Abstract
In this paper, we show that for \(T\in B({\mathcal {H}})\), if \({\mathcal {M}}\) is almost-invariant for T, then every maximal almost-invariant subspace of \({\mathcal {M}}\) is of codimension 1 in \({\mathcal {M}}\), where \({\mathcal {H}}\) is a separable, infinite-dimensional Hilbert space. We also describe the maximal hyperinvariant subspaces for normal operators with all the dimensions of eigenspaces at most 1 acting on \({\mathcal {H}}\). Our result is that for each hyperinvariant subspace, all its maximal hyperinvariant subspaces are also of codimension 1 in it.
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1 Introduction
Let \({\mathcal {H}}\) be a separable, infinite-dimensional Hilbert space and denote by \(B({\mathcal {H}})\) the set of bounded linear operators acting on \({\mathcal {H}}\). For \(T\in B({\mathcal {H}})\) a subspace \({\mathcal {M}}\) of \({\mathcal {H}}\) is called invariant for T, or T-invariant, if it is closed and \(T{\mathcal {M}}\subseteq {\mathcal {M}}\). The classical Invariant Subspace Problem, one of the most important problems in Operator Theory, is about the existence of non-trivial invariant subspaces for an operator \(T\in B({\mathcal {H}})\).
For an operator \(T\in B({\mathcal {H}})\) and an invariant subspace \({\mathcal {M}}\) for T, a T-invariant subspace \({\mathcal {N}}\) is called a maximal invariant subspace of \({\mathcal {M}}\), if \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and there is no T-invariant subspace \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\). Hedenmalm [6] obtained first the result that every maximal invariant subspace of the Bergman space is of codimension 1. For further generalizions of the Bergman space, we refer the interested readers to [1, 16]. Later, Guo et al. [5] extended the result to a much more general situation. Their result is the following.
Theorem 1.1
([5, Theorem 1.1]). Suppose T is a Fredholm operator acting on a separable Hilbert space and \(1-TT^{*} \in {\mathcal {S}}_p\) for some \(p\ge 1\). If \({\mathcal {M}}\) is an invariant subspace for T such that \( \dim {\mathcal {M}}\Theta T{\mathcal {M}}<\infty \), then every maximal invariant subspace of \({\mathcal {M}}\) is of codimension 1 in \({\mathcal {M}}\).
Here \({\mathcal {S}}_p \, (p>0)\) denotes the set of Schatten-p class operators; and for two subspaces \({\mathcal {U}}, {\mathcal {V}}\) of \({\mathcal {H}}\), \({\mathcal {U}}\Theta {\mathcal {V}}={\mathcal {U}}\cap {\mathcal {V}}^\bot \), where \({\mathcal {V}}^\bot \) denotes the orthogonal complement space of \({\mathcal {V}}\) in \({\mathcal {H}}\).
Motivated by the above work, we intend to study the maximal almost-invariant subspaces and maximal hyperinvariant subspaces.
A subspace \({\mathcal {M}}\) of \({\mathcal {H}}\) is called almost-invariant for T (or T-almost invariant) if it is closed and \(T {\mathcal {M}}\subseteq {\mathcal {M}}+{\mathcal {F}}\) for some finite-dimensional subspace \({\mathcal {F}}\) of \({\mathcal {H}}\). This concept was first introduced in [2]. The minimal dimension of such a subspace \({\mathcal {F}}\) is referred to as the defect of \({\mathcal {M}}\) for T. It is obvious that every finite-dimensional or finite-codimensional subspace is almost-invariant under T. So we only need to consider a half-space, that is, a subspace of \({\mathcal {H}}\) which is infinite-dimensional and infinite-codimensional. For further information about almost-invariant subspaces, we refer the interested readers to [2, 12,13,14].
In a similar way, we give the definition of maximal almost-invariant subspace.
Definition 1.2
For an operator \(T\in B({\mathcal {H}})\) and an almost-invariant subspace \({\mathcal {M}}\) for T, a T-almost invariant subspace \({\mathcal {N}}\) is called a maximal almost-invariant subspace of \({\mathcal {M}}\), if \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and there is no T-almost invariant subspace \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\).
We will prove that, given an operator \(T\in B({\mathcal {H}})\), for any T-almost invariant half-space \({\mathcal {M}}\) every maximal almost-invariant subspace of \({\mathcal {M}}\) is of codimension 1 in \({\mathcal {M}}\).
A subspace \({\mathcal {M}}\) of \({\mathcal {H}}\) is called hyperinvariant for T, or T-hyperinvariant, if it is closed and \(W {\mathcal {M}}\subseteq {\mathcal {M}}\) for each \(W\in \{T\}'\). Here \(\{T\}'\) denotes the commutant of T given by
There are many unsolved problems in the theory of invariant subspaces, hence these problems need close attention. In this paper, we first deal with hyperinvariant subspaces. For a further discussion about hyperinvariant subspaces, we recommend to the interested readers the recent papers [4, 7,8,9,10,11, 15].
We also define maximal hyperinvariant subspaces analogously.
Definition 1.3
For an operator \(T\in B({\mathcal {H}})\) and a hyperinvariant subspace \({\mathcal {M}}\) for T, a T-hyperinvariant subspace \({\mathcal {N}}\) is called a maximal hyperinvariant subspace of \({\mathcal {M}}\), if \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and there is no T-hyperinvariant subspace \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\).
An operator \(T\in B({\mathcal {H}})\), is said to be normal if \(T^{*} T=TT^{*}\).
Our conclusion about maximal hyperinvariant subspaces is in the setting of a Hilbert space \({\mathcal {H}}\). We will show that for a normal operator \(T\in B({\mathcal {H}})\) and a T-hyperinvariant subspace \({\mathcal {M}}\), if all the dimensions of eigenspaces of T are at most 1, then every maximal hyperinvariant subspace of \({\mathcal {M}}\) is of codimension 1 in \({\mathcal {M}}\).
Throughout the paper, for a closed subspace \({\mathcal {E}}\) of \({\mathcal {H}}\), \(P_{\mathcal {E}}\) denotes the orthogonal projection from \({\mathcal {H}}\) to \({\mathcal {E}}\) and \(T|_{\mathcal {E}}\) is the operator T restricted to \({\mathcal {E}}\).
2 Maximal almost-invariant subspaces
In this section, we give a characterization of maximal almost-invariant subspaces. The main result can be formulated as follows.
Theorem 2.1
For \(T\in B({\mathcal {H}})\), if \({\mathcal {M}}\) is a T-almost invariant half-space, then every maximal almost-invariant subspace of \({\mathcal {M}}\) is of codimension 1 in \({\mathcal {M}}\).
The first step in the proof of Theorem 2.1 is the following lemma.
Lemma 2.2
Let \(T\in B({\mathcal {H}})\). Suppose \({\mathcal {M}}\) and \({\mathcal {N}}\) are two T-almost invariant half-spaces with \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and \( \dim {\mathcal {M}}\Theta {\mathcal {N}}\ge 2\). Put \(S=P_{M\Theta N}T|_{M\Theta N}\). Then T has an almost invariant half-space \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\) if and only if there exists an S-almost invariant subspace \({\mathcal {L}}_0\) such that \({0}\subsetneq {\mathcal {L}}_0\subsetneq {\mathcal {M}}\Theta {\mathcal {N}}\).
Proof
Since \({\mathcal {M}}, {\mathcal {N}}\) are both T-almost invariant half-spaces, we write \(T{\mathcal {M}}\subseteq {\mathcal {M}}+{\mathcal {F}}_1\) and \(T{\mathcal {N}}\subseteq {\mathcal {N}}+{\mathcal {F}}_2\), where \({\mathcal {F}}_1\) and \({\mathcal {F}}_2\) are finite-dimensional subspaces of \({\mathcal {H}}\). Now, if \({\mathcal {L}}\) is almost-invariant for T with \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\), we assume that \(T{\mathcal {L}}\subseteq {\mathcal {L}}+{\mathcal {F}}_3\) for some finite-dimensional subspace \({\mathcal {F}}_3\). Put \({\mathcal {L}}_0={\mathcal {L}}\Theta {\mathcal {N}}\); then it is clear that \({0}\subsetneq {\mathcal {L}}_0\subsetneq {\mathcal {M}}\Theta {\mathcal {N}}\), and
that is \({\mathcal {L}}_0\) is S-almost invariant.
Conversely, we assume that there exists a subspace \({\mathcal {L}}_0\) with \({0}\subsetneq {\mathcal {L}}_0\subsetneq {\mathcal {M}}\Theta {\mathcal {N}}\) that is S-almost invariant by \(S {\mathcal {L}}_0\subseteq {\mathcal {L}}_0+{\mathcal {F}}_4\) for some finite-dimensional subspace \({\mathcal {F}}_4\). Put \({\mathcal {L}}={\mathcal {L}}_0+{\mathcal {N}}\), then \({\mathcal {L}}\) is half-space and \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\). Next we prove that \({\mathcal {L}}\) is T-almost invariant. Noting that \(S {\mathcal {L}}_0\subseteq {\mathcal {L}}_0+{\mathcal {F}}_4\), i.e., \(P_{{\mathcal {M}}\Theta {\mathcal {N}}}T|_{{\mathcal {M}}\Theta {\mathcal {N}}}{\mathcal {L}}_0\subseteq {\mathcal {L}}_0+{\mathcal {F}}_4\), we have \(P_{{\mathcal {N}}^\perp }T {\mathcal {L}}_0\subseteq {\mathcal {L}}_0+{\mathcal {F}}_4+{\mathcal {F}}_1\). Then \(T {\mathcal {L}}_0=P_{\mathcal {N}} T {\mathcal {L}}_0+P_{{\mathcal {N}}^\perp }T {\mathcal {L}}_0\subseteq {\mathcal {N}}+{\mathcal {L}}_0+{\mathcal {F}}_4+{\mathcal {F}}_1\), thus
for some finite-dimensional subspace \(\widehat{{\mathcal {F}}}\), that is, \({\mathcal {L}}\) is T-almost invariant. So the assertion of this lemma is proved. \(\square \)
The following lemma, proved by Popov and Tcaciuc in [14], is quite important to get the main result of this section.
Lemma 2.3
Let T be a bounded operator on an infinite-dimensional, reflexive Banach space \({\mathcal {X}}\). Then \({\mathcal {X}}\) admits an almost-invariant half-space with defect 1.
Using this lemma, we can prove the following result, which is the key idea of the proof of Theorem 2.1.
Lemma 2.4
Let \(T\in B({\mathcal {H}})\). Suppose \({\mathcal {M}}\) and \({\mathcal {N}}\) are two T-almost invariant half-spaces with \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and \( \dim {\mathcal {M}}\Theta {\mathcal {N}}\ge 2\). Then there is a T-almost invariant half-space \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\).
Proof
Set \(T{\mathcal {M}}\subseteq {\mathcal {M}}+{\mathcal {F}}_1\) and \(T{\mathcal {N}}\subseteq {\mathcal {N}}+{\mathcal {F}}_2\) as in the proof of Lemma 2.2. Firstly, assuming that \( \dim {\mathcal {M}}\Theta {\mathcal {N}}<\infty \), we can choose half-space \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\) since \( \dim {\mathcal {M}}\Theta {\mathcal {N}}\ge 2\). Moreover, for each half-space \({\mathcal {L}}\) with \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\), we have
for some finite-dimensional subspace \(\widetilde{{\mathcal {F}}}\) of \({\mathcal {H}}\) since \( \dim {\mathcal {M}}\Theta {\mathcal {N}}<\infty \).
Now we assume \( \dim {\mathcal {M}}\Theta {\mathcal {N}}=\infty \). Consider the operator \(S=P_{{\mathcal {M}}\Theta {\mathcal {N}}}T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\). Since \(S=P_{{\mathcal {M}}\Theta {\mathcal {N}}}T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\in B({\mathcal {M}}\Theta {\mathcal {N}})\) and \({\mathcal {M}}\Theta {\mathcal {N}}\) is an infinite-dimensional, reflexive Banach space, by Lemma 2.3, \({\mathcal {M}}\Theta {\mathcal {N}}\) admits an S-almost invariant half-space with defect 1. Therefore, using Lemma 2.2, there exists a T-almost invariant subspace \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\). This completes the proof. \(\square \)
Now the characterization of maximal almost-invariant subspaces is a direct consequence of this lemma.
At the end of this section, we want to pose a question to the interested readers. In the proof, the finite-dimensional subspaces making sure that \({\mathcal {N}},{\mathcal {L}},{\mathcal {M}}\) are T-almost invariant may not have the same dimension or even be the same subspace. Of course, here we mean such a finite-dimensional subspace with minimal dimension to make sure \({\mathcal {N}},{\mathcal {L}},{\mathcal {M}}\) are T-almost invariant. Hence, it is natural to ask the following question:
Question 2.5
Is there a separable, infinite-dimensional Hilbert space \({\mathcal {H}}\), and an operator \(T\in B({\mathcal {H}})\) such that there exist three half-spaces \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\) that are T-almost invariant subspaces with the same defect or even the same finite-dimensional subspace are T-almost invariant subspaces?
3 Maximal hyperinvariant subspaces
In the case when we consider the maximal hyperinvariant subspaces, we focus on the normal operators acting on a separable, infinite-dimensional Hilbert space \({\mathcal {H}}\). The main result relies on the following lemma, which is proved in a similar way to [5, Lemma 2.3].
Lemma 3.1
Let \(T\in B({\mathcal {H}})\), \({\mathcal {M}}\) and \({\mathcal {N}}\) be two T-hyperinvariant subspaces with \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and \( \dim {\mathcal {M}}\Theta {\mathcal {N}}\ge 2\). Put \(S=P_{{\mathcal {M}}\Theta {\mathcal {N}}}T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\). Then if S has a non-trivial hyperinvariant subspace then T has a hyperinvariant subspace \({\mathcal {L}}\) such that \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\).
Moreover, if in additon T is normal, then the existence of a T-hyperinvariant subspace \({\mathcal {L}}\) with \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\) implies the existence of a non-trivial hyperinvariant subspace for S.
Proof
Suppose that \({\mathcal {L}}_0\) is a nontrivial hyperinvariant subspace for S. It is clear that \({0}\subsetneq {\mathcal {L}}_0\subsetneq {\mathcal {M}}\Theta {\mathcal {N}}\). Setting \({\mathcal {L}}={\mathcal {L}}_0+{\mathcal {N}}\), we have \({\mathcal {N}}\subsetneq {\mathcal {L}}\subsetneq {\mathcal {M}}\). Next we prove that \({\mathcal {L}}\) is hyperinvariant for T. For any \(W\in \{T\}'\), we first prove \(P_{{\mathcal {M}}\Theta {\mathcal {N}}}W|_{{\mathcal {M}}\Theta {\mathcal {N}}}\in \{S\}'\). Indeed,
here we used \(P_{{\mathcal {M}}\Theta {\mathcal {N}}} W P_{\mathcal {N}} T P_{{\mathcal {M}}\Theta {\mathcal {N}}}=0\) since \({\mathcal {N}}\) is hyperinvariant for T and \(W\in \{T\}'\).
In a similar way, we can obtain
Since \(WT=TW\), thus
that is \(P_{{\mathcal {M}}\Theta {\mathcal {N}}}W|_{{\mathcal {M}}\Theta {\mathcal {N}}}\in \{S\}'\).
Since \({\mathcal {L}}_0\) is a hyperinvariant subspace for S, then \(P_{{\mathcal {M}}\Theta {\mathcal {N}}} W|_{{\mathcal {M}}\Theta {\mathcal {N}}}{\mathcal {L}}_0\subseteq {\mathcal {L}}_0\), hence we have \(P_{{\mathcal {N}}^\perp }W {\mathcal {L}}_0\subseteq {\mathcal {L}}_0\). Then it is easy to see that \(W {\mathcal {L}}_0=P_{\mathcal {N}} W {\mathcal {L}}_0+P_{{\mathcal {N}}^\perp }W {\mathcal {L}}_0\subseteq {\mathcal {N}}+{\mathcal {L}}_0={\mathcal {L}}\). Therefore, we conclude that
That is, \({\mathcal {L}}\) is hyperinvariant for T by the arbitrariness of \(W\in \{T\}'\), which proves the first assertion of this lemma.
Conversely, note that T is normal, i.e., \(T^{*} \in \{T\}'\). Therefore, \({\mathcal {M}}, {\mathcal {N}}\) are both reducing subspaces of T, so is \({\mathcal {M}}\Theta {\mathcal {N}}\). Thus we conclude that \(S=T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\). Then the operator T has the corresponding decomposition
where \(T_1\) is the restriction of T on \(({\mathcal {M}}\Theta {\mathcal {N}})^\bot \). Given an operator \(W_0\in {\mathcal {L}}({\mathcal {M}}\Theta {\mathcal {N}})\), if \(W_0S=SW_0\), set \(W=W_0\oplus I_{({\mathcal {M}}\Theta {\mathcal {N}})^\bot }\), it follows that \(W\in \{T\}'\). Thus \(W{\mathcal {L}}\subseteq {\mathcal {L}}\). It is easy to prove that \(W_0 {\mathcal {L}}_0\subseteq {\mathcal {L}}_0\), if \({\mathcal {L}}_0={\mathcal {L}}\Theta {\mathcal {N}}\). So the second assertion of the lemma is obtained. \(\square \)
Lemma 3.2
Let \(T\in B({\mathcal {H}})\) be a normal operator, and \({\mathcal {M}}\) and \({\mathcal {N}}\) two T-hyperinvariant subspaces with \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and \( \dim {\mathcal {M}}\Theta {\mathcal {N}}\ge 2\). Then \(S=P_{{\mathcal {M}}\Theta {\mathcal {N}}}T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\) is also normal.
Proof
Since \({\mathcal {M}}, {\mathcal {N}}\) are both T-hyperinvariant, and \(T^{*} \in \{T\}'\), then \({\mathcal {M}}, {\mathcal {N}}\) are both reducing subspaces of T, so is \({\mathcal {M}}\Theta {\mathcal {N}}\). Hence \(S=T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\). Next, we will show that \(T|_{{\mathcal {M}}\Theta {\mathcal {N}}}^{*} =T^{*} |_{{\mathcal {M}}\Theta {\mathcal {N}}}\). In fact,
for \(x,y\in {\mathcal {M}}\Theta {\mathcal {N}}\). Then the result follows from the normality of T. \(\square \)
The next result can be found in [3].
Lemma 3.3
A normal operator that is not a multiple of the identity has a non-trivial hyperinvariant subspace.
Using the previous lemmas, we are now ready to give the required generalization about maximal hyperinvariant subspaces.
Theorem 3.4
Let \(T\in B({\mathcal {H}})\) be a normal operator, and \({\mathcal {M}}\) be a T-hyperinvariant subspace. If all the dimensions of eigenspaces of T are at most 1, then every maximal hyperinvariant subspace of \({\mathcal {M}}\) is of codimension 1 in \({\mathcal {M}}\).
Proof
Note that the condition that all the dimensions of eigenspaces of T are at most 1 guarantees that \(P_{{\mathcal {M}}\Theta {\mathcal {N}}}T|_{{\mathcal {M}}\Theta {\mathcal {N}}}\) is not a multiple of the identity for each T-hyperinvariant subspace \({\mathcal {N}}\) with \({\mathcal {N}}\subsetneq {\mathcal {M}}\) and \( \dim {\mathcal {M}}\Theta {\mathcal {N}}\ge 2\). Then the assertion comes easily from the lemmas above. \(\square \)
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The work was supported by the National Natural Science Foundation of China (Grant No. 11771323).
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Chen, C., Zhou, ZH. & Wang, Y. A note about maximal almost-invariant subspaces and maximal hyperinvariant subspaces. Period Math Hung 79, 221–226 (2019). https://doi.org/10.1007/s10998-019-00292-3
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DOI: https://doi.org/10.1007/s10998-019-00292-3