1 Introduction
Let K be a complex Hilbert space, \({\mathbf{B}}[\mathbf{K} ]\) be the algebra of all bounded linear operators defined in K. For every N in \(\mathbf{B}[\mathbf{K}]\), denote \(\ker (N)\) as the null space and \(N^{*}\) as the adjoint of R, respectively.
The Drazin inverse of bounded linear operators on complex Banach spaces was introduced by Caradus [14] and King [26]. For more detailed study and applications of the concepts of Drazin invertibility, we invite the interested readers to refer to ([11, 12, 32]). It is well known that the Drazin inverse of the operator \(N\in{ \mathbf{B}}[\mathbf{K}]\) is the unique operator \(N^{\mathcal{D}}\in {\mathbf{B}}[\mathbf{K}]\) if it exists and satisfies the following conditions
$$ \textstyle\begin{cases} N^{\mathcal{D}} N=NN^{\mathcal{D}}, \\ (N\mathcal{^{D}} )^{2}N=N^{\mathcal{D}}, \\ N^{\nu +1}N^{\mathcal{D}}=N^{\nu}\quad \text{for some integer } \nu \geq 0. \end{cases} $$
Anzeige
We denote by \({\mathbf{B}}_{d}[\mathbf{K}]\) the set of all Drazin invertible elements of \({ \mathbf{B}}[\mathbf{K}]\).
For \(N \in {\mathbf{B}}_{d}[\mathbf{K}]\), it was observed that the Drazin inverse \(N^{\mathcal{D}}\) of N satisfies the following conditions Moreover, it was observed that if \(N \in {\mathbf{B}}_{d}[\mathbf{K}]\) and \(T \in {\mathbf{B}}[\mathbf{K}]\) is an invertible operator, then \(T^{-1}NT\in {\mathbf{B}}_{d}[\mathbf{K}]\) and \((T^{-1}NT )^{\mathcal{\mathcal{D }}}= T^{-1}N^{\mathcal{D}}T\).
$$ \textstyle\begin{cases} (N^{*} )^{\mathcal{D}} = (N^{\mathcal{D}} )^{*} \\ (N^{k} )^{\mathcal{D}}= (N^{\mathcal{D}} )^{k},\quad \forall k\in \mathbb{N}. \end{cases} $$
Lemma 1.1
(1) NT is Drazin invertible if and only if TN is Drazin invertible. Moreover
$$ (NT)^{\mathcal{D}} = N\bigl[(TN)^{\mathcal{D}}\bigr]^{\mathcal{D}}T\quad \textit{and}\quad \operatorname{ind}(NT) \leq \operatorname{ind}(TN)+ 1 $$
Anzeige
(2) If \(NT=TN\), then \((NT)^{\mathcal{D}} =T^{\mathcal{D}}N^{\mathcal{D}} = N^{\mathcal{D}}T^{ \mathcal{D}}\), \(N^{\mathcal{D}}T = TN^{\mathcal{D}}\) and \(NT^{\mathcal{D}} = T^{\mathcal{D}}N\).
(3) If \(NT = TN = 0\), then \((N+T )^{\mathcal{D}} = N^{\mathcal{D}}+T^{\mathcal{D}}\).
The success of the theory of normal operators on Hilbert spaces has led to several attempts to generalize it to large classes of operators, including normal operators.
For \(N,T \in {\mathbf{B}}[\mathbf{K}]\), we set \([N, T ]=NT-TN\). An operator \(N \in {\mathbf{B}}[\mathbf{K}]\) is called
(iii) \((n,m)\)-normal if \([N^{n}, (N^{m} )^{*} ]=0\), where n, m are two nonnegative integers ([1, 3, 4]).
These concepts of normality, studied for \(N \in \mathbf{B}[\mathbf{K}]\), have been extended to the class of Drazin inverse of bounded linear operators on K as follows: For \(R\in{ \mathbf{B}}_{d}[{ \mathbf{K}}]\), R is said to be
(iii) \((n,m)\)-power \(\mathcal{D}\)-normal if \([ (N^{\mathcal{D}} )^{n}, (N^{*} )^{m} ]=0\) for some positive integers n and m ([28]),
The study of p-tuples of operators has received great interest from many authors in recent years. Some developments in this field have been presented in [7, 9, 10, 13, 15, 16, 23, 27, 29], and further references can be found therein.
Given a p-tuple \({\mathbf{\large N}} := (N_{1}, \ldots , N_{p} )\in {\mathbf{B}}[{ \mathbf{K}}]^{p}\), we define \([{\mathbf{\large N}}^{*},{\mathbf{\large N}}]\in {\mathbf{B}}[{\mathbf{K}} \oplus \cdots \oplus{\mathbf{K}}]\) as the self-commutator of N, which is given by where \({\mathbf{\large N}}^{*}:=(N_{1}^{*},\ldots ,N_{p}^{*})\).
$$ \bigl[{\mathbf{\large N}}^{*},{\mathbf{\large N}} \bigr]_{k,l}:=\bigl[N_{l}^{*}, N_{k}\bigr]\quad \forall (k,l)\in \{1,\ldots ,p\}^{2}, $$
We shall say, following ([7, 18]), that N is jointly hyponormal if is a positive operator on \(\mathbf{K}\oplus \cdots \oplus \mathbf{K}\), or equivalently N is said to be jointly normal if N ([8]) satisfying
$$ \bigl[ {\mathbf{\large N}}^{*},{\mathbf{\large N}}\bigr]= \begin{pmatrix} [N_{1}^{*}, N_{1}] & [N_{2}^{*},N_{1}] &\cdots & [N_{p}^{*},N_{1}] \cr [N_{1}^{*}, N_{2} ]& [N_{2}^{*}, N_{2}] &\cdots & [N_{p}^{*},N_{2}] \cr \vdots & \vdots & \vdots & \vdots \cr [N_{1}^{*},N_{p}] & [N_{2}^{*}, N_{p}] &\cdots & [N_{p}^{*}, N_{p}] \cr \end{pmatrix} $$
$$ \sum_{1\leq i,k\leq p} \bigl\langle \bigl[ N_{i}^{*}\quad N_{k} \bigr]x \big|x \bigr\rangle \geq 0\quad \forall x\in {\mathbf{K}}. $$
$$ \textstyle\begin{cases} [N_{k},N_{l} ]=0,\quad k,l\in \{1,\ldots ,p\} \\ [N_{k}^{*},N_{k} ]=0,\quad k=1,\ldots ,p \end{cases} $$
Recently, in [5], the author has introduced the concept of jointly n-normal tuple as follows: \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in { \mathbf{B}}[{ \mathbf{K}}]^{p}\) is said to be joint n-normal operators if R satisfying for some positive integer n.
$$ \textstyle\begin{cases} [ N_{k},N_{l} ]=0,\quad k, l\in \{1,\ldots ,p\} \\ [N_{k}^{n},N_{k}^{*} ]=0,\quad k=1,\ldots ,p, \end{cases} $$
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in {\mathbf{B}}_{d}[\mathbf{K}]^{p}\). We set \({\mathbf{\large N}}^{\mathcal{D}}:= (N_{1}^{\mathcal{D}},\ldots ,N_{p}^{ \mathcal{D}} )\).
The present paper proposes and studies the concept of p-tuples of \((n, m)\)-\(\mathcal{D}\)-normal operators. These are natural generalizations of \(\mathcal{D}\)-normal, n-power \(\mathcal{D}\)-normal, and \((n, m)\)-power \(\mathcal{D}\)-normal single operators as done in [19, 28]. For more details on some classes of Drazin inverse operators, the reader is invited to consult [20, 21, 33].
This paper has been organized into two sections. In section two, we introduce the class of p-tuples of \((n, m)\)-\(\mathcal{D}\)-normal operators associated with Drazin invertible operators using their Drazin inverses. Some properties of this class are studied along with examples. In the third section, the tensor product of some members of this class is discussed.
2 p-tuple of \((n,m)\)-Drazin normal oreators
In this section, we introduce and study the class of jointly \((n,m)\)-power D-normal multioperators.
Definition 2.1
Let \({\mathbf{\large N}}:=(N_{1},\ldots ,N_{p}) \in {\mathbf{B}}_{d}[ \mathbf{K}]^{p}\). We said that N is p-tuple of \((n, m)\)-Drazin normal operators for some positive integers n and m if N satisfies the following conditions When \(n=m=1\), we said that N is p-tuple of \(\mathcal{D}\)-normal operators and if \(m=1\), N is p-tuple of n-\(\mathcal{D}\)-normal operators.
$$ \textstyle\begin{cases} [N_{k}, N_{l} ] =0;\quad \forall (k,l)\in \{1,\ldots ,p \}^{2} \\ [ (N_{k}^{\mathcal{D}} )^{n},N_{k}^{*m} ]=0\quad \forall k=1,\ldots ,p. \end{cases} $$
Example 2.1
Let \(N\in { \mathbf{B}}_{d}[\mathbf{K}]\) be an \((n,m)\)-\(\mathcal{D}\)-normal operator, then \({\mathbf{\large N}}=(N,\ldots ,N)\in { \mathbf{B}}_{d}[\mathbf{K}]^{p}\) is p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
Example 2.2
Let \({\mathbf{\large N}}:= (N_{1},\ldots ,N_{p} ) \in {\mathbf{B}}_{d}[ \mathbf{K}]^{p}\) be commuting operators. If each \(N_{k}\) is \((n,m)\)-\(\mathcal{D}\)-normal single operator, then N is p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
The following example shows that there exists a p-tuple of operators \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in { \mathbf{B}}_{d}({ \mathbf{K}})^{p}\) such that each \(N_{k}\) is \((n,m)\)-\(\mathcal{D}\)-normal for \(k=1,\ldots ,p\), however N is not p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. This means that the study of these concepts is not trivial.
Example 2.3
Let \({\mathbf{\large N}}=(N_{1}, N_{2})\in { {\mathbf{B}}}[{\mathbb{C}}^{4}]\) where It was observed in [19] that \(N_{1}\) and \(N_{2}\) are in \({\mathbf{B}}_{d}[\mathbb{C}^{4}]\) and It is easy to check that \([N_{1}, N_{2} ]\neq0\) and \([ (N_{j}^{\mathcal{D}} )^{n},N_{j}^{*m} ]=0\) for \(j=1,2\). This means that, each \(N_{j}\) is \((n,m)\)-power \(\mathcal{D}\)-normal, while that N is not p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
$$ N_{1}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 &1 & 1 & 0 \end{pmatrix} \quad \text{and}\quad N_{2}= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 &0 & 0 & 0 \end{pmatrix} . $$
$$ N_{1}^{\mathcal{D}}= \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 &0 & 0 & 0 \end{pmatrix} \quad \text{and} \quad N_{2}^{\mathcal{D}}= \begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 &0 & 0 & 0 \end{pmatrix} . $$
In the the following theorem we collect some properties of p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
Theorem 2.1
Let \({{\mathbf{\large N}}}=(N_{1},\ldots ,N_{p})\in {\mathbf{B}}_{d}[ \mathbf{K}]^{p}\) be p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators, then the following properties hold.
(1) N is p-tuple of \((rn,sm)\)-\(\mathcal{D}\)-normal operators for some positive integers r and s.
(2) \({\mathbf{\large{N}}}^{q}:=(N_{1}^{q_{1}},\ldots ,N_{p}^{q_{p}})\) is p-tuple of \((n,m)\)-D-normal operators for \(q=(q_{1},\ldots ,q_{p}) \in \mathbb{N}^{p}\).
(3) \({\mathbf{\large N}}^{*}=(N_{1}^{*},\ldots ,N_{p}^{*})\) is p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
(4) If V is an unitary operator, then \(V^{*}{\mathbf{\large N}}V:= (V^{*}N_{1}V,\ldots ,V^{*}N_{p}V )\) is p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
Proof
(1) Since N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators, it follows that \([N_{k},N_{l} ]=0\) for \(k,l=1,\ldots ,p\). However,
$$\begin{aligned} \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{rn}, N_{k}^{*(sm)} \bigr] =& \bigl(N_{k}^{\mathcal{D}} \bigr)^{rn}N_{k}^{*(sm)}- N_{k}^{*(sm)} N_{k}^{ \mathcal{D}} )^{rn} \\ =& \underbrace{ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} \cdots \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}}_{r- \text{times}}. \underbrace{N_{k}^{*m}\cdots N_{k}^{*m}}_{s-\text{times}}- \underbrace{N_{k}^{*m}\cdots N_{k}^{*m}}_{s-\text{times}} \underbrace{ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} \cdots \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}}_{r- \text{times}} \\ =&\underbrace{N_{k}^{*m}\cdots N_{k}^{*m}}_{s-\text{times}} \underbrace{ \bigl( N_{k}^{\mathcal{D}} \bigr)^{n} \cdots \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}}_{r- \text{times}}- \underbrace{N_{k}^{*m}\cdots N_{k}^{*m}}_{s-\text{times}} \underbrace{ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} \cdots \bigl(N_{k}^{\mathcal{\mathcal{D}}} \bigr)^{n}}_{r- \text{times}} \\ =&0. \end{aligned}$$
(2) If \(q_{k}=1\) for all \(k \in \{1,\ldots ,q\}\), then \([N_{k}^{q_{k}},N_{l}^{q_{l}}]=0\).
If \(q_{k}>1\) for all \(k \in \{1,\ldots ,p\}\), by taking into account [29, Lemma 2.1], we have Now, under the assumption that N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators, it follows that By looking that \(N_{k}\) is an \((n,m)\)-\(\mathcal{D}\)-normal, then from [28, Proposition 2.10], we obtain that \(N_{k}^{q_{k}}\) is an \((n,m)\)-\(\mathcal{D}\)-normal for all \(k \in \{1,\ldots ,q\}\). This means that \((N_{1}^{q_{1}},\ldots ,N_{p}^{q_{p}} )\) is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
$$ \bigl[N_{k}^{q_{k}},N_{l}^{q_{l}} \bigr]=\sum_{ \substack{\alpha +\alpha '=q_{k}-1 \\ \beta +\beta '=q_{l}-1}}N_{k}^{ \alpha}N_{l}^{\beta}[N_{k}, N_{l}]N_{l}^{\alpha '}N_{k}^{\beta '}. $$
$$\begin{aligned} \bigl[N_{k}^{q_{k}},N_{l}^{q_{l}} \bigr] =&\sum_{ \substack{\alpha +\alpha '=q_{k}-1 \\ \beta +\beta '=q_{l}-1}}N_{k}^{ \alpha}N_{l}^{\beta}[N_{k}, N_{l}]N_{l}^{\alpha '}N_{k}^{\beta '}=0, \quad \forall (k,l)\in \{1,\ldots ,p\}^{2}. \end{aligned}$$
(3) From Definition 2.1, we have under the condition that N is a p-tuple of \((n,m)\)-power \(\mathcal{D}\)-normal operators that is and therefore, Hence, \({\mathbf{\large N}}^{*}\) is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
$$ \textstyle\begin{cases} [N_{k}, N_{l} ]=0\quad \text{for all } (k,l)\in \{1,\ldots ,p \}^{2} \\ [ (N_{k}^{\mathcal{D}} )^{n},N_{k}^{*m} ]=0\quad \text{for } k=1,\ldots ,p. \end{cases} $$
$$ \textstyle\begin{cases} [N_{k}^{*},N_{l}^{*} ]=0 \quad \text{for all } (k,l)\in \{1, \ldots ,p\}^{2} \\ [ (N_{k}^{\mathcal{D}} )^{*n}, N_{k}^{m} ]=0 \quad \text{for } k=1,\ldots ,q. \end{cases} $$
(4) We observe that Moreover, Hence, \(V^{*}{\mathbf{\large N}}V\) is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. □
$$\begin{aligned} \bigl[V^{*}N_{k}V, V^{*}N_{l}V \bigr] =& \bigl(V^{*}N_{k}V \bigr) \bigl(V^{*}N_{l}V \bigr)- \bigl(V^{*}N_{l}V \bigr) \bigl(V^{*}N_{k}V \bigr) \\ =&V^{*}N_{k}N_{l}V-V^{*}N_{l}N_{k}V \\ =&V^{*} [N_{k},N_{l} ]V \\ =&0. \end{aligned}$$
$$\begin{aligned} \bigl[ \bigl(V^{*}N_{k}V \bigr)^{\mathcal{D}n}, \bigl( V^{*}N_{k}V \bigr)^{*m} \bigr] =& V^{*} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}V V^{*}N_{k}^{*m}V-V^{*} N_{k}^{*m}VV^{*} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}V \\ =&V^{*} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}N_{k}^{*m}V-V^{*} N_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}V \\ =&V^{*} \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n},N_{k}^{*m} \bigr]V \\ =&0. \end{aligned}$$
Proposition 2.1
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p}) \in { \mathbf{B}}_{d}[ \mathbf{K}]^{p}\). The following statements are true.
(1) If N is a p-tuple of \((n,n)\)-\(\mathcal{D}\)-normal operators then \(({\mathbf{\large N}}^{\mathcal{D}} )^{n}:= ( (N_{1}^{ \mathcal{D}} )^{n},\ldots , (N_{p}^{\mathcal{D}} )^{n} )\) is a p-tuple of normal operators.
(2) If \(({\mathbf{\large N}}^{\mathcal{D}} )^{n}\) is a p-tuple of normal operators and \(N_{k}N_{l}-N_{l}N_{k}=0\) for all \(k,l=1,\ldots ,p\), then N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
Proof
(1) If N is a p-tuple of \((n,n)\)-\(\mathcal{D}\)-normal operators. Then we get
However,
Therefore \(({\mathbf{\large N}}^{\mathcal{D}} )^{n}\) is a p-tuple of normal operators.
(2) Since \(({\mathbf{\large N}}^{\mathcal{D}} )^{n}\) is a p-tuple of normal operators, we have that Moreover, it is well known that \([N_{k}^{\mathcal{D}},N_{k} ]=0\) for each \(k=1,\ldots ,p\) and hence By taking into account the Fugled–Putnam theorem ([31]), it follows that \([ (N_{k}^{\mathcal{D}} )^{n},N_{k}^{*m} ]=0\) for each \(k=1,\ldots ,p\). Therefore N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. □
$$ \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}, \bigl(N_{k}^{\mathcal{D}}\bigr)^{*n} \bigr]=0,\quad \text{for each } k=1,\ldots ,p. $$
$$ \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}, N_{k} \bigr]=0,\quad k=1,\ldots ,p. $$
Proposition 2.2
Let \({{\mathbf{\large N}}}=(N_{1},\ldots ,N_{p})\in {\mathbf{B}}_{d}[ \mathbf{K}]^{p}\). The following assertions hold.
(1) If N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n+1,m)\)-\(\mathcal{D}\)-normal operators, then N is a p-tuple of \((n+2,m)\)-\(\mathcal{D}\)-normal operators.
(2) If N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n,m+1)\)-\(\mathcal{D}\)-normal operators, then N is a p-tuple of \((n,m+2)\)-\(\mathcal{D}\)-normal operators.
Proof
Since N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n+1,m)\)-\(\mathcal{D}\)-normal operators, we have This implies that and therefore, So, N is a p-tuple of \((n+2,m)\)-\(\mathcal{D}\)-normal operators.
$$ \textstyle\begin{cases} [N_{k},N_{l} ]=0\quad \forall k,l=1,\ldots ,p \\ [ ( N_{k}^{\mathcal{D}} )^{n},N_{k}^{*m} ]=0 \quad k=1, \ldots ,p \\ [ ( N_{k}^{\mathcal{D}} )^{n+1},N_{k}^{*m} ]=0,\quad k=1, \ldots ,p. \end{cases} $$
$$ \textstyle\begin{cases} [N_{k},N_{l} ]=0,\quad \forall k,l=1,\ldots ,p \\ (N_{k}^{\mathcal{D}} )^{n} [N_{k}^{\mathcal{D}}N_{k}^{*m}-N_{k}^{*m}N_{k}^{ \mathcal{D}} ]=0, \quad k=1,\ldots ,p, \end{cases} $$
$$ \textstyle\begin{cases} [N_{k},N_{l} ]=0\quad \forall k,l=1,\ldots ,p \\ [ (N_{k}^{\mathcal{D}} )^{n+2},N_{k}^{*m} ]=0, \quad k=1, \ldots ,p. \end{cases} $$
(2) The proof of the statement (2) is similar to the proof of statement (1), so we omit it. □
Proposition 2.3
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in {\mathbf{ B}}_{d}[\mathbf{K}]^{p}\), the following statements hold:
(1) If N is a p-tuple of \((n_{1},m)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n_{2},m)\)-\(\mathcal{D}\)-normal operators, then N is a p-tuple of \((n_{1}+n_{2},m)\)-\(\mathcal{D}\)-normal operators.
(2) If N is a p-tuple of \((n,m_{1})\)-\(\mathcal{D}\)-normal operators and a p-tuple of \((n,m_{2})\)-\(\mathcal{D}\)-normal operators, then N is a p-tuple of \((n,m_{1}+m_{2},)\)-\(\mathcal{D}\)-normal operators.
(3) If N is a p-tuple of \((n_{1},m)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n_{2},m)\)-\(\mathcal{D}\)-normal operators, then N is a p-tuple of \((rn_{1}+s n_{2},m)\)-\(\mathcal{D}\)-normal operators for \(r,s \in \mathbf{N}\).
(4) If N is a p-tuple of \((n,m_{1})\)-\(\mathcal{D}\)-normal and a p-tuple of \((n,m_{2})\)-\(\mathcal{D}\)-normal operators, then N is a p-tuple of \((n,rm_{1}+sm_{2},)\)-\(\mathcal{D}\)-normal operators for \(r,s \in \mathbb{N}\).
Proof
(1) We have \([N_{k},N_{l} ]=0\) for \(k,l=1,\ldots ,p\) and moreover for \(k=1,\ldots ,p\), (2) We have \([N_{k},N_{l} ]=0\) for \(k,l=1,\ldots ,p\) and moreover for \(k=1,\ldots ,p\), Therefore, the required results are satisfied. □
$$\begin{aligned} \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n_{1}+n_{2}}, N_{k}^{*m} \bigr] =& \bigl(N_{k}^{\mathcal{D}} \bigr)^{n_{1}+n_{2}}N_{k}^{*m}-N_{k}^{*m} \bigl(N_{k}^{ \mathcal{D}} \bigr)^{n_{1}+n_{2}} \\ =& \bigl(N_{k}^{\mathcal{D}} \bigr)^{n_{1}} \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n_{2}}, N_{k}^{*m} \bigr] \\ =&0. \end{aligned}$$
$$\begin{aligned} \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}, N_{k}^{*(m_{1}+m_{2})} \bigr] =& \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}N_{k}^{*(m_{1}+m_{2})}-N_{k}^{*(m_{1}+m_{2})} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n_{1}+n_{2}} \\ =& \bigl[ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}, N_{k}^{*m_{1}} \bigr]N_{k}^{*m_{2}} \\ =&0. \end{aligned}$$
Theorem 2.2
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in {\mathbf{ B}}_{d}[\mathbf{K}]^{p}\) such that If N is a p-tuple of \((n_{1},m)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n_{2},m)\)-\(\mathcal{D}\)-normal operators for some positive integer \(n_{1}\), \(n_{2}\) and m, then, N is a p-tuple of \((\max \{n_{1},n_{2}\}-\min \{n_{1},n_{2}\},m)\)-\(\mathcal{D}\)-normal operators. In particular, if N is jointly \((n,1)\)-\(\mathcal{D}\)-normal and a p-tuple of \((n+1,1)\)-\(\mathcal{D}\)-normal operators, then N is p-tuple of \(\mathcal{D}\)-normal operators.
$$ \ker \bigl({\mathbf{\large N^{\mathcal{D}}}}\bigr):=\bigcap _{1\leq k \leq p}\ker{N_{k}^{\mathcal{D}}}=\{0\}. $$
Proof
We have \([N_{k},N_{l} ]=0\) for all \((k,l)\in \{1,\ldots ,p\}^{2}\). Moreover, for each \(k=1,\ldots ,p\), we have Considering the case where \(n_{1}\geq n_{2}\), we get
$$ \textstyle\begin{cases} [ (N_{k}^{\mathcal{D}} )^{n_{1}}N_{k}^{*m} ]=0 \\ [ ( N_{k}^{\mathcal{D}} )^{n_{2}},N_{k}^{*m} ]=0 \end{cases} $$
and hence N is a p-tuple of \((n_{1}-n_{2},m)\)-\(\mathcal{D}\)-normal operators. □
Proposition 2.4
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p}) \in \mathbf{B}[\mathbf{K}]^{p}\) be commuting tuple of Drazin invertible operators. For \(n,m\in \mathbb{N}\), set and Then the following axioms hold.
$$ { \mathbf{\large N}}^{\prime}= \bigl(N_{1}^{\prime}, \ldots ,N_{p}^{\prime} \bigr)= \bigl( \bigl(N_{1}^{\mathcal{D}} \bigr)^{n}+N1^{*m},\ldots , \bigl(N_{p}^{ \mathcal{D}} \bigr)^{n}+N_{p}^{*m} \bigr) $$
$$ { \large \mathbf{N}}^{\prime \prime}= \bigl(N_{1}^{\prime}, \ldots ,N_{p}^{ \prime \prime} \bigr)= \bigl( \bigl(N_{1}^{\mathcal{D}} \bigr)^{n}-N_{1}^{*m}, \ldots , \bigl(N_{p}^{\mathcal{D}} \bigr)^{n}-N_{p}^{*m} \bigr). $$
(1) N is a p-tuple of n-\(\mathcal{D}\)-normal operators if and only if \([{ N}_{k}^{\prime}, { N}_{k}^{\prime \prime}]=0\) for each \(k=1,\ldots ,p\).
(2) If N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators, then \({\mathbf{\large Z}} = ( (N_{1}^{D} )^{n}N_{1}^{*m},\ldots , (N_{p}^{\mathcal{D}} )^{n}N_{p}^{*m} )\) commutes with \({\mathbf{\large N}}^{\prime}\) and \({\mathbf{\large N }}^{\prime \prime}\).
(3) N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators, if and only if \(({\mathbf{\large N}}^{\mathcal{D}} )^{n}\) commutes with \({\mathbf{\large N}}^{\prime}\).
(4) N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators if and only if \(({\mathbf{\large N}}^{\mathcal{D}} )^{n}\) commutes with \({\mathbf{\large N}}^{\prime \prime}\).
Proof
Obviously, \([N_{k},N_{l} ]=0\forall (k,l)\in \{1,\ldots ,p\}^{2}\). On the other hand, This completes the proof. □
$$\begin{aligned} & \bigl[{ N}_{k}^{\prime}, {N}_{k}^{\prime \prime} \bigr]=0 \\ &\quad \Longleftrightarrow \quad N_{k}^{\prime }N_{k}^{\prime \prime}-N_{k}^{ \prime \prime}N_{k}^{\prime}=0 \\ &\quad \Longleftrightarrow \quad \bigl( \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}+N_{k}^{*m} \bigr) \bigl( \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}-N_{k}^{*m} \bigr)- \bigl( \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}-N_{k}^{*m} \bigr) \bigl( \bigl(N_{k}^{ \mathcal{D}} \bigr)^{n}+N_{k}^{*m} \bigr)=0 \\ &\begin{aligned} \quad \Longleftrightarrow \quad &\bigl( N_{k}^{\mathcal{D}} \bigr)^{2n}- \bigl(N_{k}^{ \mathcal{D}} \bigr)^{n}N_{k}^{*m}+N_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}-N_{k}^{*2m}\\ &\quad - \bigl( \bigl(N_{k}^{\mathcal{D}} \bigr)^{2n}+ \bigl(N_{k}^{ \mathcal{D}} \bigr)^{n}N_{k}^{*m}-N_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}-N_{k}^{*2m} \bigr) =0 \end{aligned} \\ &\quad \Longleftrightarrow \quad \bigl( N_{k}^{\mathcal{D}} \bigr)^{n}N_{k}^{*m}-N_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}=0,\quad \forall k \in \{1,\ldots ,p\}. \end{aligned}$$
Theorem 2.3
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p}) \in {\mathbf{B}}_{d}[\mathbf{K}]^{p}\) be p-tuple of \((n, m)\)-\(\mathcal{D}\)-normal operators for \(n\geq m\). If each \(N_{k}^{m}\) is a partial isometry for \(k=1,\ldots ,m\), then N is a p-tuple of \((n+m,m)\)-\(\mathcal{D}\)-normal operators.
Proof
Suppose N is p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators for \(n\geq m\). It is easy to see that each \(N_{k}\) is \((n,m)\)-D-normal for \(1\leq k \leq d\). Under the hypothesis that \(N_{k}^{m}\) is a partial isometry, it follows from [28, Theorem 2.4] that \(N_{k}\) is \((n+m,m)\)-\(\mathcal{D}\)-normal operator for \(k=1,\ldots ,p\). Consequently, N is a p-tuple of \((n+m,m)\)-\(\mathcal{D}\)-normal operators. □
The following proposition shows that the class of p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators is closed subset of \({ \mathbf{B}}_{d}[{\mathbf{K}}]^{p}\).
Proposition 2.5
The class of p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators is a closed subset of \({\mathbf{B}}_{d}[\mathbf{K}]^{p}\).
Proof
Suppose that \(( {\mathbf{\large N}}_{k}= (N_{1}(k),\ldots ,N_{p}(k) ) )_{k} \in {\mathbf{B}}_{d}[\mathbf{K}]^{p}\) is a sequence of p-tuple of \((n,m)\)-power \(\mathcal{D}\)-normal operators for which where \({\mathbf{\large N}}= (N_{1},\ldots , N_{p}) \in { \mathbf{B}}_{d}[ \mathbf{K}]^{p}\). Obviously, for each \(j\in \{1,\ldots ,p\}\), we have
$$ \Vert {\mathbf{\large N}}_{k}-{\mathbf{\large N}} \Vert =\sup _{1\leq j \leq p} \bigl( \bigl\Vert N_{j}(k)-N_{j} \bigr\Vert \bigr)\longrightarrow 0,\quad \text{as } k\longrightarrow \infty , $$
$$ \lim_{k\rightarrow +\infty} \bigl\Vert N_{j}(k)-N_{j} \bigr\Vert =0. $$
(2.1)
Since \((N_{j}(k)^{\mathcal{D}} )^{n}N_{j}(k)^{*m}=N_{j}(k)^{*m} (N_{j}(k)^{ \mathcal{D}} )^{n}\) for each \(j=1,\ldots ,p\), it follows from [28, Theorem 2.4] that Moreover, for all \(i,j \in \{1,\ldots ,p\}\) and \(k\in \mathbb{N}\), we can see that Hence, in view of (2.1), we obtain On the other hand, since \(\{{\mathbf{\large N}}_{k}\}_{k}=\{ (N_{1}(k),\ldots ,N_{p}(k))\}_{k}\) is a sequence of p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators, then Therefore, we immediately get Therefore, N is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. □
$$ \bigl(N_{j}^{\mathcal{D}} \bigr)^{n}N_{j}^{*m}=N_{j}^{*m} \bigl(N_{j}^{ \mathcal{D}} \bigr)^{n},\quad \forall j \in \{1,\ldots ,p \}. $$
$$\begin{aligned} \bigl\Vert N_{i}(k)N_{j}(k)-N_{i}N_{j} \bigr\Vert =& \bigl\| N_{i}(k) \bigl(N_{j}(k)-N_{j} \bigr)+ \bigl(N_{i}(k)-N_{i}\bigr)N_{j}\bigr\| \\ \leq & \bigl\Vert N_{i}(k) \bigr\Vert \bigl\Vert N_{j}(k)-N_{j} \bigr\Vert + \bigl\Vert N_{i}(k)-N_{i} \bigr\Vert \Vert N_{j} \Vert \\ \leq & \bigl( \bigl\Vert N_{i}(k)-N_{i} \bigr\Vert + \Vert N_{i} \Vert \bigr) \bigl\Vert N_{j}(k)-N_{j} \bigr\Vert + \bigl\Vert N_{i}(k)-N_{i} \bigr\Vert \Vert N_{j} \Vert . \end{aligned}$$
$$ \bigl\Vert N_{i}(k)N_{j}(k)-N_{i}N_{j} \bigr\Vert \longrightarrow 0, \quad \text{as }k \rightarrow +\infty , \forall (i,j)\in \{1,\ldots ,q\}^{2}. $$
$$ \bigl[N_{i}(k),N_{j}(k)\bigr]=0 \quad \forall (i,j) \in \{1,\ldots ,p\}^{2} ;\text{and } k\in \mathbb{N}. $$
$$ [N_{i},N_{j}]=0\quad \forall (i,j) \in \{1,2,\ldots ,p\}^{2}. $$
Proposition 2.6
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in {\mathbf{B}}_{d}[\mathbf{K}]^{p}\) and \({\mathbf{\large S}}=(S_{1},\ldots ,S_{p})\in{\mathbf{B}}_{d}[\mathbf{K}]^{p}\) be two p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. The following statements hold.
(1) If \([N_{k},S_{l}]=0, \forall k\), \(l\in \{1,\ldots ,p\}^{2}\) and \([N_{k},S_{k}^{*}]=0\) for all \(k\in \{1,\ldots ,p\}\), then \({\mathbf{\large NS}}= (N_{1}S_{1},\ldots ,N_{p}S_{p})\) and \({\mathbf{SN}}=(S_{1}N_{1},\ldots ,S_{p}N_{p})\) are p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
(2) If \([N_{i},S_{j}]=0, \forall i\), \(j\in \{1,\ldots ,p\}\) and \(N_{k}S_{k}=N_{k}S_{k}^{*}=0\) for all \(k\in \{1,\ldots ,p\}\), then \({\mathbf{\large N}+\mathbf{S}}=(N_{1}+S_{1},\ldots ,N_{p}+S_{p})\) is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
Proof
For the statement (1), we have for all \(k,l\in \{1,\ldots ,q\}\), On the other hand, let \(k\in \{1,\ldots ,p\}\), we have This implies that NS is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. In same way, we show that SN is a p-tuple of \((n.m)\)-\(\mathcal{D}\)-normal operators.
$$\begin{aligned}{} [N_{k}S_{k},N_{l}S_{l}] =& N_{k}S_{k}N_{l}S_{l}-N_{l}S_{l}N_{k}S_{k} \\ =& N_{k}N_{l}S_{k}S_{l}-N_{l}N_{k}S_{l}S_{k} \\ =& N_{k}N_{l}S_{k}S_{l}-N_{k}N_{l}S_{l}S_{k} \\ =& N_{k}N_{l}(S_{k}S_{l}-S_{l}S_{k}) \\ =& N_{k}N_{l}[S_{k},S_{l}]=0. \end{aligned}$$
$$\begin{aligned} (N_{k} S_{k})^{*} (N_{k} S_{k} )^{\mathcal{D}n} =& S_{k}^{*} N_{k}^{*} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} \bigl( S_{k}^{\mathcal{D}} \bigr)^{n} \\ =& S_{k}^{*} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} N_{k}^{*} \bigl(S_{k}^{ \mathcal{D}} \bigr)^{n}= S_{k}^{*} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} S_{k}^{n}N_{k}^{*} \\ =& \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} \bigl(S_{k}^{\mathcal{D}} \bigr)^{n}S_{k}^{*}N_{k}^{*} \\ =& (N_{k}S_{k} )^{n\mathcal{D}} (N_{k}S_{k} )^{*}. \end{aligned}$$
(2) For all \((i,j)\in \{1,\ldots ,p\}^{2}\), we have Besides, for \(k\in \{1,2,\ldots ,p\}\), we get So, \({\mathbf{\large N}+\mathbf{S}}\) is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. □
$$\begin{aligned}{} [N_{k}+S_{k},N_{l}+S_{l} ] =& (N_{k}+S_{k} ) (N_{l}+S_{l} )- (N_{l}+S_{l}) (N_{k}+S_{k} ) \\ =& [N_{k},N_{l} ]+ [S_{k},S_{l}]+[N_{k},S_{l}]+[S_{k},N_{l}]=0. \end{aligned}$$
$$\begin{aligned} &(N_{k}+ S_{k} )^{*m} \bigl( (N_{k} +S_{k} )^{\mathcal{D}} \bigr)^{n} \\ &\quad = (N_{k}+ S_{k} )^{*m} \bigl(N_{k}^{\mathcal{D}} +S_{k}^{ \mathcal{D}} \bigr)^{n} \\ &\quad = \Biggl(\sum_{j=0}^{m} \binom{m}{j}N_{k}^{*j}S_{k}^{*m-j} \Biggr) \Biggl(\sum_{j=0}^{n} \binom{n}{k} \bigl(N_{k}^{\mathcal{D}} \bigr)^{j} \bigl(S_{k}^{\mathcal{D}} \bigr)^{n-j} \Biggr) \\ &\quad = \bigl(N_{k}^{*m}+S_{k}^{*m} \bigr) \bigl( \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}+ \bigl(S_{k}^{\mathcal{D}} \bigr)^{n} \bigr) \\ &\quad = (N_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}+N_{k}^{*m} \bigl(S_{k}^{ \mathcal{D}} \bigr)^{n} +S_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} +S_{k}^{*m} \bigl(S_{k}^{\mathcal{D}} \bigr)^{n} \\ &\quad = \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}N_{k}^{*m}+ \bigl(S_{k}^{ \mathcal{D}} \bigr)^{n}S_{k}^{*m} \\ &\quad = \bigl( \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} + \bigl(S_{k}^{\mathcal{D}} \bigr)^{n} \bigr) (N_{k}+S_{k} )^{*m} \\ &\quad = \Biggl(\sum_{j=0}^{n} \binom{n}{k} \bigl(N_{k}^{\mathcal{D}} \bigr)^{j} \bigl(S_{k}^{\mathcal{D}} \bigr)^{n-j} \Biggr) (N_{k}+ S_{k} )^{*m} \\ &\quad = \bigl( (N_{k} +S_{k} )^{\mathcal{D}} \bigr)^{n} (N_{k}+S_{k} )^{*m}. \end{aligned}$$
3 Tensor product
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in { \mathbf{B}}[\mathbf{K}]^{p}\) and \({\mathbf{\large S}}=(S_{1},\ldots ,S_{p})\in { \mathbf{B}}[\mathbf{K}]^{p}\). We denote by If \(N,S \in { \mathbf{B}}[\mathbf{K}]\), then \(N\otimes S\) is n-normal if and only if N and S are n-normal (see [6]) However, If \(N ,S\in { \mathbf{B}}_{d}[\mathbf{K}]\) such that N and S are \((n,m)\)-\(\mathcal{D}\)-normal operators, then \(N\otimes S\) is \((n,m)\)-\(\mathcal{D}\)-normal (see in [28]). The following theorem studied the tensor product of two p-tuples of \((n,m)\)-\(\mathcal{D}\)-normal operators.
$$ {\mathbf{\large N}}\otimes {\mathbf{\large S}}= (N_{1}\otimes S_{1},\ldots ,N_{p} \otimes S_{p}) $$
Theorem 3.1
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in {\mathbf{B}}_{d}[\mathbf{K}]^{p}\) and \({\mathbf{\large S}}=(S_{1},\ldots ,S_{p})\in \mathbf{B}_{d}[\mathbf{K}]^{p}\) are two p-tuples of \((n,m)\)-\(\mathcal{D}\)-normal operators, then \({\mathbf{\large N}}\otimes {\mathbf{\large S}}\) is a p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators.
Proof
Since \({\mathbf{\large N}}= (N_{1},\ldots ,N_{p} )\) and \({\mathbf{\large S}} = (S_{1},\ldots ,S_{p} )\) are p-tuples of \((n,m)\)-\(\mathcal{D}\)-normal operators, we have all \((k,l) \in \{ 1,\ldots ,p\}^{2}\) Moreover, for all \(k \in \{1,\ldots , p \}\), we have So, \({\mathbf{\large N}}\otimes {\mathbf{\large S}}\) is p-tuple of \((n,m)\)-\(\mathcal{D}\)-normal operators. □
$$\begin{aligned} & \bigl[(N_{k} \otimes S_{k}),(N_{l} \otimes S_{l} ) \bigr] \\ &\quad = \bigl[(N_{k} \otimes S_{k}) (N_{l} \otimes S_{l} )-(N_{l}\otimes S_{l}) (N_{k} \otimes S_{k}) \bigr] \\ &\quad =N_{k}N_{l}\otimes S_{k}S_{l}-N_{l}N_{k} \otimes S_{l}S_{k} \\ &\quad =N_{l}N_{k}\otimes S_{l}S_{k}-N_{l}N_{k} \otimes S_{l}S_{k} \\ &\quad =0. \end{aligned}$$
$$\begin{aligned} \bigl( (N_{k}\otimes S_{k} )^{\mathcal{D}} \bigr)^{n} (N_{k} \otimes S_{k} )^{*m} =& \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}N_{k}^{*m} \otimes \bigl( S_{k}^{\mathcal{D}} \bigr)^{n}S_{k}^{*m} \\ =&N_{k}^{*m} \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}\otimes S_{k}^{*m} \bigl(S_{k}^{D} \bigr)^{n} \\ =& (N_{k}\otimes S_{k} )^{*m} \bigl( ( N_{k}\otimes S_{k} )^{\mathcal{D}} \bigr)^{n}. \end{aligned}$$
The converse of the above theorem need not hold in general, as shown in the following example.
Example 3.1
Let
and
. A direct calculation shows that and Let \({\mathbf{\large N}}= (N_{1},N_{2} )\) and \({\mathbf{\large N}}\otimes {\mathbf{\large N}}= (N_{1}\otimes N_{1},N_{2} \otimes N_{2}) ) \). We observe that N is not 2-tuple of \((2,3)\)-\(\mathcal{D}\)-normal operators since \(N_{1}N_{2}\neq N_{2}N_{1}\). However Hence, \({\mathbf{\large N}}\otimes {\mathbf{\large N}}\) is 2-tuple of \((2,3)\)-\(\mathcal{D}\)-normal pairs.
and
. A direct calculation shows that $$ N_{1}\otimes N_{1}= \begin{pmatrix} 1 & 0 & 0 & 0 &0 & 0 & 0 & 0& 0 \\ 0 & -1 & 0 & 0 & 0& 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix} $$
$$ N_{2}\otimes N_{2}= \begin{pmatrix} 0 & 0 & 0 & 0 &0 & 0 & 0 & 0& 1 \\ 0 & 0 & 0 & 0 & 0& 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \end{pmatrix}. $$
$$\begin{aligned} &(N_{1}\otimes N_{1} )^{\mathcal{D}}= N_{1}\otimes N_{1},\qquad (N_{2}\otimes N_{2} )^{\mathcal{D}}=N_{2}\otimes N_{2} \quad \text{and}\\ & (N_{k}\otimes N_{k} )^{*3} \bigl( (N_{k}\otimes N_{k} )^{ \mathcal{D}} \bigr)^{2}= \bigl( (N_{k}\otimes N_{k} )^{\mathcal{D}} \bigr)^{2} (N_{k}\otimes N_{k} )^{*3}, \quad k\in \{1,2\}. \end{aligned}$$
In the following theorem we give the conditions under which the converse of Theorem 3.1 is true.
Theorem 3.2
Let \({\mathbf{\large N}}=(N_{1},\ldots ,N_{p})\in { \mathbf{B}}_{d}[\mathbf{K}]^{p}\) and \({\mathbf{\large S}}=(S_{1},\ldots ,S_{p})\in { \mathbf{B}}_{d}[\mathbf{K}]^{p}\) be a commuting p-tuple of operators. Then, if \({\mathbf{\large N}}\otimes {\mathbf{\large S}}\) is a p-tuple of \((n,n)\)-\(\mathcal{D}\)-normal operators, then and only then N and S are p-tuples of \((n,n)\)-\(\mathcal{D}\)-normal operators.
Proof
Assume that \({\mathbf{\large N}}\otimes {\mathbf{\large S}}\) is a p-tuple of \((n,n)\)-\(\mathcal{D}\)-normal operators. By taking into account the statement (1) of Proposition 2.1 it follows that is a p-tuple of normal operators. From which we deduce that is normal for each \(k=1,\ldots ,p\). By [19, Propositon 3.2] it is well known that However, According to [19, Propositon 3.2] it is well known that \((N_{k}^{\mathcal{D}} )^{n}\) is normal if and only if that \(N_{k}\) is n-power \(\mathcal{D}\)-normal and similarly, \((S_{k}^{ \mathcal{D}} )^{n}\) is normal if and only if that \(S_{k}\) is n-\(\mathcal{D}\)-normal. Therefore, N and S are p-tuple of \((n,n)\)-\(\mathbb{D}\)-normal operators.
$$\begin{aligned} \bigl( ({\mathbf{\large N}}\otimes {\mathbf{\large S}} )^{\mathcal{D}} \bigr)^{n}&= \bigl( \bigl( (N_{1}\otimes S_{1} )^{\mathcal{D}} \bigr)^{n}, \ldots , \bigl( (N_{p}\otimes S_{p} )^{\mathcal{D}} \bigr)^{n} \bigr)\\ &= \bigl( \bigl(N_{1}^{\mathcal{D}} \bigr)^{n}\otimes \bigl(S_{1}^{ \mathcal{D}} \bigr)^{n},\ldots \bigl(N_{p}^{\mathcal{D}} \bigr)^{n} \otimes \bigl(S_{p}^{\mathcal{D}} \bigr)^{n} \bigr) \end{aligned}$$
$$ \bigl( (N_{k}\otimes S_{k} )^{\mathcal{D}} \bigr)^{n}= \bigl(R_{k}^{ \mathcal{D}} \bigr)^{n}\otimes \bigl(S_{k}^{\mathcal{D}} \bigr)^{n}, $$
$$ \bigl(N_{k}^{\mathcal{D}} \bigr)^{n}\otimes \bigl(S_{k}^{D} \bigr)^{n} \text{ is normal if and only if } \bigl(N_{k}^{\mathcal{D}} \bigr)^{n} \text{ and } \bigl(S_{k}^{\mathcal{D}} \bigr)^{n} \text{ are normal operators}. $$
The converse follows from Theorem 3.1. □
Declarations
Competing interests
The authors declare no competing interests.
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