Fixed points of completely positive maps and their dual maps

Abstract

Let $\mathcal {A} \subset{\mathcal {B}}(\mathcal {H})$ be a row contraction and $\Phi _{\mathcal {A}}$ determined by $\mathcal {A}$ be a completely positive map on ${\mathcal {B}}(\mathcal {H})$. In this paper, we mainly consider fixed points of $\Phi _{\mathcal {A}}$ and its dual map $\Phi _{\mathcal {A}}^{\dagger}$. It is given that $\Phi _{\mathcal {A}}(X)\leq X$ (or $\Phi _{\mathcal {A}}(X)\geq X$) implies $\Phi _{\mathcal {A}}(X)= X$ and $\Phi _{\mathcal {A}}^{\dagger}(X)= X$ when $X\in {\mathcal {B}}(\mathcal {H})$ is a compact operator. Some necessary conditions of $\Phi _{\mathcal {A}}(X)= X$ and $\Phi _{\mathcal {A}}^{\dagger}(X)= X$ are given.

1 Introduction

Completely positive maps play an essential role in quantum information theory since they correspond to physical operations, see [7]. Recall that a quantum operation can be represented by a normal completely positive map, which is determined by an operator sequence, see [2, 3]. Hence, some problems about completely positive maps can be solved by researching operator sequences.

For the convenience of description, let $\mathcal {H}$ and $\mathcal {K}$ be separable Hilbert spaces and $\mathcal {B}(\mathcal {K}, {\mathcal {H}})$ be the set of all bounded linear operators from $\mathcal {K}$ into $\mathcal {H}$ and abbreviate $\mathcal {B}(\mathcal {K}, \mathcal {H})$ to $\mathcal {B}({\mathcal {H}})$ if ${\mathcal {K}}={\mathcal {H}}$. ${\mathcal {K}}(\mathcal {H})$ is the set of compact operators on $\mathcal {H}$. Denote by J a finite or infinite countable index set. Let ${\mathcal {A}}=\{A_{k}\}_{k\in J}\subset {\mathcal {B}}(\mathcal {H})$. $\mathcal {A}$ is called a row contraction if $\sum_{k\in J}A_{k}A_{k}^{*} \leq I$, where the series $\sum_{k\in J}A_{k}A_{k}^{*}$ is convergent in strong operator topology and $A_{k} ^{*}$ is the adjoint operator of $A_{k}$. We say that $\mathcal {A}$ is unital if $\sum_{k\in J}A_{k}A_{k}^{*}=I$ and trace preserving if $\sum_{k\in J}A_{k}^{*}A_{k}=I$.

To each row contraction ${\mathcal {A}}=\{A_{k}\}_{k\in J}$ one can associate a normal completely positive mapping $\Phi _{\mathcal {A}}$ on ${\mathcal {B}}(\mathcal {H})$,

$$\begin{aligned} \Phi _{\mathcal {A}}(X)= \sum_{k\in J} A_{k}XA_{k}^{*},\quad \forall X\in{ \mathcal {B}}( \mathcal {H}). \end{aligned}$$

Then, we say that $\Phi _{\mathcal {A}}$ is a quantum operation on $\mathcal {B}(\mathcal {H})$ and each $A_{k}$ is the operation element or the Kraus operator of $\Phi _{\mathcal {A}}$. $\mathcal {A}$ and $\Phi _{\mathcal {A}}$ are called self-adjoint if each $A_{k}$ is self-adjoint. If a row contraction $\mathcal {A}$ also satisfies $\sum_{k\in J}A_{k}^{*}A_{k}\leq I$, we can define a completely positive map $\Phi _{\mathcal {A}}^{\dagger}$ on ${\mathcal {B}}(\mathcal {H})$ as follows:

$$\begin{aligned} \Phi _{\mathcal {A}}^{\dagger} (X)= \sum_{k\in J} A_{k}^{*}XA_{k}, \quad\forall X\in{\mathcal {B}}( \mathcal {H}). \end{aligned}$$

The map $\Phi _{\mathcal {A}}^{\dagger}$ is well defined and is called the dual operation of $\Phi _{\mathcal {A}}$. An operator $X\in {\mathcal {B}}(\mathcal {H})$ is said to be a fixed point of $\Phi _{\mathcal {A}}$ if $\Phi _{\mathcal {A}} (X)=X$. In fact, a fixed point $\Phi _{\mathcal {A}}$ means that it is not disturbed by the action of $\Phi _{\mathcal {A}}$. Denote by ${\mathcal {B}}({\mathcal {H}})^{\Phi _{\mathcal {A}}}$ the set of fixed points of $\Phi _{\mathcal {A}}$.

Fixed points of completely positive maps were considered from different aspects since they are useful in the theory of quantum error correction, see [1, 4–6], and [8–11]. Li discussed fixed points of dual quantum operations on compact operators in [4] and given that the two fixed points sets of quantum operation and its dual operation are coincident under a certain condition. In [1], the authors noted that the positive fixed point $B\in {\mathcal {B}}({\mathcal {H}})^{\Phi _{\mathcal {A}}}$ of $\Phi _{\mathcal {A}}$ and $A_{k}$ commute if B has only discrete point spectra and $\Phi _{\mathcal {A}}$ is a self-adjoint quantum operation. However, the result does not necessary hold for a not self-adjoint quantum operation. Li generalized the result to the unital and trace-preserving quantum operation in [5], but B must be an operator when the spectra space is finite. In [4], the fixed points sets of $\Phi _{\mathcal {A}}$ and its dual map $\Phi _{\mathcal {A}}^{\dagger}$ were given by use of the properties of self-adjoint operators. It was given that the two sets were equivalent in compact operator space. Also, it was noted that $\Phi _{\mathcal {A}}(X)\geq X$ implied $\Phi _{\mathcal {A}}(X)=X$ under certain conditions. Popescu studied the inequality $\Phi _{\mathcal {A}}(X)\leq X$ and the equation $\Phi _{\mathcal {A}}(X)=X$ by use of the minimal isometric dilation and Poisson transforms in [5] and the canonical decompositions and lifting theorems were obtained to provide a description of all solutions of $\Phi _{\mathcal {A}}(X)\leq X$.

Inspired by the above results, we mainly consider fixed points of completely positive maps and their dual operations. For a given row contraction ${\mathcal {A}}$, we study the inequality $\Phi _{\mathcal {A}}(X)\leq X$ and the equation $\Phi _{\mathcal {A}}(X)=X$ on the set of all diagonalizable operators. It is given that $\Phi _{\mathcal {A}}(X)\leq X$ (or $\Phi _{\mathcal {A}}(X)\geq X$) implies $\Phi _{\mathcal {A}}(X)= X$ and $\Phi _{\mathcal {A}}^{\dagger}(X)= X$ when $X\in {\mathcal {B}}(\mathcal {H})$ is a compact operator. Simultaneously, an example is given to show that $\Phi _{\mathcal {A}}(X)= X$ does not necessarily imply $\Phi _{\mathcal {A}}^{\dagger}(X)= X$ when X is not compact. Some necessary conditions of $\Phi _{\mathcal {A}}(X)= X$ and $\Phi _{\mathcal {A}}^{\dagger}(X)= X$ are obtained.

2 Main result

In order to obtain the main results, we begin with some lemmas.

Lemma 1

([8])

Let Φ be a normal completely positive map on ${\mathcal {B}}(\mathcal {H})$ that is defined by

$$\begin{aligned} \Phi (X)=\sum_{k\in J}A_{k}XA_{k}^{*}, \quad \forall {\mathcal {B}}( \mathcal {H}). \end{aligned}$$

A positive operator $C\in {\mathcal {B}}(\mathcal {H})$ is a solution of the inequality $\Phi (X)\leq X$ (or $\Phi (X)= X$) if and only if there exists an operator sequence $\{B_{k}\}_{k\in J}\subset {\mathcal {B}}(\mathcal {H})$ such that $\sum_{k=1}B_{k}B_{k}^{*}\leq I$ (or $\sum_{k=1}B_{k}B_{k}^{*}= I$) and $A_{k} C^{\frac{1}{2}}=C^{\frac{1}{2}}B_{k}$ for any k.

Similar to Lemma 1, we give an equivalent condition of $\Phi (X)\geq X$.

Lemma 2

Let Φ be a normal completely positive map on ${\mathcal {B}}(\mathcal {H})$ that is defined by

$$\begin{aligned} \Phi (X)=\sum_{k\in J}A_{k}XA_{k}^{*}, \quad \forall {\mathcal {B}}( \mathcal {H}). \end{aligned}$$

Then, an invertible and positive operator $C\in {\mathcal {B}}(\mathcal {H})$ is a solution of the inequality $\Phi (X)\geq X$ if and only if there exists an operator sequence $\{B_{k}\}_{k\in J}\subset {\mathcal {B}}(\mathcal {H})$ such that $\sum_{k\in J} B_{k}B_{k}^{*}\geq I$ and $C^{\frac{1}{2}}B_{k}=A_{k} C^{\frac{1}{2}}$ for any k.

Proof

Suppose that C is an invertible and positive operator and also a solution of the inequality $\Phi (X)\geq X$. Define the operator $B_{k}$ by setting $B_{k}=C^{-\frac{1}{2}}A_{k} C^{\frac{1}{2}}$ for any k. By direct computing, we have

$$\begin{aligned} \sum_{k\in J} B_{k}B_{k}^{*}= C^{-\frac{1}{2}}\sum_{k\in J}A_{k} A_{k} ^{*} C^{-\frac{1}{2}}\leq C^{-1}. \end{aligned}$$

That is to say, the operator series $\sum_{k\in J} B_{k}B_{k}^{*}$ is convergent in strong operator topology. From the definition of $B_{k}$, it is easy to obtain that $C^{\frac{1}{2}} B_{k}=A_{k} C^{\frac{1}{2}}$ and

$$\begin{aligned} \Phi (C)=\sum_{k\in J}A_{k}CA_{k}^{*}=C^{\frac{1}{2}} \sum_{k\in J}B_{k}B_{k}^{*}C^{ \frac{1}{2}} \geq C. \end{aligned}$$

Thus, $C^{\frac{1}{2}}(\sum_{k\in J}B_{k}B_{k}^{*}-I)C^{\frac{1}{2}}\geq 0$ and so $\sum_{k\in J} B_{k}B_{k}^{*}\geq I$.

On the contrary, suppose that $\{B_{k}\}_{k\in J}\subset {\mathcal {B}}(\mathcal {H})$ satisfies $\sum_{k=1}B_{k}B_{k}^{*}\geq I$ and $C^{\frac{1}{2}}B_{k}=A_{k} C^{\frac{1}{2}}$ for any k, then

$$\begin{aligned} \Phi (C)=\sum_{k\in J}A_{k}CA_{k}^{*}=C^{\frac{1}{2}} \sum_{k\in J}B_{k}B_{k}^{*}C^{ \frac{1}{2}} \geq C. \end{aligned}$$

The proof is completed. □

Lemma 3

Let $\dim {\mathcal {H}}<\infty$ and ${\mathcal {A}}=\{A_{k}\}_{k\in J}\subset {\mathcal {B}}(\mathcal {H})$ be a row contraction. If $\sum_{k\in J}A_{k}A_{k}^{*} = I$ and $\sum_{k\in J}A_{k}^{*}A_{k} \leq I$, then $\sum_{k\in J}A_{k}^{*}A_{k} = I$.

Proof

Let τ be a faithful tracial state on ${\mathcal {B}}(\mathcal {H})$. This shows that $\tau ( \sum_{k\in J}A_{k}A_{k}^{*})= \tau (\sum_{k\in J}A_{k}^{*}A_{k})$. That is to say $\tau ( \sum_{k\in J}A_{k}A_{k}^{*}-\sum_{k\in J}A_{k}^{*}A_{k})=0$. This implies that $\sum_{k\in J}A_{k}^{*}A_{k} =\sum_{k\in J}A_{k}A_{k}^{*}=I$. □

Theorem 4

Let $\Phi _{\mathcal {A}}(I)= I$ and $\Phi _{\mathcal {A}}^{\dagger} (I)\leq I$. If $X \in {\mathcal {B}}(\mathcal {H})$ is a compact and self-adjoint operator that satisfies $\Phi _{\mathcal {A}}(X)\leq X$ or $\Phi _{\mathcal {A}}(X)\geq X$, then $\Phi _{\mathcal {A}}(X)=X$, $\Phi _{\mathcal {A}}^{\dagger} (X)=X$ and $X\in {\mathcal {A}}^{\prime}$.

Proof

(1) Suppose that $X \in {\mathcal {B}}(\mathcal {H})$ is a compact and self-adjoint operator with $\Phi _{\mathcal {A}}(X)\leq X$. Then, $\Phi _{\mathcal {A}}(\alpha +X)\leq \alpha +X$ holds for any real number α since $\Phi _{\mathcal {A}}(I)= I$. Without loss of generality, we may assume that X is an invertible and positive operator. According to the spectral theorem of compact normal operators, it is easy to show that the spectrum of X is at most countable and these spectral points can be arrayed as follows, $\lambda _{1}>\lambda _{2}> \cdots >\lambda _{m}$ (m is a positive integer or +∞) and the dimension of the spectral projection space associated with $\lambda _{i}$ is finite. It follows that $X={\sum_{i=1} ^{m}} \lambda _{i} P_{i}$, where $P_{i}$ is the spectral projection associated with $\lambda _{i}$. From Lemma 1, there exists an operator sequence $\{B_{k}\}_{k\in J}$ with $\sum_{k\in J}B_{k}B_{k}^{*}\leq I$ such that $B_{k}X^{\frac{1}{2}}=X^{\frac{1}{2}}A_{k}$. Denote ${\mathcal {H}}_{1}= R(P_{1})$ and ${\mathcal {H}}_{2}={\mathcal {H} }\ominus {\mathcal {H}}_{1}$. Then, $X=\lambda _{1} I_{{\mathcal {H}}_{1}}\oplus X_{1}$. It follows that $X^{\frac{1}{2}}=\lambda _{1}^{\frac{1}{2}}I_{{\mathcal {H}}_{1}} \oplus X_{1}^{\frac{1}{2}}$. $A_{k}$ and $B_{k}$ can be represented by

$$\begin{aligned} A_{k}= \begin{pmatrix} A_{11}^{k} & A_{12}^{k} \\ A_{21}^{k} & A_{22}^{k} \end{pmatrix}\quad\text{and}\quad B_{k}= \begin{pmatrix} B_{11}^{k} & B_{12}^{k} \\ B_{21}^{k} & B_{22}^{k} \end{pmatrix}, \end{aligned}$$

with respect to the space decomposition ${\mathcal {H}}={\mathcal {H}}_{1} \oplus {\mathcal {H}}_{2}$. Therefore,

$$\begin{aligned} \begin{pmatrix} \lambda _{1}^{\frac{1}{2}} B_{11}^{k} & B_{12}^{k} X_{1}^{\frac{1}{2}} \\ \lambda _{1}^{\frac{1}{2}} B_{21}^{k} & B_{22}^{k} X_{1}^{\frac{1}{2}} \end{pmatrix}= \begin{pmatrix} \lambda _{1}^{\frac{1}{2}} A_{11}^{k} & \lambda _{1}^{\frac{1}{2}} A_{12}^{k} \\ X_{1}^{\frac{1}{2}} A_{21}^{k} & X_{1}^{\frac{1}{2}} A_{22}^{k} \end{pmatrix}. \end{aligned}$$

This implies that $B_{11}^{k}=A_{11}^{k}$ and $A_{12}^{k}=\frac{1}{\sqrt{\lambda _{1}} }B_{12}^{k} X_{1}^{ \frac{1}{2}}$ hold. According to $\sum_{k\in J} A_{k}A_{k}^{*}= I$ and $\sum_{k\in J} B_{k} B_{k}^{*}\leq I$, we have

$$\begin{aligned} \sum_{k\in J} A_{11}^{k}{A_{11}^{k}}^{*}+ \sum_{k\in J} A_{12}^{k} {A_{12}^{k}}^{*}= \sum _{k\in J} A_{11}^{k}{A_{11}^{k}}^{*}+ \sum_{k\in J} \frac{1}{\lambda _{1} } B_{12}^{k} X_{1} {B_{12}^{k}}^{*}= I_{{ \mathcal {H}}_{1}} \end{aligned}$$

and

$$\begin{aligned} \sum_{k\in J} B_{11}^{k}{B_{11}^{k}}^{*}+ \sum_{k\in J}B_{12}^{k} {B_{12}^{k}}^{*} \leq I_{{\mathcal {H}}_{1}}. \end{aligned}$$

On the other hand, $0\leq {\frac{1}{\lambda _{1} }} X_{1} \leq I_{{ \mathcal {H}}_{2}}$. Hence, $\sum_{k\in J} B_{12}^{k}(I_{{\mathcal {H}}_{2}}- \frac{1}{\lambda _{1} } X_{1} ) {B_{12}^{k}}^{*}=0$, and then $B_{12}=0$. Therefore, $A_{12}^{k} =0$ and $\sum_{k\in J} A_{11}^{k}{A_{11}^{k}}^{*}=I_{{\mathcal {H}}_{1}}$. From $\Phi _{\mathcal {A}}^{\dagger} (I)\leq I$, that is, $\sum_{k\in J} A_{k}^{*}A_{k}\leq I_{\mathcal {H}}$ then $\sum_{k\in J}{ A_{11}^{k}}^{*}{A_{11}^{k}}\leq I_{{\mathcal {H}}_{1}}$. It follows from Lemma 3 that $\sum_{k\in J} { A_{11}^{k}}^{*}{A_{11}^{k}}= I_{{\mathcal {H}}_{1}}$ since ${\mathcal {H}}_{1}$ is a finite-dimensional space. Thus, $\sum_{k\in J} {A_{21}^{k} }^{*}{A_{21}^{k}}= 0$ and then $A_{21}^{k}=0$. This shows that $A_{k} P_{1}=P_{1}A_{k}$, $\Phi _{ \mathcal {A}}(P_{1})=P_{1}$, $\Phi _{ \mathcal {A}}^{\dagger}(P_{1})=P_{1}$ and $\Phi _{\mathcal {A}}(X)=\lambda _{1}\Phi _{\mathcal {A}}(P_{1})\oplus \Phi _{\mathcal {A}}(X_{1})\leq \lambda _{1}P_{1}\oplus X_{1}$. Therefore, $\Phi _{\mathcal {A}}(X_{1})\leq X_{1}$. By induction, $X\in {\mathcal {A}}^{\prime}$, $\Phi _{\mathcal {A}}(X)=X$ and $\Phi _{\mathcal {A}}^{\dagger} (X)=X$.

(2) If $\Phi _{\mathcal {A}}(X)\geq X$, the process is as above, the result holds by Lemma 2. The proof is completed. □

Similar to the proof of Theorem 4, we have the following result.

Theorem 5

([4])

Let $\Phi _{\mathcal {A}}(I)\leq I$ and $\Phi _{\mathcal {A}}^{\dagger} (I)\leq I$. If $X \in {\mathcal {K}}(\mathcal {H})$ satisfies $\Phi _{\mathcal {A}}(X)\geq X\geq 0$, then $\Phi _{\mathcal {A}}(X)=X$ and $X\in {\mathcal {A}}^{\prime}$ hold.

Corollary 6

([1])

Let $\dim \mathcal {H}<\infty$ and ${\mathcal {A}}\subset {\mathcal {B}}(\mathcal {H})$ be a unital and trace-preserving row contraction. Then, ${\mathcal {B}}{(\mathcal {H})}^{\Phi _{\mathcal {A}}}= {\mathcal {A}}^{ \prime}$.

Proof

As $\mathcal {A}$ is unital, it is natural that ${\mathcal {A}}^{\prime} \subset {\mathcal {B}}({\mathcal {H}})^{\Phi _{ \mathcal {A}}}$ holds. We need only to prove that ${\mathcal {B}}({\mathcal {H}})^{\Phi _{\mathcal {A}}}\subset \mathcal {A}^{ \prime}$. For any $X\in {\mathcal {B}}({\mathcal {H}})^{\Phi _{\mathcal {A}}}$, then $X^{*} \in {\mathcal {B}}({\mathcal {H}})^{\Phi _{\mathcal {A}}}$. Hence, we can assume that X is self-adjoint. Denote ${\mathcal {H}}_{1}= P^{X} (0,\|X\|]$ and ${\mathcal {H}}_{2}=[-\|X\|, 0]$. Then, ${\mathcal {H}}={\mathcal {H}}_{1} \oplus {\mathcal {H}}_{2}$ and X has the representation $X=X^{+} \oplus ( -X^{-})$, where $X^{+}$ is invertible in ${\mathcal {B}}({\mathcal {H}}_{1})$. With respect to the space decomposition as above, the operator $A_{k}$ can be expressed as $A_{k}=(A_{ij}^{k})_{2\times 2}$ and then ${A_{k}}^{*}=({A_{ji}^{k}}^{*})_{2\times 2}$. It follows that

$$\begin{aligned} A_{k}XA_{k}^{*}= \begin{pmatrix} A_{11}^{k} X^{+}{A_{11}^{k}}^{*}-A_{12}^{k} X^{-} {A_{12}^{k}}^{*} & A_{11}^{k} X^{+}{A_{21}^{k}}^{*}-A_{12}^{k} X^{-} {A_{22}^{k}}^{*} \\ A_{21}^{k} X^{+}{A_{11}^{k}}^{*}-A_{22}^{k} X^{-} {A_{12}^{k}}^{*} & A_{21}^{k} X^{+}{A_{21}^{k}}^{*}-A_{22}^{k} X^{-} {A_{22}^{k}}^{*} \end{pmatrix}. \end{aligned}$$

From $\Phi _{\mathcal {A}}(X)=X$, we obtain

$$\begin{aligned} \textstyle\begin{cases} \sum_{k\in J} A_{11}^{k} X^{+}{A_{11}^{k}}^{*}-\sum_{k\in J} A_{12}^{k} X^{-} {A_{12}^{k}}^{*}=X^{+}, \\ \sum_{k\in J} A_{21}^{k} X^{+}{A_{21}^{k}}^{*}-\sum_{k\in J} A_{22}^{k} X^{-} {A_{22}^{k}}^{*} = -X^{-}, \end{cases}\displaystyle \end{aligned}$$

whereas, $\sum_{k\in J} A_{12}^{k} X^{-} {A_{12}^{k}}^{*}\geq 0$, so $\sum_{k\in J} A_{11}^{k} X^{+}{A_{11}^{k}}^{*}\geq X^{+}$. Combining this with Theorem 5, we have $X^{+}\in {\{A_{11}^{k}\}_{k\in J}}^{\prime}$ and $\sum_{k\in J} A_{11}^{k} X^{+} {A_{11}^{k}}^{*}=X^{+}$. Furthermore, $\sum_{k\in J} A_{11}^{k} {A_{11}^{k}}^{*}=I_{{\mathcal {H}}_{1}}$ holds. Moreover, $\sum_{k\in J} A_{k} {A_{k}}^{*}=I_{\mathcal {H}}$ implies $\sum_{k\in J} A_{12}^{k} {A_{12}^{k}}^{*}=0$ and hence $A_{12}^{k}=0$. From Lemma 3 and $\sum_{k\in J} A_{k} ^{*} {A_{k}}=I_{\mathcal {H}}$, we have $A_{21}^{k}=0$ for any k. Hence, $\sum_{k\in J} A_{22}^{k} {A_{22}^{k}}^{*}=I_{{ \mathcal {H}}_{2}}$. Combining $\sum_{k\in J} A_{22}^{k} X^{-} {A_{22}^{k}}^{*} \geq X^{-}$ with Theorem 4, it is easy to obtain $X^{-}\in {\{A_{22}^{k}\}_{k\in J}}^{\prime}$, and then $X\in \mathcal {A}^{\prime}$. The proof is completed. □

In Theorem 4, the result does not necessarily hold if X is not a compact operator.

Example 7

Let $\{e_{1},e_{2},\ldots \}$ be a basis of an infinite Hilbert space $\mathcal {H}$ and S be the unilateral operator on $\mathcal {H}$. Then, $Se_{i}=e_{i+1}, \forall i \geq 1$. Suppose that $\mathcal {K}=\mathcal {H} \oplus \mathcal {H} \oplus \mathcal {H}$. Define an operator A as follows,

$$\begin{aligned} A= \begin{pmatrix} S^{*} & 0 & 0 \\ \frac{1}{\sqrt{2}}(I-SS^{*}) & SS^{*} & \frac{1}{\sqrt{2}}(I-SS^{*}) \\ 0 & 0 &S^{*} \end{pmatrix}. \end{aligned}$$

Then,

$$\begin{aligned} A^{*}= \begin{pmatrix} S & \frac{1}{\sqrt{2}}(I-SS^{*}) & 0 \\ 0 & SS^{*} & 0 \\ 0 & \frac{1}{\sqrt{2}}(I-SS^{*}) &S \end{pmatrix}. \end{aligned}$$

By direct computing, it is easy to obtain that $AA^{*}=I_{\mathcal {K}}$ and $A^{*}A\leq I_{\mathcal {K}}$. Assume that $X\in \mathcal {B}(\mathcal {K})$ has the following matrix form,

$$\begin{aligned} X= \begin{pmatrix} I_{\mathcal {H}} & 0 & I_{\mathcal {H}} \\ 0 & \frac{3}{2}I_{\mathcal {H}} & 0 \\ 0 & 0 & I_{\mathcal {H}} \end{pmatrix}. \end{aligned}$$

According to the matrix forms of $A, A^{*}, X$, $AXA^{*}=X$ holds, whereas,

$$\begin{aligned} AX= \begin{pmatrix} S^{*} & 0 & 0 \\ \frac{1}{\sqrt{2}}(I-SS^{*}) & \frac{3}{2} SS^{*} & {\sqrt{2}}(I-SS^{*}) \\ 0 & 0 &S^{*} \end{pmatrix},\\ XA= \begin{pmatrix} S^{*} & 0 &S^{*} \\ \frac{3}{\sqrt{2}} (I-SS^{*}) & \frac{3}{2} SS^{*} & \frac{3}{\sqrt{2}}(I-SS^{*}) \\ 0 & 0 &S^{*} \end{pmatrix}. \end{aligned}$$

These show that $AX\neq XA$ and $A^{*}XA\neq X$.

Proposition 8

Let $\Phi _{\mathcal {A}}(I)\leq I$ and $\Phi _{\mathcal {A}}^{\dagger} (I)\leq I$. Suppose that X is a positive operator with only at most a countable set of distinct eigenvalues $\{\lambda _{i}\}$ such that $X =\sum_{i}\lambda _{i}P_{i}$, where $P_{i}P_{j}=P_{j}P_{i}=0$ and $\lambda _{i}$ is strictly decreasing. If $\Phi _{\mathcal {A}}(X)\geq X$ and $\Phi _{\mathcal {A}}^{\dagger} (X)\geq X$, then $X\in {\mathcal {A}}^{\prime}$ and $\Phi _{\mathcal {A}}(X)=\Phi _{ \mathcal {A}}^{\dagger}(X)= X$.

Proof

Suppose that X is a positive operator with $X =\sum_{i}\lambda _{i}P_{i}$ and $\lambda _{i}$ is strictly decreasing. Denote ${\mathcal {H}}_{1}= P^{X}\{\lambda _{1}\} \mathcal {H}$ and ${\mathcal {H}}_{2}={\mathcal {H} }\ominus {\mathcal {H}}_{1}$. Then, $X=\lambda _{1} I_{{\mathcal {H}}_{1}}\oplus X_{1}$. $A_{k}$ and $A_{k}^{*}$ have the following matrix forms,

$$\begin{aligned} A_{k}= \begin{pmatrix} A_{11}^{k} & A_{12}^{k} \\ A_{21}^{k} & A_{22}^{k} \end{pmatrix}\quad\text{and}\quad A_{k}^{*}= \begin{pmatrix} {A_{11}^{k}}^{*} & {A_{21}^{k}}^{*} \\ {A_{12}^{k}}^{*} & {A_{22}^{k}}^{*} \end{pmatrix}. \end{aligned}$$

Therefore,

$$\begin{aligned} A_{k}XA_{k}^{*}= \begin{pmatrix} \lambda _{1} A_{11}^{k} {A_{11}^{k}}^{*}+A_{12}^{k} X_{1} {A_{12}^{k}}^{*} & \lambda _{1} A_{11}^{k} {A_{21}^{k}}^{*}+A_{12}^{k} X_{1} {A_{22}^{k}}^{*} \\ \lambda _{1} A_{21}^{k} {A_{11}^{k}}^{*}+A_{22}^{k} X_{1} {A_{12}^{k}}^{*} & \lambda _{1} A_{21}^{k} {A_{21}^{k}}^{*}+A_{22}^{k} X_{1} {A_{22}^{k}}^{*} \end{pmatrix}. \end{aligned}$$

From $\sum_{k\in J} A_{k}XA_{k}^{*}\geq X$, we have

$$\begin{aligned} \lambda _{1} I_{{\mathcal {H}}_{1}}\leq \sum_{k\in J} \lambda _{1} A_{11}^{k} {A_{11}^{k}}^{*}+ \sum_{k\in J}A_{12}^{k} X_{1} {A_{12}^{k}}^{*} \leq \lambda _{1} \biggl(\sum_{k\in J}A_{11}^{k} {A_{11}^{k}}^{*}+\sum _{k \in J}A_{12}^{k} {A_{12}^{k}}^{*} \biggr)\leq \lambda _{1}{{\mathcal {H}}_{1}}. \end{aligned}$$

If $X_{1}=0$, then $\sum_{k\in J}A_{11}^{k} {A_{11}^{k}}^{*}=I_{{\mathcal {H}}_{1}}$. It follows that $\sum_{k\in J}A_{12}^{k} {A_{12}^{k}}^{*}=0$, hence $A_{12}^{k}=0$. If $X_{1}\neq 0$, then $X_{1}<\lambda _{1}I_{{\mathcal {H}}_{2}}$, which means $\lambda _{1}I_{{\mathcal {H}}_{2}}-X_{1}$ is a positive and invertible operator. Therefore, $\sum_{k\in J}A_{12}^{k} {A_{12}^{k}}^{*}=0$ and so $A_{12}^{k}=0$. On the other hand, from $\Phi _{\mathcal {A}}^{\dagger}(X)\geq X$, we can obtain $A_{21}^{k}=0$. That is, $A_{k}P_{1}=P_{1}A_{k}$, $\Phi _{\mathcal {A}}(P_{1})=P_{1}$ and $\Phi _{\mathcal {A}}^{\dagger}(P_{1})=P_{1}$. Meanwhile, $\Phi _{\mathcal {A}}(0\oplus X_{1})\geq 0\oplus X_{1}$, $\Phi _{\mathcal {A}}^{\dagger} (0\oplus X_{1})\geq 0\oplus X_{1}$. Continuing the above process, the result holds. The proof is completed. □

Theorem 9

Let $\Phi _{\mathcal {A}}(I)\leq I$ and $\Phi _{ \mathcal {A}}^{\dagger} (I)\leq I$. Suppose that X is a self-adjoint operator with only at most a countable set of distinct eigenvalues $\{\lambda _{i}\}$ and $|\lambda _{i}|$ can be arranged in decreasing order, where $|\lambda _{i}|$ means the absolute value of $\lambda _{i}$. If $\Phi _{\mathcal {A}}(X)=X$ and $\Phi _{\mathcal {A}}^{\dagger}(X)= X$, then $X\in {\mathcal {A}}^{\prime}$.

Proof

Let ${\mathcal {H}}_{1}=P^{X} [-\|X\|, 0), {\mathcal {H}}_{2}=P^{X} \{ 0\}$ and ${\mathcal {H}}_{3}=P^{X} (0,\|X\|]$, where $P^{X} (\cdot )$ is the spectral measure of X. Then, ${\mathcal {H}}={\mathcal {H}}_{1} \oplus { \mathcal {H}}_{2} \oplus {\mathcal {H}}_{3}$. X has the matrix form $X=X_{1} \oplus 0 \oplus (-X_{3})$, where $X_{1}$ and $X_{3}$ are injective and have dense ranges. Denote $A_{k}=(A_{ij}^{k})_{3\times 3}$, then ${A_{k}}^{*}=({A_{ji}^{k}}^{*})_{3\times 3}$. By direct computing, we have

$$\begin{aligned} A_{k}XA_{k}^{*}= \begin{pmatrix} A_{11}^{k} X_{1} A_{11}^{k} *-A_{13}^{k} X_{3} {A_{13}^{k}}^{*} & A_{11}^{k} X_{1}{A_{21}^{k}}^{*}-A_{13}^{k} X_{3} {A_{23}^{k}}^{*} & A_{11}^{k} X_{1}{A_{31}^{k}}^{*}-A_{13}^{k} X_{3} {A_{33}^{k}}^{*} \\ A_{21}^{k} X_{1}{A_{11}^{k}}^{*}-A_{23}^{k} X_{3} {A_{13}^{k}}^{*} & A_{21}^{k} X_{1}{A_{21}^{k}}^{*}-A_{23}^{k} X_{3} {A_{23}^{k}}^{*} & A_{21}^{k} X_{1}{A_{31}^{k}}^{*}-A_{23}^{k} X_{3} {A_{33}^{k}}^{*} \\ A_{31}^{k} X_{1}{A_{11}^{k}}^{*}-A_{33}^{k} X_{3} {A_{13}^{k}}^{*} & A_{31}^{k} X_{1}{A_{11}^{k}}^{*}-A_{33}^{k} X_{3} {A_{13}^{k}}^{*} & A_{31}^{k} X_{1}{A_{31}^{k}}^{*}-A_{33}^{k} X_{3} {A_{33}^{k}}^{*} \end{pmatrix}. \end{aligned}$$

From $\Phi _{\mathcal {A}}(X)=X$, it is easy to see that

$$\begin{aligned} &\sum_{k\in J} A_{11}^{k} X_{1}{A_{11}^{k}}^{*}-\sum _{k\in J} A_{13}^{k} X_{3} {A_{13}^{k}}^{*}=X_{1}, \end{aligned}$$

(1)

$$\begin{aligned} &\sum_{k\in J} A_{31}^{k} X_{1}{A_{31}^{k}}^{*}-\sum _{k\in J} A_{33}^{k} X_{3} {A_{33}^{k}}^{*} = -X_{3}, \end{aligned}$$

(2)

whereas,

$$\begin{aligned} A_{k}^{*}XA_{k}= \begin{pmatrix} {A_{11}^{k}}^{*}X_{1}A_{11}^{k} - {A_{31}^{k}}^{*}X_{3} A_{31}^{k} & {A_{11}^{k}}^{*}X_{1}A_{12}^{k} - {A_{31}^{k}}^{*}X_{3} A_{32}^{k} & {A_{11}^{k}}^{*}X_{1}A_{13}^{k} - {A_{31}^{k}}^{*}X_{3} A_{33}^{k} \\ {A_{12}^{k}}^{*}X_{1}A_{11}^{k} -{A_{32}^{k}}^{*}X_{3} A_{31}^{k} & {A_{12}^{k}}^{*}X_{1}A_{12}^{k} - {A_{32}^{k}}^{*}X_{3} A_{32}^{k} & {A_{12}^{k}}^{*}X_{1}A_{13}^{k} -X_{3} {A_{32}^{k}}^{*} A_{33}^{k} \\ {A_{13}^{k}}^{*}X_{1}A_{11}^{k} -{A_{33}^{k}}^{*}X_{3} A_{31}^{k} & {A_{13}^{k}}^{*}X_{1}A_{12}^{k} -{A_{33}^{k}}^{*}X_{3} A_{32}^{k} & {A_{13}^{k}}^{*}X_{1} A_{13}^{k} - {A_{33}^{k}}^{*}X_{3} A_{33}^{k} \end{pmatrix}. \end{aligned}$$

From $\Phi _{\mathcal {A}}^{\dagger}(X)= X$, we can obtain

$$\begin{aligned} &\sum_{k\in J} {A_{11}^{k}}^{*}X_{1}A_{11}^{k} - \sum_{k\in J}{A_{31}^{k}}^{*}X_{3} A_{31}^{k}=X_{1}, \end{aligned}$$

(3)

$$\begin{aligned} &\sum_{k\in J} {A_{13}^{k}}^{*}X_{1}A_{13}^{k} - \sum_{k\in J}{A_{33}^{k}}^{*}X_{3} A_{33}^{k}= -X_{3}. \end{aligned}$$

(4)

As $\sum_{k\in J} A_{13}^{k} X_{3} {A_{13}^{k}}^{*}\geq 0$ and $\sum_{k\in J} A_{31}^{k} X_{1}{A_{31}^{k}}^{*}\geq 0$, combining Eq. (1) with Eq. (2), we have

$$\begin{aligned} &\sum_{k\in J} A_{11}^{k} X_{1}{A_{11}^{k}}^{*}\geq X_{1}, \end{aligned}$$

(5)

$$\begin{aligned} &\sum_{k\in J} {A_{33}^{k}}^{*}X_{3}A_{33}^{k} \geq X_{3}. \end{aligned}$$

(6)

Similarly, combining $\sum_{k\in J}{A_{13}^{k}}^{*} X_{1} A_{13}^{k}\geq 0$, $\sum_{k\in J}{A_{31}^{k}}^{*} X_{3} A_{31}^{k}\geq 0$ with Eqs. (3) and (4), the following equations hold,

$$\begin{aligned} &\sum_{k\in J} {A_{11}^{k}}^{*} X_{1} A_{11}^{k} \geq X_{1}, \end{aligned}$$

(7)

$$\begin{aligned} &\sum_{k\in J} {A_{33}^{k}}^{*}X_{3} A_{33}^{k} \geq X_{3}. \end{aligned}$$

(8)

It follows from Proposition 8, Eqs. (5), (7), (6), and (8) that

$$\begin{aligned} X_{1} \in \bigl\{ A_{11}^{k}, {A_{11}^{k}}^{*} \bigr\} ^{\prime}\quad\text{and}\quad X_{3} \in \bigl\{ A_{33}^{k}, {A_{33}^{k}}^{*} \bigr\} ^{\prime}, \end{aligned}$$

(9)

and

$$\begin{aligned} \sum_{k\in J} A_{11}^{k} X_{1}{A_{11}^{k}}^{*}= X_{1},\qquad \sum_{k \in J} A_{33}^{k} X_{3}{A_{33}^{k}}^{*}=X_{3}.~ \end{aligned}$$

Hence, $\sum_{k\in J} A_{13}^{k} X_{3} {A_{13}^{k}}^{*}=0$. As $X_{3}$ is positive, injective, and also has dense range, hence $A_{13}^{k}=0$. Similarly, $A_{31}^{k}=0$. The operator $X_{1}$ is also a positive and injective operator with dense range, $\sum_{k\in J} A_{11}^{k} {A_{11}^{k}}^{*}=I_{{\mathcal {H}}_{1}}$ from Eq. (5). According to $\sum_{k\in J} A_{k} {A_{k}}^{*}\leq I_{\mathcal {H} }$, then $\sum_{k \in J} A_{12}^{k} {A_{12}^{k}}^{*}=0$, and so $A_{12}^{k}=0$ for any k. Similarly, $A_{21}^{k}=0$. This shows that $A_{k}= A_{11}^{k}\oplus A_{22}^{k}\oplus A_{33}^{k}$. Combining Eq. (9) and the matrix forms of X and $A_{k}$, we have $A_{k}X=XA_{k}$ for any k. The proof is completed. □

If X has only two spectral points, we have the following result.

Theorem 10

Let $\mathcal {A}$ be a unital operator sequence and X be a self-adjoint operator with only two spectral points. If $\Phi _{\mathcal {A}}(X)=X$, then $X\in {\mathcal {A}}^{\prime}$.

Proof

Let $\lambda _{1}, \lambda _{2}$ be the two spectral points of X. Without loss of generality, suppose that $\lambda _{1}>\lambda _{2}>0$ since $\Phi _{\mathcal {A}}(I)=I$. Denote ${\mathcal {H}}_{1}=P^{X}\{\lambda _{1}\}{\mathcal {H}}$ and ${\mathcal {H}}_{2}=P^{X}\{\lambda _{2}\}{\mathcal {H}}$, then ${\mathcal {H}}_{1} \oplus {\mathcal {H}}_{2}=\mathcal {H}$. Hence, $X=\lambda _{1}I_{{\mathcal {H}}_{1}}\oplus \lambda _{2} I_{{\mathcal {H}}_{2}}$. Assume that $A_{k}$ has the matrix form $A_{k}=(A_{ij}^{k})_{2\times 2}$ with respect to the space decomposition ${\mathcal {H}} ={\mathcal {H}}_{1} \oplus {\mathcal {H}}_{2}$. From $\Phi _{\mathcal {A}}(X)=X$, we have

$$\begin{aligned} &\sum_{k\in J} A_{k}XA_{k}^{*} \\ &\quad= \begin{pmatrix} \lambda _{1}\sum_{k\in J} A_{11}^{k} {A_{11}^{k}}^{*}+\lambda _{2} \sum_{k\in J}A_{12}^{k} {A_{12}^{k}}^{*} & \lambda _{1} \sum_{k \in J}A_{11}^{k} {A_{21}^{k}}^{*}+\lambda _{2} \sum_{k\in J} A_{12}^{k} {A_{22}^{k}}^{*} \\ \lambda _{1} \sum_{k\in J} A_{21}^{k} {A_{11}^{k}}^{*}+\lambda _{2} \sum_{k\in J}A_{22}^{k} {A_{12}^{k}}^{*} & \lambda _{1} \sum_{k \in J}A_{21}^{k} {A_{21}^{k}}^{*}+ \lambda _{2} \sum_{k\in J}A_{22}^{k} {A_{22}^{k}}^{*} \end{pmatrix} \\ & \quad= \begin{pmatrix} \lambda _{1} & 0 \\ 0 & \lambda _{2} \end{pmatrix}. \end{aligned}$$

This shows that $\lambda _{1}\sum_{k\in J} A_{11}^{k} {A_{11}^{k}}^{*}+\lambda _{2} \sum_{k\in J}A_{12}^{k} {A_{12}^{k}}^{*} =\lambda _{1} I_{{ \mathcal {H}}_{1}}$. That is,

$$\begin{aligned} \sum_{k\in J} A_{11}^{k} {A_{11}^{k}}^{*}+ \frac{\lambda _{2}}{\lambda _{1}} \sum _{k\in J}A_{12}^{k} {A_{12}^{k}}^{*} =I_{{\mathcal {H}}_{1}}. \end{aligned}$$

As $\mathcal {A}$ is unital, it is easy to obtain that

$$\begin{aligned} \sum_{k\in J} A_{11}^{k} {A_{11}^{k}}^{*}+ \sum _{k\in J}A_{12}^{k} {A_{12}^{k}}^{*} =I_{{\mathcal {H}}_{1}}. \end{aligned}$$

Hence, $\sum_{k\in J}A_{12}^{k} {A_{12}^{k}}^{*}=0$ and then $A_{12}^{k}=0$. Similarly, according to

$$\begin{aligned} \lambda _{1} \sum_{k\in J}A_{21}^{k} {A_{21}^{k}}^{*}+ \lambda _{2} \sum _{k\in J}A_{22}^{k} {A_{22}^{k}}^{*}=\lambda _{2}I_{{\mathcal {H}}_{2}} \end{aligned}$$

and

$$\begin{aligned} \sum_{k\in J}A_{21}^{k} {A_{21}^{k}}^{*}+ \sum _{k\in J}A_{22}^{k} {A_{22}^{k}}^{*}=I_{{ \mathcal {H}}_{2}}, \end{aligned}$$

we have $A_{21}^{k}=0$. Hence $X\in \mathcal {A}^{\prime}$. The proof is completed. □