T-BT Inverse and T-GC Partial Order via the T-Product

Abstract

In this paper, we extend the BT inverse to the set of third-order tensors, and we call it the T-BT inverse. We give characterizations and properties of the inverse by applying tensor decomposition. Based on the inverse, we introduce a new binary relation: T-BT order. Furthermore, by applying the T-BT order, we introduce a generalized core partial order (called T-GC partial order).

1. Introduction

The tensor $\mathcal{A}=\left(a_{i_1{i}_2...i_N}\right)\in {\mathbb{C}}^{I_1\times I_2\times \cdots \times I_N}$ is a higher-order extension of the matrix, in which $\mathbb{C}$ stands for the field of complex numbers. Write N as the order of the tensor $\mathcal{A}$. The tensors are linear representations of linear transformations in nth-order space. Obviously, the first-order tensor is a vector, the second-order tensor is a matrix, and tensors of the third or greater order are called high-order tensors. Higher-order tensors are applied in numerical linear algebra, data mining, graph analysis, etc.

There are many definitions of tensor multiplication, such as Einstein product and T-product. It is known that there are some studies on tensors, for example tensor decomposition, eigenvalues of tensor, etc. In [1], Sun, Zheng, Bu and Wei define the Moore–Penrose inverse of tensors with the Einstein product and the explicit formulas of the Moore–Penrose inverse of some block tensors. The general solutions of some multilinear systems are given, as well as the minimum-norm least-square solution of some multilinear systems using the Moore–Penrose inverse of tensors. Furthermore, in [2], Ratikanta Behera and Debasisha Mishra elaborate on this theory by producing a few characterizations of different generalized inverses of tensors. A new method, the Moore–Penrose inverse of tensors, is proposed. Reverse order laws for several generalized inverses of tensors are also presented. In addition, they discuss general solutions of multilinear systems of tensors using such theories. In [3], Kilmer and Martin introduce the concept of the tensor T-product. The tensor multiplication is converted into matrix multiplication by using block-cyclic matrix theory and discrete Fourier transform. Based on the T-product, Miao, Qi and Wei [4] obtain the T-Schur decomposition and T-core-nilpotent decomposition, obtain characterizations of the T-Drazin inverse, and propose the T-characteristic polynomial and corresponding Cayley–Hamilton theorem. Jin, Bai, Benítez and Liu [5] introduce the T-Moore–Penrose inverse based on the T-product. The fast Fourier transform algorithm is used to prove that the T-Moore–Penrose inverse of tensors $\mathcal{A}$ exists and is unique. Some properties of the tensor are given, and the representations of the {1}-inverse, {1,3}-inverse, {1,4}-inverse of the tensor are established. An algorithm for computing the T-Moore–Penrose inverse of tensors is constructed. Furthermore, Jin, Bai, Benítez and Liu [5] show how the T-Moore–Penrose inverse can be applied to the high-order Gauss-Markov theorem.

The T-Moore–Penrose inverse is denoted by ${\mathcal{A}}^{†}$ of $\mathcal{A}\in {\mathbb{C}}^{m\times n\times p}$, which is the unique tensor satisfying the following tensor equations:

$$\begin{array}{c}\hfill \mathcal{A}\ast {\mathcal{A}}^{†}\ast \mathcal{A}=\mathcal{A},{\mathcal{A}}^{†}\ast \mathcal{A}\ast {\mathcal{A}}^{†}={\mathcal{A}}^{†},{(\mathcal{A}\ast {\mathcal{A}}^{†})}^{\ast }=\mathcal{A}\ast {\mathcal{A}}^{†},{({\mathcal{A}}^{†}\ast \mathcal{A})}^{\ast }={\mathcal{A}}^{†}\ast \mathcal{A}.\end{array}$$

If $\mathcal{A}$ is nonsingular, then ${\mathcal{A}}^{†}={\mathcal{A}}^{-1}$. Let ${\mathcal{P}}_{\mathcal{A}}=\mathcal{A}\ast {\mathcal{A}}^{†}$, ${\mathcal{R}}_{\mathcal{A}}={\mathcal{A}}^{†}\ast \mathcal{A}$.

In [6], Baksalary and Trenkler introduce the $\mathrm{core}$ inverse: let $A\in {\mathbb{C}}^{n\times n}$ and $\mathrm{Ind}\left(A\right)=1$, then there exists the unique matrix X satisfying $AX=A{A}^{†}$, $\mathcal{R}\left(X\right)\subseteq \mathcal{R}\left(A\right)$, which is denoted as $X=A^{\#}$. Subsequently, several generalized core inverses are introduced. For example, letting $A\in {\mathbb{C}}^{n\times n}$ with $\mathrm{Ind}\left(A\right)=k$, Manjunatha Prasad and Mohana [7] introduce the $\mathrm{core}$-$\mathrm{EP}$ inverse $A^{†}=A^k{\left({\left(A^{\ast }\right)}^k{A}^{k+1}\right)}^{-}{\left(A^{\ast }\right)}^k$. Malik and Thome [8] introduce the DMP inverse $A^{d,†}=A^{D}A{A}^{†}$. Baksalary and Trenkler introduce [9] the BT inverse $A^{\diamond }={\left(A{P}_A\right)}^{†}$. When $\mathrm{Ind}\left(A\right)=1$, it is obvious that $A^{d,†}=A^{†}=A^{\diamond }=A^{\#}$. Furthermore, let the Hartwig–Spindelböck decomposition of A be as in [10]:

$$\begin{array}{c}\hfill A=Q\left[\begin{array}{cc}EK& EL\\ O& O\end{array}\right]Q^{\ast },\end{array}$$

(1)

where Q is a unitary matrix, $K{K}^{\ast }+L{L}^{\ast }=I$, and E is a diagonal positive definite matrix. Then

$$\begin{array}{c}\hfill A^{\diamond }=Q\left[\begin{array}{cc}{\left(EK\right)}^{†}& O\\ O& O\end{array}\right]Q^{\ast }.\end{array}$$

(2)

It is important to note that compared to other generalized core inverses (e.g., core-EP inverse and DMP inverses), there are few research results on the BT inverse. As we all know, the core, core-EP and DMP inverses of tensors are defined, but there are no studies of the T-BT inverse on third-order tensors. Therefore, it is meaningful and challenging to study the T-BT inverse of tensors.

In this paper, the BT inverse is extended from matrices to tensors using tools such as T-Hartwig–Spindelböck decomposition and discrete Fourier transform based on the T-product. Based on the inverse, a new binary relationship “$\stackrel{\diamond }{⪯}$” is introduced. It does not satisfy the antisymmetry and transitivity, so “$\stackrel{\diamond }{⪯}$” is not a partial order. Then, we consider the relationship between the new binary relationship and T-Core, $\mathrm{T}$-minus order. Finally, we add some conditions to “$\stackrel{\diamond }{⪯}$” to make a new partial order ($\mathrm{T}$-$\mathrm{GC}$ partial order).

2. Preliminaries

Suppose that $\mathcal{A}\in {\mathbb{C}}^{m\times n\times p}$ is a third-order tensor. We write the i-th front slice of $\mathcal{A}$ as $A^{\left(i\right)}\in {\mathbb{C}}^{m\times n}$ ($i=1,\dots ,p$). Then

$$\mathcal{A}=\left[\begin{array}{cccc}{A}^{\left(1\right)}& A^{\left(2\right)}& \cdots & A^{\left(p\right)}\end{array}\right].$$

The operations bcirc, unfold and fold are defined as follows [3,11,12,13]:

$$\begin{array}{c}\hfill bcirc\left(\mathcal{A}\right):=\left[\begin{array}{ccccc}{A}^{\left(1\right)}& A^{\left(p\right)}& A^{(p-1)}& \cdots & A^{\left(2\right)}\\ A^{\left(2\right)}& A^{\left(1\right)}& A^{\left(p\right)}& \cdots & A^{\left(3\right)}\\ ⋮& ⋮& ⋮& & ⋮\\ A^{\left(p\right)}& A^{(p-1)}& A^{(p-2)}& \cdots & A^{\left(1\right)}\end{array}\right],unfold\left(\mathcal{A}\right):=\left[\begin{array}{c}{A}^{\left(1\right)}\\ A^{\left(2\right)}\\ ⋮\\ A^{\left(p\right)}\end{array}\right],\end{array}$$

and $fold\left(unfold\right(\mathcal{A}\left)\right):=\mathcal{A}$ means that “ fold ” is the inverse operator of “unfold”. Meanwhile, Miao, Qi and Wei [4] define the inverse operation $bcir{c}^{-1}:{\mathbb{C}}^{mp\times np}\to {\mathbb{C}}^{m\times n\times p}$ such that $bcir{c}^{-1}\left(bcirc\left(\mathcal{A}\right)\right)=\mathcal{A}$.

The definition of tensor T-product is given.

Definition 1

([3], T-product). Let $\mathcal{A}\in {\mathbb{C}}^{m\times n\times p}$, $\mathcal{B}\in {\mathbb{C}}^{n\times s\times p}$. Then

$$\mathcal{A}\ast \mathcal{B}:=fold\left(bcirc\right(\mathcal{A})\ast unfold(\mathcal{B}\left)\right).$$

Definition 2

([4], Conjugate transpose). Let ${\mathcal{A}}^{\ast }$ be the conjugate transpose of $\mathcal{A}\in {\mathbb{C}}^{m\times n\times p}$ and

$$\begin{array}{c}\hfill {\mathcal{A}}^{\ast }=fold\left(\left[\begin{array}{c}{\left(A^{\left(1\right)}\right)}^{\ast }\\ {\left(A^{\left(p\right)}\right)}^{\ast }\\ {\left(A^{(p-1)}\right)}^{\ast }\\ ⋮\\ {\left(A^{\left(2\right)}\right)}^{\ast }\end{array}\right]\right).\end{array}$$

In the T-range space of $\mathcal{A}$,

$\mathcal{R}\left(\mathcal{A}\right)=Ran\left((F_p\otimes I_m)bcric\left(\mathcal{A}\right)(F_p^{\ast }\otimes I_n)\right)$, and “Ran” means the range space; in the T-null space of $\mathcal{A}$, $\mathcal{N}\left(\mathcal{A}\right)=Null\left((F_p\otimes I_m)bcric\left(\mathcal{A}\right)(F_p^{\ast }\otimes I_n)\right)$, “Null” represents the null space ([4]), where “⊗” and $F_p$ respectively represent the Kronecker product and discrete Fourier transform matrix.

The T-product has the following basic properties.

Lemma 1

([5,13,14]). If $\mathcal{A}$, $\mathcal{B}$, $\mathcal{C}$ are tensors of appropriate sizes, then

(1)

$(\mathcal{A}\ast \mathcal{B})\ast \mathcal{C}=\mathcal{A}\ast (\mathcal{B}\ast \mathcal{C})$;

(2)

$bcirc(\mathcal{A}\ast \mathcal{B})=bcirc\left(\mathcal{A}\right)bicrc\left(\mathcal{B}\right)$;

(3)

$bcirc{\left(\mathcal{A}\right)}^j=bcirc\left({\mathcal{A}}^j\right)$, for arbitrary $j=0,1,2,\dots$;

(4)

$bcirc{\left(\mathcal{A}\right)}^{\ast }=bcirc\left({\mathcal{A}}^{\ast }\right)$;

(5)

$bcirc{\left(\mathcal{A}\right)}^{†}=bcirc\left({\mathcal{A}}^{†}\right)$.

Definition 3

([4], Identity tensor and inverse tensor). Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ and $\mathcal{B}\in {\mathbb{C}}^{n\times n\times p}$. If $\mathcal{A}\ast \mathcal{B}=\mathcal{B}\ast \mathcal{A}={\mathcal{I}}_{nnp}$, then $\mathcal{A}$ is invertible and $\mathcal{B}$ $(\mathcal{B}={\mathcal{A}}^{-1})$ is the inverse tensor of $\mathcal{A}$, where ${\mathcal{I}}_{nnp}$ is the $n\times n\times p$ identity tensor, the first frontal slice of ${\mathcal{I}}_{nnp}$ is the $n\times n$ identity matrix, and its other frontal slices are all zeros.

The circular matrix can be transformed into a diagonal form by discrete Fourier transform [15]. For the block circulant matrix $bcirc\left(\mathcal{A}\right)$, we can diagonalize its blocks through discrete Fourier transform [16,17]. Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$, then

$$\begin{array}{c}\hfill bcirc\left(\mathcal{A}\right)=(F_p^{\ast }\otimes I_n)Diag(A_1,...,A_p)(F_p\otimes I_n),\end{array}$$

where $A_1,\cdots ,A_p\in {\mathbb{C}}^{n\times n}$. Furthermore,

$${\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}=\left\{\mathcal{A}:\mathrm{rank}\left(A_1\right)=\cdots =\mathrm{rank}\left(A_p\right),\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}\right\}.$$

On the basis of the above block diagonalization, Miao, Qi and Wei [4] introduce the definitions of the $\mathrm{T}$-index and $\mathrm{T}$-rank.

Definition 4

([4], T-index and T-rank). Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$. Then

$${\mathrm{rank}}_T\left(\mathcal{A}\right)=\mathrm{rank}\left(bcirc\left(\mathcal{A}\right)\right)=\sum _{i=1}^p(\mathrm{rank}\left(A_i\right)).$$

and

$${\mathrm{Ind}}_T\left(\mathcal{A}\right)=\mathrm{Ind}\left(bcirc\left(\mathcal{A}\right)\right)=\underset{1\le i\le p}{max}\left\{\mathrm{Ind}\left(A_i\right)\right\},$$

in which $\mathrm{Ind}\left(A_i\right)$ satisfies the minimum positive integer k of $\mathrm{rank}\left(A_i^k\right)=\mathrm{rank}\left(A_i^{k+1}\right)$, $i=1,\dots ,p$.

Denote

$${\mathbb{C}}_{\mathcal{T}}^{\mathrm{CM}}=\left\{\mathcal{A}:{\mathrm{Ind}}_T\left(\mathcal{A}\right)=1,\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}\right\}.$$

In [18], Liu and Ma give the following decomposition.

Theorem 1

([18]). Let $\mathcal{A}=\left[\begin{array}{cccc}{A}^{\left(1\right)}& A^{\left(2\right)}& \cdots & A^{\left(p\right)}\end{array}\right]\in {\mathbb{C}}^{n\times n\times p}$. Then

$$\begin{array}{c}\hfill \mathcal{A}=\widehat{\mathcal{Q}}\ast \mathcal{S}\ast {\widehat{\mathcal{Q}}}^{\ast },\end{array}$$

(3)

in which

$$\begin{array}{cc}\hfill \widehat{\mathcal{Q}}& =bcir{c}^{-1}\left((F_p^{\ast }\otimes I_n)Diag({\widehat{Q}}_1,...,{\widehat{Q}}_p)(F_p\otimes I_n)\right),\hfill \\ \hfill \mathcal{S}& =bcir{c}^{-1}\left((F_p^{\ast }\otimes I_n)Diag(S_1,...,S_p)(F_p\otimes I_n)\right),\hfill \\ \hfill & =bcir{c}^{-1}\left((F_p^{\ast }\otimes I_n)\left[\begin{array}{ccc}\left[\begin{array}{cc}{E}_1{K}_1& E_1{L}_1\\ O& O\end{array}\right]& & \\ & \ddots & \\ & & \left[\begin{array}{cc}{E}_p{K}_p& E_p{L}_p\\ O& O\end{array}\right]\end{array}\right](F_p\otimes I_n)\right),\hfill \end{array}$$

${\widehat{Q}}_i$ is unitary, $E_i\in {\mathbb{C}}^{r_i\times r_i}$, $r_i=\mathrm{rank}\left(A^{\left(i\right)}\right)$, $K_i{K}_i^{\ast }+L_i{L}_i^{\ast }=I_{r_i}$, $i=1,\dots ,p$.

Based on the above decomposition, Liu and Ma consider the T-Moore–Penrose inverse of the third-order tensor under the T-product.

Theorem 2

([18]). Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ and its decomposition be given as in(3). Then

$$\begin{array}{c}\hfill {\mathcal{A}}^{†}=\widehat{\mathcal{Q}}\ast {\mathcal{S}}^{†}\ast {\widehat{\mathcal{Q}}}^{\ast },\end{array}$$

in which

$$\begin{array}{cc}\hfill {\mathcal{S}}^{†}& =bcir{c}^{-1}\left((F_p^{\ast }\otimes I_n)\left[\begin{array}{ccc}\left[\begin{array}{cc}{K}_1^{\ast }{E}_1^{-1}& O\\ L_1^{\ast }{E}_1^{-1}& O\end{array}\right]& & \\ & \ddots & \\ & & \left[\begin{array}{cc}{K}_p^{\ast }{E}_p^{-1}& O\\ L_p^{\ast }{E}_p^{-1}& O\end{array}\right]\end{array}\right](F_p\otimes I_n)\right).\hfill \end{array}$$

Theorem 3

([5]). Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$, $\mathcal{B}\in {\mathbb{C}}^{n\times n\times p}$ and $\mathcal{C}\in {\mathbb{C}}^{n\times n\times p}$. Then the tensor equation $\mathcal{A}\ast \mathcal{X}\ast \mathcal{B}=\mathcal{C}$ is consistent if and only if there exist ${\mathcal{A}}^{\left(1\right)}\in \mathcal{A}\left\{1\right\}$ and ${\mathcal{B}}^{\left(1\right)}\in \mathcal{B}\left\{1\right\}$ such that $\mathcal{A}\ast {\mathcal{A}}^{\left(1\right)}\ast \mathcal{C}\ast {\mathcal{B}}^{\left(1\right)}\ast \mathcal{B}=\mathcal{C},$ and the general solution is $\mathcal{X}={\mathcal{A}}^{\left(1\right)}\ast \mathcal{C}\ast {\mathcal{B}}^{\left(1\right)}+\mathcal{Z}-{\mathcal{A}}^{\left(1\right)}\ast \mathcal{A}\ast \mathcal{Z}\ast \mathcal{B}\ast {\mathcal{B}}^{\left(1\right)},$ in which $\mathcal{Z}\in {\mathbb{C}}^{n\times n\times p}$ is arbitrary.

Theorem 4

([14], Tensor block multiplication under T-Product). Let ${\mathcal{A}}_1\in {\mathbb{C}}^{n_1\times m_1\times p}$, ${\mathcal{B}}_1\in {\mathbb{C}}^{n_1\times m_2\times p}$, ${\mathcal{C}}_1\in {\mathbb{C}}^{n_2\times m_1\times p}$, ${\mathcal{D}}_1\in {\mathbb{C}}^{n_2\times m_2\times p}$, ${\mathcal{A}}_2\in {\mathbb{C}}^{m_1\times r_2\times p}$, ${\mathcal{B}}_2\in {\mathbb{C}}^{m_1\times r_2\times p}$, ${\mathcal{C}}_2\in {\mathbb{C}}^{m_2\times r_2\times p}$ and ${\mathcal{D}}_2\in {\mathbb{C}}^{m_2\times r_2\times p}$. Then

$$\begin{array}{c}\hfill \left[\begin{array}{cc}{\mathcal{A}}_1& {\mathcal{B}}_1\\ {\mathcal{C}}_1& {\mathcal{D}}_1\end{array}\right]\ast \left[\begin{array}{cc}{\mathcal{A}}_2& {\mathcal{B}}_2\\ {\mathcal{C}}_2& {\mathcal{D}}_2\end{array}\right]=\left[\begin{array}{cc}{\mathcal{A}}_1\ast {\mathcal{A}}_2+{\mathcal{B}}_1\ast {\mathcal{C}}_2& {\mathcal{A}}_1\ast {\mathcal{B}}_2+{\mathcal{B}}_1\ast {\mathcal{D}}_2\\ {\mathcal{C}}_1\ast {\mathcal{A}}_2+{\mathcal{D}}_1\ast {\mathcal{C}}_2& {\mathcal{C}}_1\ast {\mathcal{B}}_2+{\mathcal{D}}_1\ast {\mathcal{D}}_2\end{array}\right].\end{array}$$

(4)

In [19], Zhang and Ma introduce the T-Hartwig–Spindelböck decomposition for tensor and the definition of the T-core inverse. By applying the decomposition, Zhang and Ma obtain representations of the T-core inverse and T-Moore–Penrose inverse under the T-product.

Theorem 5

([19]). Let $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$. Then, there is a unitary tensor $\mathcal{Q}\in {\mathbb{C}}^{n\times n\times p}$ satisfying

$$\begin{array}{c}\hfill \mathcal{A}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\end{array}$$

(5)

in which $\mathcal{E}\in {\mathbb{C}}^{t\times t\times p}$ is an invertible diagonal tensor, $\mathcal{K}\in {\mathbb{C}}^{t\times t\times p}$, $\mathcal{L}\in {\mathbb{C}}^{t\times (n-t)\times p}$ and $\mathcal{K}\ast {\mathcal{K}}^{\ast }+\mathcal{L}\ast {\mathcal{L}}^{\ast }={\mathcal{I}}_t$.

Furthermore,

$$\begin{array}{c}\hfill {\mathcal{A}}^{†}=\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{K}}^{\ast }\ast {\mathcal{E}}^{-1}& \mathcal{O}\\ {\mathcal{L}}^{\ast }\ast {\mathcal{E}}^{-1}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\end{array}$$

(6)

Definition 5

([19], T-core inverse). Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ and ${\mathrm{Ind}}_T\left(\mathcal{A}\right)=1$. If $\mathcal{X}$ satisfies

$$\begin{array}{c}\hfill \left(1\right)\mathcal{A}\ast \mathcal{X}\ast \mathcal{A}=\mathcal{A},\left(2^{{}^{\prime }}\right)\mathcal{A}\ast {\mathcal{X}}^2=\mathcal{X},\left(3\right){(\mathcal{A}\ast \mathcal{X})}^{\ast }=\mathcal{A}\ast \mathcal{X},\end{array}$$

then $\mathcal{X}={\mathcal{A}}^{\#}$, and $\mathcal{X}$ is called the T-core inverse of $\mathcal{A}$.

Theorem 6

([19]). Let $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$ have the form in(5) and ${\mathrm{Ind}}_T\left(\mathcal{A}\right)=1$. Then $\mathcal{K}$ is invertible and

$$\begin{array}{c}\hfill {\mathcal{A}}^{\#}=\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{K}}^{-1}\ast {\mathcal{E}}^{-1}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\end{array}$$

(7)

3. T-BT Inverse

Firstly, we introduce the T-BT inverse:

Definition 6

([10], T-BT inverse). Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$. Denote

$$\begin{array}{c}\hfill {\mathcal{A}}^{\diamond }={(\mathcal{A}\ast {\mathcal{P}}_{\mathcal{A}})}^{†}.\end{array}$$

(8)

We call ${\mathcal{A}}^{\diamond }$ the T-BT inverse of $\mathcal{A}$.

It is well known that the T-Moore–Penrose inverse of any $\mathcal{A}$ always exists and is unique, then the T-BT inverse of any $\mathcal{A}$ always exists and is unique.

Theorem 7.

Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ and its decomposition be given as in(3). Then, the T-BT inverse of $\mathcal{A}$ is

$$\begin{array}{c}\hfill {\mathcal{A}}^{\diamond }=\widehat{\mathcal{Q}}\ast {\mathcal{S}}^{\diamond }\ast {\widehat{\mathcal{Q}}}^{\ast },\end{array}$$

in which

$$\begin{array}{cc}\hfill {\mathcal{S}}^{\diamond }& =bcir{c}^{-1}\left((F_p^{\ast }\otimes I_n){\left(Diag(S_1,...,S_p)\right)}^{\diamond }(F_p\otimes I_n)\right),\hfill \\ \hfill & =bcir{c}^{-1}\left((F_p^{\ast }\otimes I_n)\left[\begin{array}{ccc}\left[\begin{array}{cc}{\left(E_1{K}_1\right)}^{†}& O\\ O& O\end{array}\right]& & \\ & \ddots & \\ & & \left[\begin{array}{cc}{\left(E_p{K}_p\right)}^{†}& O\\ O& O\end{array}\right]\end{array}\right](F_p\otimes I_n)\right).\hfill \end{array}$$

(9)

Proof. 

Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ and its decomposition be given as in (3). Then

$$\begin{array}{cc}\hfill bcirc\left({\mathcal{P}}_{\mathcal{S}}\right)& =bcirc(\mathcal{S}\ast {\mathcal{S}}^{†})=(F_p^{\ast }\otimes I_n)\left[\begin{array}{ccc}\left[\begin{array}{cc}{I}_{\mathrm{rank}\left(S_1\right)}& O\\ O& O\end{array}\right]& & \\ & \ddots & \\ & & \left[\begin{array}{cc}{I}_{\mathrm{rank}\left(S_p\right)}& O\\ O& O\end{array}\right]\end{array}\right](F_p\otimes I_n),\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill bcirc(\mathcal{S}\ast {\mathcal{P}}_{\mathcal{S}})& =(F_p^{\ast }\otimes I_n)\left[\begin{array}{ccc}\left[\begin{array}{cc}{E}_1{K}_1& O\\ O& O\end{array}\right]& & \\ & \ddots & \\ & & \left[\begin{array}{cc}{E}_p{K}_p& O\\ O& O\end{array}\right]\end{array}\right](F_p\otimes I_n),\hfill \end{array}$$

(11)

and

$$\begin{array}{cc}\hfill bcirc\left({\mathcal{S}}^{\diamond }\right)& =bcirc{(\mathcal{S}\ast {\mathcal{P}}_{\mathcal{S}})}^{†}\hfill \\ \hfill & =(F_p^{\ast }\otimes I_n)\left[\begin{array}{ccc}\left[\begin{array}{cc}{\left(E_1{K}_1\right)}^{†}& O\\ O& O\end{array}\right]& & \\ & \ddots & \\ & & \left[\begin{array}{cc}{\left(E_p{K}_p\right)}^{†}& O\\ O& O\end{array}\right]\end{array}\right](F_p\otimes I_n).\hfill \end{array}$$

(12)

According to Theorem 1, (8), and ${\mathcal{P}}_{\mathcal{A}}=\mathcal{A}\ast {\mathcal{A}}^{†}$, we have

$$\begin{array}{c}\hfill {\mathcal{A}}^{\diamond }={(\mathcal{A}\ast \mathcal{A}\ast {\mathcal{A}}^{†})}^{†}={({\mathcal{A}}^2\ast {\mathcal{A}}^{†})}^{†}={(\widehat{\mathcal{Q}}\ast {\mathcal{S}}^2\ast {\mathcal{S}}^{†}\ast {\widehat{\mathcal{Q}}}^{\ast })}^{†}.\end{array}$$

(13)

The “bcirc” discrete Fourier transform operation is performed on both sides of $\left(\right)$:

$$\begin{array}{cc}\hfill bcirc\left({\mathcal{A}}^{\diamond }\right)& =bcirc{({\mathcal{A}}^2\ast {\mathcal{A}}^{†})}^{†}=bcirc{(\widehat{\mathcal{Q}}\ast {\mathcal{S}}^2\ast {\mathcal{S}}^{†}\ast {\widehat{\mathcal{Q}}}^{\ast })}^{†}=bcirc{(\widehat{\mathcal{Q}}\ast \mathcal{S}\ast {\mathcal{P}}_{\mathcal{S}}\ast {\widehat{\mathcal{Q}}}^{\ast })}^{†}.\hfill \end{array}$$

(14)

Furthermore, applying (10), (11), (12) and (14) we obtain (9). □

Corollary 1.

Let $\mathcal{A},\mathcal{B},{\mathcal{Q}}_1\in {\mathbb{C}}^{n\times n\times p}$ and $\mathcal{B}={\mathcal{Q}}_1\ast \mathcal{A}\ast {{\mathcal{Q}}_1}^{\ast }$, in which ${\mathcal{Q}}_1$ is a unitary tensor. Then ${\mathcal{B}}^{\diamond }={\mathcal{Q}}_1\ast {\mathcal{A}}^{\diamond }\ast {{\mathcal{Q}}_1}^{\ast }$.

Proof. 

According to $\left(\right)$, we have

$$\begin{array}{cc}\hfill {\mathcal{B}}^{\diamond }& ={(\mathcal{B}\ast {\mathcal{P}}_{\mathcal{B}})}^{†}={(\mathcal{B}\ast \mathcal{B}\ast {\mathcal{B}}^{†})}^{†}={({\mathcal{Q}}_1\ast \mathcal{A}\ast \mathcal{A}\ast {{\mathcal{Q}}_1}^{\ast }\ast {({\mathcal{Q}}_1\ast \mathcal{A}\ast {{\mathcal{Q}}_1}^{\ast })}^{†})}^{†}={({\mathcal{Q}}_1\ast \mathcal{A}\ast \mathcal{A}\ast {\mathcal{A}}^{†}\ast {{\mathcal{Q}}_1}^{\ast })}^{†}\hfill \\ & ={\mathcal{Q}}_1\ast {(\mathcal{A}\ast \mathcal{A}\ast {\mathcal{A}}^{†})}^{†}\ast {{\mathcal{Q}}_1}^{\ast }={\mathcal{Q}}_1\ast {(\mathcal{A}\ast {\mathcal{P}}_{\mathcal{A}})}^{†}\ast {{\mathcal{Q}}_1}^{\ast }={\mathcal{Q}}_1\ast {\mathcal{A}}^{\diamond }\ast {{\mathcal{Q}}_1}^{\ast }.\hfill \end{array}$$

□

Theorem 8.

Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$. Then ${\mathcal{A}}^{\diamond }$ is the $\left\{2\right\}$-inverse of $\mathcal{A}$, that is, ${\mathcal{A}}^{\diamond }\ast \mathcal{A}\ast {\mathcal{A}}^{\diamond }={\mathcal{A}}^{\diamond }$.

Proof. 

Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ and its decomposition be given as in Theorem 1. By Theorem 7, we have

$$\begin{array}{cc}\hfill & bcirc({\mathcal{A}}^{\diamond }\ast \mathcal{A}\ast {\mathcal{A}}^{\diamond })=bcirc\left({\mathcal{A}}^{\diamond }\right)bcirc\left(\mathcal{A}\right)bcirc\left({\mathcal{A}}^{\diamond }\right)\hfill \\ & =(F_p^{\ast }\otimes I_n)\left(\left[\begin{array}{ccc}{Q}_1\left[\begin{array}{cc}{\left(E_1{K}_1\right)}^{†}{E_1{K}_1\left(E_1{K}_1\right)}^{†}& O\\ O& O\end{array}\right]Q_1^{\ast }& & \\ & \ddots & \\ & & Q_p\left[\begin{array}{cc}{\left(E_p{K}_p\right)}^{†}{E_p{K}_p\left(E_p{K}_p\right)}^{†}& O\\ O& O\end{array}\right]Q_p^{\ast }\end{array}\right]\right)(F_p\otimes I_n)\hfill \\ & =(F_p^{\ast }\otimes I_n)\left(\left[\begin{array}{ccc}{Q}_1\left[\begin{array}{cc}{\left(E_1{K}_1\right)}^{†}& O\\ O& O\end{array}\right]Q_1^{\ast }& & \\ & \ddots & \\ & & Q_p\left[\begin{array}{cc}{\left(E_p{K}_p\right)}^{†}& O\\ O& O\end{array}\right]Q_p^{\ast }\end{array}\right]\right)(F_p\otimes I_n)\hfill \\ & =bcirc\left({\mathcal{A}}^{\diamond }\right).\hfill \end{array}$$

Therefore, ${\mathcal{A}}^{\diamond }\ast \mathcal{A}\ast {\mathcal{A}}^{\diamond }=\mathcal{A}$. □

Remark 1.

Since ${\mathcal{A}}^{\diamond }$ is the $\left\{2\right\}$-inverse of $\mathcal{A}$, ${(\mathcal{A}\ast {\mathcal{A}}^{\diamond })}^2=\mathcal{A}\ast {\mathcal{A}}^{\diamond }$ and ${({\mathcal{A}}^{\diamond }\ast \mathcal{A})}^2={\mathcal{A}}^{\diamond }\ast \mathcal{A}$, then $\mathcal{A}\ast {\mathcal{A}}^{\diamond }$ and ${\mathcal{A}}^{\diamond }\ast \mathcal{A}$ are idempotent tensors.

Based on Theorem 5, we have the following characterization of the T-BT inverse.

Theorem 9.

Let $\mathcal{A}\in {\mathbb{C}}^{n\times n\times p}$ have the form in(5). Then

$$\begin{array}{c}\hfill {\mathcal{A}}^{\diamond }=\mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\end{array}$$

(15)

Proof. 

According to (5) and (6), we can obtains

$$\begin{array}{cc}\hfill {\mathcal{P}}_{\mathcal{A}}& =\mathcal{A}\ast {\mathcal{A}}^{†}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{I}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\mathcal{A}\ast {\mathcal{P}}_{\mathcal{A}}=\mathcal{A}\ast \mathcal{A}\ast {\mathcal{A}}^{†}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\hfill \end{array}$$

It follows from (8) that we obtain (15). □

4. T-BT Order

Let A and B be complex matrices of the same order. The well-known star partial order is defined as follows: $A\stackrel{\ast }{\le }B$: $A{A}^{†}=B{A}^{†}$ and $A^{†}A=A^{†}B$. Similarly, we write $\mathcal{A}\stackrel{\ast }{⪯}\mathcal{B}:\mathcal{A}\ast {\mathcal{A}}^{†}=\mathcal{B}\ast {\mathcal{A}}^{†},{\mathcal{A}}^{†}\ast \mathcal{A}={\mathcal{A}}^{†}\ast \mathcal{B}$, where $\mathcal{A}$ and $\mathcal{B}$ are tensors of the same order. In this section, we consider the binary relation “$\stackrel{\diamond }{⪯}$”:

$$\begin{array}{c}\hfill \mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B}\iff \mathcal{A}\ast {\mathcal{A}}^{\diamond }=\mathcal{B}\ast {\mathcal{A}}^{\diamond },{\mathcal{A}}^{\diamond }\ast \mathcal{A}={\mathcal{A}}^{\diamond }\ast \mathcal{B},\end{array}$$

(16)

where $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$ and $\mathcal{B}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$. We call “$\stackrel{\diamond }{⪯}$” the T-BT order.

Theorem 10.

Let $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$, $\mathcal{B}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$ and $\mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B}$. Then

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathcal{A}& =\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\hfill \\ \hfill \mathcal{B}& =\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{B}}_1& \mathcal{E}\ast \mathcal{L}+(\mathcal{I}-\mathcal{E}\ast \mathcal{K}\ast {(\mathcal{E}\ast \mathcal{K})}^{†})\ast \mathcal{Y}\\ \mathcal{W}\ast (\mathcal{I}-{(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{K})& {\mathcal{B}}_4\end{array}\right]\ast {\mathcal{Q}}^{\ast },\hfill \end{array}\end{array}$$

(17)

where $\mathcal{Q}$, $\mathcal{L}$, $\mathcal{E}$ and $\mathcal{K}$ are of the forms as in Theorem 5, ${\mathcal{B}}_1\in {\mathbb{C}}^{t\times t\times p}$; ${\mathcal{B}}_4\in {\mathbb{C}}^{(n-t)\times (n-t)\times p}$, $\mathcal{E}\ast \mathcal{K}\stackrel{\ast }{⪯}{\mathcal{B}}_1$, $\mathcal{Y}$ and $\mathcal{W}$ are arbitrary tensors with appropriate orders.

Proof. 

Let $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$ have the form in (5), then the T-BT inverse of $\mathcal{A}$ is of the form (15). It follows that

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill \mathcal{A}\ast {\mathcal{A}}^{\diamond }& =\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\hfill \\ \hfill {\mathcal{A}}^{\diamond }\ast \mathcal{A}& =\mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{K}& {(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\hfill \end{array}\end{array}$$

(18)

Write

$$\begin{array}{c}\hfill {\mathcal{Q}}^{\ast }\ast \mathcal{B}\ast \mathcal{Q}=\left[\begin{array}{cc}{\mathcal{B}}_1& {\mathcal{B}}_2\\ {\mathcal{B}}_3& {\mathcal{B}}_4\end{array}\right],\end{array}$$

(19)

where ${\mathcal{B}}_1\in {\mathbb{C}}^{t\times t\times p}$. Applying (15) and (19) gives

$$\begin{array}{c}\hfill \begin{array}{cc}\hfill & \mathcal{B}\ast {\mathcal{A}}^{\diamond }=\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{B}}_1\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\\ {\mathcal{B}}_3\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\hfill \\ \hfill & {\mathcal{A}}^{\diamond }\ast \mathcal{B}=\mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^{†}\ast {\mathcal{B}}_1& {(\mathcal{E}\ast \mathcal{K})}^{†}\ast {\mathcal{B}}_2\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\hfill \end{array}\end{array}$$

(20)

From (16), (18) and (20), we obtain that $\mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B}$ if and only if

$$\begin{array}{cc}\hfill \mathcal{E}\ast \mathcal{K}\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& ={\mathcal{B}}_1\ast {(\mathcal{E}\ast \mathcal{K})}^{†},\hfill \end{array}$$

(21)

$$\begin{array}{cc}\hfill {(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{K}& ={(\mathcal{E}\ast \mathcal{K})}^{†}\ast {\mathcal{B}}_1,\hfill \end{array}$$

(22)

$$\begin{array}{cc}\hfill \mathcal{O}& ={\mathcal{B}}_3\ast {(\mathcal{E}\ast \mathcal{K})}^{†},\hfill \end{array}$$

(23)

$$\begin{array}{cc}\hfill {(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{L}& ={(\mathcal{E}\ast \mathcal{K})}^{†}\ast {\mathcal{B}}_2.\hfill \end{array}$$

(24)

From (21) and (), we obtain $\mathcal{E}\ast \mathcal{K}\stackrel{\ast }{⪯}{\mathcal{B}}_1$. From () and (), we obtain ${\mathcal{B}}_3=\mathcal{W}\ast (\mathcal{I}-{(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{K})$ and ${\mathcal{B}}_2=\mathcal{E}\ast \mathcal{L}+(\mathcal{I}-\mathcal{E}\ast \mathcal{K}\ast {(\mathcal{E}\ast \mathcal{K})}^{†})\ast \mathcal{Y}$, where $\mathcal{Y}$, $\mathcal{W}$ are arbitrary. □

Remark 2.

The T-BT order is not antisymmetric.

Example 1.

Let $\mathcal{A}\in {\mathbb{C}}^{3\times 3\times 2}$, $\mathcal{B}\in {\mathbb{C}}^{3\times 3\times 2}$: $\mathcal{A}=\left[\begin{array}{cc}{A}^{\left(1\right)}& A^{\left(2\right)}\end{array}\right]$, $\mathcal{B}=\left[\begin{array}{cc}{B}^{\left(1\right)}& B^{\left(2\right)}\end{array}\right]$, with frontal slices

$$\begin{array}{c}\hfill A^{\left(1\right)}=\left[\begin{array}{ccc}1& -1& 0\\ 1& -1& 0\\ 0& 0& 1\end{array}\right],A^{\left(2\right)}=\left[\begin{array}{ccc}2& -2& 0\\ 2& -2& 0\\ 0& 0& 2\end{array}\right],B^{\left(1\right)}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 1\end{array}\right],B^{\left(2\right)}=\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 2\end{array}\right].\end{array}$$

Hence,

$$\begin{array}{cc}\hfill {\mathcal{A}}^{†}& =\left[\begin{array}{cccccc}-\frac{1}{12}& -\frac{1}{12}& 0& \frac{1}{6}& \frac{1}{6}& 0\\ \frac{1}{12}& \frac{1}{12}& 0& -\frac{1}{6}& -\frac{1}{6}& 0\\ 0& 0& -\frac{1}{3}& 0& 0& \frac{2}{3}\end{array}\right],\hfill \\ \hfill {\mathcal{A}}^2& =\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 5& 0& 0& 4\end{array}\right],{\mathcal{A}}^2\ast {\mathcal{A}}^{†}=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& -\frac{1}{3}& 0& 0& \frac{2}{3}\end{array}\right],\hfill \\ \hfill \mathcal{A}\ast {\mathcal{A}}^{\diamond }& =\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\end{array}\right],\mathcal{B}\ast {\mathcal{A}}^{\diamond }=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{A}}^{\diamond }\ast \mathcal{A}& =\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\end{array}\right],{\mathcal{A}}^{\diamond }\ast \mathcal{B}=\left[\begin{array}{cccccc}0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0\end{array}\right].\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \mathcal{A}& \stackrel{\diamond }{⪯}\mathcal{B}\iff \mathcal{A}\ast A^{\diamond }=\mathcal{B}\ast {\mathcal{A}}^{\diamond },{\mathcal{A}}^{\diamond }\ast \mathcal{A}={\mathcal{A}}^{\diamond }\ast \mathcal{B},\hfill \\ \hfill \mathcal{B}& \stackrel{\diamond }{⪯}\mathcal{A}\iff \mathcal{B}\ast {\mathcal{B}}^{\diamond }=\mathcal{A}\ast {\mathcal{B}}^{\diamond },{\mathcal{B}}^{\diamond }\ast \mathcal{B}={\mathcal{B}}^{\diamond }\ast \mathcal{A}.\hfill \end{array}$$

Since $\mathcal{A}\ne \mathcal{B}$, the T-BT order is not antisymmetric.

Remark 3.

The T-BT order is not transitive.

Example 2.

Let $\mathcal{A}\in {\mathbb{C}}^{3\times 3\times 2}$, $\mathcal{B}\in {\mathbb{C}}^{3\times 3\times 2}$, $\mathcal{C}\in {\mathbb{C}}^{3\times 3\times 2}$: $\mathcal{A}=\left[\begin{array}{cc}{A}^{\left(1\right)}& A^{\left(2\right)}\end{array}\right]$, $\mathcal{B}=\left[\begin{array}{cc}{B}^{\left(1\right)}& B^{\left(2\right)}\end{array}\right]$, $\mathcal{C}=\left[\begin{array}{cc}{C}^{\left(1\right)}& C^{\left(2\right)}\end{array}\right]$, with frontal slices

$$\begin{array}{c}\hfill A^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 2\\ 0& 0& 0\end{array}\right],A^{\left(2\right)}=\left[\begin{array}{ccc}2& 2& 2\\ 0& 0& 4\\ 0& 0& 0\end{array}\right],B^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],\\ \hfill B^{\left(2\right)}=\left[\begin{array}{ccc}2& 2& 2\\ 0& 0& 0\\ 0& 0& 0\end{array}\right],C^{\left(1\right)}=\left[\begin{array}{ccc}1& 1& 1\\ 0& 0& 0\\ 0& 1& 1\end{array}\right],C^{\left(2\right)}=\left[\begin{array}{ccc}2& 2& 2\\ 0& 0& 0\\ 0& 2& 2\end{array}\right].\end{array}$$

Hence,

$$\begin{array}{cc}\hfill {\mathcal{A}}^{†}& =\left[\begin{array}{cccccc}-\frac{1}{6}& \frac{1}{12}& 0& \frac{1}{3}& -\frac{1}{6}& 0\\ -\frac{1}{6}& \frac{1}{12}& 0& \frac{1}{3}& -\frac{1}{3}& 0\\ 0& -\frac{1}{6}& 0& 0& \frac{1}{3}& 0\end{array}\right],{\mathcal{B}}^{†}=\left[\begin{array}{cccccc}-\frac{1}{9}& 0& 0& \frac{2}{9}& 0& 0\\ -\frac{1}{9}& 0& 0& \frac{2}{9}& 0& 0\\ -\frac{1}{9}& 0& 0& \frac{2}{9}& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{A}}^2& =\left[\begin{array}{cccccc}5& 5& 15& 4& 4& 12\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],{\mathcal{A}}^2\ast {\mathcal{A}}^{†}=\left[\begin{array}{cccccc}1& 1& 0& 2& 2& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{A}}^{\diamond }& =\left[\begin{array}{cccccc}-\frac{1}{6}& 0& 0& \frac{1}{3}& 0& 0\\ -\frac{1}{6}& 0& 0& \frac{1}{3}& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\mathcal{A}\ast {\mathcal{A}}^{\diamond }=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill \mathcal{B}\ast {\mathcal{A}}^{\diamond }& =\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],{\mathcal{A}}^{\diamond }\ast \mathcal{A}=\left[\begin{array}{cccccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& 0& 0& 0\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{A}}^{\diamond }\ast \mathcal{B}& =\left[\begin{array}{cccccc}\frac{1}{2}& \frac{1}{2}& \frac{1}{2}& 0& 0& 0\\ \frac{1}{2}& \frac{1}{2}& \frac{1}{2}& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{B}}^2& =\left[\begin{array}{cccccc}5& 5& 5& 4& 4& 4\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],{\mathcal{B}}^2\ast {\mathcal{B}}^{†}=\left[\begin{array}{cccccc}1& 0& 0& 2& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{B}}^{\diamond }& =\left[\begin{array}{cccccc}-\frac{1}{3}& 0& 0& \frac{1}{6}& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\mathcal{B}\ast {\mathcal{B}}^{\diamond }=\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill \mathcal{C}\ast {\mathcal{B}}^{\diamond }& =\left[\begin{array}{cccccc}1& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],{\mathcal{B}}^{\diamond }\ast \mathcal{B}=\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\hfill \\ \hfill {\mathcal{B}}^{\diamond }\ast \mathcal{C}& =\left[\begin{array}{cccccc}1& 1& 1& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\end{array}\right],\mathcal{C}\ast {\mathcal{A}}^{\diamond }=\left[\begin{array}{cccccc}\frac{1}{2}& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0\\ \frac{1}{2}& 0& 0& 0& 0& 0\end{array}\right].\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill \mathcal{A}& \stackrel{\diamond }{⪯}\mathcal{B}\iff \mathcal{A}\ast A^{\diamond }=\mathcal{B}\ast {\mathcal{A}}^{\diamond },{\mathcal{A}}^{\diamond }\ast \mathcal{A}={\mathcal{A}}^{\diamond }\ast \mathcal{B},\hfill \\ \hfill \mathcal{B}& \stackrel{\diamond }{⪯}\mathcal{A}\iff \mathcal{B}\ast {\mathcal{B}}^{\diamond }=\mathcal{A}\ast {\mathcal{B}}^{\diamond },{\mathcal{B}}^{\diamond }\ast \mathcal{B}={\mathcal{B}}^{\diamond }\ast \mathcal{A}.\hfill \end{array}$$

Because $\mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B}$, $\mathcal{B}\stackrel{\diamond }{⪯}\mathcal{A}$ and $\mathcal{C}\ast {\mathcal{A}}^{\diamond }\ne \mathcal{A}\ast {\mathcal{A}}^{\diamond }$, the T-BT order is not transitive.

5. T-GC Partial Order

Let $\mathcal{A},\mathcal{B}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$. We determine that $\mathcal{A}$ is below $\mathcal{B}$ under the partial order “⪯”. If the binary relation “⪯” satisfies the following, we obtain the subsequent results: (1) reflexivity: $\mathcal{A}⪯\mathcal{A}$; (2) antisymmetry: if $\mathcal{A}⪯\mathcal{B}$ and $\mathcal{B}⪯\mathcal{A}$, then $\mathcal{A}=\mathcal{B}$; and (3) transitivity: if $\mathcal{A}⪯\mathcal{B}$ and $\mathcal{B}⪯\mathcal{C}$, then $\mathcal{A}⪯\mathcal{C}$. From Remarks 2 and 3, we see that the binary relation “$\stackrel{\diamond }{⪯}$” is not a partial order. In this section, we consider the binary relation “$\stackrel{▵}{⪯}$”:

$$\begin{array}{c}\hfill \mathcal{A}\stackrel{▵}{⪯}\mathcal{B}:\mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B},{\mathcal{P}}_{\mathcal{A}}\ast (\mathcal{B}-\mathcal{A})\ast (\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})=\mathcal{O},(\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})\ast \mathcal{B}\ast {\mathcal{P}}_{\mathcal{A}}=\mathcal{O}\\ \hfill and{\mathrm{rank}}_T(\mathcal{B}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{B}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right).\end{array}$$

(25)

in which $\mathcal{A},\mathcal{B}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$.

Theorem 11.

Let $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$, $\mathcal{B}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$ and ${\mathrm{rank}}_T(\mathcal{B}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{B}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right)$. The following conditions are equivalent:

(1)

$\mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B}$ and ${\mathcal{P}}_{\mathcal{A}}\ast (\mathcal{B}-\mathcal{A})\ast (\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})=\mathcal{O}$, $(\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})\ast \mathcal{B}\ast {\mathcal{P}}_{\mathcal{A}}=\mathcal{O}$.

(2)

Then there is a unitary tensor $\mathcal{Q}\in {\mathbb{C}}^{n\times n\times p}$ such that

$$\begin{array}{c}\hfill \mathcal{A}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\mathcal{B}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& {\mathcal{B}}_4\end{array}\right]\ast {\mathcal{Q}}^{\ast },\end{array}$$

(26)

where $\mathcal{L}$, $\mathcal{E}$ and $\mathcal{K}$ are of the forms as in Theorem5, and ${\mathcal{B}}_4\in {\mathbb{C}}_{\mathcal{I}}^{(n-t)\times (n-t)\times p}$.

(3)

${\mathcal{A}}^{†}\ast \mathcal{A}={\mathcal{A}}^{†}\ast \mathcal{B}$ and ${\mathcal{A}}^2=\mathcal{B}\ast \mathcal{A}$.

Proof. 

$\left(1\right)\Rightarrow \left(2\right)$ Let the forms of $\mathcal{A}$ and $\mathcal{B}$ be as in (17). According to ${\mathcal{P}}_{\mathcal{A}}\ast (\mathcal{B}-\mathcal{A})\ast (\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})=\mathcal{O}$, $(\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})\ast \mathcal{B}\ast {\mathcal{P}}_{\mathcal{A}}=\mathcal{O}$, we have $(\mathcal{I}-\mathcal{E}\ast \mathcal{K}\ast {(\mathcal{E}\ast \mathcal{K})}^{†})\ast \mathcal{Y}=\mathcal{O}$, $\mathcal{W}\ast (\mathcal{I}-{(\mathcal{E}\ast \mathcal{K})}^{†}$$\ast \mathcal{E}\ast \mathcal{K})=\mathcal{O}$, then ${\mathcal{B}}_2=\mathcal{E}\ast \mathcal{L}$ and ${\mathcal{B}}_3=\mathcal{O}$. It follows from ${\mathrm{rank}}_T(\mathcal{B}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{B}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right)$, we get ${\mathcal{B}}_1=\mathcal{E}\ast \mathcal{K}$. Thus, $\left(2\right)$ is established.

$\left(2\right)\Rightarrow \left(1\right)$ Let the forms of $\mathcal{A}$, $\mathcal{B}$ be as in (26). Applying Theorem 9, it is easy to check that

$$\begin{array}{cc}\hfill \mathcal{A}\ast {\mathcal{A}}^{\diamond }& =\mathcal{Q}\ast \left[\begin{array}{cc}(\mathcal{E}\ast \mathcal{K})\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast }=\mathcal{B}\ast {\mathcal{A}}^{\diamond },\hfill \\ \hfill {\mathcal{A}}^{\diamond }\ast \mathcal{A}& =\mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^{†}\ast (\mathcal{E}\ast \mathcal{K})& {(\mathcal{E}\ast \mathcal{K})}^{†}\ast (\mathcal{E}\ast \mathcal{L})\\ \mathcal{O}& \mathcal{O}\end{array}\right]={\mathcal{A}}^{\diamond }\ast \mathcal{B}.\hfill \end{array}$$

(27)

Therefore, we obtain $\mathcal{A}\stackrel{\diamond }{⪯}\mathcal{B}$. Since $\mathcal{A}\ast {\mathcal{A}}^{\diamond }=\mathcal{A}\ast {\mathcal{A}}^{†}$, applying (26) and (27) gives

$$\begin{array}{cc}\hfill {\mathcal{P}}_{\mathcal{A}}\ast (\mathcal{B}-\mathcal{A})\ast (\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})& =\mathcal{Q}\ast \left[\begin{array}{cc}(\mathcal{E}\ast \mathcal{K})\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast \left[\begin{array}{cc}\mathcal{O}& \mathcal{O}\\ \mathcal{O}& {\mathcal{B}}_4\end{array}\right]\ast \left[\begin{array}{cc}\mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{I}\end{array}\right]\ast {\mathcal{Q}}^{\ast }=\mathcal{O},\hfill \\ \hfill (\mathcal{I}-{\mathcal{P}}_{\mathcal{A}})\ast \mathcal{B}\ast {\mathcal{P}}_{\mathcal{A}}& =\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{I}\end{array}\right]\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& {\mathcal{B}}_4\end{array}\right]\ast \left[\begin{array}{cc}(\mathcal{E}\ast \mathcal{K})\ast {(\mathcal{E}\ast \mathcal{K})}^{†}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast \mathcal{Q}=\mathcal{O}.\hfill \end{array}$$

$\left(2\right)\Rightarrow \left(3\right)$ Let the forms of $\mathcal{A}$, $\mathcal{B}$ be as in $\left(\right)$. By applying Theorem 10, we have

$$\begin{array}{c}\hfill {\mathcal{A}}^{†}\ast \mathcal{A}=\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{K}}^{\ast }\ast \mathcal{K}& {\mathcal{K}}^{\ast }\ast \mathcal{L}\\ {\mathcal{L}}^{\ast }\ast \mathcal{K}& {\mathcal{L}}^{\ast }\ast \mathcal{L}\end{array}\right]{\mathcal{Q}}^{\ast }=\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{K}}^{\ast }\ast {\mathcal{E}}^{-1}& \mathcal{O}\\ {\mathcal{L}}^{\ast }\ast {\mathcal{E}}^{-1}& \mathcal{O}\end{array}\right]\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& {\mathcal{B}}_4\end{array}\right]{\mathcal{Q}}^{\ast }={\mathcal{A}}^{†}\ast \mathcal{B},\\ \hfill {\mathcal{A}}^2=\mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^2& \mathcal{E}\ast \mathcal{K}\ast \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]{\mathcal{Q}}^{\ast }=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& {\mathcal{B}}_4\end{array}\right]\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]{\mathcal{Q}}^{\ast }=\mathcal{B}\ast \mathcal{A}.\end{array}$$

$\left(3\right)\Rightarrow \left(2\right)$ Let $\mathcal{A}$ have the form in $\left(\right)$. Write

$$\begin{array}{c}\hfill {\mathcal{Q}}^{\ast }\ast \mathcal{B}\ast \mathcal{Q}=\left[\begin{array}{cc}{\mathcal{B}}_1& {\mathcal{B}}_2\\ {\mathcal{B}}_3& {\mathcal{B}}_4\end{array}\right],\end{array}$$

(28)

where ${\mathcal{B}}_1\in {\mathbb{C}}^{t\times t\times p}$ and ${\mathcal{B}}_4\in {\mathbb{C}}^{(n-t)\times (n-t)\times p}$. Since ${\mathcal{A}}^{†}\ast \mathcal{A}={\mathcal{A}}^{†}\ast \mathcal{B}$, we obtain

$$\begin{array}{c}\hfill \mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{K}}^{\ast }\ast \mathcal{K}& {\mathcal{K}}^{\ast }\ast \mathcal{L}\\ {\mathcal{L}}^{\ast }\ast \mathcal{K}& {\mathcal{L}}^{\ast }\ast \mathcal{L}\end{array}\right]{\mathcal{Q}}^{\ast }=\mathcal{Q}\ast \left[\begin{array}{cc}{\mathcal{K}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_1& {\mathcal{K}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_2\\ {\mathcal{L}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_1& {\mathcal{L}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_2\end{array}\right]{\mathcal{Q}}^{\ast }.\end{array}$$

Therefore,

$$\begin{array}{cc}\hfill {\mathcal{K}}^{\ast }\ast \mathcal{K}& ={\mathcal{K}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_1,{\mathcal{K}}^{\ast }\ast \mathcal{L}={\mathcal{K}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_2,\hfill \\ \hfill {\mathcal{L}}^{\ast }\ast \mathcal{K}& ={\mathcal{L}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_1,{\mathcal{L}}^{\ast }\ast \mathcal{L}={\mathcal{L}}^{\ast }\ast {\mathcal{E}}^{-1}\ast {\mathcal{B}}_2.\hfill \end{array}$$

From $\mathcal{K}\ast {\mathcal{K}}^{\ast }+\mathcal{L}\ast {\mathcal{L}}^{\ast }={\mathcal{I}}_r$ and ${\mathcal{A}}^2=\mathcal{B}\ast \mathcal{A}$, it follows that

$$\begin{array}{c}\hfill {\mathcal{B}}_1=\mathcal{E}\ast \mathcal{K},{\mathcal{B}}_2=\mathcal{E}\ast \mathcal{L},\end{array}$$

(29)

and

$$\begin{array}{cc}\hfill \mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^2& \mathcal{E}\ast \mathcal{K}\ast \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]{\mathcal{Q}}^{\ast }& =\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \ast \mathcal{E}\ast \mathcal{L}\\ {\mathcal{B}}_3& {\mathcal{B}}_4\end{array}\right]\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]{\mathcal{Q}}^{\ast }\hfill \\ & =\mathcal{Q}\ast \left[\begin{array}{cc}{(\mathcal{E}\ast \mathcal{K})}^2& \mathcal{E}\ast \mathcal{K}\ast \mathcal{E}\ast \mathcal{L}\\ {\mathcal{B}}_3\ast \mathcal{E}\ast \mathcal{K}& {\mathcal{B}}_3\ast \mathcal{E}\ast \mathcal{L}\end{array}\right]{\mathcal{Q}}^{\ast }.\hfill \end{array}$$

Hence, ${\mathcal{B}}_3\ast \mathcal{E}\ast \mathcal{K}=\mathcal{O}$ and ${\mathcal{B}}_3\ast \mathcal{E}\ast \mathcal{L}=\mathcal{O}$. Since $\mathcal{E}$ is invertible and $\mathcal{K}\ast {\mathcal{K}}^{\ast }+\mathcal{L}\ast {\mathcal{L}}^{\ast }={\mathcal{I}}_r$, then ${\mathcal{B}}_3=\mathcal{O}$ and ${\mathcal{B}}_4\in {\mathbb{C}}_{\mathcal{I}}^{(n-t)\times (n-t)\times p}$. Therefore, $\left(2\right)$ is established. □

In [9], Baksalary and Trenkler introduce the $\mathrm{BT}$ order “$\stackrel{\diamond }{\le }$” on ${\mathbb{C}}^{n\times n}$:

$$\begin{array}{c}\hfill A\stackrel{\diamond }{\le }B:A{A}^{\diamond }=B{A}^{\diamond }and{A}^{\diamond }A=A^{\diamond }B.\end{array}$$

(30)

By applying Theorems 10 and 11, we can obtain relevant conclusions on ${\mathbb{C}}^{n\times n}$.

Corollary 2.

Let $A,B\in {\mathbb{C}}^{n\times n}$, $P_A(B-A)(I-P_A)=O$, $(I-P_A)B{P}_A=O$, $A\stackrel{\diamond }{\le }B$ and the Hartwig–Spindelböck decomposition of A be as in(1). Then

$$\begin{array}{c}\hfill B=Q\left[\begin{array}{cc}{B}_1& EL\\ O& B_4\end{array}\right]Q^{\ast },\end{array}$$

where $B_4\in {\mathbb{C}}^{\left(n-r\right)\times \left(n-r\right)}$, $r=\mathrm{rank}\left(A\right)$ and $EK\stackrel{\ast }{\le }{B}_1$.

Lemma 2

Let $\mathcal{A}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$, $\mathcal{B}\in {\mathbb{C}}_{\mathcal{I}}^{n\times n\times p}$ and $\mathcal{A}\stackrel{▵}{⪯}\mathcal{B}$. Then

$$\begin{array}{cc}\hfill \mathcal{B}& =\mathcal{A}+\left(\mathcal{I}-\mathcal{A}\ast {\mathcal{A}}^{†}\right)\ast \mathcal{B}\hfill \end{array}$$

(31)

$$\begin{array}{cc}\hfill & =\mathcal{A}\ast \mathcal{A}\ast {\mathcal{A}}^{†}+\mathcal{B}\ast \left(\mathcal{I}-\mathcal{A}\ast {\mathcal{A}}^{†}\right),\hfill \end{array}$$

(32)

$$\begin{array}{cc}\hfill \mathcal{O}& ={\mathcal{A}}^{†}\ast \left(\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}\right)\hfill \end{array}$$

(33)

$$\begin{array}{cc}\hfill & =\left(\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}\right)\ast \mathcal{A}.\hfill \end{array}$$

(34)

Proof. 

Let $\mathcal{A}\stackrel{▵}{⪯}\mathcal{B}$ and $\mathcal{A}$ be of the form (5). Applying Theorem 5 gives the T-Hartwig–Spindelböck decomposition of ${\mathcal{B}}_4$,

$$\begin{array}{c}\hfill {\mathcal{B}}_4={\mathcal{Q}}_B\ast \left[\begin{array}{cc}{\mathcal{E}}_2\ast {\mathcal{K}}_2& {\mathcal{E}}_2\ast {\mathcal{L}}_2\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_B^{\ast },\end{array}$$

(35)

in which ${\mathcal{E}}_2\in {\mathbb{C}}^{t_2\times t_2\times p}$ is an invertible diagonal tensor, ${\mathcal{K}}_2\in {\mathbb{C}}^{t_2\times t_2\times p}$, $\mathcal{L}\in {\mathbb{C}}^{t_2\times (n-t_2)\times p}$ and ${\mathcal{K}}_2\ast {{\mathcal{K}}_2}^{\ast }+{\mathcal{L}}_2\ast {{\mathcal{L}}_2}^{\ast }={\mathcal{I}}_{t_2}$. Write

$$\begin{array}{c}\hfill {\mathcal{Q}}_1=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{I}& \mathcal{O}\\ \mathcal{O}& {\mathcal{Q}}_B\end{array}\right]\ast {\mathcal{Q}}^{\ast },{\mathcal{L}}_1{\mathcal{Q}}_B=\left[\begin{array}{cc}{\mathcal{L}}_{11}& {\mathcal{L}}_{12}\end{array}\right],\end{array}$$

where ${\mathcal{L}}_{11}\in {\mathbb{C}}^{t_1\times t_2\times p}$, $r_A={\mathrm{rank}}_T\left(\mathcal{A}\right)$ and ${r_B=\mathrm{rank}}_T\left(\mathcal{B}\right)$. Then

$$\begin{array}{cc}\hfill \mathcal{A}& ={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast {\mathcal{L}}_{11}& \mathcal{E}\ast {\mathcal{L}}_{12}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast },\mathcal{B}={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast {\mathcal{L}}_{11}& \mathcal{E}\ast {\mathcal{L}}_{12}\\ \mathcal{O}& {\mathcal{E}}_2\ast {\mathcal{K}}_2& \mathcal{E}{\ast }_2{\mathcal{L}}_2\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast }.\hfill \end{array}$$

It follows that

$$\begin{array}{cc}\hfill \left(\mathcal{I}-\mathcal{A}\ast {\mathcal{A}}^{†}\right)\ast \mathcal{B}& ={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{I}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{I}\end{array}\right]\ast \left[\begin{array}{ccc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast {\mathcal{L}}_{11}& \mathcal{E}\ast {\mathcal{L}}_{12}\\ \mathcal{O}& {\mathcal{E}}_2\ast {\mathcal{K}}_2& \mathcal{E}{\ast }_2{\mathcal{L}}_2\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast }\hfill \\ \hfill & ={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& {\mathcal{E}}_2\ast {\mathcal{K}}_2& \mathcal{E}{\ast }_2{\mathcal{L}}_2\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast },\hfill \\ \hfill \mathcal{A}\ast \mathcal{A}\ast {\mathcal{A}}^{†}& ={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{E}\ast \mathcal{K}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast },\hfill \\ \hfill \mathcal{B}\ast \left(\mathcal{I}-\mathcal{A}\ast {\mathcal{A}}^{†}\right)& ={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{O}& \mathcal{E}\ast {\mathcal{L}}_{11}& \mathcal{E}\ast {\mathcal{L}}_{12}\\ \mathcal{O}& {\mathcal{E}}_2\ast {\mathcal{K}}_2& \mathcal{E}{\ast }_2{\mathcal{L}}_2\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast },\hfill \end{array}$$

then we get (31) and ().

By applying

$$\begin{array}{c}\hfill \mathcal{A}\ast {\mathcal{A}}^{†}={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{I}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast },\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}={\mathcal{Q}}_1\ast \left[\begin{array}{ccc}\mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{O}\\ \mathcal{O}& \mathcal{O}& \mathcal{I}\end{array}\right]\ast {\mathcal{Q}}_1^{\ast }\end{array}$$

we can obtain () and (). □

In the following Theorem 12, we prove that the binary relationship “$\stackrel{▵}{⪯}$” is a partial order. We call it the T-GC partial order.

Theorem 12.

The T-GC order is a partial order.

Proof. 

Reflexivity is clearly established.

Now, we prove the antisymmetry: according to $\mathcal{A}\stackrel{▵}{⪯}\mathcal{B}$, we have ${\mathrm{rank}}_T\left(\mathcal{A}\right)\le {\mathrm{rank}}_T\left(\mathcal{B}\right)$. According to $\mathcal{B}\stackrel{▵}{⪯}\mathcal{A}$, we have ${\mathrm{rank}}_T\left(\mathcal{B}\right)\le {\mathrm{rank}}_T\left(\mathcal{A}\right)$ and ${\mathrm{rank}}_T(\mathcal{B}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{B}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right)={\mathrm{rank}}_T\left({\mathcal{B}}_4\right)=0$, then ${\mathrm{rank}}_T\left(\mathcal{B}\right)={\mathrm{rank}}_T\left(\mathcal{A}\right)$. Hence, antisymmetry is established.

Finally, we prove the transitivity: let $\mathcal{A}\stackrel{▵}{⪯}\mathcal{B}$ and $\mathcal{B}\stackrel{▵}{⪯}\mathcal{C}$.

By ${\mathrm{rank}}_T(\mathcal{B}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{B}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right)$ and ${\mathrm{rank}}_T(\mathcal{C}-\mathcal{B})={\mathrm{rank}}_T\left(\mathcal{C}\right)-{\mathrm{rank}}_T\left(\mathcal{B}\right)$, then

$$\begin{array}{c}\hfill {\mathrm{rank}}_T(\mathcal{C}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{C}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right).\end{array}$$

(36)

Applying Lemma 2 to $\mathcal{B}\stackrel{▵}{⪯}\mathcal{C}$, we have $\mathcal{C}=\mathcal{B}+\left(\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}\right)\ast \mathcal{C}$. Because of $\mathcal{A}\stackrel{▵}{⪯}\mathcal{B}$ and (31), we obtain

$$\begin{array}{c}\hfill {\mathcal{A}}^{†}\ast \mathcal{C}={\mathcal{A}}^{†}\ast \left(\mathcal{B}+\left(\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}\right)\ast \mathcal{C}\right)={\mathcal{A}}^{†}\ast \mathcal{B}.\end{array}$$

(37)

Applying Lemma 2 to $\mathcal{B}\stackrel{▵}{⪯}\mathcal{C}$, we have $\mathcal{C}=\mathcal{B}\ast \mathcal{B}\ast {\mathcal{B}}^{†}+\mathcal{C}\ast \left(\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}\right)$. By (), we give $\mathcal{A}=\mathcal{B}\ast {\mathcal{B}}^{†}\ast \mathcal{A}$ and

$$\begin{array}{c}\hfill \mathcal{C}\ast \mathcal{A}=\left(\mathcal{B}\ast \mathcal{B}\ast {\mathcal{B}}^{†}+\mathcal{C}\ast \left(\mathcal{I}-\mathcal{B}\ast {\mathcal{B}}^{†}\right)\right)\ast \mathcal{A}=\mathcal{B}\ast \mathcal{B}\ast {\mathcal{B}}^{†}\ast \mathcal{A}=\mathcal{B}\ast \mathcal{A}.\end{array}$$

(38)

Because of $\mathcal{A}\stackrel{▵}{⪯}\mathcal{B}$, we have ${\mathcal{A}}^{†}\ast \mathcal{A}={\mathcal{A}}^{†}\ast \mathcal{B}$ and ${\mathcal{A}}^2=\mathcal{B}\ast \mathcal{A}$. It follows from (37) and (38) that

$$\begin{array}{c}\hfill {\mathcal{A}}^{†}\ast \mathcal{A}={\mathcal{A}}^{†}\ast \mathcal{C},{\mathcal{A}}^2=\mathcal{C}\ast \mathcal{A}.\end{array}$$

(39)

Therefore, applying (36) and (39), we obtain $\mathcal{A}\stackrel{▵}{⪯}\mathcal{C}$, so the transmissibility is established.

To sum up, the binary relation “$\stackrel{▵}{⪯}$ " is a partial order. □

Applying (25), Theorems 11 and 12, we obtain relevant conclusions in the complex field:

Definition 7

(Generalized core partial order). Let $A\in {\mathbb{C}}^{n\times n}$, $B\in {\mathbb{C}}^{n\times n}$. If they satisfy

$$\begin{array}{c}\hfill A\stackrel{▵}{\le }B:A\stackrel{\diamond }{\le }B,P_A(B-A)(I-P_A)=O,(I-P_A)B{P}_A=O\\ \hfill and\mathrm{rank}(B-A)=\mathrm{rank}\left(B\right)-\mathrm{rank}\left(A\right).\end{array}$$

Then the binary relation “$\stackrel{▵}{\le }$” is called the generalized core partial order (GC partial order for short).

Remark 4.

If $A,B\in {\mathbb{C}}^{\mathrm{CM}}$, it is easy to check that the generalized $\mathrm{core}$ partial order and the core partial order coincide.

Theorem 13.

Let $A,B\in {\mathbb{C}}^{n\times n}$. The following conditions are equivalent:

(1)

$A\stackrel{▵}{\le }B$: $A\stackrel{\diamond }{\le }B$, $P_A(B-A)(I-P_A)=O$, $(I-P_A)B{P}_A=O$ and $\mathrm{rank}(B-A)=\mathrm{rank}\left(B\right)-\mathrm{rank}\left(A\right)$.

(2)

Then there is a unitary matrix $Q\in {\mathbb{C}}^{n\times n}$ such that

$$\begin{array}{c}\hfill A=Q\left[\begin{array}{cc}EK& EL\\ O& O\end{array}\right]Q^{\ast },B=Q\left[\begin{array}{cc}EK& EL\\ O& B_4\end{array}\right]Q^{\ast },\end{array}$$

where L, E and K are of the forms as in (1), $B_4\in {\mathbb{C}}^{(n-t)\times (n-t)}$ and $\mathrm{rank}\left(B_4\right)=\mathrm{rank}\left(B\right)-\mathrm{rank}\left(A\right)$.

(3)

$A^{†}A=A^{†}B$, $A^2=BA$ and $\mathrm{rank}(B-A)=\mathrm{rank}\left(B\right)-\mathrm{rank}\left(A\right)$.

Example 3.

Let

$$\begin{array}{c}\hfill A=\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right],B=\left[\begin{array}{cccc}1& 1& 0& 2\\ 0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 0\end{array}\right].\end{array}$$

It is obvious that

$$\begin{array}{c}\hfill A^{†}A=A^{†}B=\left[\begin{array}{cccc}\frac{1}{2}& \frac{1}{2}& 0& 0\\ \frac{1}{2}& \frac{1}{2}& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 1\end{array}\right],A^2=AB=\left[\begin{array}{cccc}1& 1& 0& 3\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right]\end{array}$$

and

$$\begin{array}{c}\hfill \mathrm{rank}(B-A)=\mathrm{rank}\left(B\right)-\mathrm{rank}\left(A\right).\end{array}$$

Therefore, $A\stackrel{▵}{\le }B$.

When $\mathcal{A},\mathcal{B}\in {\mathbb{C}}_{\mathcal{T}}^{\mathrm{CM}}$, by applying (2) to (16) we denote

$$\begin{array}{c}\hfill \mathcal{A}\stackrel{\#}{⪯}\mathcal{B}:\mathcal{A}\ast {\mathcal{A}}^{\#}=\mathcal{B}\ast {\mathcal{A}}^{\#},{\mathcal{A}}^{\#}\ast \mathcal{A}={\mathcal{A}}^{\#}\ast \mathcal{B}.\end{array}$$

(40)

When ${\mathrm{Ind}}_T\left(\mathcal{A}\right)=1$, since $K_i$ is invertible, applying (21) and () gives ${\mathcal{B}}_1=\mathcal{E}\ast \mathcal{K}$, $\mathcal{I}-{(\mathcal{E}\ast \mathcal{K})}^{†}\ast \mathcal{E}\ast \mathcal{K}=\mathcal{O}$ and $\mathcal{I}-\mathcal{E}\ast \mathcal{K}\ast {(\mathcal{E}\ast \mathcal{K})}^{†}=\mathcal{O}$. Applying (17), it follows that

$$\begin{array}{c}\hfill \begin{array}{c}\hfill \mathcal{A}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& \mathcal{O}\end{array}\right]\ast {\mathcal{Q}}^{\ast },\mathcal{B}=\mathcal{Q}\ast \left[\begin{array}{cc}\mathcal{E}\ast \mathcal{K}& \mathcal{E}\ast \mathcal{L}\\ \mathcal{O}& {\mathcal{B}}_4\end{array}\right]\ast {\mathcal{Q}}^{\ast }.\end{array}\end{array}$$

Therefore, ${\mathrm{rank}}_T(\mathcal{B}-\mathcal{A})={\mathrm{rank}}_T\left(\mathcal{B}\right)-{\mathrm{rank}}_T\left(\mathcal{A}\right)$. Furthermore, applying (16) and (25) gives that “$\stackrel{\diamond }{⪯}$” is the equivalent to “$\stackrel{▵}{⪯}$” on ${\mathbb{C}}_{\mathcal{T}}^{\mathrm{CM}}$.

By applying Theorem 12, we obtain the following Theorem 14.

Theorem 14.

The binary relation “$\stackrel{\#}{⪯}$” is a partial order on ${\mathbb{C}}_{\mathcal{T}}^{\mathrm{CM}}$. We call it the T-core partial order.

6. Conclusions

In this paper, the characterization, properties and corresponding partial order of the T-BT inverse of the third-order tensor with the T-product are studied. It provides a feasible scheme for the BT inverse of other types of products, such as the Einstein product. These results also lay a foundation for subsequent research on tensor equations and other issues.

In [20,21], we see that the BT inverse can be used to characterize the EP matrix and study matrix equations. Based on results of this paper, we will discuss characterizations of special tensors and tensor equations. In particular, based on [22,23,24], we will discuss the Sylvester-type tensor equations via the T-product.