k E k Therefore, we conclude µ A V = ( p k (λ) ) e k Now, it is straightforward to see that V = E k because otherwise one would get a contradiction to µ A = χ A (choose any complementary subspace of V and consider a corresponding bloxk matrix representation of A Ek ) Finally, we have to show that x K n is cyclic if and only is π k (x) E k \ E k for all k = 1,, r = : Assume without loss of generaliy π 1 (x) E 1 Then the subspace span{x, Ax,, A n 1 x} is contained in the A-invariant subspace E 1 E 2 E r and thus x is not cyclic =: Let x K n with π k (x) E k \ E k for all k = 1,, r and V := span{x, Ax,, An 1 x} Certainly, V is an A-invariant subspace of K n and the minimal polynomial µ A V is a factor of

10 µ A By the hypothesis π k (x) E k \ E k for all k = 1,, r, we conclude2 that µ A V = µ A Hence the assumption µ A = χ A implies V = E 1 E r = K n Proof of part (a): By the above Lemma 1, we have to show that there exists u K m such that π k (Bu) E k \ E k for all k = 1,, r To this end, consider for k = 1,, r the subspaces U k := (π B) 1 (E k) Then, none of them is equal to U = K m, becasue if for examle U 1 = U were true then imb would be contained in the A-invariant subspace E 1 E 2 E n contradicting the controllability assumption on (A, B) Therefore, by the Lemma 2 (see below) we conclude r U k K m and thus there exists u 0 K m such that π k (Bu 0 ) E k \ E k for all k = 1,, r Hence, (A, b := Bu 0 ) satisfies the desired controllability property Lemma 2: Let K be an infinite field and let V be any K-vector space Then the union of finitely many, proper subspaces of V never coincides with V Proof of the Lemma 2: Suppose the above statement were false Hence, there exists a finite number U 0, U 1,, U n of proper subspaces such that n V = k=0 holds Moreover, assume without loss of generality that none of the subspaces is completely contained in the union of the remaining ones (otherwise, just remove those!) Now, choose y X \ U 0 and x 0 U 0 such that x 0 n U k Moreover, construct (by induction) a sequence (λ i ) i N of scalars which are mutually distinct and different from zero, ie λ i λ j for i j and λ i 0, and consider the linear combinations U k x i := x 0 + λ i y, i N Clearly, one has x i U 0 for all i N, because if x i := x 0 + λ i y would be in U 0 then one could conclude λ 1 y = x i x 0 U 0 But this contradicts the assumption y U 0 Hence, all x i have to be contained in n i=1 U i and thus there exists a subspace U k, k {1,, n} which contains at least to different vectors of the above form, ie x i, x j U k for i j Hence, one has (λ i λ j )x 0 = λ i x j λ j x i U k and thus the contradiction x 0 U k follows This completes the proof (b): Choose A := ( ) A with A := and B := (e 1, e n ) Then a straightforward calculation show that (A, B) is contrallable but (A, Bu) is never controllable (eg due to Hautus) 2 Note that f l (A) r g k (A)x l = x l and thus r g k(a)x l = 0 g l (A)x l k l

11 Aufgabe 27 Zeigen, dass Kontrollierbarkeit von (A, B) asymptotische Kontrollierbarkeit (zum Ursprung) impliziert Proof: Since (A, B) is controllable we can find a T 0 and a control u such that Φ(T, 0, x, u) = 0 Switching to the zero-control for t > T one obtains Φ(t, 0, x, u) = 0 for all t T and therefore (A, B) is asymptotically controllable to the origin Alternative Proof: Apply simply the result of Exercise 28 Remark: Note that, in general, controllability of (A, B) does not imply asymptotic controllability of (A, B) to any state x 0 Consider the following counter-example: ( ) ( ) ( ) A :=, b :=, and x := Then (A, b) is certainly controllable But, if we assume that Φ(t, 0, 0, u) is closed to x then we obtain x Φ(t + 1, 0, 0, u) = e 1 AΦ(t, 0, 0, u) u(t)e 1 (1 u)(t)e 1 e 2 1 Hence, there does not exist a T 0 such that Φ(t, 0, 0, u) stays in a small neighbourhood of x for all t T

12 Aufgabe 28 Zeigen Sie, dass (A, B) C n n C n m genau dann asymptotisch kontrollierbar ist, wenn das folgende modifizierte Hautus-Kriterium gilt: rk ( A λ B ) = n for all λ C with λ 1 Proof: = : Let (A, B) C n n C n m be asymptotically controllable (to the origin) and assume without loss of generality that (A, B) is in Kalman s controllability form, ie ( ) A11 A A = 12 and B = ( B 0 A 1 0 ) 22 with (A 11, B 1 ) controllable By Theorem???, we know that the spectrum of the A 22 is contained in the open unit disc Hence, for λ C with λ 1, we obtain im ( A λ B ) ( ) ( ) ( ) A11 λ A = im 12 B 1 A11 λ B = im 1 A12 + im 0 A 22 λ A 22 λ { ( ) x { = 1 x1 C 1} ( ) n A + 12 (A 22 λ) 1 x 2 x2 C 2} n = C n 0 x 2 and thus rk ( A λ B ) = n for all λ C with λ 1 =: Now, suppose that rk ( A λ B ) = n holds for all λ C with λ 1 Thus, assuming Kalman s controllability form yields ( ) A11 λ A n = rk 12 B 1 0 A 22 λ 0 for all λ C with λ 1 Therefore, we conclude that the spectrum of A 22 has to be contained in the open unit disc and hence Theorem??? implies that (A, B) is asymptotically controllable (to the origin)

13 Aufgabe 29 (a) Seien L F (A, B, C) und L F,0 (A, B, C) wie in Abschnitt IV1 definiert Zeigen Sie die folgenden Aussagen: (i) (x, v, y) L F (A, B, C) (x, F y + v, y) L(A, B, C) (ii) L F,0 (A, B, C) = {(x, u, y) L(A, B, C) u = F y} (b) Zeigen Sie, dass für D 0 Identität (ii) aus Teil (a) im Allgemeinen falsch ist (a): (b):