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# Generalizations of some classical theorems to D-normal operators on Hilbert spaces

## Abstract

We say that a Drazin invertible operator T on Hilbert space is of class $$[DN]$$ if $$T^{D}T^{*} = T^{*}T^{D}$$. The authors in (Oper. Matrices 12(2):465–487, 2018) studied several properties of this class. We prove the Fuglede–Putnam commutativity theorem for D-normal operators. Also, we show that T has the Bishop property $$(\beta)$$. Finally, we generalize a very famous result on products of normal operators due to I. Kaplansky to D-normal matrices.

## Introduction

Let $$\mathcal{H}$$ be a complex Hilbert space. By $$\mathcal{B}(\mathcal {H})$$ we denote the space of all bounded linear operators on $$\mathcal {H}$$ and by $$I = I_{\mathcal{H}}$$ the identity operator. If $$T \in \mathcal{B}(\mathcal{H})$$, then $$T^{*}$$ denotes the adjoint of T. By $$\mathcal{N}(T)$$, $$\mathcal{R}(T)$$, and $$\sigma(T)$$ we denote the null space, the range, and the spectrum of T, respectively. For convenience, we write $$T - \lambda$$ instead of $$T - \lambda I$$.

Property $$(\beta)$$ has been introduced by Bishop  and is defined as follows.

### Definition 1.1

An operator $$T \in\mathcal{B}(\mathcal{H})$$ is said to have the Bishop property $$(\beta)$$ (shortly, property $$(\beta)$$) if for every open set D of $$\mathbb{C}$$ and every sequence of analytic functions $$f_{n} : D \longrightarrow\mathcal{H}$$ such that $$(T - \mu )f_{n}(\mu) \longrightarrow0$$ uniformly on all compact subsets of D, then also $$f_{n}(\mu) \longrightarrow0$$, again locally uniformly on D.

It is well known that every normal operator has property $$(\beta)$$. The study of operators satisfying property $$(\beta)$$ is of significant interest and is currently being done by a number of mathematicians around the world (see [3, 12]).

### Definition 1.2

Let $$T \in\mathcal{B}(\mathcal{H})$$. The operator T is said to have the single-valued extension property at $$\lambda\in\mathbb{C}$$ (abbreviated SVEP at λ) if for every neighborhood D of λ, the only analytic function $$f : D \longrightarrow\mathcal{H}$$ that satisfies the equation

$$(T - \mu) f (\mu) = 0$$

is the constant function $$f \equiv0$$.

The operator T is said to have the SVEP if T has the SVEP at every $$\lambda\in\mathbb{C}$$.

The quasinilpotent part and the analytic core of $$(T - \lambda)$$ are, respectively, defined by

$$H_{0}(T - \lambda) = \Bigl\{ x \in \mathcal{H}: \lim _{n \longrightarrow \infty} \bigl\Vert (T - \lambda)^{n}x \bigr\Vert ^{\frac{1}{n}} = 0 \Bigr\}$$

and

\begin{aligned} K_{0}(T - \lambda) ={}& \bigl\{ x \in \mathcal{H}: \text{there exist a sequence } (x _{n}) \subset\mathcal{H} \text{ and a constant } \delta> 0 \\ &\text{such that } (T - \lambda)x_{1} = x, (T - \lambda)x_{n+1} = x_{n}, \text{and } \Vert x_{n} \Vert \leq\delta^{n} \Vert x \Vert \\ &\text{for all } n\in \mathbb{N} \bigr\} .\end{aligned}

The subspace $$C(T )$$ in purely algebraic terms was introduced by Saphar .

### Definition 1.3

Let T be a linear operator on $$\mathcal{H}$$. The algebraic core $$C(T )$$ is is the greatest subspace $$\mathcal{M}$$ of $$\mathcal{H}$$ for which $$T (\mathcal {M}) =\mathcal{M}$$.

For bounded linear operators, the Drazin inverse was introduced and studied by Caradus . It is shown that the Drazin inverse is helpful in analyzing Markov chains, difference equation, differential equations, Cauchy problems, and iterative procedures [2, 5].

### Definition 1.4

For $$T \in\mathcal{B}(\mathcal{H})$$, suppose that there exists an operator $$T^{D} \in\mathcal{B}(\mathcal{H})$$ satisfying the following three operator equations:

$$TT^{D} = T^{D}T, \qquad T^{D}TT^{D} = T^{D}, \qquad T^{k+1}T^{D} = T^{k},$$

where $$k = \operatorname{ind} (T)$$, the index of T, is the smallest nonnegative integer for which $$\mathcal{R}(T^{k}) = \mathcal{R}(T^{k + 1})$$ and $$\mathcal{N}(T^{k}) = \mathcal{N}(T^{k + 1})$$. Then $$T^{D}$$ is called a Drazin inverse of T.

In particular, when $$\operatorname{ind}(T) = 1$$, the operator $$T^{D}$$ is called the group inverse of T and is denoted by $$T^{\sharp}$$. Clearly, $$\operatorname{ind}(T) = 0$$ if and only if T is invertible, and in this case, $$T^{D} = T^{-1}$$.

### Remark 1.5

Let T be Drazin invertible.

1. 1.

The spectral idempotent $$T^{\pi}$$ of T corresponding to $$\{0\}$$ is given by $$T^{\pi}= I - TT^{D}$$. The operator matrix form of T with respect to the space decomposition $$\mathcal{H} = \mathcal{N}(T^{\pi}) \oplus\mathcal{R}(T^{\pi})$$ is given by $$T = T_{1} \oplus T_{2}$$, where $$T_{1}$$ is invertible, and $$T_{2}$$ is nilpotent.

2. 2.
• $$H_{0}(T) = \mathcal{R}(T^{\pi}) = \mathcal{N}(T^{D}) = \mathcal{N}(T^{k})$$,

• $$K_{0}(T) = \mathcal{N}(T^{\pi}) = \mathcal{R}(T^{D}) = \mathcal{R}(T^{k})$$,

where $$k = \operatorname{ind}(T)$$.

For $$T \in\mathcal{B}(\mathcal{H})$$, it is well known that the Drazin inverse $$T^{D}$$ of T is unique if it exists, and then $$(T^{*})^{D} = (T^{D})^{*}$$.

### Lemma 1.6

()

Let$$S, T \in\mathcal{B}(\mathcal{H})$$be Drazin invertible. Then

1. (i)

STis Drazin invertible if and only ifTSis Drazin invertible, $$\operatorname{ind}(ST) \leq \operatorname{ind}(TS) + 1$$, and$$(ST)^{D} = S[(TS)^{D}]^{2}T$$.

2. (ii)

IfSis idempotent, then$$S^{D} = S$$.

3. (iii)

If$$ST = TS$$, then$$(ST)^{D} = T^{D}S^{D} = S^{D}T^{D}$$, $$S^{D}T = TS^{D}$$, and$$ST^{D} = T^{D}S$$.

### Definition 1.7

()

Let $$T \in\mathcal{B}(\mathcal{H})$$ be Drazin invertible. T is called a D-normal operator if

$$T^{D}T^{*} = T^{*}T^{D}.$$

The class of all D-normal operators is denoted by $$[DN]$$.

### Proposition 1.8

Let$$T \in\mathcal{B}(\mathcal{H})$$be Drazin invertible. ThenTisD-normal if and only if$$T^{D}$$is normal.

### Proof

Let T be D-normal. Then $$T^{D}T^{*} = T^{*}T^{D}$$ and, by Lemma 1.6(3), $$T^{D}(T^{*})^{D} = (T^{*})^{D}T^{D}$$. Since $$(T^{*})^{D} = (T^{D})^{*}$$, $$T^{D}$$ is normal. Now let $$T^{D}$$ be normal. Since $$T^{D}T = TT^{D}$$, by the Fuglede theorem, $$T^{D}T^{*} = T^{*}T^{D}$$. Therefore T is D-normal. □

D-normal operators were introduced and studied by Dana and Yousefi . The authors in [8, 9] studied several properties of this class.

## Fuglede–Putnam theorem for D-normal operators

The Fuglede–Putnam theorem is a very useful tool when dealing with products (and even sums) involving normal operators. As an application of this theorem, we can name Kaplansky theorem . Many mathematicians attempt to extend this theorem to nonnormal operators (see ).

The Hilbert–Schmidt operators in $$\mathcal{H}$$ form an ideal $$\mathbb {H}$$ in the algebra $$\mathcal{B}(\mathcal{H})$$ of all operators in $$\mathcal{H}$$. The ideal $$\mathbb{H}$$ itself is a Hilbert space with inner product

$$\langle X, Y \rangle= \sum\langle Xe_{i}, Ye_{i} \rangle= \operatorname{tr} \bigl(Y^{*}X \bigr) = tr \bigl(XY^{*} \bigr),$$

where $$\{e_{i} \}$$ is any orthonormal basis of $$\mathcal{H}$$. For each pair of operators $$S, T \in \mathcal{B}(\mathcal{H})$$, there is an operator Γ defined on $$\mathcal{B}(\mathcal{H})$$ by the formula $$\varGamma X = SXT$$ as in . The adjoint and the Drazin inverse of Γ are given by the formulas

$$\varGamma^{*}X = S^{*}XT^{*} \quad\mbox{and} \quad \varGamma^{D}X = S^{D}XT^{D}.$$

We say that normal operators $$S, T$$ satisfy the Fuglede–Putnam theorem if $$SX = XT$$ implies $$S^{*}X = XT^{*}$$. The aim of this section is to show that if $$S, T$$ are of class $$[DN]$$ and T is invertible, then for a Hilbert–Schmidt operator X,

$$SX = XT \quad\text{implies} \quad S^{*}X = XT^{*}.$$

### Theorem 2.1

Let$$S, T, X \in \mathcal{B}(\mathcal{H})$$be such thatSandTare Drazin invertible. If$$SX = XT$$, then$$S^{D}X = X T^{D}$$.

### Proof

There exists a scalar polynomial g such that $$(S \oplus T)^{D} = g(S \oplus T)$$ . This implies that $$S^{D} = g(S)$$ and $$T^{D} = g(T)$$. Hence $$S^{D}X = g(S) X = X g(T) = X T^{D}$$. □

### Lemma 2.2

If$$S, T \in[DN]$$, then the operatorΓis of class$$[DN]$$.

### Proof

By hypothesis, $$S^{D}S^{*} = S^{*}S^{D}$$ and $$T^{D}T^{*} = T^{*}T^{D}$$. For any pair $$S, T \in\mathcal{B}(\mathcal{H})$$,

\begin{aligned} \bigl(\varGamma^{*}\varGamma^{D} - \varGamma^{D} \varGamma^{*} \bigr)X &= \varGamma ^{*} \varGamma^{D}X - \varGamma^{D}\varGamma^{*}X \\ &= \varGamma^{*} \bigl(S^{D}XT^{D} \bigr) - \varGamma^{D} \bigl(S^{*}XT^{*} \bigr) \\ &= S^{*} \bigl(S^{D}XT^{D} \bigr)T^{*} - S^{D} \bigl(S^{*}XT^{*} \bigr)T^{D} \\ &= 0, \end{aligned}

which implies that Γ is of class $$[DN]$$. □

### Theorem 2.3

Let$$S, T \in[DN]$$nr such thatTis invertible, and letXbe a Hilbert–Schmidt operator. If$$SX = XT$$, then$$S^{*}X = XT^{*}$$.

### Proof

Let Γ be the Hilbert–Schmidt operator defined by $$\varGamma Y = SYT^{-1}$$, where $$Y \in\mathcal{B}(\mathcal{H})$$. Since S, T are of class $$[DN]$$, by Lemma 2.2, Γ is of class $$[DN]$$. The hypothesis $$SX = XT$$ implies that $$\varGamma X = X$$ and $$\varGamma^{D}X = X$$ and also

\begin{aligned} \bigl\Vert \varGamma^{*}X \bigr\Vert ^{2} &= \bigl\langle \varGamma^{*}X , \varGamma^{*}X \bigr\rangle \\ &= \bigl\langle \varGamma^{*} \bigl(\varGamma^{D} \bigr)^{2}X, \varGamma^{*} \bigl(\varGamma ^{D} \bigr)^{2}X \bigr\rangle \\ &= \bigl\langle \varGamma{ \bigl(\varGamma^{D} \bigr)^{2}}^{*} \varGamma^{*} \bigl(\varGamma ^{D} \bigr)^{2}X, X \bigr\rangle \\ &= \bigl\langle \varGamma^{D}X, \varGamma^{D}X \bigr\rangle \\ & = \Vert X \Vert ^{2}. \end{aligned}

On the other hand,

\begin{aligned} \bigl\langle \varGamma^{*}X , X \bigr\rangle &= \bigl\langle \varGamma^{*}X , \bigl(\varGamma^{D} \bigr)^{2}X \bigr\rangle \\ &= \bigl\langle { \bigl(\varGamma^{D} \bigr)^{2}}^{*} \varGamma^{*}X, X \bigr\rangle \\ &= \bigl\langle {\varGamma^{D}}^{*}X, X \bigr\rangle \\ &= \bigl\langle X, \varGamma^{D}X \bigr\rangle \\ &= \langle X, X \rangle. \end{aligned}

So we have

\begin{aligned} \bigl\Vert \varGamma^{*}X - X \bigr\Vert ^{2} &= \bigl\langle \varGamma^{*}X - X , \varGamma^{*}X - X \bigr\rangle \\ &= \bigl\langle \varGamma^{*}X , \varGamma^{*}X \bigr\rangle - \bigl\langle \varGamma ^{*}X , X \bigr\rangle - \bigl\langle X , \varGamma^{*}X \bigr\rangle + \langle X , X \rangle \\ &= \bigl\Vert \varGamma^{*}X \bigr\Vert ^{2} - \bigl\langle \varGamma^{*}X , X \bigr\rangle - \bigl\langle X , \varGamma^{*}X \bigr\rangle + \Vert X \Vert ^{2} \\ &= 0. \end{aligned}

Therefore $$\varGamma^{*}X = X$$, and hence $$S^{*}X = XT^{*}$$. □

Here we give an example that if $$X \in\mathcal{B}(\mathcal{H})$$ and $$S, T \in[DN]$$ satisfy $$SX = XT$$, then we cannot get $$S^{*}X = XT^{*}$$. Just consider the operator $S=X=(0010)$ and $$T = 0$$. Then $$SX =XT$$, but $S∗X=(1000)$ and $XT∗=(0000)$.

## Bishop property for D-normal operators

We start this section with the matrix representation for $$T \in [DN]$$.

### Lemma 3.1

If$$T \in[DN]$$, then$$\mathcal{R}(T^{D})$$reducesT.

### Proof

Since $$T \in[DN]$$, $$T^{D}T^{*} = T^{*}T^{D}$$. Obviously, $$\mathcal {R}(T^{D})$$ is invariant under T. We will show that $$\mathcal{R}(T^{D})$$ is invariant under $$T^{*}$$. Let $$x \in \mathcal{R}(T^{D})$$. Then $$x = T^{D}y$$ for some $$y \in\mathcal{H}$$, and $$T^{*}x = T^{*} T^{D}y = T^{D}T^{*} y \in \mathcal{R}(T^{D})$$. Thus $$\mathcal{R}(T^{D})$$ is invariant under $$T^{*}$$, and $$\mathcal{R}(T^{D})$$ reduces T. □

### Theorem 3.2

IfTis of class$$[DN]$$, thenThas the following matrix representation: $T=(T100T2)$on$$\mathcal{H} = \mathcal{R}(T^{D}) \oplus\mathcal{N}(T^{D})$$, where$$T_{1} = T |_{\mathcal{R}(T^{D})}$$is also of class$$[N]$$, and$$T_{2}$$is a nilpotent operator with nilpotency$$\operatorname{ind}(T)$$. Furthermore, $$\sigma(T) = \sigma(T_{1}) \cup\{0\}$$.

### Proof

By Lemma 3.1, $$\mathcal{R}(T^{D})$$ reduces T. Hence T has the matrix representation $T=(T100T2)$ on $$\mathcal{H} = \mathcal{R}(T^{D}) \oplus {\mathcal{N}(T^{*}}^{D})$$. Note that since $$T \in[DN]$$, $$\mathcal{N}(T^{D}) = {\mathcal{N}(T^{*}}^{D})$$. Let P be the orthogonal projection onto $$\mathcal{R}(T^{D})$$. Then

$$\left ( \textstyle\begin{array}{c@{\quad}c} T_{1} & 0 \\ 0 & 0 \end{array}\displaystyle \right ) = TP = PT = PTP.$$

Hence

$$P \bigl( T^{D}T^{*} \bigr)P = \left ( \textstyle\begin{array}{c@{\quad}c} T_{1}^{D}T_{1}^{*} & 0 \\ 0 & 0 \end{array}\displaystyle \right )$$

and

$$P \bigl(T^{*} T^{D} \bigr)P = \left ( \textstyle\begin{array}{c@{\quad}c} T_{1}^{*} T_{1}^{D} & 0 \\ 0 & 0 \end{array}\displaystyle \right ).$$

Since $$T \in[DN]$$, $$P(T^{*} T^{D} )P = P(T^{D}T^{*} )P$$, implying $$T_{1}^{*}T_{1}^{D} = T_{1}^{D}T_{1}^{*}$$. Hence $$T_{1} \in[DN]$$. On the other hand, by Remark 1.5, $$T_{1}$$ is invertible. So $$T_{1} \in[N]$$.

For any $z=(z1z2)∈H$,

\begin{aligned} \bigl\langle T_{2}^{D}z_{2} , z_{2}\bigr\rangle &= \bigl\langle T^{D}(I - P)z , (I - P)z \bigr\rangle \\ & = \bigl\langle (I - P)z , \bigl({T^{D}} \bigr)^{*}(I - P)z \bigr\rangle \\ &= 0. \end{aligned}

Therefore $$T_{2}^{D} = 0$$. Then $$T_{2}$$ is a nilpotent operator. Since $$\mathcal{R}(T^{D})$$ reduces T, $$\sigma(T) = \sigma(T_{1}) \cup \sigma (T_{2}) = \sigma(T_{1}) \cup\{0\}$$. □

### Theorem 3.3

If$$T \in [DN]$$, thenThas property$$(\beta)$$.

### Proof

If $$D \subset\mathbb{C}$$ is an open neighborhood of $$\lambda\in \mathbb{C}$$ and $$f_{m}$$ ($$m = 1, 2, \ldots$$) are vector-valued analytic functions on D such that $$(T - \mu)f_{m}(\mu) \longrightarrow0$$ uniformly on every compact subset of D, then we decompose $$\mathcal{H}$$ as $$\mathcal{H} =\mathcal{R}(T^{D}) \oplus \mathcal{N}(T^{D})$$, and by Theorem 3.2$T=(T100T2)$ where $$T_{1} \in[N]$$, and $$T_{2}$$ is a nilpotent operator with nilpotency $$\operatorname{ind}(T)$$. The convergence $$(T - \mu)f_{m}(\mu) \longrightarrow0$$ implies

$$\left ( \textstyle\begin{array}{c@{\quad}c} T_{1} - \mu & 0 \\ 0 & T_{2} - \mu \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{c} f_{m_{1}}(\mu) \\ f_{m_{2}}(\mu) \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} (T_{1} - \mu) f_{m_{1}}(\mu) \\ (T_{2} - \mu) f_{m_{2}}(\mu) \end{array}\displaystyle \right )$$

Since $$T_{2}$$ is nilpotent, it has property $$(\beta)$$, and therefore $$f_{m_{2}}(\mu) \longrightarrow0$$. Also, since $$T_{1}$$ is normal, it has property $$(\beta)$$. So by Theorem 3.39 in , T has property $$(\beta)$$. □

From the theorem we immediately have the following:

### Corollary 3.4

If$$T \in [DN]$$, thenThas the SVEP.

The following example shows that for a D-normal operator T, the corresponding eigenspaces need not be reducing subspaces of T.

### Example 3.5

$T=(0100)$. Clearly, T is a D-normal operator, and the eigenspace of T is $(x0)$, but it is not a reducing subspace of T.

### Theorem 3.6

Suppose that$$T \in [DN]$$. Then$$C(T^{D})$$is invariant under$$T^{*}$$.

### Proof

By the definition of algebraic core of $$T^{D}$$, $$T^{D} (C(T^{D})) = C(T^{D})$$. Since $$T \in [DN]$$, $$T^{*}T^{D} = T^{D}T^{*}$$. So we have $$T^{*}T^{D}C(T^{D}) = T^{D}T^{*}C(T^{D} )$$. This implies $$T^{*}C(T^{D}) = T^{D}T^{*} C(T^{D})$$. Now, since $$C(T^{D})$$ is the greatest subspace satisfying $$T^{D} (C(T^{D})) = C(T^{D})$$, we have $$T^{*}C(T^{D}) \subseteq C(T^{D})$$. Thus $$C(T^{D})$$ is invariant under $$T^{*}$$. □

### Theorem 3.7

If$$T \in[DN]$$, then the following properties hold:

1. 1.

$$H_{0}(T^{D} - \lambda)$$is a reducing subspace ofT.

2. 2.

$$x \in H_{0}(T)$$if and only if$$T^{*}x \in H_{0}(T)$$.

3. 3.

$$H_{0}(T^{D} - \lambda) = \mathcal{N}(T^{D} - \lambda) = \mathcal{N}(T^{D} - \lambda)^{*}$$. In particular, $$H_{0}(T) = \mathcal {N}(T^{D}) = \mathcal{N}((T^{D})^{*})$$.

4. 4.

If$$\mathcal{M}$$is an invariant subspace ofTand$$T_{1} = T \vert_{\mathcal{M}}$$on$$\mathcal{H} = \mathcal{M} \oplus\mathcal {M}^{\perp}$$, then$$H_{0}(T_{1}^{D} - \lambda) = \mathcal{N}(T_{1}^{D} - \lambda) = \mathcal{N}(T_{1}^{D} - \lambda)^{*}$$

### Proof

1. Since $$T \in[DN]$$, $$(T^{D} - \lambda) T^{*} = T^{*}(T^{D} - \lambda)$$,and hence for $$x \in H_{0}(T^{D} - \lambda)$$, we have

\begin{aligned} \lim_{n \rightarrow\infty} \bigl\Vert \bigl(T^{D} - \lambda \bigr) ^{n}T^{*}x \bigr\Vert ^{\frac{1}{n}} &= \lim _{n \rightarrow\infty} \bigl\Vert T^{*} \bigl(T^{D} - \lambda \bigr) ^{n}x \bigr\Vert ^{\frac{1}{n}} \\ &\leq \lim_{n \rightarrow\infty} \bigl\Vert T^{*} \bigr\Vert ^{\frac {1}{n}} \bigl\Vert \bigl(T^{D} - \lambda \bigr) ^{n}x \bigr\Vert ^{\frac{1}{n}} \\ &= 0. \end{aligned}

Hence $$T^{*}x \in H_{0}(T^{D} - \lambda)$$. It is easy to see that $$Tx \in H_{0} (T^{D} - \lambda)$$.

2. We have $$H_{0}(T) = \mathcal{N}(T^{D})$$. On the other hand, $$(T^{D})^{D} = T^{2}T^{D}$$. It is clear that $$\mathcal{N}(T^{D}) = \mathcal{N}( T^{2}T^{D})$$. So, $$H_{0}(T) = H_{0}(T^{D})$$.

If $$x \in H_{0}(T) = H_{0}(T^{D})$$, then we easily get that $$T^{*}x \in H_{0}(T)$$. To prove the converse, let $$T^{*}x \in H_{0}(T)$$. For every $$n > 1$$, we have

\begin{aligned} \bigl\Vert \bigl(T^{D} \bigr)^{n}T^{*}x \bigr\Vert ^{2} &= \bigl\langle \bigl(T^{D} \bigr)^{n}T^{*}x, \bigl(T^{D} \bigr)^{n}T^{*}x \bigr\rangle \\ &= \bigl\langle T^{*} \bigl(T^{D} \bigr)^{n}x, T^{*} \bigl(T^{D} \bigr)^{n}x \bigr\rangle \\ &= \bigl\langle \bigl({T^{*}}^{D} \bigr)^{n}TT^{*} \bigl(T^{D} \bigr)^{n}x, x \bigr\rangle \\ &= \bigl\langle T \bigl({T^{*}}^{D} \bigr)^{n}T^{*} \bigl(T^{D} \bigr)^{n}x, x \bigr\rangle \\ &= \bigl\langle T \bigl(T^{D} \bigr)^{n} \bigl({T^{*}}^{D} \bigr)^{n}T^{*}x, x \bigr\rangle \\ &= \bigl\langle \bigl({T^{*}}^{D} \bigr)^{n - 1}x, \bigl({T^{*}}^{D} \bigr)^{n - 1}x \bigr\rangle \\ &= \bigl\Vert \bigl({T^{*}}^{D} \bigr)^{n - 1}x \bigr\Vert ^{2} \\ &= \bigl\Vert \bigl(T^{D} \bigr)^{n - 1}x \bigr\Vert ^{2}. \end{aligned}

So, for every $$n > 1$$,

$$\bigl\Vert \bigl(T^{D} \bigr)^{n}T^{*}x \bigr\Vert ^{2} = \bigl\Vert \bigl(T^{D} \bigr)^{n - 1}x \bigr\Vert ^{2},$$
(3.1)

and for $$n = 1$$,

$$\bigl\Vert T^{D}T^{*}x \bigr\Vert ^{2} = \bigl\Vert TT^{D}x \bigr\Vert ^{2}.$$

According to (3.1),

$$\lim_{n \rightarrow\infty} \bigl\Vert \bigl(T^{D} \bigr)^{n - 1}x \bigr\Vert ^{\frac {1}{n - 1}} = \lim _{n \rightarrow\infty} \bigl( \bigl\Vert \bigl(T^{D} \bigr)^{n}T^{*}x \bigr\Vert ^{{\frac{1}{n}}} \bigr)^{\frac{n}{n - 1}} = 0.$$

Thus $$x \in H_{0}(T^{D} ) = H_{0}(T)$$.

3. Notice that for a totally paranormal operator T, $$H_{0} (T - \lambda) = \mathcal{N}(T - \lambda)$$ for every $$\lambda\in \mathbb {C}$$ . The class of totally paranormal operators includes the class of hyponormal operators and hence normal operators. In view of normality $$T^{D}$$, we have

$$H_{0} \bigl(T^{D} - \lambda \bigr) = \mathcal{N} \bigl(T^{D} - \lambda \bigr) = \mathcal {N} \bigl(T^{D} - \lambda \bigr)^{*}.$$

For $$\lambda= 0$$, $$H_{0}(T^{D}) = \mathcal{N}(T^{D}) = \mathcal {N}(T^{D})^{*}$$.

3. By Proposition 2.6 of , $$T_{1}^{D} = T^{D} \vert_{\mathcal{M}}$$ is hyponormal, and hence $$H_{0}(T_{1}^{D} - \lambda) = \mathcal {N}(T_{1}^{D} - \lambda) = \mathcal{N}(T_{1}^{D} - \lambda)^{*}$$. □

## Generalization of Kaplansky theorem for D-normal matrices

Let $$\mathcal{M}_{n}( \mathbb{C})$$ be the set of $$n \times n$$ complex matrices. In this section, we are mainly interested in generalizing the following famous result on products of normal operators, due to I. Kaplansky, to D-normal matrices.

### Theorem 4.1

()

LetAandBbe two bounded operators on a Hilbert space such thatABandAare normal. ThenBcommutes with$$AA^{*}$$iffBAis normal.

### Proposition 4.2

Let$$A, B \in \mathcal{M}_{n}( \mathbb{C})$$be such thatABisD-normal. Then

$$A^{*}AB = BAA^{*} \quad\Longrightarrow \quad BA \textit{ is }D\textit{-normal}.$$

### Proof

Let $$A = UP$$, where P is positive, and U is unitary. Note that there exists a positive semidefinite $$K \in\mathcal{M}_{n}( \mathbb{C})$$ such that $$A = KU$$. We obtain

\begin{aligned} P^{2}B & = A^{*}AB \\ &= (BA) A^{*} \\ & = BK^{2}. \end{aligned}

Hence, since P and K are positive semidefinite, $$PB = BK$$. Then $$PBU = BKU$$. So $$PBU = BUP$$. Thus

$$U^{*}ABU = U^{*}UPBU = PBU = BA.$$

Hence BA is unitary equivalent to a D-normal operator, and thus by [7, Proposition 2.6], it is D-normal itself. □

### Remark 4.3

Using a similar method as in Proposition 4.2, we can show that for $$A, B \in \mathcal {M}_{n}( \mathbb{C})$$, by the Kaplansky theorem the condition that A is normal is superfluous.

### Proposition 4.4

Let$$A, B \in \mathcal{M}_{n}( \mathbb{C})$$be such thatABisD-normal. Then

$$A^{*}(AB)^{D} = (BA)^{D}A^{*} \quad\Longleftrightarrow\quad BA \textit{ is }D\textit{-normal}.$$

### Proof

Let AB and BA be D-normal matrices. Then by Lemma 1.6(i)

\begin{aligned} A(BA)^{D}&= AB \bigl((AB)^{2} \bigr)^{D}A \\ &= (AB)^{D}A. \end{aligned}

Hence, by the Fuglede–Putnam theorem,

$$A \bigl((BA)^{D} \bigr)^{*} = \bigl((AB)^{D} \bigr)^{*}A.$$

So,

$$A^{*}A \bigl((BA)^{D} \bigr)^{2}B = B \bigl((AB)^{D} \bigr)^{2}AA^{*}.$$

Hence

$$A^{*}(AB)^{D} = (BA)^{D}A^{*}.$$

Conversely, if $$A^{*}(AB)^{D} = (BA)^{D}A^{*}$$, then $$A^{*}A((BA)^{D})^{2}B = B((AB)^{D})^{2}AA^{*}$$. Let $$A = UP$$, where P is positive, and U is unitary. Note that there exists a positive semidefinite $$K \in\mathcal{M}_{n}( \mathbb{C})$$ such that $$A = KU$$. So $$P^{2}((BA)^{D})^{2}B = B((AB)^{D})^{2}K^{2}$$. Hence, since P and K are positive semidefinite, $$P((BA)^{D})^{2}B = B((AB)^{D})^{2}K$$. So we have

$$P \bigl((BA)^{D} \bigr)^{2}BU = B \bigl((AB)^{D} \bigr)^{2}KU.$$
(4.1)

Now

\begin{aligned} U^{*}(AB)^{D}U &= U^{*}A \bigl((BA)^{D} \bigr)^{2}BU \quad(\text{by Lemma 1.6}) \\ &= U^{*}UP \bigl((BA)^{D} \bigr)^{2}BU \\ &= B \bigl((AB)^{D} \bigr)^{2}KU\quad (\text{by (4.1)}) \\ &= B \bigl((AB)^{D} \bigr)^{2}A \\ &= (BA)^{D}. \end{aligned}

Hence $$(BA)^{D}$$ is unitary equivalent to a normal operator and thus is normal itself. □

## References

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### Acknowledgements

The authors would like to thank the anonymous referees for careful reading and helpful comments and suggestions for improving this paper.

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