Reducibility of Matrix Weights

Transcription

1 Reducibility of Matrix Weights Juan Tirao V Encuentro Iberoamericano de Polinomios Ortogonales y sus Aplicaciones Instituto de Matemáticas CU, UNAM 8-12 de junio de 2015

2 Outline 1. Matrix Weights 2. The Commuting Space 3. Complete Reducibility 4. Orthogonal Polynomials Typeset by FoilTEX 1

3 Weight Functions Let V be a finite dimensional complex inner product vector space. A weight function W = W (x) of linear operators on V is an integrable function on an interval (a, b), such that W (x) is positive semi-definite (resp. positive definite) for all (resp. almost all) x 2 (a, b), andhasfinite moments of all orders: for all n 2 N 0 we have M n = Z b a x n W (x) dx 2 End(V ). If {e 1,...,e N } is an orthonormal basis in V and we represent each W (x) by a matrix of size N we obtain a matrix weight on the real line. Typeset by FoilTEX 2

4 Let S be an arbitrary set of elements. According to C. Chevalley an S-module on a field K is pair (W, V ) formed by a finite dimensional vector space V over K and a mapping W which assigns to every x 2 S alinear operator W (x) on V. Thus an S-module is an additive group with two domains of operators, the field K and the set S. In particular, if W = W (x) is a weight of linear operators on V,then (W, V ) is an S-module, where S is the support of W. Two weights W and W 0 of linear operators on V and V 0, respectively, defined on the same interval are equivalent, W W 0, if there exists an isomorphism T of V onto V 0 such that W 0 (x) =TW(x)T for all x. Notice that the equivalence of weights of linear operators does not coincide with the notion of isomorphism among S-modules. Typeset by FoilTEX 3

5 If W = W (x) is a weight of linear operators on V,wesaythatW is the orthogonal direct sum W = W 1 W d if V is the orthogonal direct sum V = V 1 V d where each subspace V i is W (x)-invariant for all x and W i (x) is the restriction of W (x) to V i. We say that a weight W = W (x) of linear operators on V reduces to scalar weights, ifw is equivalent to a direct sum W 0 = W 0 1 W 0 N of orthogonal one dimensional weights. We also say that an operator weight is irreducible when it is not equivalent to a direct sum of matrix weights of smaller size. Theorem. A weight W of linear operators on V reduces to scalar weights if and only if there is a positive definite operator P such that for all x, y W (x)pw(y) =W (y)pw(x). (1) Typeset by FoilTEX 4

6 Proof. Suppose that W (x) =M (x)m, = (x) a diagonal weight. Then P =(MM ) 1 is a positive definite operator such that W (x)(mm ) 1 W (y) =(M (x)m )(MM ) 1 (M (y)m ) = M (x) (y)m = W (y)(mm ) 1 W (x). Conversely, assume (1). Let x 0 be such that W (x 0 ) is nonsingular. Let A be a positive definite operator such that A 2 = W (x 0 ). By taking W 0 (x) =A 1 W (x)a 1 and P 0 = AP A one sees that we may assume that W (x 0 )=I. Hence W (x)p = PW(x) for all x. Let E be any eigenspace of P. Then E is W (x)-invariant. Hence W (x) and W (y) restricted to E commute by (1) and are self-adjoint. Therefore W (x) restricted to E are simultaneously diagonalizable through a unitary operator. Since this happens for all eigenspaces of P and they are orthogonal we have proved that W (x) is unitarily equivalent to a direct sum of orthogonal one dimensional weights. Typeset by FoilTEX 5

7 Two weights W and W 0 of linear operators on V and V 0 defined on the same interval are unitarily equivalent, W W 0, if there is a unitary isomorphism U of V onto V 0 such that W 0 (x) =UW(x)U for all x. Theorem. The following conditions are equivalent: (i) A weight W of linear operators on V is unitarily equivalent to a direct sum W 0 = W1 0 WN 0 of orthogonal one dimensional weights. (ii) For all x, y we have W(x)W(y)=W(y)W(x); (iii) There exists a positive definite operator P such that for all x, y we have W (x)pw(y) =W (y)pw(x) and W (x)p = PW(x). Corollary. Let W = W (x) be a weight of linear operators on V such that W (x 0 )=I for some x 0 in the support of W. Then W reduces to scalar weights if and only if W is unitarily equivalent to a direct sum W 0 = W1 0 WN 0 of orthogonal one dimensional weights. Typeset by FoilTEX 6

8 In particular, if W (x 0 )=I then W is equivalent to a direct sum of orthogonal one dimensional weights if and only if W (x)w (y) =W (y)w (x) for all x, y (Durán-Grünbaum). x Example. Let W (x) = 2 + x x supported in (0, 1). Then x x W (x) = 1 1 x x 1 1 Therefore W reduces to scalar weights and W (x)pw(y) =W (y)pw(x) 1 1 for all x, y 2 (0, 1) with P =. But W is not unitarily 1 2 equivalent to a diagonal weight. In fact W (x)w (y) 6= W (y)w (x). Typeset by FoilTEX 7

9 x x 1 Example. Let W (x) = supported in (1, 2). Let P = x Then P is positive definite and for all x, y 1 2 W (x)pw(y) =W (y)pw(x). Since W (1) = I, W is unitarily equivalent to a diagonal weight. In fact, x x 1 p x 1 x p2 1 1 = 2x Typeset by FoilTEX 8

10 The Commuting Space Let W be a weight of linear operators on the inner product space V. We define the commuting space C of W by C = {T 2 End(V ):TW(x) =W (x)t for all x}. The commuting space of W is a real vector space that contains valuable information on the extend W is equivalent to a direct sum of orthogonal weights. Another important properties of C are the following ones: if T 2Cand p 2 R[x] then p(t ) 2C;ifM : V! Ṽ is a linear isomorphism of inner product spaces, then C(MWM )=MC(W)M 1. Typeset by FoilTEX 9

11 Main Theorem. Let W = W (x) be a weight of linear operators on the inner product space V. Then the following conditions are equivalent: (i) W is equivalent to a direct sum of operator weights of smaller size; (ii) there is a non trivial idempotent Q 2C; (iii) C > RI. Proof. (i) implies (ii) Suppose that W (x) =MWM = W 1 W2, Ṽ = Ṽ1 Ṽ 2. Let P be the orthogonal projection onto Ṽ1. Then P W (x) = W (x)p. Hence P (MW(x)M )=(MW(x)M )P and (M 1 PM)W (x) =W (x)(m P (M ) 1 )=W (x)(m 1 PM). (ii) implies (iii) The implication is obvious. Typeset by FoilTEX 10

12 (iii) implies (i) Our first observation is that if T 2Cthen its eigenvalues are real. In fact, we may assume that W is a matrix weight function of size n, and by changing W by an equivalent weight we may assume that T is in Jordan canonical form. Thus T is the direct sum of elementary Jordan matrices J i of size d i with characteristic value i of the form J i = 0 1 i i 0 0. B 0 0 i 0 A i Let us write W (x) as an s s-matrix of blocks W ij (x) of d i -rows and d j -columns. Then, by hypothesis, we have J i W ii (x) =W ii (x)j i. Typeset by FoilTEX 11

13 Thus, if 1 apple k apple n is the index corresponding to the first row of W ii (x), we have iw kk (x) =w kk (x) i. But w kk (x) > 0 for almost all x in the support of W.Therefore i = i. Let S and N be, respectively, the semi-simple and the nilpotent parts of T. The minimal polynomial of T is (x 1) r 1 (x j) r j where 1,..., j are the di erent eigenvalues of T and r i is the greatest dimension of the Jordan blocks with eigenvalue i. The proof of the Jordan canonical form reveals that S and N are real polynomials in T. Thus S, N 2C. The minimal polynomial of S is p =(x 1) (x j). Let us consider the Lagrange polynomials p k = Y (x i) ( i k ). i6=k Since p k ( i )= ik,ands = S it follows that P i = p i (S) is the orthogonal projection of C n onto the i-eigenspace of S. ThereforeP i 2C. Typeset by FoilTEX 12

14 Now N is the direct sum of the nilpotent parts N i of each elementary Jordan block J i,i.e. N i = B A Since N 2Cwe have N i W ii (x) =W ii (x)n i. If N i were a matrix of size larger than one, and if 1 apple k apple n is the index corresponding to the first row of W ii (x), we would have w kk (x) =0for all x, which is a contradiction. Therefore all elementary Jordan matrices of T are one dimensional, hence N =0and T = S. Typeset by FoilTEX 13

15 If C > RI, then there exists T 2C with j > 1 di erent eigenvalues. Let M 2 GL(V ) such that S = MTM 1 be a diagonal matrix and let P i, 1 apple i apple j be the orthogonal projections onto the eigenspaces of S. Let W (x) =MW(x)M.ThenI = P P j, P r P s = P s P r = rs P r, 1 apple r, s apple j and P r = P r, P r W (x) = W (x)pr for all 1 apple r apple j. Therefore, W (x) =(P P j ) W (x)(p P j )= X 1appler,sapplej P r W (x)ps = P 1 W (x)p1 + + P j W (x)pj = W 1 (x) Wj (x), completing the proof that (iii) implies (i). Hence the theorem is proved. Corollary. Let W = W (x) be an operator weight function. Then W is irreducible if and only if its commuting space C = RI. Typeset by FoilTEX 14

16 Complete Reducibility Let (W, V ) be an abstract S-module. Then (W, V ) is said to be simple if it is of positive dimension and if the only invariant subspaces of V are {0} and V.AnS-module is called semi-simple if it can be represented as a sum of simple submodules. Asemi-simpleS-module (W, V ) can be represented as the direct sum V = V 1 V j of a collection ={V i } of simple S-submodules. Moreover, if we have a representation of this kind, and if V 0 is any invariant subspace of V, then there exists a subcollection 0 of such that V is the direct sum of V 0 and of the sum of the submodules belonging to 0. Conversely, let V be an S-module which has the following property: if V 0 is any invariant subspace of V,thereexistsaninvariantsubspaceV 00 such that V = V 0 V 00.ThenV is semi-simple (Chevalley). Typeset by FoilTEX 15

17 Proposition. Let W be a weight of linear operators on V with support S. Then the S-module (W, V ) is semi-simple. Proof. Let V 0 be any invariant subspace of V and let V 00 be its orthogonal complement. Then, since W (x) is self-adjoint for all x, V 00 is invariant. Therefore the S-module (W, V ) is semi-simple. Let V = V 1 V j = V1 0 Vj 0 0 be two representations of an abstract semi-simple S-module V as a direct sum of simple submodules. Then we have j = j 0 and there exists a permutation of the set {1,...,j} such that V i is isomorphic to V 0 (i) for all 1 apple i apple j (Chevalley). Clearly, this uniqueness result can be generalized to the following one: if V = V 1 V j and V 0 = V1 0 Vj 0 0 are two isomorphic S-modules represented as direct sums of simple submodules, then we have j = j 0 and there exists a permutation of the set {1,...,j} such that V i is isomorphic to V 0 (i) for all 1 apple i apple j. Typeset by FoilTEX 16

18 This last statement is not true for operator weights on an inner product space V, because, as we pointed out at the beginning, the equivalence among weights has a di erent meaning than the isomorphism of S-modules. Example. The matrix weight W (x) = x 2 + x x x x has no invariant subspace of C n, i.e. it is simple, but it is equivalent to W 0 (x) = x x which is the direct sum of two scalar weights. Typeset by FoilTEX 17

19 Theorem. Let W = W 1 W j and W 0 = W1 0 Wj 0 0 be representations of the operator weights W and W 0 as orthogonal direct sums of simple (irreducible) weights. If W and W 0 are unitarily equivalent, then we have j = j 0 and there exists a permutation of the set {1,...,j} such that W i is unitarily equivalent to W 0 (i) for all 1 apple i apple j. Proof. It is enough to consider the case W = W 0. We shall construct the permutation. Suppose that (i) is already defined for i<k(1apple k apple j) and has the following properties: a) (i) 6= (i 0 ) for i<i 0 <k; b) Vi 0 V 0 (i) for i<k; c) we have the orthogonal direct sum V = M i<k V 0 (i) M V i. i k Typeset by FoilTEX 18

20 Let E = L i<k V 0 L (i) i>k V i.thene? is the direct sum of a certain number of the spaces Vi 0. On the other hand E? is unitarily isomorphic to V/E, i.etov k. It follows that E? is simple and therefore E? is one of the Vi 0,sayE? = Vi 0 0.SinceV 0 (i) E for i<k,wehavei 0 6= (i) for i<k. We define (k) =i 0. It is clear that the function (i) now defined for i<k+1, satisfies conditions a), b), c) with k replaced by k +1.Because is injective on the set {1,...,j} we must have j 0 j. Since the two decompositions play symmetric roles, we also have j j 0.Hencej = j 0. Typeset by FoilTEX 19

21 If we come back to our previous example we realize that a matrix weight may not be expressible as direct sum of irreducible matrix weights. But such a weight is equivalent to one that is the direct sum of two irreducible weights. Taking into account this fact and our Main Theorem we make the following definition. Definition. We say that an operator weight is completely reducible if it is equivalent to an orthogonal direct sum of irreducible operators weight. Observe that the Main Theorem implies that every operator weight is completely reducible. Theorem.Let W = W 1 W j and W 0 = W1 0 Wj 0 0 be representations of the operator weights W and W 0 as orthogonal direct sums of irreducible weights. If W and W 0 are equivalent, then we have j = j 0 and there exists a permutation of the set {1,...,j} such that W i is equivalent to W 0 (i) for all 1 apple i apple j. Typeset by FoilTEX 20

22 Proof. Modulo unitary equivalence we may assume from the beginning that W and W 0 are matrix weights of the same size. Let W 0 (x) =MW(x)M for all x, with M a nonsingular matrix. We may write M = UP in auniquewaywithu unitary and P positive definite, and P = VDV where V is unitary and D is a positive diagonal matrix. Then W 0 (x) = (UV)D(V W (x)v )D(UV). Modulo unitary equivalences we may assume that W 0 (x) =DW(x)D. If we write D = D 1 D j where D i is a diagonal matrix block of the same size as the matrix block W i. Then W 0 =(D 1 W 1 D 1 ) (D j W j D j ). From the hypothesis we also have the representation of W 0 = W1 0 Wj 0 0 as an orthogonal direct sum of irreducible weights. Now we are ready to apply our previous theorem to conclude that j = j 0 and that there exists a permutation such that D i W i D i W 0 (i) for all 1 apple i apple j. HenceW i W 0 (i) for all 1 apple i apple j. Typeset by FoilTEX 21

23 Theorem. Let W = W (x) be an operator weight equivalent to an orthogonal direct sum W 1 W d of irreducible weights. If j(t ) is the number of distinct eigenvalues of T 2C, then d = max{j(t ):T 2C}. Moreover, W is equivalent to a matrix weight W 0 = W1 0 Wd 0 where Wi 0 is the restriction of W 0 to one of the eigenspaces of a diagonal matrix D 2C. Besides j(t ) is the degree of the minimal polynomial of T. Proof. Suppose that W = W 1 W d and let P 1,...,P d be the corresponding orthogonal projections. Define T = 1 P d P d with 1,..., d all di erent. Then clearly T 2C.Henced apple max{j(t ):T 2C}. Conversely, let T 2Csuch that j(t ) = max{j(t ):T 2C}. Modulo unitary equivalence we may assume that W is a matrix weight. In the proof of the Main Theorem we established that T is semi-simple. Thus we may write T = A 1 DA with D a diagonal matrix. Let W 0 (x) =AW (x)a. Then W 0 W and D 2C(W 0 ).WemayassumethatD = D 1 D j(t ) where D i is the Jordan diagonal block corresponding to the eigenvalue i of Typeset by FoilTEX 22

24 D. ThenW 0 = W1 0 Wj(T 0 ) where the size of the block W i 0 is equal to the size of the block D i,forall1 apple i apple j(t ), becausedw 0 (x) =W 0 (x)d. If some Wi 0 were not irreducible we could replace it, modulo equivalence, by a direct sum of matrix irreducible weights. Thus there exists a matrix weight W 00 W 0 such that W 00 = W1 00 Wj 00 is a direct sum of matrix irreducible weights with j(t ) apple j. By hypothesis W = W 1 W d. Hence the previous theorem implies that d = j. Therefored = j(t ) and the Wi 0 are in fact irreducible. Moreover there exists a permutation of the set {1,...,d} such that Wi 0 = W (i). This completes the proof. We include the following instructive example to get a better understanding of some concepts and proofs given. Typeset by FoilTEX 23

25 Example. Let W be the matrix weight with support S =[1, 2] given by 0 W (x) 2x 1 x 1 1 x 2 1A. x 1 x The S-module (W, C 3 ) is simple i.e. there are no proper invariant subspaces. In this case this is equivalent to showing that there is no one dimensional invariant subspace. This follows easily. An easy computation leads to: T 2C if and only if 0 T s 0 t s 1 0 t 0 A, s,t 2 R. 0 0 t Typeset by FoilTEX 24

26 The characteristic polynomial of T is det( T )=( s)( t) 2 and the minimal polynomial p( ) is: p( )=( s)( t) if s 6= t and p( ) =( t) when s = t. Thus every T 2Cis diagonalizable, as expected, and max{j(t ):T 2C}=2. Therefore we know that W is equivalent to a direct sum of two irreducible matrix weights, one of size one and the other of size two. Take T = A The eigenvalues of T are 1 and 2 and the corresponding eigenspaces are: V 1 = Ce 1 and V 2 = Ce 2 C(e 1 + e 3 ). Let A be the matrix which changes the basis {e 1, e 2, e 1 + e 3 } into the basis {e 1, e 2, e 3 }. Typeset by FoilTEX 25

27 Let D = diag(1, 2, 2). Then, A = A and T = A 1 DA. From the previous theorem we know that W 0 = AW A is represented as a direct sum of irreducible matrix weights. In fact W (x) W 0 (x) = x x 2 1A. 0 1 x Typeset by FoilTEX 26

28 Orthogonal Polynomials Let W be weight function of linear operators on V. In End(V )[x] we introduce the following sesquilineal Hermitian form: (P, Q) = Z b a P (x)w (x)q(x) dx. (ap + bq, R) =a(p, R)+b(Q, R), (TP,Q)=T(P, Q), (P, Q) =(Q, P ), (P, P) 0; if (P, P) =0then P =0. Typeset by FoilTEX 27

29 Proposition. Let V n = {F 2 End(V )[x] :degf apple n} for all n 0, V 1 =0and V n? 1 = {H 2 V n :(H, F) =0 for all F 2 V n 1 }. Then V n? 1 is a free End(V )-left module of dimension one and (i) V n = V n 1 V? n 1 para todo n 0. (ii) There is a unique monic polynomial P n in V? n 1, anddeg(p n )=n. Corollary. Every sequence {Q n } of orthogonal polynomials in End(V )[x] is of the form Q n = A n P n where A n 2 GL N (C). Typeset by FoilTEX 28

30 To round o our exposition on reducibility of an operator weight W on the real line, we give an alternative description of the commuting space C. Whenever one considers equality among integrable or measurable functions, f = g means f(x) = g(x) for almost every x (for a.e. x). In particular, the precise definition of the commuting space of a matrix weight W is C = {T : TW(x) =W (x)t for a.e. x}. Theorem. If W is an operator weight on (a, b), let {P n } n 0 be the sequence of monic orthogonal polynomials and let M 0 be the moment of order zero of W. Then C = {T : TM 0 = M 0 T,TP n = P n T for all n 0}. Typeset by FoilTEX 29

31 Proof. If T 2C, then TW(x) =W (x)t for a.e. x 2 (a, b). Since P 0 = I it is obvious that TP 0 = P 0 T. Now assume that n 1 and that TP j (x) =P j (x)t for all x and all 0 apple j apple n 1. Then (P n T,P j )= Z b a P n (x)tw(x)p j (x)dx = Z b =(P n,p j T )=(P n,tp j )=(P n,p j )T =0. a P n (x)w (x)t P j (x)dx Hence P n (x)t = AP n (x) for some A 2 Mat N (C) and all x. Since P n is monic it follows that T = A. Therefore by induction on n we have proved that P n T = TP n for all n 0. Moreover it is clear that TM 0 = M 0 T. Therefore C {T : TM 0 = M 0 T,TP n = P n T for all n 0}. Typeset by FoilTEX 30

32 To prove the reverse inclusion we start with a matrix T such that TM 0 = M 0 T and TP n = P n T for all n 0. We will first prove by induction on n 0 that TM n = M n T.Thusassumethatn 1 and that TM j = M j T for all 0 apple j apple n 1. IfwewriteP n = x n I + x n 1 A n A 0,then by hypothesis TA j = A j T for all 0 apple j apple n 1. Now (P n,p 0 )=0is equivalent to M n + A n 1 M n A 0 M 0 =0. Hence TM n = M n T.Therefore Z b a x n (TW(x) W (x)t ) dx = for all n 0, which is equivalent to TW(x) =W (x)t for almost every x 2 (a, b). Thus {T : TM 0 = M 0 T,TP n = P n T for all n 0} C, completing the proof of the theorem. Typeset by FoilTEX 31