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. □