Revisiting of the BT-inverse of matrices

1.   Introduction

For many different generalized inverses such as $A^{\dagger}, A^{D}$, , , $A^{D, \dagger}, A^{(B, C)}$, below can all be characterized by several equations respectively, while there is no such equations to define $A^{\diamondsuit}$. Our main aim is to develop some necessary and sufficient conditions for a matrix to be the BT-inverse by equations and derive some properties of the BT-inverse.

Throughout this paper, we denote the set of $m\times n$ complex matrices by $\mathbb{C}^{m\times n}$. We denote the identity matrix of order $n$ by $I_{n}$, the range space, the null space, the conjugate transpose and the rank of the matrix $A\in\mathbb{C}^{m\times n}$ by $\mathcal{R}(A)$, $\mathcal{N}(A)$, $A^{*}$ and $r(A)$, respectively. The index of $A\in\mathbb{C}^{n\times n}$, denoted by ${\rm Ind}(A)$, is the smallest nonnegative integer $k$ such that $r(A^{k}) = r(A^{k+1})$. $P_{\mathcal{L}, \mathcal{M}}$ stands for the projector (idempotent) on the space $\mathcal{L}$ along the $\mathcal{M}$. For $A\in\mathbb{C}^{m\times n}$, $P_{A}$ represents the orthogonal projection onto $\mathcal{R}(A)$, i.e. $P_{A} = P_{\mathcal{R}(A)} = AA^{\dagger}$.

For the readers' convenience, we will first recall the definitions of some generalized inverses. For $A\in\mathbb{C}^{m\times n}$, the Moore-Penrose inverse $A^{\dagger}$ of $A$ is the unique matrix $X\in\mathbb{C}^{n\times m}$ satisfying the following four Penrose equations ^[1]:

A matrix $X\in\mathbb{C}^{n\times m}$ that satisfies condition (1) above is called an inner inverse of $A$ and is denoted by $A^{(1)}$. A matrix $X\in\mathbb{C}^{n\times m}$ that satisfies condition (2) above is called an outer inverse of $A$ and is denoted by $A^{(2)}$. A matrix $X\in\mathbb{C}^{n\times m}$ that satisfies condition (1) and condition (3) above is denoted by $A^{(1, 3)}$. The symbol $A\{1\}$, $A\{1, 3\}$ stand for the set of all $A^{(1)}$, $A^{(1, 3)}$ respectively. Let $A\in\mathbb{C}^{m\times n}$ be of rank $r$, and $\mathcal{T}$, $\mathcal{S}$ be a subspace of $\mathbb{C}^{n}, \mathbb{C}^{m}$ where $\mathcal{T}$, $\mathcal{S}$ is of dimension $t$ $(\leqslant r)$, $m-t$, respectively. Then a matrix $X$ satisfies $X = XAX$, $\mathcal{R}(X) = \mathcal{T}$ and $\mathcal{N}(X) = \mathcal{S}$ if and only if $A\mathcal{T}\oplus \mathcal{S} = \mathbb{C}^{m}$, and in this case $X$ denoted by $A^{(2)}_{\mathcal{T}, \mathcal{S}}$ is unique.

The Drazin inverse of $A\in\mathbb{C}^{n\times n}$ with ${\rm Ind}(A) = k$, denoted by $A^{D}$ ^[2], is the unique matrix $X\in\mathbb{C}^{n\times n}$ satisfying:

Especially, if ${\rm Ind}(A) = 1$, then the Drazin inverse of $A$ is called the group inverse of $A$ and is denoted by $A^{\#}$.

Baksalary and Trenkler ^[3] introduced the core inverse on the $\mathbb{C}^{\rm CM}_{n}$ $\big(\mathbb{C}^{\rm CM}_{n} = \{A|A\in\mathbb{C}^{n\times n}, r(A) = r(A^{2})\}\big)$: the core inverse of $A\in\mathbb{C}^{\rm CM}_{n}$ is defined to be the unique matrix $X\in\mathbb{C}^{n\times n}$ such that

and denoted by (see ^[3,4,5,6]).

Moreover, three kinds of generalizations of the core inverse were given for $n\times n$ complex matrices, called core-EP inverse, DMP-inverse and BT-inverse, respectively.

Firstly, for $A\in\mathbb{C}^{n\times n}$ with ${\rm Ind}(A) = k$, the unique matrix $X\in\mathbb{C}^{n\times n}$ satisfying:

is called the Core-EP inverse of $A$ written as (see ^[7,8,9,10]). Moreover, it is seen that $= (A^{k+1}(A^{k})^{\dagger})^{\dagger}$ (see [7,Theorem 2.7]).

Secondly, the DMP-inverse of $A\in\mathbb{C}^{n\times n}$ with ${\rm Ind}(A) = k$, written by $A^{D, \dagger}$ ^[11,12], is defined as the unique matrix $A\in\mathbb{C}^{n\times n}$ satisfying:

Moreover, it was proved that $A^{D, \dagger} = A^{D}AA^{\dagger}$. Also, the dual DMP inverse of $A$ was introduced in ^[12], namely $A^{\dagger, D} = A^{\dagger}AA^{D}$.

Thirdly, the BT-inverse of $A\in\mathbb{C}^{n\times n}$, denoted by $A^{\diamondsuit}$^[13], is defined as

In recent years, some new generalized inverses are introduced. The (B, C)-inverse of $A\in\mathbb{C}^{m\times n}$, denoted by $A^{(B, C)}$ ^[14,15], is the unique matrix $X\in\mathbb{C}^{n\times m}$ satisfying:

where $B, C\in\mathbb{C}^{n\times m}.$

In ^[16], Wang and Chen introduced a new generalized inverse called the weak group inverse of $A\in\mathbb{C}^{n\times n}$, denoted by . It is defined as the unique matrix $X\in\mathbb{C}^{n\times n}$ satisfying:

Moreover, it is proved that = ()$^{2}A.$

While the authors in ^[13] introduced the BT-inverse defined as $A^{\diamondsuit} = (AP_{A})^{\dagger}$, the characterizations of how a matrix is $A^{\diamondsuit}$, however, seldom gave. In this paper, we concern more on the necessary and sufficient conditions for a matrix to be $A^{\diamondsuit}$ and characterize the relationships between $A^{\diamondsuit}$ and other generalized inverses. The research is as follows. In Section 2, some indispensable matrix classes and lemmas are given. In Section 3, some characterizations of $A^{\diamondsuit}$ are given too. In Section 4, we first derive a canonical form of $A^{\diamondsuit}$ by Core-EP decomposition and verify the validity of it by Example 1. By the canonical form of $A^{\diamondsuit}$ and Core-EP decomposition, we obtain the relationships between $A^{\diamondsuit}$ and other generalized inverses and some properties of $A^{\diamondsuit}$ when $A^{\diamondsuit}$ or $A$ belongs to some special matrix classes. In Section 5, we extend the representation $A^{\diamondsuit} = (AP_{A})^{\dagger}$ to a more general one by the maximal classes of matrices.

2.   Preliminaries

For convenience, some matrix classes will be given as follows.

These symbols $\mathbb{C}^{\rm CM}_{n}$, $\mathbb{C}^{\rm P}_{n}$, $\mathbb{C}^{\rm OP}_{n}$ and $\mathbb{C}^{\rm EP}_{n}$ will stand for the subsets of $\mathbb{C}^{n\times n}$ consisting of core matrices, projectors (idempotent matrices), orthogonal projectors (Hermitian idempotent matrices) and EP (Range-Hermitian) matrices, respectively, i.e.,

In order to present some characterizations and properties of $A^{\diamondsuit}$, we need to introduce the following lemmas.

Lemma 2.1. ^[17] Let $A\in\mathbb{C}^{n\times n}$, $r(A) = r$. Then we have

where $U\in\mathbb{C}^{n\times n}$ is unitary, $\Sigma = diag(\sigma_{1}, \sigma_{2}, \ldots, \sigma_{r})$ is the diagonal matrix of singular values of $A$, $\sigma_{i} > 0(i = 1, 2, \cdots, r)$ and $K\in\mathbb{C}^{r\times r}$, $L\in\mathbb{C}^{r\times(n-r)}$ satisfy

Moreover, from $(2.1)$, it follows that

By ^[12,13], we obtain that

and

The lemma below gives the Core-EP decomposition introduced by Wang which plays an important role in this paper.

Lemma 2.2. ^[9] Let $A\in\mathbb{C}^{n\times n}$ with ${\rm Ind}(A) = k$. Then there exists a unitary matrix $U\in\mathbb{C}^{n\times n}$ such that

where $T\in\mathbb{C}^{t\times t}$ is nonsingular with $t = r(T) = r(A^{k})$ and $N$ is nilpotent of index $k$.

Lemma 2.3. [18,Lemma 6] Let $A\in\mathbb{C}^{n\times n}$ with ${\rm Ind}(A) = k$ be the form of $(2.7)$. Then

where $N$ is not necessary nilpotent, $\triangle = (TT^{*}+S(I_{n-t}-N^{\dagger}N)S^{*})^{-1}$, $t = r(A^{k})$.

From $(2.7)$ and $(2.8)$, a straightforward computation shows that

Lemma 2.4. [13,Theorem 1] Let $A\in\mathbb{C}^{n\times n}$. Then

3.   Different characterizations about BT-inverse

It is well-known that some of generalized inverses such as MP-inverse, Drazin inverse, DMP-inverse, etc. can be presented as an outer inverse under the condition of prescribed range and null space. Therefore, we will prove that the same holds in the case of BT-inverse as follows. In the following theorem, we show the other characterizations of BT-inverse by the fact that $A^{\diamondsuit}AA^{\diamondsuit} = A^{\diamondsuit}.$

Theorem 3.1. Let $A, X\in\mathbb{C}^{n\times n}$. Then the following conditions are equivalent:

$(a)$ $X = A^{\diamondsuit};$

$(b)$ $XAX = X,$ $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*})$ and $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*}),$ i.e., $X = A^{(2)}_{\mathcal{R}(P_{A}A^{*}), \mathcal{N}(P_{A}A^{*})};$

$(c)$ $XAX = X,$ $AX = A(AP_{A})^{\dagger}$ and $XA = (AP_{A})^{\dagger}A;$

$(d)$ $XAX = X,$ $AX = P_{AP_{A}}$ and $XA = (AP_{A})^{\dagger}A.$

Proof. $(a)\Rightarrow(b).$ From the definition of BT-inverse and Lemma 2.4, we derive that

moreover

From the definition of BT-inverse and (3.2), it follows that

$(b)\Rightarrow(c).$ From [19,Remark 3.1], we have that $A^{(2)}_{\mathcal{R}(A^{\diamondsuit}), \mathcal{N}(A^{\diamondsuit})}$ exits. It is easy to check that $A^{\diamondsuit} = A^{(2)}_{\mathcal{R}((AP_{A})^{\dagger}), \mathcal{N}((AP_{A})^{\dagger})} = A^{(2)}_{\mathcal{R}(P_{A}A^{*}), \mathcal{N}(P_{A}A^{*})}$. Since $X = A^{(2)}_{\mathcal{R}(P_{A}A^{*}), \mathcal{N}(P_{A}A^{*})}$ and the uniqueness of $X$, we obtain that $X = A^{\diamondsuit}.$ Then the rest of proof is trivial.

$(c)\Rightarrow(d).$ Since $AX = A(AP_{A})^{\dagger}$, by (3.1), we obtain that $AX = AP_{A}(AP_{A})^{\dagger} = P_{AP_{A}}.$

$(d)\Rightarrow(a).$ By the condition, we conclude that

In the following theorem, we present a connection between (B, C)-inverse and BT-inverse showing that a BT-inverse of a matrix $A\in\mathbb{C}^{n\times n}$ is its $(P_{A}A^{*}, P_{A}A^{*})$-inverse.

Theorem 3.2. Let $A\in\mathbb{C}^{n\times n}$. Then $A^{\diamondsuit} = A^{(P_{A}A^{*}, P_{A}A^{*})}.$

Proof. From the definition of BT-inverse and (3.1), it follows that

Hence $A^{\diamondsuit} = A^{(P_{A}A^{*}, P_{A}A^{*})}.$

According to the fact that $\mathcal{R}(A^{\diamondsuit}) = \mathcal{R}(P_{A}A^{*})$ and $\mathcal{N}(A^{\diamondsuit}) = \mathcal{N}(P_{A}A^{*})$, there are several different characterizations of BT-inverse as follows.

Theorem 3.3. Let $A, X\in\mathbb{C}^{n\times n}$. Then the following conditions are equivalent:

$(a)$ $X = A^{\diamondsuit};$

$(b)$ $AX = A(AP_{A})^{\dagger},$ $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*});$

$(c)$ $AX = P_{AP_{A}},$ $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*});$

$(d)$ $P_{A}X = (AP_{A})^{\dagger},$ $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*});$

$(e)$ $A^{\dagger}X = A^{\dagger}(AP_{A})^{\dagger},$ $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*});$

$(f)$ $XA = (AP_{A})^{\dagger}A,$ $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*});$

$(g)$ $XA = P_{\mathcal{R}(P_{A}A^{*}), \mathcal{N}((AP_{A})^{\dagger}A)},$ $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*}).$

Proof. That $(a)$ implies all other items $(b),$ $(c),$ $(d),$ $(e),$ $(f)$ and $(g)$ can be checked directly by Theorem 3.1, the definition of BT-inverse and Lemma 2.4.

$(b)\Rightarrow(a).$ By $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*})$, we have $X = (AP_{A})^{\dagger}T$ for some $T\in\mathbb{C}^{n\times n}.$ By (3.2), then

$(c)\Rightarrow(b).$ Since $AX = P_{AP_{A}}$, by (3.1), we obtain that $AX = P_{AP_{A}} = AP_{A}(AP_{A})^{\dagger} = A(AP_{A})^{\dagger}$.

$(d)\Rightarrow(a).$ By $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*})$, we get $X = (AP_{A})^{\dagger}T$ for some $T\in\mathbb{C}^{n\times n}.$ By (3.2), then

$(e)\Rightarrow(d).$ Premultiplying $A^{\dagger}X = A^{\dagger}(AP_{A})^{\dagger}$ by $A$, we obtain that $P_{A}X = P_{A}(AP_{A})^{\dagger} = (AP_{A})^{\dagger}.$

$(f)\Rightarrow(a).$ By $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*})$, we obtain $X = K(AP_{A})^{\dagger}$ for some $K\in\mathbb{C}^{n\times n}.$ By (3.2), then

$(g)\Rightarrow(a).$ Since $XA = P_{\mathcal{R}(P_{A}A^{*}), \mathcal{N}((AP_{A})^{\dagger}A)} = P_{\mathcal{R}((AP_{A})^{\dagger}), \mathcal{N}((AP_{A})^{\dagger}A)}$, we get $XA(AP_{A})^{\dagger} = (AP_{A})^{\dagger}.$ By $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*})$, we have $X = K(AP_{A})^{\dagger}$ for some $K\in\mathbb{C}^{n\times n}.$ Then

Remark 3.4. Notice that the condition $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*})$ in items $(b),$ $(c),$ $(d)$ and $(e)$ of Theorem 3.3 can be replaced by $\mathcal{R}(X)\subseteq\mathcal{R}(P_{A}A^{*})$. Also the condition $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*})$ in items $(f), (g)$ of Theorem 3.3 can be replaced by $\mathcal{N}(P_{A}A^{*})\subseteq\mathcal{N}(X)$.

Theorem 3.5. Let $A, X\in\mathbb{C}^{n\times n}.$ Then the following conditions are equivalent:

$(a)$ $X = A^{\diamondsuit};$

$(b)$ $r(X) = r(A^{2}),$ $XA(AP_{A})^{*} = (AP_{A})^{*}$ and $AX = A(AP_{A})^{\dagger};$

$(c)$ $r(X) = r(A^{2}),$ $(AP_{A})^{*}AX = (AP_{A})^{*}$ and $XA = A(AP_{A})^{\dagger}A;$

$(d)$ $r(X) = r(A^{2})$, $XA(AP_{A})^{*} = (AP_{A})^{*}$ and $AX = P_{AP_{A}};$

$(e)$ $r(X) = r(A^{2})$, $XA(AP_{A})^{*} = (AP_{A})^{*}$ and $P_{A}X = (AP_{A})^{\dagger};$

$(f)$ $r(X) = r(A^{2})$, $XA(AP_{A})^{*} = (AP_{A})^{*}$ and $A^{\dagger}X = A^{\dagger}(AP_{A})^{\dagger};$

$(g)$ $r(X) = r(A^{2}),$ $(AP_{A})^{*}AX = (AP_{A})^{*}$ and $XA = P_{\mathcal{R}(P_{A}A^{*}), \mathcal{N}(P_{A}A^{*})}.$

Proof. $(a)\Rightarrow(b).$ For $X = A^{\diamondsuit}$, we get that $r(A^{\diamondsuit}) = r(AP_{A})$. For $\mathcal{R}(A^{2}) = \mathcal{R}(AP_{A}A)\subseteq\mathcal{R}(AP_{A})\subseteq\mathcal{R}(A^{2}),$ then we get that $\mathcal{R}(AP_{A}) = \mathcal{R}(A^{2}),$ hence $r(A^{\diamondsuit}) = r(AP_{A}) = r(A^{2})$. From the definition of BT-inverse and the latter half of (2.11), we derive that $A^{\diamondsuit}A(AP_{A})^{*} = (AP_{A})^{*}$ and $AA^{\diamondsuit} = A(AP_{A})^{\dagger}.$

That $(a)$ implies all other items $(c),$ $(d),$ $(e),$ $(f)$ and $(g)$ can be similarly proved.

$(b)\Rightarrow(a).$ Combining $r(X) = r(A^{2}) = r(AP_{A})$ with $XA(AP_{A})^{*} = (AP_{A})^{*}$, we obtain $\mathcal{R}(X) = \mathcal{R}(P_{A}A^{*})$. Hence it follows from $(b)$ of Theorem 3.3 that $X = A^{\diamondsuit}$.

$(c)\Rightarrow(a).$ From $r(X) = r(A^{2}) = r(AP_{A})$ and $(AP_{A})^{*}AX = (AP_{A})^{*}$, we get $\mathcal{N}(X) = \mathcal{N}(P_{A}A^{*})$. Hence we get $X = A^{\diamondsuit}$ by $(f)$ of Theorem 3.3.

The proofs of $(d)\Rightarrow(a)$, $(e)\Rightarrow(a)$ and $(f)\Rightarrow(a)$ are analogous to that of $(b)\Rightarrow(a).$ Also $(g)\Rightarrow(a)$ follows similarly as in the part $(c)\Rightarrow(a).$

4.   Canonical form and some properties of BT-inverse

In this section, we first give the canonical form of BT-inverse by using Core-EP decomposition. Then some properties of BT-inverse will be given by utilizing the definition and the canonical form of BT-inverse.

Theorem 4.1. Let $A\in\mathbb{C}^{n\times n}$ be of the form $(2.7)$. Then

where $\triangle = [TT^{*}+S(P_{N}-P_{N^{\diamondsuit}})S^{*}]^{-1}.$

Proof. By (2.9) of Lemma 2.3, we get that

From (2.8) of Lemma 2.3, we have that

where $\triangle = [TT^{*}+S(P_{N}-P_{N}P_{N^{\diamondsuit}})S^{*}]^{-1}.$

It is easy to check that $P_{N}N^{\diamondsuit} = N^{\diamondsuit}$ by (2.3) and (2.5). Hence

where $\triangle = [TT^{*}+S(P_{N}-P_{N}P_{N^{\diamondsuit}})S^{*}]^{-1} = [TT^{*}+S(P_{N}-P_{N^{\diamondsuit}})S^{*}]^{-1}.$

Next, we will verify the correctness of the expression (4.1) as follows.

Example 1. Given matrix

By the definition of BT-inverse, it turns out that

Assume that $A$ is of the form (2.7), we obtain that

According to (4.1), a straightforward computation shows that

Let $\parallel \cdot \parallel$ be the Frobenius norm, then it follows that

which implies the validity of the representation (4.1).

Lemma 4.2. ^[20] Let $A\in\mathbb{C}^{n\times n}$ written as in $(2.7)$. Then

where $\tilde{T} = \sum\limits_{j = 0}^{k-1}T^{j}SN^{k-1-j}$.

In ^[13], the necessary and sufficient conditions for $A^{\diamondsuit} = A^{\dagger},$ were given by using the Hartwig-Spindelböck decomposition in Lemma 2.1. We will prove the conditions that $A^{\diamondsuit} = A^{D}$, $A^{\diamondsuit} = A^{\dagger, D}$ and $A^{\diamondsuit} =$ are equivalent by utilizing Core-EP decomposition as follows.

Theorem 4.3. Let $A\in\mathbb{C}^{n\times n}$ be decomposed by $(2.7)$. Then the following statements are equivalent:

$(a)$ $S = 0$ and $N^{2} = 0;$

$(b)$ $A^{\diamondsuit} = A^{D};$

$(c)$ $A^{2}\in\mathbb{C}^{\rm EP}_{n};$

$(d)$ $A^{\diamondsuit} = A^{\dagger, D};$

$(e)$ $A^{\diamondsuit} =$.

Proof. $(a)\Longleftrightarrow(b).$ It follows from the definition of $A^{\diamondsuit}$, Lemma 2.3 and (4.2).

$(a)\Longleftrightarrow(c).$ From (2.7) and (2.8), we can calculate that

where $\triangle' = (T^{2}(T^{2})^{*}+(TS+SN)(I_{n-t}-(N^{2})^{\dagger}N^{2})(TS+SN)^{*})^{-1}.$

Then it follows that

$(d)\Longrightarrow (a).$ We can get $AA^{\diamondsuit} = AA^{D}$ by $A^{\diamondsuit} = A^{\dagger, D}.$ From (2.1), (2.4) and (2.5), $AA^{\diamondsuit} = AA^{D}$ is equivalent to

Thus $\Sigma K(\Sigma K)^{\dagger} = \Sigma K(\Sigma K)^{D}$. Then we have $\Sigma K = (\Sigma K)^{2}(\Sigma K)^{D}$ which implies ${\rm Ind}(\Sigma K)\leq1,$ moreover ${\rm Ind}(A)\leq2.$

Then let $A$ be the form of (2.7). For ${\rm Ind}(A)\leq2,$ we obtain $N^{2} = 0.$ Representations (4.1) and (4.2) directly lead to

Hence we get $\widetilde{T} = 0$ which implies $S = 0.$

$(a)\Longrightarrow (d).$ It can be directly checked.

$(a)\Longleftrightarrow(e).$ From the definition of $A^{\diamondsuit}$ and together with Lemma 2.3, it follows that

From ^[7], it is shown that $A^{\diamondsuit} =$ is equivalent to $A^{\diamondsuit} = A^{D, \dagger}$ by using the Hartwig-Spindelböck decomposition. Now we can verify the equivalence of $A^{\diamondsuit} =$ and $A^{\diamondsuit} = A^{D, \dagger}$ by Core-EP decomposition.

Theorem 4.4. Let $A\in\mathbb{C}^{n\times n}$ be decomposed by $(2.7)$. Then the following statements are equivalent:

$(a)$ $A^{\diamondsuit} =$ ;

$(b)$ $SN = 0$ and $N^{2} = 0;$

$(c)$ $A^{\diamondsuit} = A^{D, \dagger}.$

Proof. $(a)\Longleftrightarrow(b).$ According to Corollary 3.3 in ^[9], we have that

From the definition of $A^{\diamondsuit}$, and (2.9) together with the equation above, it follows that

$(b)\Longleftrightarrow(c).$ From the definition of $A^{\diamondsuit}$ and $A^{D, \dagger}$ together with (4.2), by using Lemma 2.3, it follows that

where $\tilde{T} = \sum\limits_{j = 0}^{k-1}T^{j}SN^{k-1-j}$.

Remark 4.5. If $A$ of the form $(2.7)$ is nilpotent, it follows that $A = UNU^{*}$. Then the $(a)$ of Theorem $4.3$ and the $(b)$ of the Theorem $4.4$ are equivalent to $N^{2} = 0$. In other words, if $A$ is nilpotent, then it follows that the conditions $A^{\diamondsuit} = A^{D}, A^{\diamondsuit} =$ , $A^{\diamondsuit} = A^{D, \dagger}, A^{\diamondsuit} = A^{\dagger, D}$ and $A^{\diamondsuit} =$ are equivalent.

In [13,Theorem 4], the author gave some equivalent conditions for $A^{\diamondsuit}\in\mathbb{C}^{\rm EP}_{n}.$ Then we will give some necessary and sufficient conditions for $A^{\diamondsuit}$ which belongs to some special matrix classes by using Core-EP decomposition.

Theorem 4.6. Let $A\in\mathbb{C}^{n\times n}$ be the form of $(2.7)$. Then,

$(a)$ $A^{\diamondsuit}\in\mathbb{C}^{\rm CM}_{n}\Longleftrightarrow N^{2} = 0;$

$(b)$ $A^{\diamondsuit}\in\mathbb{C}^{\rm P}_{n}\Longleftrightarrow N^{2} = 0$ and $T = TT^{*}+SP_{N}S^{*};$

$(c)$ $A^{\diamondsuit}\in\mathbb{C}^{\rm OP}_{n}\Longleftrightarrow$ $T = I_{t}, SN = 0$ and $N^2 = 0$ $($or $A^2 = A_{1}.$ where $A_{1}$ is presented in Lemma 2.2.$)$

Proof. $(a).$ From the definition of BT-inverse, it follows that

By (2.7) and (2.9), we obtain that

Thus $A^{\diamondsuit}\in\mathbb{C}^{\rm CM}_{n}\Longleftrightarrow N^{2}N^{\dagger} = 0\Longleftrightarrow N^{2} = 0$ which establishes point $(a)$ of the theorem.

$(b).$ For $A^{\diamondsuit}\in\mathbb{C}^{\rm P}_{n}\subseteq \mathbb{C}^{\rm CM}_{n},$ we have $N^{2} = 0$. From (4.1), now we have that

where $\triangle = (TT^{*}+SP_{N}S^{*})^{-1}.$

Since $A^{\diamondsuit}\in\mathbb{C}^{\rm P}_{n},$ we get that $T^{*}\triangle = I_{t},$ hence $T = (\triangle^{*})^{-1} = \triangle^{-1}.$ The sufficient condition of $(b)$ can be directly checked, therefore point $(b)$ of the theorem holds.

$(c).$ It can be directly checked that $A^2 = A_{1}$ is equivalent to $T = I_{t}, SN = 0$ and $N^2 = 0$ by Core-EP decomposition. For $A^{\diamondsuit}\in\mathbb{C}^{\rm OP}_{n}\subseteq \mathbb{C}^{\rm P}_{n},$ we have $N^{2} = 0$ and $T = \triangle^{-1}.$ From (4.1), we have

where $\triangle = (TT^{*}+SP_{N}S^{*})^{-1}.$

Since $A^{\diamondsuit}\in\mathbb{C}^{\rm OP}_{n},$ we get that $SP_{N} = 0$ which implies $T = I_{t}, SN = 0.$ The sufficient condition of $(c)$ can be directly checked, therefore point $(c)$ of the theorem holds.

Remark 4.7. If $A$ of the form $(2.7)$ is nilpotent which implies $A = UNU^{*}$, then $A^{\diamondsuit}\in\mathbb{C}^{\rm CM}_{n}$ or $\mathbb{C}^{\rm P}_{n}$ or ${\mathbb{C}^{\rm OP}_{n}}$ is equivalent to $A^{2} = 0$ $($or $N^{2} = 0$$)$.

From ^[13], it is known that $A^{\diamondsuit}A = AA^{\diamondsuit}$ and $(A^{\diamondsuit})^{\dagger} = (A^{\dagger})^{\diamondsuit}$ are both satisfied when $A\in\mathbb{C}^{\rm EP}_{n}$, but we can't conclude $A\in\mathbb{C}^{\rm EP}_{n}$ when $A^{\diamondsuit}A = AA^{\diamondsuit}$ or $(A^{\diamondsuit})^{\dagger} = (A^{\dagger})^{\diamondsuit}$ holds. How to establish an equivalence relation between them, the following theorem will give.

Theorem 4.8. Let $A\in\mathbb{C}^{n\times n}$ written as in $(2.1)$. Then the following statements are equivalent:

$(a)$ $A\in\mathbb{C}^{\rm EP}_{n};$

$(b)$ $AA^{\diamondsuit} = A^{\diamondsuit}A$ and $A\in\mathbb{C}^{\rm CM}_{n};$

$(c)$ $(A^{\diamondsuit})^{\dagger} = (A^{\dagger})^{\diamondsuit}$ and $A\in\mathbb{C}^{\rm CM}_{n};$

$(d)$ $(A^{\diamondsuit})^{m} = (A^{\dagger})^{m}$ for some $m\geq2$ and $A\in\mathbb{C}^{\rm CM}_{n}.$

Proof. That $(a)$ implies items $(b), (c)$ and $(d)$ can be checked directly by the definition of $A^{\diamondsuit}.$

$(b)\Rightarrow(a).$ For $A\in\mathbb{C}^{\rm CM}_{n},$ we get that $K$ is nonsingular. By (2.5) and (2.6), we get that $A^{\diamondsuit} =$ and $A$ = $A$. Hence it follows that $A\in\mathbb{C}^{\rm EP}_{n}$ by [3,Theorem 3].

$(c)\Rightarrow(a).$ This follows similarly as in the part $(b)\Rightarrow(a).$

$(d)\Rightarrow(a).$ It is known that $A\in\mathbb{C}^{\rm EP}_{n}$ is equivalent to $L = 0$. Combining (2.3), (2.5) with $(A^{\diamondsuit})^{m} = (A^{\dagger})^{m}$ leads to $L = 0$ which means $A\in\mathbb{C}^{\rm EP}_{n}$.

5.   Representations of BT-inverse by the maximal classes

Finally, we study the representations for the BT-inverse. In ^[4], let $A\in\mathbb{C}^{\rm CM}_{n}$. While $= A^{\#}AA^{\dagger}$ or $(A^{2}A^{\dagger})^{\dagger}$, the author gave new representations by the maximal matrix classes such as $= XAY$ or $(A^{2}Z)^{\dagger}$ where $\mathcal{R}(XA)\subseteq\mathcal{R}(A)$ and $Y\in A\{1, 3\}$ or $Z\in A\{1, 3\}$. Similarly, the author in ^[21] gave the representations of , $A^{D, \dagger}$ by the maximal classes. Now, we will derive the representations of BT-inverse by the maximal classes. We first give the important lemma as follows.

Lemma 5.1. ^[22] Let $A, B, C\in\mathbb{C}^{n\times n}$. Then the matrix equation $AXB = C$ is consistent if and only if for some $A^{(1)}\in A\{1\},$ $B^{(1)}\in B\{1\},$

in which case the general solution is

for arbitrary $Z\in\mathbb{C}^{n\times n}.$

Theorem 5.2. Let $A\in\mathbb{C}^{n\times n}$ of rank $r$ has the form $(2.1)$. Then the following conditions are equivalent:

$(a)$ $A^{\diamondsuit} = (A^{2}X)^{\dagger};$

$(b)$ $A^{2}X = AP_{A};$

$(c)$ $X = P_{(A^{2})^{\dagger}}A^{\dagger}+(I_{n}-P_{(A^{2})^{\dagger}})Z,$ for arbitrary $Z\in\mathbb{C}^{n\times n};$

$(d)$ $X$ can be expressed as

where $R = PP^{*}+QQ^{*},$ $P = (\Sigma K)^{2}$ and $Q = \Sigma K\Sigma L,$ for arbitrary $Z_{1}, Z_{2}, Z_{3}, Z_{4}.$

Proof. $(a)\Rightarrow(b).$ Since $A^{\diamondsuit} = (AP_{A})^{\dagger} = (A^{2}X)^{\dagger},$ we have $A^{2}X = AP_{A}.$

$(b)\Rightarrow(c).$ It is evident that $P_{(A^{2})^{\dagger}}A^{\dagger}$ satisfies the equation

Applying Lemma 5.1 to this equation, the general solution of (4.3) is given by

for arbitrary $Z\in\mathbb{C}^{n\times n}.$

$(c)\Longleftrightarrow(d).$ From (2.1), it follows that

and applying [23,Lemma 1] to (5.2), we obtain that

where $R = PP^{*}+QQ^{*},$ $P = (\Sigma K)^{2}$ and $Q = \Sigma K\Sigma L.$ Next, partitioning accordingly

a straightforward computation shows that $X = P_{(A^{2})^{\dagger}}A^{\dagger}+(I_{n}-P_{(A^{2})^{\dagger}})Z$ is equivalent to

where $R = PP^{*}+QQ^{*},$ $P = (\Sigma K)^{2}$ and $Q = \Sigma K\Sigma L,$ for arbitrary $Z_{1}, Z_{2}, Z_{3}, Z_{4}.$

$(c)\Rightarrow(a).$ By a direct calculation, we have that $A^{2}X = A^{2}A^{\dagger}.$ Therefore

Theorem 5.3. Let $A\in\mathbb{C}^{n\times n}$ be of the form $(2.1)$, $X, Y\in AP_{A}\{1\}.$ Then the following conditions are equivalent:

$(a)$ $A^{\diamondsuit} = XAP_{A}Y;$

$(b)$ $XAP_{A} = P_{(A^{2})^{\dagger}}$ and $AP_{A}Y = A(AP_{A})^{\dagger};$

$(c)$ $X = (AP_{A})^{\dagger}+Z(I_{n}-P_{AP_{A}})$ and $Y = (AP_{A})^{\dagger}+(I_{n}-P_{(AP_{A})^{\dagger}})W,$ for arbitrary $Z, W\in\mathbb{C}^{n\times n};$

$(d)$ $X, Y$ can be expressed as

for arbitrary $Z_{1}, Z_{2}, Z_{3}, Z_{4};$

for arbitrary $W_{1}, W_{2}, W_{3}, W_{4}.$

Proof. $(a)\Rightarrow(b).$ Postmultiplying $A^{\diamondsuit} = XAP_{A}Y$ by $AP_{A}.$ For $Y\in AP_{A}\{1\}$, it follows that $XAP_{A} = P_{(A^{2})^{\dagger}}.$ Premultiplying $A^{\diamondsuit} = XAP_{A}Y$ by $AP_{A}$. Since $X\in AP_{A}\{1\},$ it follows that $AP_{A}Y = AP_{A}A^{\diamondsuit} = AA^{\diamondsuit}.$

$(b)\Rightarrow(c).$ Applying Lemma 5.1 to two equations $XAP_{A} = (AP_{A})^{\dagger}AP_{A}$ and $AP_{A}Y = A(AP_{A})^{\dagger}$ respectively, the general solutions are given by $X = (AP_{A})^{\dagger}+Z(I_{n}-P_{AP_{A}})$ for arbitrary $Z\in\mathbb{C}^{n\times n}$ and $Y = (AP_{A})^{\dagger}+(I_{n}-P_{(AP_{A})^{\dagger}})W$ for arbitrary $W\in\mathbb{C}^{n\times n}.$

$(c)\Rightarrow(d).$ Assume that $A$ has the form given in (2.1), we have

Next, partitioning accordingly

a straightforward shows that $X = (AP_{A})^{\dagger}+Z(I_{n}-P_{AP_{A}})$ is equivalent to

for arbitrary $Z_{1}, Z_{2}, Z_{3}, Z_{4}$. $Y = (AP_{A})^{\dagger}+(I_{n}-P_{(AP_{A})^{\dagger}})W$ is equivalent to

for arbitrary $W_{1}, W_{2}, W_{3}, W_{4}.$

$(d)\Rightarrow(a).$ According to (5.4) and (5.5), a straightforward computation shows that

6.   Conclusions

In this work, different characteristics of the BT-inverse of a square matrix have been developed. Some necessary and sufficient conditions for a matrix to be the BT-inverse have been derived. The Core-EP decomposition is efficient for investigating the relationships between the BT-inverse and other generalized inverses. The expression of BT-inverse has been extended to more general ones by the maximal classes of matrices.

Acknowledgement

The authors are thankful to four anonymous referees for their careful reading, detailed corrections, insightful comments and pertinent suggestions on the first version of the paper, which enhance the presentation of the results distinctly. This research is supported by the Natural Science Foundation of China under Grants 11961076.

Conflict of interest

The authors declare no conflict of interest.