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 (HölderMcCarthy'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 HölderMcCarthy'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 HölderMcCarthy'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 selfadjoint 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.