In the following lemma, Tanahashi, Jeon, Kim, and Uchiyama studied the matrix representation of a
-quasiclass A operator with respect to the direct sum of
and its orthogonal complement.
Lemma 2.1 (see [21]).
Let
be a
-quasiclass A operator for a positive integer
and let
on
be
matrix expression. Assume that ran
is not dense, then
is a class A operator on
and
. Furthermore,
.
Proof.
Consider the matrix representation of
with respect to the decomposition
:
Let
be the orthogonal projection of
onto
. Then
. Since
is a
-quasiclass A operator, we have
Then
by Hansen's inequality [22]. On the other hand
Hence
That is,
is a class A operator on
.
For any
,
which implies
.
Since
, where
is the union of the holes in
which happen to be subset of
by [23, Corollary 7], and
and
has no interior points, we have
.
Theorem 2.2.
Let
be a
-quasiclass A operator for a positive integer
. Then the following assertions hold.
for all
and all positive integers
.
If
for some positive integer
, then
.
for all positive integers
, where
denotes the spectral radius of
.
To give a proof of Theorem 2.2, the following famous inequality is needful.
Lemma 2.3 (Hölder-McCarthy's inequality [24]).
Let
. Then the following assertions hold.
for
and all
.
for
and all
.
Proof of Theorem 2.2.
-
(1)
Since it is clear that
-quasiclass A operators are
-quasiclass A operators, we only need to prove the case
. Since
by Hölder-McCarthy's inequality, we have
for
is a
-quasiclass A operator.
-
(2)
If
, it is obvious that
. If
, then
by (
). The rest of the proof is similar.
-
(3)
We only need to prove the case
, that is,
If
for some
, then
by (2) and in this case
. Hence (3) is clear. Therefore we may assume
for all
. Then
by (
), and we have
Hence
By letting
, we have
that is,
Lemma 2.4 (see [21]).
Let
be a
-quasiclass A operator for a positive integer
. If
and
for some
, then
.
Proof.
We may assume that
. Let
be a span of
. Then
is an invariant subspace of
and
Let
be the orthogonal projection of
onto
. It suffices to show that
in (2.14). Since
is a
-quasiclass A operator, and 
, we have
We remark
Then by Hansen's inequality and (2.15), we have
Hence we may write
We have
This implies
and
. On the other hand,
Hence
and
. Since
is a
-quasiclass A operator, by a simple calculation we have
Recall that
if and only if
and
for some contraction
. Thus we have
. This completes the proof.
Lemma 2.5 (see [25]).
If
satisfies
for some complex number
, then
for any positive integer
.
Proof.
It suffices to show
by induction. We only need to show
since
is clear. In fact, if
, then we have
by hypothesis. So we have
, that is,
. Hence
.
An operator is said to have finite ascent if
for some positive integer
.
Theorem 2.6.
Let
be a
-quasiclass A operator for a positive integer
. Then
has finite ascent for all complex number
.
Proof.
We only need to show the case
because the case
holds by Lemmas 2.4 and 2.5.
In the case
, we shall show that
. It suffices to show that
since
is clear. Now assume that
. We may assume
since if
, it is obvious that
. By Hölder-McCarthy's inequality, we have
So we have
, which implies
. Therefore
.
In the following lemma, Tanahashi, Jeon, Kim, and Uchiyama extended the result (1.3) to
-quasiclass A operators in the case
.
Lemma 2.7 (see [21]).
Let
be a
-quasiclass A operator for a positive integer
. Let
be an isolated point of
and
the Riesz idempotent for
. Then the following assertions hold.
If
, then
is self-adjoint and
If
, then
.
An operator
is said to be isoloid if every isolated point of
is an eigenvalue of
.
Theorem 2.8.
Let
be a
-quasiclass A operator for a positive integer
. Then
is isoloid.
Proof.
Let
be an isolated point. If
, by (
) of Lemma 2.7,
for
. Therefore
is an eigenvalue of
. If
, by (
) of Lemma 2.7,
for
. So we have
. Therefore
is an eigenvalue of
. This completes the proof.
Let
denote the tensor product on the product space
for nonzero
,
. The following theorem gives a necessary and sufficient condition for
to be a
-quasiclass A operator, which is an extension of [20, Theorem 4.2].
Theorem 2.9.
Let
,
be nonzero operators. Then
is a
-quasiclass A operator if and only if one of the following assertions holds
(1)
or
.
(2)
and
are
-quasiclass A operators.
Proof.
It is clear that
is a
-quasiclass A operator if and only if
Therefore the sufficiency is clear.
To prove the necessary, suppose that
is a
-quasiclass A operator. Let
,
be arbitrary. Then we have
It suffices to prove that if (
) does not hold, then (
) holds. Suppose that
and
. To the contrary, assume that
is not a
-quasiclass A operator, then there exists
such that
From (2.25) we have
that is,
for all
. Therefore
is a
-quasiclass A operator. As the proof in Theorem 2.2 (
), we have
So we have
for all
by (2.28). Because
is a
-quasiclass A operator, from Lemma 2.1 we can write
on
, where
is a class A operator (hence it is normaloid). By (2.30) we have
So we have
where equality holds since
is normaloid.
This implies that
. Since
for all
, we have
. This contradicts the assumption
. Hence
must be a
-quasiclass A operator. A similar argument shows that
is also a
-quasiclass A operator. The proof is complete.