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.

1 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 [4] 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 [13].

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 [6]. 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. 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.

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

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

  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

([5])

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

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

• (ii) IfSis idempotent, then$S^{D} = S$.

• (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

([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 [7]. The authors in [8, 9] studied several properties of this class.

2 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 [10]. Many mathematicians attempt to extend this theorem to nonnormal operators (see [14]).

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 [3]. 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 $(S \oplus T)^{D} = g(S \oplus T)$ [5]. 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 $S=X=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right)$ and $T = 0$. Then $SX =XT$, but $S^{\ast }X=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$ and $X{T}^{\ast }=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$.

3 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: $T=\left(\begin{array}{cc}{T}_1& 0\\ 0& T_2\end{array}\right)$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 $T=\left(\begin{array}{cc}{T}_1& 0\\ 0& T_2\end{array}\right)$ 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 $z=\left(\begin{array}{c}{z}_1\\ z_2\end{array}\right)\in \mathcal{H}$,

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

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$T=\left(\begin{array}{cc}{T}_1& 0\\ 0& T_2\end{array}\right)$ 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 [11], 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

$T=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Clearly, T is a D-normal operator, and the eigenspace of T is $\left(\begin{array}{c}x\\ 0\end{array}\right)$, 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. $H_{0}(T^{D} - \lambda)$is a reducing subspace ofT.

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

• 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. 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 $\lambda\in \mathbb {C}$ [1]. 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 [7], $T_{1}^{D} = T^{D} \vert_{\mathcal{M}}$ is hyponormal, and hence $H_{0}(T_{1}^{D} - \lambda) = \mathcal {N}(T_{1}^{D} - \lambda) = \mathcal{N}(T_{1}^{D} - \lambda)^{*}$. □

4 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

([10])

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. □