On normal operator logarithms

0.14 MB·

by  Eduardo Chiumiento

On normal operator logarithms
Eduardo Chiumiento ∗
Abstract
3
1 Let X,Y be normal bounded operators on a Hilbert space such that eX = eY. If the
0
spectraofX andY arecontainedinthestripS ofthecomplexplanedefinedby|Im(z)|≤π,
2
we show that |X| = |Y|. If Y is only assumed to be bounded, then |X|Y = Y|X|. We
n
give a formula for X −Y in terms of spectral projections of X and Y provided that X,Y
a
J are normal and eX = eY. If X is an unbounded self-adjoint operator, which does not have
(2k+1)π, k ∈ Z, as eigenvalues, and Y is normal with spectrum in S satisfying eiX = eY,
4
then Y ∈ {eiX}′′. We give alternative proofs and generalizations of results on normal

] operator exponentials proved by Ch. Schmoeger.
A
AMS classification: 47B15; 47A60.
F
.
h Keywords: Exponential map, normal operator, spectral theorem.
t
a
m 1 Introduction
[
Solutions to the equation eX = eY were studied by E. Hille [1] in the general setting of

1 unital Banachalgebras. Under the assumption that the spectrum σ(X) of X is incongruent
v
(mod2πi),whichmeansthatσ(X)∩σ(X+2kπi)=∅forallk =±1,±2,...,he provedthat
7
XY =YX and there exist idempotents E ,E ,...,E commuting with X and Y such that
9 1 2 n
7
n n
0
X−Y =2πi k E , E =I, E E =δ ,
. j j j i j ij
1 X X
j=1 j=1
0
3 wherek ,k ,...,k aredifferentintegers. Ifthehypothesisonthespectrumisremoved,itis
1 2 n
1
possible to find non commuting logarithms (see e.g. [1, 6]). In the setting of Hilbert spaces,
:
v when X is a normal operator, the above assumption on the spectrum can be weakened. In
Xi fact, Ch. Schmoeger[5]provedthatX belongsto the doublecommutantofY providedthat
E (σ(X)∩σ(X+2kπi))=0, k =1,2,..., where E is the spectralmeasure of X. We also
r X X
a refer to [3] for a generalization of this result by F. C. Paliogiannis.
In this paper, we study the operator equation eX = eY in the setting of Hilbert spaces
under the assumption that the spectra of X and Y belong to a non-injective domain of the
complex exponential map. Our results include the relation between the modulus of X and
Y (Theorem 3.1), a formula for the difference of two normal logarithms in terms of their
spectral projections (Theorem 4.1) and commutation relations when X is a skew-adjoint
unbounded operator (Theorem 5.1). The proofs of these results are elementary. In fact,
they rely on the spectral theorem for normal operators. This approach allows us to give a
generalization(Corollary 4.2) and an alternative proof (Corollary 3.2) of two results by Ch.
Schmoeger (see [6]).
∗Partially supported by InstitutoArgentino deMatema´tica ‘Alberto P. Caldero´n’ and CONICET
1
2 Notation and preliminaries
Let (H,h·, ·i) be a complex Hilbert space and B(H) be the algebra of bounded operators
on H. The spectrum of an operator X is denoted by σ(X), and the set of eigenvalues of X
is denoted by σ (X). The real part of X ∈B(H) is Re(X)= 1(X +X∗) and its imaginary
p 2
part is Im(X)= 1(X −X∗).
2
If X is a bounded or unbounded normal operator on H, we denote by E the spectral
X
measure of X. Recall that E is defined on the Borel subsets of σ(X), but we may think
X
thatE isdefinedonallthe BorelsubsetsofC. Indeed,wecansetE (Ω)=E (Ω∩σ(X))
X X X
for every Borel set Ω ⊆ C. Our first lemma is a generalized version of [4, Ch. XII Ex. 25],
where the normal operator can now be unbounded.
Lemma 2.1. Let X be a (possibly unbounded) normal operator on H and f a bounded Borel
function on σ(X). Then
E (Ω)=E (f−1(Ω)),
f(X) X
for every Borel set Ω⊆C.
Proof. WedefineaspectralmeasurebyE′(Ω)=E (f−1(Ω)),whereΩisanyBorelsubsetof
X
C. We aregoingtoshowthatE′ =E . Sincef isbounded, itfollowsthatf(X)∈B(H).
f(X)
Moreover,the operator f(X) is given by
hf(X)ξ,ηi= f(z)dE (z),
Z Xξ,η
C
where ξ,η ∈H and E is the complex measure defined by E (Ω)=hE (Ω)ξ,ηi (see
Xξ,η Xξ,η X
[4, Theorem 12.21]). By the change of measure principle ([4, Theorem 13.28]), we have
zE′ (z)= f(z)E (z).
Z ξ,η Z Xξ,η
C C
ThereforeE′ satisfiestheequation zE′ (z)=hf(X)ξ,ηi,whichuniquelydeterminesthe
C ξ,η
spectral measure of f(X) (see [4, TRheorem 12.23]). Hence E′ =E .
f(X)
The following lemma was first proved in [6, Corollary 2]. See also [3, Corollary 3] for
another proof. We give below a proof for the sake of completeness, which does not depend
on further results of these articles.
Lemma 2.2. Let X and Y be normal operators in B(H). If eX =eY, then Re(X)=Re(Y).
Proof. The following computation was done in [6]:
eX+X∗ =eXeX∗ =eX(eX)∗ =eY(eY)∗ =eYeY∗ =eY+Y∗,
wherethe firstandlastequalitiesholdbecauseX andY arenormal. Now wemayfinishthe
proof in a different fashion: note that the exponential map, restricted to real axis, has an
inverselog:R →R. Sinceσ(X+X∗)⊆Randσ(eX+X∗)⊆R ,wecanusethecontinuous
+ +
functional calculus to get X +X∗ =log(eX+X∗)=log(eY+Y∗)=Y +Y∗.
Throughout this paper, we use the following notation for subsets of the complex plane:
• Ω +iΩ ={x+iy : x∈Ω , y ∈Ω }, where Ω , i=1,2, are subsets of R.
1 2 1 2 i
• For short, we write R+ia for the set R+i{a}.
• We write S for the complex strip {z ∈C : −π ≤ Im(z)≤π}, and S◦ for the interior
of S.
2
Lemma 2.3. Let X,Y be normal operators in B(H) such that σ(X) ⊆ S and σ(Y) ⊆ S.
Then eX =eY if and only if the following conditions hold:
i) E (Ω)=E (Ω) for all Borel subsets Ω of S◦.
X Y
ii) Re(X)=Re(Y).
Proof. Suppose that eX = eY. Let Ω be a Borel measurable subset of S◦. By the spectral
mapping theorem,
σ(eX)={eλ : λ∈σ(X)}={eµ : µ∈σ(Y)}=σ(eY).
It is well-known that the restriction of the complex exponential map exp|S◦ is bijective.
Therefore we have σ(X)∩Ω=σ(Y)∩Ω, and by Lemma 2.1,
E (Ω)=E (Ω∩σ(X))=E (exp−1(exp(Ω∩σ(X))))
X X X
=EeX(exp(Ω∩σ(X)))=EeY(exp(Ω∩σ(Y)))=EY(Ω),
which proves i). On the other hand, ii) is proved in Lemma 2.2.
To prove the converse assertion, we first note that
E (R−iπ)+E (R+iπ)=I −E (S◦)=I−E (S◦)
X X X Y
=E (R−iπ)+E (R+iπ),
Y Y
sinceσ(X)⊆S,σ(Y)⊆S andE (S◦)=E (S◦). DuetothefactthatE andE coincide
X Y X Y
on Borel subsets of S◦, we find that
ezdE (z)= ezdE (z).
Z X Z Y
S◦ S◦
Hence we get
eX = ezdE (z)=− eRe(z)dE (z)− eRe(z)dE (z)+ ezdE (z)
Z X Z X Z X Z X
S R+iπ R−iπ S◦
=−eRe(X)(E (R+iπ)+E (R−iπ))+ ezdE (z)
X X Z X
S◦
=−eRe(Y)(E (R+iπ)+E (R−iπ))+ ezdE (z)=eY.
Y Y Z Y
S◦
Remark 2.4. We have shownthat E (R−iπ)+E (R+iπ)=E (R−iπ)+E (R+iπ),
X X Y Y
whenever X,Y are normalbounded operatorssuch that σ(X)⊆S, σ(Y)⊆S and eX =eY.
Theorem 2.5. (S. Kurepa [2])LetX ∈B(H)suchthateX =N isanormaloperator. Then
X =N +2πiW,
0
where N =log(N) and log is the principal (or any) branch of the logarithm function. The
0
bounded operator W commutes with N and there exists a bounded and regular, positive def-
0
inite self-adjoint operator Q such that W =Q−1WQ is a self-adjoint operator the spectrum
0
of which belongs to the set of all integers.
3
3 Modulus and square of logarithms
Now we show the relation between the modulus of two normal logarithms with spectra
contained in S.
Theorem 3.1. Let X be a normal operator in B(H). Assume that σ(X)⊆S and eX =eY.
i) If Y is normal in B(H) and σ(Y)⊆S, then |X|=|Y|.
ii) If Y ∈B(H), then |X|Y =Y|X|.
Proof. i) We will prove that the spectral measures of |Im(X)| and |Im(Y)| coincide. Let us
set A = Im(X) and B = Im(Y). Given Ω ⊆ [0,π), put Ω′ = {x∈ R : |x| ∈ Ω}. Note that
R+iΩ′ ⊆S◦. As an application of Lemma 2.1 and Lemma 2.3, we see that
E (Ω)=E (Ω′)=E (R+iΩ′)=E (R+iΩ′)=E (Ω′)=E (Ω).
|A| A X Y B |B|
By Remark 2.4, we have
E ({π})=E ({−π, π})=E (R−iπ)+E (R+iπ)
|A| A X X
=E (R−iπ)+E (R+iπ)=E ({π}).
Y Y |B|
Thus, we have proved E = E , which implies that |A| = |B|. On the other hand, by
|A| |B|
Lemma 2.3, we know that Re(X)=Re(Y). Therefore
|X|2 =Re(X)2+|A|2 =Re(Y)2+|B|2 =|Y|2.
Hence |X|=|Y|, and the proof is complete.
ii) Since X is a normal operator, eX = eY is also a normal operator. Then by a result
by S. Kurepa (see Theorem 2.5), there exist operators N and W such that N is normal,
0 0
eX = eN0, W commutes with N and Y = N +2πiW. In fact, N can be defined using
0 0 0
the Borel functional calculus by N = log(eX), where log is the principal branch of the
0
logarithm. In particular, this implies that σ(N ) ⊆ S. Now we can apply i) to find that
0
|N | = |X|. Since N W = WN , we have |N |W = W|N |, and this gives W|X| = |X|W.
0 0 0 0 0
Hence |X|Y =Y|X|.
Followingsimilararguments,wecangiveanalternativeproofofaresultbyCh. Schmoeger
([6, Theorem 3]). This result was originally proved using inner derivations. Note that a mi-
nor improvement on the assumption on σ(X) over the boundary ∂S of the strip S can now
be done. Given a set Ω⊆C, we denote by Ω¯ the set {x−iy : x+iy ∈Ω}.
Corollary 3.2. Let X be a normal operator in B(H), σ(X) ⊆S, Y ∈ B(H) and eX = eY.
Suppose that for every Borel subset Ω⊆∂S\{−iπ, iπ}, it holds that E (Ω¯)=0, whenever
X
E (Ω)6=0. Then X2Y =YX2.
X
Proof. We will show that EX2(Ω0) commutes with Y for every Borel subset Ω0 ⊆ σ(X2).
From the equation eX =eY, we have eXY =YeX, and thus, EeX(Ω)Y =YEeX(Ω) for any
Borel set Ω. Since the set Ω is arbitrary,by Lemma 2.1 we get
1. E (Ω′)Y =YE (Ω′) for every subset Ω′ ⊆S◦.
X X
2. (E (Ω′)+E (Ω¯′))Y =Y(E (Ω′)+E (Ω¯′)), whenever Ω′ ⊆∂S.
X X X X
On the other hand, the image of S by the analytic map f(z)=z2 is given by
f(S)={u±i2t u+t2 : u∈[−π2,∞), u+t2 ≥0}.
p
4
Let us write f−1(Ω ) = (−Ω′)∪Ω′, where Ω′ is a subset of the half-plane Re(z) ≥ 0 and
0
−Ω′ denotes the set {−z : z ∈ Ω}. We need to consider three cases. In the case in which
Ω0 ⊆ f(S)◦, then f−1(Ω0) ⊆ S◦. It follows that EX2(Ω0) = EX(−Ω′)+EX(Ω′), and by
the item 1. above we have EX2(Ω0)Y = YEX2(Ω0). In the case where Ω0 ⊆∂f(S)\{π2},
we have either E (Ω′) = 0 or E (Ω¯′) = 0 by our assumption on the spectral measure of
X X
X. Similarly, it must be either E(−Ω′) = 0 or E (−Ω¯′) = 0. Therefore item 2. above
X
reducestothedesiredconclusion,i.e. EX2(Ω0)Y =YEX2(Ω0). Finally,ifΩ0 ={−π2},then
EX2(Ω0) = EX({−iπ})+EX({iπ}) commutes with Y by item 2., and this concludes the
proof.
4 Difference of logarithms
Let X,Y be normal operators and k ∈ Z. In order to avoid lengthly formulas, let us fix a
notation for some special spectral projections of these operators:
• P =E (R+i((2k−1)π,(2k+1)π));
2k+1 X
• Q =E (R+i((2k−1)π,(2k+1)π));
2k+1 Y
• E =E (R+i(2k+1)π);
2k+1 X
• F =E (R+i(2k+1)π).
2k+1 Y
As we havepointed out inthe introduction, E.Hille showedthat the difference betweentwo
logarithms in Banach algebras may be expressed as the sum of multiples of projections (see
[1, Theorem 4]). In order to prove that result, the spectrum of one of the logarithms must
be incongruent (mod 2πi). In the case where X and Y are both normal logarithms on a
Hilbert space, the spectral theorem can be used to provide a more general formula.
Theorem 4.1. Let X and Y be normal operators in B(H) such that eX =eY. If σ(X) and
σ(Y) are contained in R+i[(2k +1)π,(2k +1)π] for some k ,k ∈Z, then
0 1 0 1
k1
X −Y = 2kπi(P −Q )+(2k+1)πi(E −F ).
2k+1 2k+1 2k+1 2k+1
X
k=k0
Proof. We first suppose that σ(X) and σ(Y) are contained in the strip S. Then we have
Im(X) = Im(X)(E (S◦)+E (R+iπ)+E (R−iπ)) = Im(X)P +πE −πE . Anal-
X X X 1 1 −1
ogously, Im(Y) = Im(Y)Q +πF −πF . By Lemma 2.3, we know that Re(X) = Re(Y)
1 1 −1
and E (Ω)=E (Ω) for every Borel subset Ω of S◦. It follows that
X Y
Im(X)P = Im(z)dE (z)= Im(z)dE (z)=Im(Y)Q ,
1 Z X Z Y 1
S◦ S◦
which implies
X −Y =πi(E −F )−πi(E −F ). (1)
1 1 −1 −1
Thus, we have proved the formula in this case. For the general case, without restrictions on
spectrum of X and Y, we need to consider the following Borel measurable function
k1
f(t)= (t−2kπ)χ (t),
((2k−1)π,(2k+1)π]
X
k=k0−1
where χ (t) is the characteristic function of the interval I. Set A=Im(X) and B =Im(Y).
I
ByLemma2.2,Re(X)=Re(Y),andsincetherealandimaginarypartofX andY commute
because X andY arenormal, eiA =eXe−Re(X) =eYe−Re(Y) =eiB. The function f satisfies
5
eif(t) =eit, which implies that eif(A) =eiA =eiB =eif(B). Since σ(f(A)) and σ(f(B)) are
contained in [−π,π], we can replace in equation (1) to find that
f(A)−f(B)=π(E ({π})−E ({π}))
f(A) f(B)
k1
=π E ({(2k+1)π})−E ({(2k+1)π})
A B
X
k=k0−1
k1
=π E −F . (2)
2k+1 2k+1
X
k=k0
Here we have used Lemma 2.1 to express E ,E and E ,E in terms of E and E
f(A) A f(B) B X Y
respectively. Inparticular,notethatE ({−π})=E ({−π})=0. Ontheotherhand,
f(A) f(B)
we have
k1 k1
1. f(A)= (A−2kπ)χ (A)=A− 2kπ(P +E ),
((2k−1)π,(2k+1)π] 2k+1 2k+1
X X
k=k0−1 k=k0
k1
2. f(B)=B− 2kπ(Q +F ).
2k+1 2k+1
X
k=k0
Therefore
X −Y =i(A−B)
k1
=i(f(A)−f(B))+ 2kπi(P −Q )+2kπi(E −F ).
2k+1 2k+1 2k+1 2k+1
X
k=k0
Combining this with the expression in (2), we get the desired formula.
Below we give a generalizationof another result due to Ch. Schmoeger (see [6, Theorem
5]). The assumptions on the spectrum of X and Y were more restrictive in [6]: kXk ≤ π,
kYk≤πandeither−iπoriπdoesnotbelongtothepointspectrumofoneoftheseoperators.
However,thesehypothesiswerenecessarytoconcludethatX−Y isamultipleofaprojection;
meanwhile XY =YX canbe obtainedunder moregeneralassumptions(see [6, Theorem3],
[5, Theorem 1.4] and [3, Theorem 9]).
Corollary 4.2. Let X,Y be normal operators in B(H). Assume that σ(X)⊆S, σ(Y)⊆S
and eX =eY. The following assertions hold:
i) If E =0, then XY =YX and X −Y =−2πiF .
1 1
ii) If E =0, then XY =YX and X −Y =2πiF .
−1 −1
iii) If E =E =0, then X =Y.
1 −1
Proof. i) Under these assumptions on the spectra of X and Y, we have established that
E +E = F +F in Remark 2.4. On the other hand, by equation (1) in the proof of
−1 1 −1 1
Theorem 4.1, we know that X −Y =πi(E −F )−πi(E −F ). Since E =0, we have
1 1 −1 −1 1
E = F +F . It follows that X = −2πiF +Y. Hence X and Y commute. We can
−1 1 −1 1
similarly conclude that ii) holds true. To prove iii), note that E = E = 0 implies that
1 −1
F +F =0, and consequently, F =F =0. Hence we get X =Y.
1 −1 1 −1
6
5 Unbounded logarithms
Let X be a self-adjoint unbounded operator on H. As before, E denotes the spectral
X
measure of X. In item i) of our next result, we will give a version of [5, Theorem 1.4]
for unbounded operators (see also [3, Theorem 9]). To this end, we extend the definition
given in [5] for bounded operators: a self-adjoint unbounded operator X is generalized 2π-
congruence-free if
E (σ(X)∩σ(X +2kπ))=0, k =±1,±2,....
X
Given Y ∈B(H), the commutant of Y is the set
{Y }′ ={Z ∈B(H) : ZY =YZ}.
The double commutant of Y is defined by
{Y }′′ ={W ∈B(H) : WZ =ZW, for all Z ∈{Y }′}.
If X is a self-adjoint unbounded operator and Y ∈ B(H), recall that XY = YX, that is X
commutes with Y, if YE (Ω) = E (Ω)Y for every Borel subset Ω ⊆ R. Recall that the
X X
exponential eiX of a self-adjoint unbounded operator X is a unitary operator, which can be
defined via the Borel functional calculus (see e.g. [4]).
Theorem 5.1. Let X be a self-adjoint operator on H and Y ∈B(H) such that eiX =eY.
i) If X is generalized 2π-congruence-free, then E (Ω)∈{Y }′′ for all Borel subsets Ω of
X
R. In particular, XY =YX.
ii) If {(2k+1)π : k ∈ Z}∩σ (X) has at most one element and Y is normal in B(H)
p
such that σ(Y)⊆S, then XY =YX.
iii) If (2k+1)π ∈/ σ (X) for all k ∈Z and Y is normal in B(H) such that σ(Y)⊆S, then
p
Y ∈{eiX}′′.
Proof. i) Let Z ∈ B(H) such that ZY = YZ. It follows that ZeY = eYZ. Then we have
ZeiX = eiXZ, and by Lemma 2.1, ZE (exp−1(Ω)) = E (exp−1(Ω))Z for every Ω ⊆ T. If
X X
Ω′ =exp−1(Ω)∩[−π,π], then
E (exp−1(Ω))= E (Ω′+2kπ),
X X
Xk∈Z
where this series converges in the strong operator topology. Suppose now that there is
some k ∈ Z such that E (Ω′ +2kπ) 6= 0. It follows that σ(X)∩ (Ω′ + 2kπ) 6= ∅, and
X
(Ω′+2lπ)∩σ(X)⊆σ(X)∩σ(X+2(l−k)π)foralll ∈Z. Bytheassumptiononthespectral
measure of X, E (Ω′+2lπ) ≤ E (σ(X)∩σ(X +2(l−k)π)) = 0 for l 6= k. Therefore for
X X
eachΩ,the aboveseriesreducestoonlyonespectralprojectioncorrespondingtoasetofthe
form Ω′+2kπ. Hence Z commutes with all the spectral projections of X.
ii) We need to consider the Borel measurable function f defined in the proof of Theorem
4.1. Since eiX = eY, we have that eif(X) = eY. Recall that E ({(2k+1)π}) 6= 0 if and
X
only if (2k+1)π ∈σ (X) ([4, Theorem 12.19]). By the hypothesis on the eigenvalues of X,
p
there is at most one n ∈Z such that E ({(2n +1)π})6=0. According to Lemma 2.1, we
0 X 0
get
E ({π})= E ({(2k+1)π})=E ({(2n +1)π}).
f(X) X X 0
Xk∈Z
Ontheotherhand,E ({−π})=0forallk ∈Zbydefinitionofthefunctionf. According
f(X)
to Corollary 4.2 ii), it follows that f(X)=Y +2πiF . By Remark 2.4, we also know that
−1
7
E ({(2n +1)π}) = F +F . In order to show that Y commutes with all the spectral
X 0 −1 1
projections of X, we divide into two cases. If Ω ⊆ C\{(2k +1)π : k ∈ Z}, note that
E (Ω)F =0 because F ≤E ({(2n +1)π}). Hence we get
X −1 −1 X 0
E (Ω)Y =E (Ω)(f(X)−2πiF )=E (Ω)f(X)=f(X)E (Ω)=YE (Ω).
X X −1 X X X
If Ω ⊆ {(2k +1)π : k ∈ Z}, we only need to prove that E ({(2n +1)π}) commutes
X 0
with Y. This follows immediately, because E ({(2n +1)π}) is the sum of two spectral
X 0
projections of Y.
iii) As in the proof of ii), we have eif(X) = eY. Now by the assumption on the eigenvalues
of X, it follows that
E ({−π, π})= E ({(2k+1)π})=0. (3)
f(X) X
Xk∈Z
Applying Corollary 4.2 iii), we get f(X) = Y. In particular, Y is a self-adjoint operator
such that σ(Y)⊆[−π,π].
Let Z ∈ B(H) such that ZeiX = eiXZ. Then we have ZEeiX(Ω) = EeiX(Ω)Z for every
Borel set Ω ⊆ T. We are going to show that ZE (Ω′) = E (Ω′)Z for every Ω′ ⊆ [−π,π].
Y Y
We needto considertwocases. IfΩ′ ⊆(−π,π), there existsa unique setΩ⊆T\{−1}such
that exp−1(Ω)∩[−π,π]=Ω′. Therefore
EY(Ω′)=Ef(X)(Ω′)= EX(Ω′+2kπ)=EX(exp−1(Ω))=EeiX(Ω).
Xk∈Z
If Ω′ ⊆ {−π, π}, by equation (3) we find that E (Ω′) = E (Ω′) = 0. Hence we obtain
Y f(X)
thatZ commuteswitheveryspectralprojectionofY. Thelatterisequivalenttosayingthat
Z commute with Y, and this concludes the proof.
Acknowledgment
I wouldliketothankEstebanAndruchowandGabrielLarotondaforsuggestingmetoprove
Theorem 3.1 i). I am also grateful to them for several helpful conversations.
References
[1] E.Hille, On roots and logarithms of elements of a complex Banach alge-
bra, Math. Ann. 136 (1958), 46-57.
[2] S. Kurepa, A note on logarithms of normal operators, Proc. Amer. Math.
Soc. 13 (1962), 307-311.
[3] F. C. Paliogiannis, On commuting operator exponentials, Proc. Amer. Math.
Soc. 131 (2003), no. 12, 3777-3781.
[4] W. Rudin, Functional Analysis, McGraw-Hill, New York (1974).
[5] Ch. Schmoeger,On normal operator exponentials, Proc.Amer.Math.Soc.130
(2001), no. 3, 697-702.
[6] Ch. Schmoeger, On the operator equation eA = eB, Linear Algebra Appl. 359
(2003), 169-179.
Departamento de de Matema´tica, FCE-UNLP,Calles50 y 115, (1900) La Plata,
Argentina and Instituto Argentino de Matema´tica, ‘Alberto P. Caldero´n’,
CONICET, Saavedra 15 3er. piso, (1083) Buenos Aires, Argentina.
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