juan tirao - unam · juan tirao v encuentro iberoamericano de polinomios ortogonales y sus...

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

Outline

1. Matrix Weights

2. The Commuting Space

3. Complete Reducibility

4. Orthogonal Polynomials

– Typeset by FoilTE

X – 1

Weight Functions

Let V be a finite dimensional complex inner product vector space. Aweight function W = W (x) of linear operators on V is an integrablefunction 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), and has finitemoments of all orders: for all n 2 N0 we have

Mn =

Z b

axnW (x) dx 2 End(V ).

If {e1, . . . , eN} 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.


X – 2

Let S be an arbitrary set of elements. According to C. Chevalley anS-module on a field K is pair (W,V ) formed by a finite dimensional vectorspace V over K and a mapping W which assigns to every x 2 S a linearoperator W (x) on V . Thus an S-module is an additive group with twodomains 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 0of 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 0such that W 0

(x) = TW (x)T ⇤for all x.

Notice that the equivalence of weights of linear operators does notcoincide with the notion of isomorphism among S-modules.


X – 3

If W = W (x) is a weight of linear operators on V , we say that W is theorthogonal direct sum W = W1 � · · · �Wd if V is the orthogonal directsum V = V1 � · · ·� Vd where each subspace Vi is W (x)-invariant for all xand Wi(x) is the restriction of W (x) to Vi.

We say that a weight ˜W =

˜W (x) of linear operators on V reduces to

scalar weights, if W is equivalent to a direct sum W 0= W 0

1 � · · ·�W 0N of

orthogonal one dimensional weights.

We also say that an operator weight is irreducible when it is notequivalent 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)


X – 4

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⇤)

�1W (y) = (M⇤(x)M⇤)(MM⇤

)

�1(M⇤(y)M⇤

)

= M⇤(x)⇤(y)M⇤= W (y)(MM⇤

)

�1W (x).

Conversely, assume (1). Let x0 be such that W (x0) is nonsingular. LetA be a positive definite operator such that A2

= W (x0). By takingW 0

(x) = A�1W (x)A�1 and P 0= APA one sees that we may assume that

W (x0) = I. Hence W (x)P = PW (x) for all x. Let E be any eigenspaceof P. Then E is W (x)-invariant. Hence W (x) and W (y) restricted toE commute by (1) and are self-adjoint. Therefore W (x) restricted toE are simultaneously diagonalizable through a unitary operator. Sincethis happens for all eigenspaces of P and they are orthogonal we haveproved that W (x) is unitarily equivalent to a direct sum of orthogonal onedimensional weights.


X – 5

Two weights W and W 0of 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 0such 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= W 0

1 � · · ·�W 0N 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, ywe 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 (x0) = I for some x0 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= W 0

1 � · · ·�W 0N of orthogonal one dimensional weights.


X – 6

In particular, if W (x0) = I then W is equivalent to a direct sum oforthogonal one dimensional weights if and only if W (x)W (y) = W (y)W (x)for all x, y (Duran-Grunbaum).

Example. Let W (x) =

✓x2

+ x xx x

◆supported in (0, 1). Then

W (x) =

✓1 1

0 1

◆✓x2

0

0 x

◆✓1 0

1 1

◆.

Therefore W reduces to scalar weights and W (x)PW (y) = W (y)PW (x)

for all x, y 2 (0, 1) with P =

✓1 �1

�1 2

◆. But W is not unitarily

equivalent to a diagonal weight. In fact W (x)W (y) 6= W (y)W (x).


X – 7

Example. Let W (x) =

✓x x� 1

x� 1 1

◆supported in (1, 2). Let P =

✓2 1

1 2

◆. Then P is positive definite and for all x, y

W (x)PW (y) = W (y)PW (x).

Since W (1) = I, W is unitarily equivalent to a diagonal weight. In fact,

1p2

✓1 1

1 �1

◆✓x x� 1

x� 1 x

◆1p2

✓1 1

1 �1

◆=

✓2x� 1 0

0 1

◆.


X – 8

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 valuableinformation on the extend W is equivalent to a direct sum of orthogonalweights.

Another important properties of C are the following ones: if T 2 C andp 2 R[x] then p(T ) 2 C; if M : V ! ˜V is a linear isomorphism of innerproduct spaces, then C(MWM⇤

) = MC(W )M�1.


X – 9

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 2 C;

(iii) C > RI.

Proof.

(i) implies (ii) Suppose that ˜W (x) = MWM⇤=

˜W1� ˜W2, ˜V =

˜V1� ˜V2.Let P be the orthogonal projection onto ˜V1. Then P ˜W (x) =

˜W (x)P .Hence P (MW (x)M⇤

) = (MW (x)M⇤)P and

(M�1PM)W (x) = W (x)(M⇤P (M⇤)

�1) = W (x)(M�1PM)

⇤.

(ii) implies (iii) The implication is obvious.


X – 10

(iii) implies (i) Our first observation is that if T 2 C then its eigenvaluesare real. In fact, we may assume that W is a matrix weight function of sizen, and by changing W by an equivalent weight we may assume that T isin Jordan canonical form. Thus T is the direct sum of elementary Jordanmatrices Ji of size di with characteristic value �i of the form

Ji =

0

BBBBBBBB@

�i 0 · · · 0 0

1 �i · · · 0 0

· · · ·· · · ·· · · ·0 0 · · · �i 0

0 0 · · · 1 �i

1

CCCCCCCCA

.

Let us write W (x) as an s ⇥ s-matrix of blocks Wij(x) of di-rows anddj-columns. Then, by hypothesis, we have JiWii(x) = Wii(x)J

⇤i .


X – 11

Thus, if 1 k n is the index corresponding to the first row of Wii(x),we have �iwkk(x) = wkk(x)¯�i. But wkk(x) > 0 for almost all x in thesupport of W . Therefore �i =

¯�i.

Let S and N be, respectively, the semi-simple and the nilpotent parts ofT . The minimal polynomial of T is (x��1)

r1 · · · (x��j)rj where �1, . . . ,�j

are the di↵erent eigenvalues of T and ri is the greatest dimension of theJordan blocks with eigenvalue �i. The proof of the Jordan canonical formreveals that S and N are real polynomials in T . Thus S,N 2 C. Theminimal polynomial of S is p = (x � �1) · · · (x � �j). Let us consider theLagrange polynomials

pk =

Y

i 6=k

(x� �i)

(�i � �k).

Since pk(�i) = �ik, and S = S⇤ it follows that Pi = pi(S) is the orthogonalprojection of Cn onto the �i-eigenspace of S. Therefore Pi 2 C.


X – 12

Now N is the direct sum of the nilpotent parts Ni of each elementaryJordan block Ji, i.e.

Ni =

0

BBBBBBBB@

0 0 · · · 0 0

1 0 · · · 0 0

· · · ·· · · ·· · · ·0 0 · · · 0 0

0 0 · · · 1 0

1

CCCCCCCCA

.

Since N 2 C we have NiWii(x) = Wii(x)N⇤i . If Ni were a matrix of size

larger than one, and if 1 k n is the index corresponding to the first rowof Wii(x), we would have wkk(x) = 0 for all x, which is a contradiction.Therefore all elementary Jordan matrices of T are one dimensional, henceN = 0 and T = S.


X – 13

If C > RI, then there exists T 2 C with j > 1 di↵erent eigenvalues.Let M 2 GL(V ) such that S = MTM�1 be a diagonal matrix and let Pi,1 i j be the orthogonal projections onto the eigenspaces of S. Let˜W (x) = MW (x)M⇤. Then I = P1 + · · ·+ Pj, PrPs = PsPr = �rsPr, 1 r, s j and P ⇤

r = Pr, Pr˜W (x) = ˜W (x)Pr for all 1 r j. Therefore,

˜W (x) = (P1 + · · ·+ Pj)˜W (x)(P1 + · · ·+ Pj) =

X

1r,sj

Pr˜W (x)Ps

= P1˜W (x)P1 + · · ·+ Pj

˜W (x)Pj =˜W1(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.


X – 14

Complete Reducibility

Let (W,V ) be an abstract S-module. Then (W,V ) is said to be simpleif it is of positive dimension and if the only invariant subspaces of V are{0} and V . An S-module is called semi-simple if it can be represented as asum of simple submodules.

A semi-simple S-module (W,V ) can be represented as the direct sumV = V1 � · · · � Vj of a collection � = {Vi} of simple S-submodules.Moreover, if we have a representation of this kind, and if V 0 is any invariantsubspace of V , then there exists a subcollection �0 of � such that V is thedirect 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: ifV 0 is any invariant subspace of V , there exists an invariant subspace V 00

such that V = V 0 � V 00. Then V is semi-simple (Chevalley).


X – 15

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 orthogonalcomplement. 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 = V1 � · · · � Vj = V 01 � · · · � V 0

j0 be two representations of anabstract semi-simple S-module V as a direct sum of simple submodules.Then we have j = j0 and there exists a permutation � of the set {1, . . . , j}such that Vi is isomorphic to V 0

�(i) for all 1 i j (Chevalley). Clearly, thisuniqueness result can be generalized to the following one: if V = V1�· · ·�Vj

and V 0= V 0

1 � · · ·�V 0j0 are two isomorphic S-modules represented as direct

sums of simple submodules, then we have j = j0 and there exists apermutation � of the set {1, . . . , j} such that Vi is isomorphic to V 0

�(i) forall 1 i j.


X – 16

This last statement is not true for operator weights on an inner productspace V , because, as we pointed out at the beginning, the equivalenceamong weights has a di↵erent meaning than the isomorphism of S-modules.

Example. The matrix weight

W (x) =

✓x2

+ x xx x

◆

has no invariant subspace of Cn, i.e. it is simple, but it is equivalent to

W 0(x) =

✓x2

0

0 x

◆

which is the direct sum of two scalar weights.


X – 17

Theorem. Let W = W1 � · · · � Wj and W 0= W 0

1 � · · · � W 0j0

be representations of the operator weights W and W 0as orthogonal

direct sums of simple (irreducible) weights. If W and W 0are unitarily

equivalent, then we have j = j0 and there exists a permutation � of the set

{1, . . . , j} such that Wi is unitarily equivalent to W 0�(i) for all 1 i j.

Proof. It is enough to consider the case W = W 0. We shall construct thepermutation �. Suppose that �(i) is already defined for i < k (1 k j)and has the following properties: a) �(i) 6= �(i0) for i < i0 < k; b)V 0i ⇡ V 0

�(i) for i < k; c) we have the orthogonal direct sum

V =

M

i<k

V 0�(i) �

M

i�k

Vi.


X – 18

Let E =

Li<k V

0�(i)�

Li>k Vi. Then E? is the direct sum of a certain

number of the spaces V 0i . On the other hand E? is unitarily isomorphic to

V/E, i.e to Vk. It follows that E? is simple and therefore E? is one of theV 0i , say E?

= V 0i0. Since V 0

�(i) ⇢ E for i < k, we have i0 6= �(i) for i < k.We define �(k) = i0. It is clear that the function �(i) now defined fori < k + 1, satisfies conditions a), b), c) with k replaced by k + 1. Because� is injective on the set {1, . . . , j} we must have j0 � j. Since the twodecompositions play symmetric roles, we also have j � j0. Hence j = j0.


X – 19

If we come back to our previous example we realize that a matrix weightmay not be expressible as direct sum of irreducible matrix weights. Butsuch a weight is equivalent to one that is the direct sum of two irreducibleweights. Taking into account this fact and our Main Theorem we make thefollowing 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 iscompletely reducible.

Theorem.Let W = W1 � · · · � Wj and W 0= W 0

1 � · · · � W 0j0 be

representations of the operator weights W and W 0as orthogonal direct

sums of irreducible weights. If W and W 0are equivalent, then we have

j = j0 and there exists a permutation � of the set {1, . . . , j} such that

Wi is equivalent to W 0�(i) for all 1 i j.


X – 20

Proof. Modulo unitary equivalence we may assume from the beginning thatW 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 ina unique way with U unitary and P positive definite, and P = V DV ⇤

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 assumethat W 0

(x) = DW (x)D. If we write D = D1 � · · · � Dj where Di isa diagonal matrix block of the same size as the matrix block Wi. ThenW 0

= (D1W1D1) � · · · � (DjWjDj). From the hypothesis we also havethe representation of W 0

= W 01 � · · · � W 0

j0 as an orthogonal direct sumof irreducible weights. Now we are ready to apply our previous theoremto conclude that j = j0 and that there exists a permutation � such thatDiWiDi ⇡ W 0

�(i) for all 1 i j. Hence Wi ⇠ W 0�(i) for all 1 i j.


X – 21

Theorem. Let W = W (x) be an operator weight equivalent to an

orthogonal direct sum W1� · · ·�Wd of irreducible weights. If j(T ) is thenumber of distinct eigenvalues of T 2 C, then d = max{j(T ) : T 2 C}.Moreover, W is equivalent to a matrix weight W 0

= W 01� · · ·�W 0

d where

W 0i is the restriction of W 0

to one of the eigenspaces of a diagonal matrix

D 2 C. Besides j(T ) is the degree of the minimal polynomial of T .

Proof. Suppose that W = W1 � · · · � Wd and let P1, . . . , Pd be thecorresponding orthogonal projections. Define T = �1P1 + · · · + �dPd with�1, . . . ,�d all di↵erent. Then clearly T 2 C. Hence d max{j(T ) : T 2 C}.

Conversely, let T 2 C such that j(T ) = max{j(T ) : T 2 C}. Modulounitary equivalence we may assume that W is a matrix weight. In the proofof the Main Theorem we established that T is semi-simple. Thus we maywrite T = A�1DA with D a diagonal matrix. Let W 0

(x) = AW (x)A⇤.Then W 0 ⇠ W and D 2 C(W 0

). We may assume that D = D1�· · ·�Dj(T )

where Di is the Jordan diagonal block corresponding to the eigenvalue �i of


X – 22

D. Then W 0= W 0

1� · · ·�W 0j(T ) where the size of the block W 0

i is equal tothe size of the block Di, for all 1 i j(T ), because DW 0

(x) = W 0(x)D.

If some W 0i were not irreducible we could replace it, modulo equivalence,

by a direct sum of matrix irreducible weights. Thus there exists a matrixweight W 00 ⇠ W 0 such that W 00

= W 001 � · · ·�W 00

j is a direct sum of matrixirreducible weights with j(T ) j.

By hypothesis W = W1� · · ·�Wd. Hence the previous theorem impliesthat d = j. Therefore d = j(T ) and theW 0

i are in fact irreducible. Moreoverthere exists a permutation � of the set {1, . . . , d} such that W 0

i = W�(i).This completes the proof.

We include the following instructive example to get a betterunderstanding of some concepts and proofs given.


X – 23

Example. Let W be the matrix weight with support S = [1, 2] given by

W (x) =

0

@2x 1 x1 x2

1

x 1 x

1

A .

The S-module (W,C3) 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 2 C if and only if

T =

0

@s 0 t� s0 t 0

0 0 t

1

A , s, t 2 R.


X – 24

The characteristic polynomial of T is det(� � T ) = (� � s)(� � t)2

and the minimal polynomial p(�) is: p(�) = (� � s)(� � t) if s 6= t andp(�) = (� � t) when s = t. Thus every T 2 C is diagonalizable, as

expected, and max{j(T ) : T 2 C} = 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 =

0

@1 0 1

0 2 0

0 0 2

1

A .

The eigenvalues of T are 1 and 2 and the corresponding eigenspaces are:

V1 = Ce1 and V2 = Ce2�C(e1+e3). Let A be the matrix which changes

the basis {e1, e2, e1 + e3} into the basis {e1, e2, e3}.


X – 25

Let D = diag(1, 2, 2). Then,

A =

0

@1 0 �1

0 1 0

0 0 1

1

A

and T = A�1DA. From the previous theorem we know that W 0= AWA⇤

is represented as a direct sum of irreducible matrix weights. In fact

W (x) ⇠ W 0(x) =

0

@x 0 0

0 x21

0 1 x

1

A .


X – 26

Orthogonal Polynomials

Let W be weight function of linear operators on V . In End(V )[x] weintroduce the following sesquilineal Hermitian form:

(P,Q) =

Z b

aP (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 ) = 0 then P = 0.


X – 27

Proposition. Let Vn = {F 2 End(V )[x] : degF n} for all n � 0,

V�1 = 0 and V ?n�1 = {H 2 Vn : (H,F ) = 0 for all F 2 Vn�1}. Then

V ?n�1 is a free End(V )-left module of dimension one and

(i) Vn = Vn�1 � V ?n�1 para todo n � 0.

(ii) There is a unique monic polynomial Pn in V ?n�1, and deg(Pn) = n.

Corollary. Every sequence {Qn} of orthogonal polynomials in End(V )[x]is of the form Qn = AnPn where An 2 GLN(C).


X – 28

To round o↵ our exposition on reducibility of an operator weight W onthe real line, we give an alternative description of the commuting space C.

Whenever one considers equality among integrable or measurablefunctions, 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 matrixweight W is

C = {T : TW (x) = W (x)T ⇤ for a.e. x}.

Theorem. If W is an operator weight on (a, b), let {Pn}n�0 be the

sequence of monic orthogonal polynomials and let M0 be the moment of

order zero of W . Then

C = {T : TM0 = M0T⇤, TPn = PnT for all n � 0}.


X – 29

Proof. If T 2 C, then TW (x) = W (x)T ⇤ for a.e. x 2 (a, b). SinceP0 = I it is obvious that TP0 = P0T . Now assume that n � 1 and thatTPj(x) = Pj(x)T for all x and all 0 j n� 1. Then

(PnT, Pj) =

Z b

aPn(x)TW (x)P ⇤

j (x)dx =

Z b

aPn(x)W (x)T ⇤P ⇤

j (x)dx

= (Pn, PjT ) = (Pn, TPj) = (Pn, Pj)T⇤= 0.

Hence Pn(x)T = APn(x) for some A 2 MatN(C) and all x. Since Pn ismonic it follows that T = A. Therefore by induction on n we have provedthat PnT = TPn for all n � 0. Moreover it is clear that TM0 = M0T

⇤.Therefore

C ✓ {T : TM0 = M0T⇤, TPn = PnT for all n � 0}.


X – 30

To prove the reverse inclusion we start with a matrix T such that TM0 =

M0T⇤ and TPn = PnT for all n � 0. We will first prove by induction on

n � 0 that TMn = MnT⇤. Thus assume that n � 1 and that TMj = MjT

⇤

for all 0 j n� 1. If we write Pn = xnI + xn�1An�1 + · · ·+ A0, thenby hypothesis TAj = AjT for all 0 j n� 1.

Now (Pn, P0) = 0 is equivalent to Mn+An�1Mn�1+ · · ·+A0M0 = 0.Hence TMn = MnT

⇤. Therefore

Z b

axn

(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 : TM0 = M0T⇤, TPn = PnT for all n � 0} ✓ C,

completing the proof of the theorem.


X – 31

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