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
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1
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), and has finite
moments of all orders: for all n 2 N0 we have
Mn =
Z
b
a
xnW (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.
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2
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 a linear
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) = T W (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.
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If W = W (x) is a weight of linear operators on V , we say that W is the
orthogonal direct sum W = W1 · · · Wd if V is the orthogonal direct
sum V = V1 · · · Vd where each subspace Vi is W (x)-invariant for all x
and Wi(x) is the restriction of W (x) to Vi.
We say that a weight W̃ = W̃ (x) of linear operators on V reduces to
scalar weights, if W is equivalent to a direct sum W 0 = W10 · · · WN0 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)P W (y) = W (y)P W (x).
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(1)
4
Proof. Suppose that W (x) = M ⇤(x)M ⇤, ⇤ = ⇤(x) a diagonal weight.
Then P = (M M ⇤) 1 is a positive definite operator such that
W (x)(M M ⇤)
1
W (y) = (M ⇤(x)M ⇤)(M M ⇤)
1
(M ⇤(y)M ⇤)
= M ⇤(x)⇤(y)M ⇤ = W (y)(M M ⇤)
1
W (x).
Conversely, assume (1). Let x0 be such that W (x0) is nonsingular. Let
A be a positive definite operator such that A2 = W (x0). By taking
W 0(x) = A 1W (x)A 1 and P 0 = AP A one sees that we may assume that
W (x0) = I. Hence W (x)P = P W (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.
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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) = U W (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 = W10 · · · WN0 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)P W (y) = W (y)P W (x) and W (x)P = P W (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 = W10 · · · WN0 of orthogonal one dimensional weights.
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In particular, if W (x0) = 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).
✓ 2
◆
x +x x
Example. Let W (x) =
supported in (0, 1). Then
x
x
W (x) =
✓
◆✓ 2
◆✓
◆
1 1
x 0
1 0
.
0 1
0 x
1 1
Therefore W reduces to scalar weights
and ◆
W (x)P W (y) = W (y)P W (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).
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✓
◆
x
x 1
Example. Let W (x) =
supported in (1, 2). Let P =
x
1
1
✓
◆
2 1
. Then P is positive definite and for all x, y
1 2
W (x)P W (y) = W (y)P W (x).
Since W (1) = I, W is unitarily equivalent to a diagonal weight. In fact,
✓
1 1
p
2 1
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◆✓
1
1
◆
✓
x
x 1 1 1
p
x 1
x
2 1
1
1
◆
=
✓
2x
◆
1 0
.
0
1
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 ) : T W (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 2 C and
p 2 R[x] then p(T ) 2 C; if M : V ! Ṽ is a linear isomorphism of inner
product spaces, then C(M W M ⇤) = M C(W )M 1.
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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 W̃ (x) = M W M ⇤ = W̃1 W̃2, Ṽ = Ṽ1 Ṽ2.
Let P be the orthogonal projection onto Ṽ1. Then P W̃ (x) = W̃ (x)P .
Hence P (M W (x)M ⇤) = (M W (x)M ⇤)P and
(M
1
P M )W (x) = W (x)(M ⇤P (M ⇤)
1
) = W (x)(M
1
P M )⇤ .
(ii) implies (iii) The implication is obvious.
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(iii) implies (i) Our first observation is that if T 2 C then 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 Ji of size di with characteristic value i of the form
0
i
B1
B
B·
B
Ji = B
B·
B·
B
@0
0
0
i
·
·
·
0
0
· · ·
· · ·
· · ·
· · ·
0
0
·
·
·
i
1
1
0
0C
C
·C
C
·C
C.
·C
C
0A
i
Let us write W (x) as an s ⇥ s-matrix of blocks Wij (x) of di-rows and
dj -columns. Then, by hypothesis, we have JiWii(x) = Wii(x)Ji⇤.
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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 the
support of W . Therefore i = ¯ i.
Let S and N be, respectively, the semi-simple and the nilpotent parts of
rj
r1
T . The minimal polynomial of T is (x
1 ) · · · (x
j ) where 1 , . . . , j
are the di↵erent eigenvalues of T and ri 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 2 C. The
minimal polynomial of S is p = (x
1 ) · · · (x
j ). Let us consider the
Lagrange polynomials
Y (x
i)
pk =
.
( i
k)
i6=k
Since pk ( i) = ik , and S = S ⇤ it follows that Pi = pi(S) is the orthogonal
projection of Cn onto the i-eigenspace of S. Therefore Pi 2 C.
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Now N is the direct sum of the nilpotent parts Ni of each elementary
Jordan block Ji, i.e.
0
0
B1
B
B·
B
Ni = B
B·
B·
B
@0
0
0
0
·
·
·
0
0
1
· · · 0 0
· · · 0 0C
C
· ·C
C
· ·C
C.
· ·C
C
· · · 0 0A
· · · 1 0
Since N 2 C we have NiWii(x) = Wii(x)Ni⇤. If Ni were a matrix of size
larger than one, and if 1  k  n is the index corresponding to the first row
of 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, hence
N = 0 and T = S.
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If C > RI, then there exists T 2 C with j > 1 di↵erent eigenvalues.
Let M 2 GL(V ) such that S = M T M 1 be a diagonal matrix and let Pi,
1  i  j be the orthogonal projections onto the eigenspaces of S. Let
W̃ (x) = M W (x)M ⇤. Then I = P1 + · · · + Pj , Pr Ps = PsPr = rsPr , 1 
r, s  j and Pr⇤ = Pr , Pr W̃ (x) = W̃ (x)Pr for all 1  r  j. Therefore,
W̃ (x) = (P1 + · · · + Pj )W̃ (x)(P1 + · · · + Pj ) =
= P1W̃ (x)P1 + · · · + Pj W̃ (x)Pj = W̃1(x)
X
Pr W̃ (x)Ps
1r,sj
···
W̃j (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.
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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 . An S-module is called semi-simple if it can be represented as a
sum of simple submodules.
A semi-simple S-module (W, V ) can be represented as the direct sum
V = 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 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 , there exists an invariant subspace V 00
such that V = V 0 V 00. Then V is semi-simple (Chevalley).
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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 = V1 · · · Vj = V10 · · · Vj00 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 Vi is isomorphic to V 0 (i) for all 1  i  j (Chevalley). Clearly, this
uniqueness result can be generalized to the following one: if V = V1 · · · Vj
and V 0 = V10 · · · Vj00 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 Vi is isomorphic to V 0 (i) for
all 1  i  j.
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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) =
✓
2
x +x x
x
x
◆
has no invariant subspace of Cn, i.e. it is simple, but it is equivalent to
W 0(x) =
✓
2
x
0
0
x
◆
which is the direct sum of two scalar weights.
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Theorem.
Let W = W1
···
Wj and W 0 = W10
···
Wj00
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 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 the
permutation . 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)
Vi0 ⇡ V 0 (i) for i < k; c) we have the orthogonal direct sum
V =
M
i<k
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V
0
(i)
M
Vi .
i k
18
L
L
?
Let E = i<k V (i)
V
.
Then
E
is the direct sum of a certain
i
i>k
number of the spaces Vi0. 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 the
Vi0, say E ? = Vi00 . 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 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. Hence j = j 0.
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0
19
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 = W1
···
Wj and W 0 = W10
···
Wj00 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
Wi is equivalent to W 0 (i) for all 1  i  j.
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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) = M W (x)M ⇤
for all x, with M a nonsingular matrix. We may write M = U P in
a 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) =
(U V )D(V ⇤W (x)V )D(U V )⇤. Modulo unitary equivalences we may assume
that W 0(x) = DW (x)D. If we write D = D1 · · · Dj where Di is
a diagonal matrix block of the same size as the matrix block Wi. Then
W 0 = (D1W1D1) · · · (Dj Wj Dj ). From the hypothesis we also have
the representation of W 0 = W10 · · · Wj00 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
DiWiDi ⇡ W 0 (i) for all 1  i  j. Hence Wi ⇠ W 0 (i) for all 1  i  j.
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Theorem. Let W = W (x) be an operator weight equivalent to an
orthogonal direct sum W1 · · · Wd of irreducible weights. If j(T ) is the
number 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 = W10 · · · Wd0 where
Wi0 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 the
corresponding 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}. 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 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
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0
0
D. Then W 0 = W10 · · · Wj(T
) where the size of the block Wi is equal to
the size of the block Di, for all 1  i  j(T ), because DW 0(x) = W 0(x)D.
If some Wi0 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 = W100 · · · Wj00 is a direct sum of matrix
irreducible weights with j(T )  j.
By hypothesis W = W1 · · · Wd. Hence the previous theorem implies
that d = j. Therefore d = j(T ) and the Wi0 are in fact irreducible. Moreover
there exists a permutation of the set {1, . . . , d} such that Wi0 = W (i).
This completes the proof.
We include the following instructive example to get a better
understanding of some concepts and proofs given.
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Example. Let W be the matrix weight with support S = [1, 2] given by
0
1
2x 1 x
W (x) = @ 1 x2 1 A .
x 1 x
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
0
1
s 0 t s
0 A,
T = @0 t
0 0
t
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s, t 2 R.
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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 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
0
1
1 0 1
T = @0 2 0A .
0 0 2
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}.
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Let D = diag(1, 2, 2). Then,
0
1 0
A = @0 1
0 0
1
1
0A
1
and T = A 1DA. From the previous theorem we know that W 0 = AW A⇤
is represented as a direct sum of irreducible matrix weights. In fact
0
1
x 0 0
W (x) ⇠ W 0(x) = @ 0 x2 1 A .
0 1 x
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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
P (x)W (x)Q(x)⇤ dx.
a
(aP + bQ, R) = a(P, R) + b(Q, R),
(T P, Q) = T (P, Q),
(P, Q)⇤ = (Q, P ),
(P, P )
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0;
if (P, P ) = 0 then P = 0.
27
Proposition. Let Vn = {F 2 End(V )[x] : deg F  n} for all n
0,
V 1 = 0 and Vn? 1 = {H 2 Vn : (H, F ) = 0 for all F 2 Vn 1}. Then
Vn? 1 is a free End(V )-left module of dimension one and
(i) Vn = Vn
1
Vn? 1 para todo n
0.
(ii) There is a unique monic polynomial Pn in Vn? 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).
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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 : T W (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 : T M0 = M0T ⇤, T Pn = PnT for all n
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0}.
29
Proof. If T 2 C, then T W (x) = W (x)T ⇤ for a.e. x 2 (a, b). Since
P0 = I it is obvious that T P0 = P0T . Now assume that n 1 and that
T Pj (x) = Pj (x)T for all x and all 0  j  n 1. Then
(PnT, Pj ) =
Z
b
a
Pn(x)T W (x)Pj⇤(x)dx =
Z
b
a
Pn(x)W (x)T ⇤Pj⇤(x)dx
= (Pn, Pj T ) = (Pn, T Pj ) = (Pn, Pj )T ⇤ = 0.
Hence Pn(x)T = APn(x) for some A 2 MatN (C) and all x. Since Pn is
monic it follows that T = A. Therefore by induction on n we have proved
that PnT = T Pn for all n 0. Moreover it is clear that T M0 = M0T ⇤.
Therefore
C ✓ {T : T M0 = M0T ⇤, T Pn = PnT for all n
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0}.
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To prove the reverse inclusion we start with a matrix T such that T M0 =
M0T ⇤ and T Pn = PnT for all n 0. We will first prove by induction on
n 0 that T Mn = MnT ⇤. Thus assume that n 1 and that T Mj = Mj T ⇤
for all 0  j  n 1. If we write Pn = xnI + xn 1An 1 + · · · + A0, then
by hypothesis T Aj = Aj T for all 0  j  n 1.
Now (Pn, P0) = 0 is equivalent to Mn + An
Hence T Mn = MnT ⇤. Therefore
Z
b
xn(T W (x)
W (x)T ⇤) dx =
1 M n 1 + · · · + A0 M 0
for all
n
= 0.
0,
a
which is equivalent to T W (x) = W (x)T ⇤ for almost every x 2 (a, b). Thus
{T : T M0 = M0T ⇤, T Pn = PnT for all n
0} ✓ C,
completing the proof of the theorem.
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