NEW ZEALAND JOURNAL OF MATHEMATICS
Volume 44 (2014), 103-106
THE WEIGHTED MEAN OPERATOR ON `2 WITH THE
WEIGHT SEQUENCE wn = n + 1 IS HYPONORMAL
H. C. Rhaly Jr. and B. E. Rhoades
(Received 19 January, 2013)
Abstract. Posinormality is used to demonstrate that the weighted mean matrix
whose weight sequence is the sequence of positive integers is a hyponormal
operator on `2 .
1. Introduction
In this paper our attention will be focused on an example of a weighted mean
matrix that does not satisfy the sufficient conditions for hyponormality given in [8].
If B(H) denotes the set of all bounded linear operators on a Hilbert space H, then
A ∈ B(H) is said to be posinormal (see [2], [5]) if AA∗ = A∗ P A for some positive
operator P ∈ B(H), called the interrupter, and A is hyponormal if A∗ A − AA∗ ≥ 0.
Hyponormal operators are necessarily posinormal.
A lower triangular infinite matrix M = [mij ], acting through multiplication to
give a bounded linear operator on `2 , is factorable if its entries are
ai cj , if j ≤ i
mij =
0,
if j > i
where ai depends only on i and cj depends only on j; the matrix M is terraced
(see [3], [4]) if cj = 1 for all j. A weighted mean matrix is a lower triangular
matrix with entries wj /Wi , where {wj } is a nonnegative sequence with w0 > 0,
Pi
and Wi = j=0 wj . A weighted mean matrix is factorable, with ai = 1/Wi and
cj = wj for all i,j.
The Cesaro matrix C is the terraced matrix that occurs when ai = 1/(i + 1) for
all i, and it is also a weighted mean matrix with weight sequence wj = 1 for all j.
In [1] C was shown to be a hyponormal operator on `2 . In [8] an example was given
of a hyponormal weighted mean matrix with weight sequence (strictly) increasing
to a finite limit. Also in [8] it was seen that the weighted mean matrix associated
with wn = q n for q > 1 is not hyponormal, and neither is the matrix associated
with wn = q −n for 0 < q < 1. These are known as the q-Cesàro operators, and
clearly they satisfy wn → ∞. In this paper we will present an example satisfying
wn → ∞ which is a hyponormal operator on `2 .
Previously, attempts have been made to follow the procedure of [6] for terraced
matrices in order to obtain analogous sufficient conditions for factorable matrices
to be hyponormal, but those attempts have so far met with limited success; see,
2010 Mathematics Subject Classification Primary 47B99.
Key words and phrases: posinormal operator, hyponormal operator, factorable matrix, weighted
mean matrix.
104
H. C. RHALY JR. and B. E. RHOADES
Pn
e.g., [9]. It is clear, however, that the sequence {( k=0 c2k )an /cn } plays a significant role in the calculations for factorable matrices, similar to the role which
the sequence {(n + 1)an } played for the terraced matrices. For the weighted mean
matrix associated with the weight sequence wn = n + 1, one can easily verify that
n+1
X
(
c2k )an+1 /cn+1 − (
k=0
n
X
c2k )an /cn = −(1/6)an ,
k=0
and that is what helps to make the calculations in the next section manageable.
As an aside, P
we note that a diagonal interrupter P results from the case where
n
the sequence {( k=0 c2k )an /cn } is constant; that case has been studied in [7].
2. Main Result
As mentioned previously, our example will be the weighted mean matrix M
associated with the weight sequence wn = n + 1. Although w0 > 0 and {wn }
is increasing, this example fails to satisfy the displayed condition in the following
result from [8].
Theorem 1. If w0 > 0, wn ≤ wn+1 for all n ≥ 0, and
n−1
X
wn+1 (
k=0
n+1
X
wk )(
wk ) ≤ wn (
n
X
wk )2 for all n ≥ 1,
k=0
k=0
then the associated weighted mean matrix (whenever bounded) is hyponormal.
However, this example survives the following necessary condition for hyponormality, which follows from [8, Corollary 2].
Theorem 2. If the weighted mean matrix associated with the sequence {wn } is to
Pn−1
be hyponormal, it is necessary that {wn / k=0 wk }n≥1 ∈ `2 .
Taking encouragement from this, we set out to prove that M is hyponormal. The
next theorem will provide us with our main tool, an expression for the interrupter
P associated with the matrix M . For notational convenience, we define
n
X
tn :≡ (
c2k )an /cn
k=0
for each nonnegative integer n.
Theorem 3. Suppose M = [ai cj ] is a lower triangular factorable matrix that acts
as a bounded operator on `2 and that the following conditions are satisfied:
(1) both {an } and {an /cn } are positive decreasing sequences that converge to 0, and
(2) the matrix B = [bij ] defined by

 ci {1/cj − aj+1 /(cj+1 aj )} if i ≤ j;
−aj+1 /aj
if i = j + 1;
bij =

0
if i > j + 1.
is a bounded operator on `2 . Then M is posinormal with interrupter P = B ∗ B,
and the entries of P = [pij ] are given by

Pj
 a2j+1 /a2j + ( k=0 c2k )(1/cj − aj+1 /[cj+1 aj ])2 if i = j;
pij =
(a /[c a ] − 1/ci )(tj+1 − tj )/aj
if i > j;
 i+1 i+1 i
(aj+1 /[cj+1 aj ] − 1/cj )(ti+1 − ti )/ai
if i < j.
Proof. See [7].
THE WEIGHTED MEAN OPERATOR WITH wn = n + 1 IS HYPONORMAL
105
The following lemma will help with our example.
Lemma 1. Suppose that the factorable matrix M satifies the hypotheses of Theorem
3, and assume that, for all n, tn+1 − tn = γan for some constant γ 6= 0. Then the
N th finite section QN of Q :≡ I − P can be reduced to the following tridiagonal
form:


d0 s0 0 . . .
0
0
 s0 d1 s1 . . .
0
0 



 0 s1 d2 . . .
.
0


,
YN :≡  .
.
.
.
.
.
..
..
..
..
.. 

 ..


 0 0
. . . . dN −1 sN −1 
0 0 0 . . . sN −1
dN
where dn = 2 − cn+1 an+1 /[cn an ] − cn+2 an+2 /[cn+1 an+1 ]
+γ{1/cn − an+1 /[cn+1 an ] − 1/cn+1 + an+2 /[cn+2 an+1 ]}
and sn = −(cn+1 an+1 − cn+2 an+2 )/(cn+1 an+1 ) when 0 ≤ n ≤ N − 1; and dN =
1 − pN N .
Proof. We note that, because of the assumption regarding tn+1 − tn , QN is almost reverse L-shaped, and this simplifies the reduction process. For columns
n = 1, 2, ..., N , we subtract the nth column of QN from the (n − 1)st column. Call
the new matrix Q0N . We then work with the rows of Q0N . For m = 1, 2, ..., N , we
subtract the mth row of Q0N from the (m − 1)st row. This leads to the tridiagonal
form YN .
We exclude the case γ = 0 from Lemma 1 because Q is already a diagonal matrix
in that case.
As we have already seen, when M is the weighted mean matrix associated with
wn = n + 1, the hypothesis of Lemma 1 is satisfied with γ = −1/6. We are now
ready to proceed with our main result.
Theorem 4. The weighted mean matrix M associated with the weight sequence
wn = n + 1 is hyponormal.
Proof. One easily verifies that the weighted mean matrix M associated with wn =
n + 1 satisfies the hypotheses of Theorem 3. For M to be hyponormal, we must
have
h(M ∗ M − M M ∗ )f, f i = h(M ∗ M − M ∗ P M )f, f i = h(I − P )M f, M f i ≥ 0
for all f in `2 . Consequently, we can conclude that M will be hyponormal when
Q :≡ I − P ≥ 0. We note that the range of M contains all of the en ’s from the
standard orthonormal basis for `2 .
We refer now to Lemma 1. In order to show that Q is positive, it suffices to
show that QN has a positive determinant for each positive integer N . Note that
det YN = det Q0N = det QN . Substituting in the appropriate expressions for our
example, we obtain the tridiagonal matrix YN with entries
dn = (2n2 + 9n + 6)/{(n + 1)(n + 3)(n + 4)}
and sn = −1/(n + 4) when 0 ≤ n ≤ N − 1; and
dN = (6N 2 + 15N + 7)/{6(N + 1)(N + 2)(N + 3)}.
106
H. C. RHALY JR. and B. E. RHOADES
Next we transform YN into a triangular matrix with the same determinant, and we
find that the new matrix has diagonal entries δn which are given by the recursion
formula: δ0 = d0 , δn = dn − s2n−1 /δn−1 (1 ≤ n ≤ N ). An induction argument
shows that
δn = (n + 2)/{(n + 1)(n + 4)} > 0
for 0 ≤ n ≤ N − 1. Since dN departs from the pattern set by the earlier dn ’s, δN
must be handled separately:
δN = (6N 2 + 15N + 7)/{6(N + 1)(N + 2)(N + 3)} − N/{(N + 1)(N + 3)}
= (3N + 7)/{6(N + 1)(N + 2)(N + 3)} > 0.
So det QN =
QN
n=0 δn
> 0, and the proof is complete.
We observe that this example illustrates that it is possible
Pn for a factorable matrix
M ∈ B(`2 ) to be hyponormal even if the sequence {( k=0 c2k )an /cn } is strictly
decreasing.
In closing, we note that when M is the weighted mean matrix associated with
the weight sequence given by wn = (n + 1)2 , a straightforward computation demonstrates that M does not satisfy the hypothesis of Lemma 1.
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H. C. Rhaly Jr.,
1081 Buckley Drive,
Jackson, MS 39206, U.S.A.
rhaly@member.ams.org
B. E. Rhoades
Indiana University,
Department of Mathematics,
Bloomington, IN 47405, U.S.A.
rhoades@indiana.edu