ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Séria I: PRACE MATEMATYCZNE XXX (1991)
Ng u y e n Va n Ma u
(Warszawa)
Conditions for polynomials in right inverses with stationary and algebraic coefficients
to be Yolterra operators
Abstract. The conditions for polynomials in right inverses with constant coefficients were obtained by Przeworska-Rolewicz and von Trotha (see Th. 2.4.1 in [6]). In the present note we generalize this result to the case of polynomials with algebraic and stationary coefficients.
Moreover, we also give a necessary and sufficient condition for
R1 + R2, R\Ri
to be Volterra operators, provided R 2, R 2 are Volterra right inverses of a right invertible operator D. The method used here is essentially based on the properties of generalized algebraic operators.0. Let X be a linear space over C. Denote by L(X) the set of all linear operators with domains and ranges in X. Let L0(X) = {A eL(X): domzl = X}.
Denote by R{X) the set of all right invertible operators belonging to L(X). For a DeR(X), we write
= { R e L
0(X): DR = I},
= {.F
eL 0(X): F
2= F, F X = kerD, 3Re@ D FR = 0}.
If F e and FR = 0 for an R e &D then F is called an initial operator corresponding to R. In the sequel, we assume kerD Ф 0.
Following [6] we say that an operator A e L
0(X) is a Volterra operator if for every scalar t the operator I — tR is invertible. We denote by V(X) the set of all Volterra operators from L
0(X).
De f in it io n
1 [8]. An operator A e L
0(X) is said to be stationary if DA = AD on domD, RA = AR.
For given D e R (X ) and R
e@
dwe denote by SDtR the set of all stationary operators.
De f in it io n
2 [7]. We say that an operator A e L
0(X) is algebraic if there exists a non-zero polynomial
P{t) =
P 0 + P i t + . . . + P Nt N,p
0, . . . , p NeC,
such that P(A) = 0.
Without loss of generality we can assume that pN = 1, i.e. P(t) is unitary.
We say that the algebraic operator A is of order N if no unitary polynomial (2(0 of degree m < N such that Q(A) = 0 on X exists. The minimal unitary polynomial annihilating A is called the characteristic polynomial of A and will be denoted by PA(t). Denote by sé the set of all algebraic operators.
De f in it io n
3 [1]. Let 38 c L0(X) be a commutative algebra. An operator S
eL
q(X) is said to be generalized algebraic over 38 if there exists a polynomial
Af(0 = A 0 + A l t + ... + Antn, An Ф 0,
with A0, ..., A„e38, AjS = SAj (j = 0, ..., n) such that M(S) = 0.
The following results were obtained by Przeworska-Rolewicz and von Trotha (see [6], Th. 2.4.1).
Th e o r e m
I. Write
(1) Q(t, s) = £ è (0 = Q(t, 1), P(t) = tMQ(t),
k = 0
where q0, ..., q^-x eC , qN = 1, M is a non-negative integer.
(i) (Przeworska-Rolewicz). I f there exists an Re3êDn V(X) (Le. R is a Volterra right inverse of D) then P(D)e R(X) and
(2) i^o = ^ M+iV[ 0 ( / , ^ ) ] _1 is a Volterra inverse of P(D).
(ii) (von Trotha). I f R 0 of the form (2) is a Volterra operator then R is a Volterra operator.
In the present note we generalize Theorem I to the case when Q(t, s) is a polynomial with algebraic and stationary coefficients (Theorems 1, 2 and 3).
Moreover, we give conditions for R l + R 2, R l R 2 to be Volterra operators provided that R lt R 2 are Volterra right inverses of a D
eR(X) (Theorems 9,10).
The method used here is essentially based on the properties of generalized algebraic operators obtained in [l]-[5 ].
Th e o r e m
II [2]. Let S be an algebraic operator with characteristic polynomial of the form
p s(t) = f t ( t - t y
i = 1
Ch # tj for i r1+ ... +r„ = N 0, t j c C j = 1 ,..., и), and m
V:= V(S)= ^ AjSm- j
7 = 1
where A
jE38 (see Definition 3), AjS = SAj, j = 1, ..., m. Then V is a generalized
algebraic operator.
Theorem
III
[3].Let A and В be commuting algebraic operators with characteristic polynomials
П
PA(t) = П ui ^ u j f ° r i # j ( i j = !>•••, n),
i= 1
m
р в( 0 = П ^ vi for {i, j m),
j= i
respectively. Then A + B is an algebraic operator with characteristic roots belonging to the set
{ щ + Vf.i =
1,n; j =
1,. . . m}.
1. Let D
eR{X), R e M D. Let A0, A
n ej^ n S DtR be mutually com
muting and let A
n= J.
Write
(4) Q(t,
S)= £ 0 (0 = Q(t, 1), p(t) =
7 = 0
where M is a non-negative integer.
Theorem
1. I f R
eV(X) then the operator Q(I, R) is invertible and
(5) R
0= R N+MIQ(I, V(X).
P ro o f. Write
^ 0 = Нп{Я*} (k = 0, 1,...).
Then J '
qc= L0(X) is a commutative algebra. Hence, by Theorem II, Q(I, R) is a generalized algebraic operator with characteristic roots belonging to the set
N
№) {/+ I k = 1 , ,
k= 1
where { ^ _ м , • ••, ^-к,гк -к)к
=1are the characteristic roots of the operators
^N-k (k = 1 ,..., ЛГ), respectively.
Theorem I implies that every operator I + Y
j!=
itN- ktjRk is invertible. It follows that Q(I, R) is invertible (cf. [2]).
Now we prove that R ± = R N[Q(I, Я)]-1 is a right inverse of Q(D). Indeed, Q(D)R
1= Q(D)RN[Q(I, Я )]"1 = £ Я)]-1 = /.
k = 0
Consequently, the operator P(D) = DMQ(D) is also right invertible and has a right inverse of the form (5).
To finish the proof, we have to check that R
0eV(X). For tE C we
obtain I - t R
0= [Q(I, P )]_1H(P), where H(R) = YJk=oAkRN~k- t R N+M. By
Theorem II, H(R) is a generalized algebraic operator with characteristic roots
belonging to ,the set
(7) {I + tR”+M + £ tN. k,jRk; j = 1 ,.... rN_t}.
k = 0
Theorem I shows that every operator I + tRN + M + YJk=otN-kjRk is invertible.
Hence, H(.R) is invertible (cf. [2]). We conclude that I — tR0, as a superposition of invertible operators, is again invertible, i.e.
,R0 e F (X ).Theorem 2.
Suppose that R e 0lD
nV
(X)for a D eR(X). Suppose, moreover, that Aestf n SD R. Then AR e V (X).
P ro o f. By Theorem II, I-{-AAR is a generalized algebraic operator over
= lin{Rk} (к = 0, 1, ...) with characteristic roots of the form {/ + 2uff?: i
= 1 ,..., n}. Since every operator / + Au}R is invertible for all А e C we conclude that I + AAR is invertible for all ДеС, i.e. AR is a Volterra operator.
The converse of Theorem 1 is given by the following
Theorem 3.
Let DeR(X), R
eMd. Let A 0, ..., ANe j / n S DtR be mutually commuting and let AN = I. Suppose that Q{t, s), Q(t), P(t) are of the form (4). I f
Q(i, R) is invertible then
(8) R 0 = RM+Nm i , R ) T 1^^PiD).
Moreover, if R 0eV (X ) then ReV(X).
P ro o f. It is enough to check that R e V(X), provided that the operator R0 of the form (8) is a Volterra operator. Fix ДеС. Write A = ()(Д). Then A is an algebraic operator (cf. [6]) and AP(D) = P(D)A, AR0 = R 0A. By our assump
tion, R0 is a Volterra operator. This implies that so is AR 0 (Theorem 2), and I — A R 0 is invertible.
On the other hand,
I - A R 0 = I - A R M+N = IQ(I, R ) - A R M+N][Q(I, R )YK From the last equality we obtain
/ = ( I - A R or l m , Д)]-1 [0 (/, R ) - A R M+*-]
= IQ(I, R ) - A R M+NW - A R or 1lQ(I, R ) T \ i.e. Q(I, R) — ARM+N is invertible.
Write
HA(t, s) = Q(t, s ) - A s M+N, HA(t) = HA(t, 1).
Then H
a(A) = HA(Д, 1) = Q(A, 1) — A = Q(A) — A = 0 and HA(I, R) is invertible for HA(I, R) = Q(I, R ) ~ A R m+n.
Hence
HA(t) = (t-AI)Q A(t), where QA(t) = £ Bjtj
j = о
(9)
and B jeS D>R n jrf(j = 0, . . N — 1) are mutually commuting. From (9) we get HA(t, s) = (t-Às)QA(t, s),
where
Q
a&
s)= V B j t W - 1 4 QA(t) = 0 Ж 1).
7=i
Thus HA(I, R) = R), i.e.
I = ( / - А К Ш / , Л)[ЯА(/, Я )]"1 = [Я А(/, я ) Г ^ ( J , Я)(1-ЛЯ).
This shows that I — XR is invertible for all XeC, i.e. ReV(X).
Co r o l l a r y 1.
Suppose that DeR(X), R e $ Dn V ( X ) and A e S DyR
nsf.
Then every solution of the equation
(10) (D — A)x = y, y e X
is of the form
x = (I — AR)~1(Ry + z), where zekerD is arbitrary.
In particular, if A has the characteristic polynomial of the form П
PA(t) = П (f - ^ ) (*« ф h f ° r
1 7=1t/геп the solutions of (10) are of the form П
x — (/ — tjR)~
1Pj(Ry + z), zekerD , 7=i
where
P ,= П ] = 1 ,.... n.
k= l,k¥>j
Co r o l l a r y
2. dim ker 0(D) = iV dim kerf).
Indeed,
Q(D) = DN ^ AjRN~j = DNQ(I, R).
7 = 0
By Theorem 1, the operator Q(I, R) is invertible. Hence dim ker Q(D)
= dim ker DN = N dim ker D.
As a corollary, we obtain the formula
dim DMQ{D) = (M + AO dim ker D.
R em ark 1. In general, the converse to Theorem 2 is not true. Further
more, for every D
eR { X ) ,R
eM
dthere exists an Aestf n S DyR such that
12 - Comment. Math. 30.2
AR e V (X), i.e. AR e V (X) does not imply R e V (X). Indeed, if A e stf n SDtR and A 2 — 0 then I — XAR is invertible for all Xe C and {I — XAR)"1 = I+ XAR. This implies A R e V ( X ) for every Re@tD. However, we have the following
Corollary 3.
Suppose that DeR(X), Re£%D. Suppose, moreover, that A e s / n S DiR and A is invertible. I f AR is a Volterra operator then so is R.
Indeed, by Theorem 2, A ~ 1(AR) = R is Volterra, provided that AReV(X).
2. Algebraic exponentials.
Definition 4.
If A e jrf n SD<R
andО ф xA e
ker(D—A)
thenx A
is said to be analgebraic exponential element corresponding to A.
It is easy to check that if О Ф x^eker (I — AR) for an A e S D>Rn sé then xAeker(D — A), i.e. x A is an algebraic exponential and FxA = 0, where F is an initial operator for D corresponding to the right inverse R. By Theorems 1 and 3, in that case, R is not a Volterra operator.
Theorem 4.
Suppose that {An} c SDfR n sé is a sequence of algebraic operators such that Aj, А { — А-} are invertible for i Ф j . Then for an arbitrary positive integer n the algebraic exponentials xAl, ..., x An are linearly independent over the set srf r\S D>R.
P ro o f. By Definition 4, x Al Ф 0. Suppose that for a fixed к (к ^ 1) the algebraic exponentials x Al, ..., x Ak are linearly independent over sé n S DtR. If x Ak+1 is linearly dependent on {x^, •••,xAk} then there exists an operator 0 ф dk + 1
gsé n SDtR such that
This implies (
11)
fc + i
E
dj x Aj> dj<=jrf r \ S DtR.j = i
k+ 1
0 = D (E ^
j x A j)» d j e s / r\SDtR,
j = i
and
(12)fe + i
° = Z dJÂ l XAj- 7=1
The equalities (11) and (12) together imply
0 = Z dM j - A i)*
a, = Z d№ j ~ A i)x Aj-
j = 1 J = 2
By our assumption, dj(Aj — A 1) = 0 (j = 2, . . . , k + \ ) and the operators
Aj — A iij = 2, ..., k+1) are invertible. Hence dj = 0 {j = 2, ..., k+ 1), which
contradicts our assumption. We therefore conclude that the set {xAl, ..., xAk+1} is linearly independent over sé r\S DtR for an arbitrary k e N.
Theorem 5.
I f RDn V
(X)for a D e R(X), A e s é
nSDtR and
dim kerA —
0,then the operator eA = (I — AR)~i has no eigenvectors.
P ro o f. Suppose that there exists а и ф 0 such that for a t ф 0 we have eAu = tu. This implies
(13) [(\ — t)I + tRA~\u = 0.
By Theorem 2, AR e V(X). Hence, by (13), и — 0 for t Ф 1. This contradicts the assumption. If t — 1 then from (13) we get ARu = 0. Our assumption, that dim ker A = 0 implies и = 0, again yields a contradiction.
Definition 5.
If D
eR(X) and there is an R e 01D n V
(X),then every operator eA = (I — AR)- 1 , where A e SDtR n stf, is called an algebraic exponen
tial operator.
Theorem6
. I f R e f%D
nV
(X),A e SDtR
ns f and F is an initial operator for D corresponding to R, then the algebraic exponentials eA(z) are uniquely determined by their initial values:
eA(z) = ( I - A R ) ~
1F(eA{z)), i.e. F(eA(z)) = z for zeker D.
P ro o f. Suppose that zekerD . Then (I — AR)eA{z) = z, i.e. eA(z)
= z + AReA{z) and DeA(z) = AeA(z). This means that eA(z) is an algebraic eigenvector of D corresponding to A. Thus
FeA(z) = (I-R D )e A(z) = eA(z)-R D eA(z)
= eA( z )-R A e A{z) = (I - A R ) e A(z), i.e. F(eA{z)) = z.
Corollary 4.
I f D eR(X), R e MD
пV
(X),then for every operator 0
té A e S DR n sé there exist non-trivial exponentials.
Corollary 5.
Suppose that DeR(X), {Rv}veIo <= n
F(X )and A s s # n SD'R. Then
(Fv- F J e A(z) = AFvR/ieA(z) for
zek erZ ),D, v, p e l 0.
Indeed,
AFvR„eAz) = р Л цЛеА(2) = FvR»DeA(Z)
= F f l - F J e A z ) = (Fv — F VF f)eA(z) = {Fy- F f}eA{z).
Definition
6. If D eR(X ) and there is an R e ^ Dn V(X), then the operators
£ a ^ ( e i A F e —î a) , s a 2 j ( @ i A F e —i A ) ,
where Aestf n S D R, are called the algebraic cosine and sine operator, respec
tively. The elements cA(z) and sA(z), where zekerZ), are called the algebraic cosine and sine elements, respectively.
Theorem 7.
The algebraic cosine and sine operators have the following properties:
c a
= {I + A 2R 2)~1,
sa= A(I + A 2R 2)~1R, c2 A + s2 A = eiAe - iA, (14)
casb+
cbsa= i(eiAeiB- e - iAe - iB),
CA CB ~ S A S B = 2.(e i A e iB + e - i A e ~ i B ) -
P r o o f is the same as for “scalar” cosine and sine operators (cf. [6], p. 99).
Corollary
6. DcA = —A
sa,
Dsa= A
ca.
Theorem 8.
Let F be an initial operator for a D
eR(X) corresponding to R e & Dr\V(X). Then for every A e s / n S D<R and O ^ z e k e r D we have F
sa(
z) = 0 and cA(z) ф 0.
P ro o f. By (14),
F
sa(
z) = FA(I + A 2R2)~1R
z= FRA{I + A 2R 2)~1z = 0.
If cA(z) = 0 then z = (/ + A 2R 2)(I + A 2R 2)~1z = (I+
A 2R 2)ca(z)= 0. which contradicts our assumption that z ф 0.
3. Operations on Volterra right inverses. Let De
R{X), R r, R 2
eM
dn V(X). It is well known that R XR 2 Ф P 2^ i (cf- [6]). The following question arises: Are R
i_R2 and R 1 + R 2 Volterra operators, provided that R1? R 2 are Volterra operators? In general, the answer in negative. In order for R
i+ R 2, ^
1^2to be Volterra operators, we obtain the following conditions:
Theorem 9.
Let D
eR{X), R x, R 2
e$ D. Then R2 is a Volterra operator if and only if so is R 2R 1.
P ro o f. Suppose that R 1R 2
eV(X). Write
C
rxR
z= У ^
1^
2) ’ E = I t R 2eRlR2R l , tE C.
Then E is well defined on X and
(15) (I tR 1R 2)E = (I tR2R l)(I + tR2eBiB2R 1)
— I — tR2R l + t(I — tR2R 1)R2eRiR2R 1
= / — tR2R 1 + tR2{I — tR^R2)eRlRiR^ — I — tR2R^-\-tR2Ri = /.
Similarly, we find
(16)
(I — tR2R})E = /.
Thus I~ tR
2R , is invertible for all te C , i.e. R
2R i is a Volterra operator.
Th e o r e m
10. Suppose that D eR(X) and R t , R
2are Volterra right inverses of D. Then a necessary and sufficient condition for R t R
2to be a Volterra operator is that
(17) F2(I — tR2)~1z
Ф0, 0 ^ zekerD , where F
2is an initial operator for D corresponding to R 2.
P ro o f. Note that R
1R
2e& D2and R
2is a Volterra right inverse of D2.
Hence, if R i R
2has an eigenvector then it must be of the form (18) q = (I — t R l y
1z, zekerD 2, z ф 0.
If v
Фt, veC then
и — (I — vR t R 2)(I — tR2) ~1 z = U - t R 2 l + R
1{tRl - v R
2f t ( I - t R 2l)~l z
= z + R^tRj^ — vR2)(I — tRl)~l z.
Hence D2u = (t — v)(I — tR2)~1z ф 0 for all 0 ^ zek erD 2, i.e. R t R
2has no eigenvectors.
If v = teC , then we have и = (/ — tR
1R 2)(I — tRj)~
1z. Consider two cases:
(i) zekerD and (ii) zek e rD 2\kerD.
(i) In this case, we get
Fxu = F l ( I - t R l ) ~ 1z = l I - t R l + R ^ t R ^ D m i - t R l y ' z
= z + R y t R ^ - D y i - t R l y ' z .
Since domD = kerD + /?1X (cf. [6]), we conclude that F
jM # 0 for 0
ФzekerD . Thus
и Ф0, i.e. R t R
2has no eigenvector in kerD.
(ii) Let z = R
1z
1+ z2, where z1? z2ekerD , z l
Ф0. It is easy to check that и = ( I - t R ^ J q = (I — R l R
2D2)(I — tR2)~l z
= z + tR
1F
2R 1(I — t R l y
1z.
If z2 Ф 0 then F^u = F
1R l z
1+ F
1z
2— z
2ф 0, which implies и ф 0. If
z2— 0, i.e. z = R l zi , then
u = R
1z
1+ tR
1F
2R
21( I - t R l y
1z l = R
1z
1+ R l F
2[I—(I—tRlJ] (I — t R j y
1Zl
= R
1zl + R
1F2(I — tR
2)~
1zl — R
1F
2z l = R
1F2(I — tR
2)~
1z
1Ф0.
Hence, и
Ф 0for all v eC and zek erD 2 if and only if the condition (17) is
satisfied, which was to be proved. »
Similarly, we obtain the following
Th e o r e m
10'. I f R 1 and R 2 are Volterra right inverses of D
eR(X) then a necessary and sufficient condition for R X R 2 to be a Volterra operator is that (19) FX(I — tR2)~l z ф 0 for all te C , zekerT),
where Fx is an initial operator for D corresponding to R x.
Ex a m p l e
1. Let X = të([0, 1], where F = C or F — R, and let
d x x
D
= —, R t = $, R 2 =
J ,x x Ф x 2i x x,
x2g[0 , 1].XI X 2
It is easy to check that
(7 — tR2 ) ~1 с = c cos y / —t(x — Xj) for
c eF, j = l , 2. Hence
F 2(I — tR2)~1 c = ccoSyf--t(x2 — x 1).
If t0 = — ^(
x2 —
x x)~2
k2 then F2(I — toR 2)~1c = 0. Theorem 10 shows that R XR2 is not a Volterra operator.
Th e o r e m 11.
Suppose that D
eR(X) and R x, R 2
en V
(X ).Then a neces
sary and sufficient condition for R x + R 2 to be a Volterra operator is that (20) (7 — tjR1)_1z + (7 — tR2)~l z Ф 0 for all zeker7)\{0}.
P ro o f. Let R = %(RX+ R 2). Then R
eMd. Hence every non-trivial eigen
vector of R (if exists) must be of the form
q = (I — tRf)~x z, 0 # z e k e r D . If v
eC and v ф t then
и = (I — vR)q = I - ^ v ( R x + R 2) ( I - t R x)~xz
= I — tR x + (t — %v)(Rx — jv R 2)(I — tR x)~l z
= z + ((t — v/2)R1—(v/2)R2)(I — tR 1)~1z.
This implies that
Du = (t — v){I — tRx)~xz # 0 for 0 # zekerD . If v = t
eC, then
и = (I — tR)q = z + (t/2)(Rx—R 2)(I — tR x)~xz = ^z + ^(I — tR2)(I — tR x)~xz.
Hence
2 и = z + (7 — tR2)(I — zR x)~x z, i.e.
2(1 — tR2)~xu = (I — tR x)~x z + (I — tR2)~x z .
From this equality we conclude that и Ф 0 if and only if the right hand side of this last equality is different from zero, i.e. we get the condition (20).
Ex a m p l e
2. Let X , D, R lt R
2be the same as in Example 1. It is easy to check that
(/ — = cexp[t(x — x})~\ f o r c e d ,
j= 1,2.
Hence
u(x) = (I — tR 1)~1c + (I — tR2)~1c = cetx[e~tXl +etX2].
This equality implies that u(x) Ф 0 for all t e R. By Theorem 11, ^(R
1+ R 2) is a Volterra operator in the space X = ^([0, 1], R).
Note that in the space X = ^ [ [ 0 , 1], C ], u(x) = 0 for t = (x2 — x ^ 1^ . Hence, by Theorem 11, R t + R
2is not a Volterra operator.
References
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INSTITUTE OF MATHEMATICS, TECHNICAL UNIVERSITY OF WARSAW PL. POLITECHNIKI 1, 00-661 WARSZAWA, POLAND