RACSAM https://doi.org/10.1007/s13398-017-0486-4 ORIGINAL PAPER

The pseudo core inverse of a lower triangular matrix Yuefeng Gao1

· Jianlong Chen1

Received: 10 May 2017 / Accepted: 13 December 2017 © Springer-Verlag Italia S.r.l., part of Springer Nature 2017

Abstract The notion of core inverse was introduced by Baksalary and Trenkler for a complex matrix of index one in 2010, and then it was generalized to an arbitrary ∗-ring case by Raki´c, Dinˇci´c and Djordjevi´c in 2014. Later, the notion of pseudo core inverse extended the notion of core inverse to an element of an arbitrary index in ∗-rings; meanwhile, it generalized the notion of core-EP inverse, introduced by Manjunatha Prasad and Mohana for complex matrices in 2014, to the case of ∗-ring. In this paper, we characterize the pseudo core invertibility of a matrix in product form under some prescribed conditions. As applications, the existence and expression for the pseudo core inverse of a lower triangular matrix are considered. Keywords Pseudo core inverse · Core inverse · Core-EP inverse · Lower triangular matrix Mathematics Subject Classification 15A09 · 16W10 · 16S50

1 Introduction Throughout this paper, R is a unital ring with involution, i.e., a ∗-ring and R m×n denotes the set of all m × n matrices over R. For any matrix A = (ai j ) ∈ R m×n , we let A∗ = (a ∗ji ) ∈ R n×m . Let A ∈ R m×n . We say that A is (von Neumann) regular if A has an inner inverse A− ∈ R n×m , i.e., A A− A = A. Such an inner inverse of A may not be unique if it exists. We use A{1} to denote the set of all the inner inverses of A. A {1, 3}-inverse of A is a solution to the equations AX A = A and (AX )∗ = AX ; A {1, 4}-inverse of A is a solution to the equations AX A = A and (X A)∗ = X A; such a {1, 3}-inverse (resp. {1, 4}-inverse) of A may not be unique if it exists. We use A{1, 3}

B

Jianlong Chen [email protected] Yuefeng Gao [email protected]

1

School of Mathematics, Southeast University, Nanjing 210096, China

Y. Gao, J. Chen

(resp. A{1, 4}) to denote the set of all the {1, 3}-inverses (resp. {1, 4}-inverses) of A. The Moore-Penrose inverse of A is denoted, as usual, by A† . Let A ∈ R n×n , the Drazin inverse (resp. group inverse) and Drazin index of A are denoted, as usual, by A D (resp. A# ) and i(A) respectively. Baksalary and Trenkler [1] introduced the concept of core inverse for a complex matrix in 2010. In 2014, Raki´c et al. [2] extended this notion to an arbitrary ∗-ring. Then, Xu et al. [3] characterized the core (resp. dual core) invertible elements in ∗-rings by three equations, which can be used, without modification, to characterize the existence of core inverses of matrices over ∗-rings. The core inverse of A, denoted by A# , is the unique solution (if exists) to the equations X A2 = A, AX 2 = X, (AX )∗ = AX. The dual core inverse of A, denoted by A# , is the unique solution (if exists) to the equations A2 X = A, X 2 A = X, (X A)∗ = X A. We refer readers to [4,5] for a deep study of the core inverse. In [6], the authors introduced the notions of pseudo core inverse and dual pseudo core inverse in ∗-rings. D The pseudo core inverse of A, denoted by A , is the unique solution (if exists) to the equations X Am+1 = Am for some positive integer m, AX 2 = X and (AX )∗ = AX. The smallest positive integer m satisfying the above equations is called the pseudo core index of A. If A is pseudo core invertible, then it must be Drazin invertible, and the pseudo core index coincides with the Drazin index [6]. So here and subsequently, we denote the pseudo core index of A by i(A). The pseudo core inverse is an outer inverse, i.e., we always have D D D A A A = A [6]. It is obvious that if the pseudo core index equals one, then the pseudo core inverse of A is the core inverse of A. The pseudo core inverse also extends the core-EP inverse [7] from complex matrices to ∗-rings (see [6]). Dually, the dual pseudo core inverse of A, if exists, is the unique matrix A D satisfying m for some positive integer m, (A )2 A = A and (A A)∗ the equations Am+1 A D = A D D D    = A D A. The smallest positive integer m satisfying the above equations is called the dual pseudo core index of A, denoted by i(A) as well (see [6]). Gouveia et al. [8], Puystjens et al. [9] and Mary et al. [10] investigated the group invertibility of a product; Chen [11] and Puystjens et al. [12] gave characterizations for a product to have Drazin inverse; Gouveia et al. [8] and Patrício [13] studied the Moore– Penrose invertibility of a product; Zhu et al. [14] and Castro-González et al. [15] presented characterizations for a product to have {1, 3}-inverse; Ke et al. [16] gave necessary and sufficient conditions for a product to have core inverse (resp. dual core inverse). Inspired by the above papers, in Sect. 2, under some prescribed conditions, we obtain existence criteria and formulae for the pseudo core inverse of a product. As applications, in Sect. 3, we investigated the pseudo core invertibility of a lower triangular matrix.

2 Pseudo core inverses and dual pseudo core inverses of a product In this section, we consider matrices over R unless otherwise indicated. Our main goal is to propose different existing criteria and formulae for the pseudo core inverse of the matrix T expressed as a specific product P AQ, depending on the existence of specific inverses of A. I denotes the identity matrix. We begin with some auxiliary lemmas.

The pseudo core inverse of a lower triangular matrix

Lemma 1 [11, Theorem 2.1] Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with A{1}  = ∅ and T k = P AQ for some positive integer k. Let A− ∈ A{1}. Then the following are equivalent: (1) T D exists with i(T ) ≤ k, and P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s . (2) A− AQT P A + I − A− A is invertible. In this case, T D = P A(A− AQT P A + I − A− A)−1 Q. Lemma 2 [11, Theorem 3.2] Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with T = P AQ. Suppose P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s . (1) If A{1, 4}  = ∅. Let A(1,4) ∈ A{1, 4}. Then T {1, 3}  = ∅ if and only if (P A)∗ P A + I − A(1,4) A is invertible. In this case, Q  A(1,4) A[(P A)∗ P A + I − A(1,4) A]−1 (P A)∗ ∈ T {1, 3}. (2) If A{1, 3}  = ∅. Let A(1,3) ∈ A{1, 3}. Then T {1, 4}  = ∅ if and only if AQ(AQ)∗ + I − A A(1,3) is invertible. In this case, (AQ)∗ [AQ(AQ)∗ + I − A A(1,3) ]−1 A A(1,3) P  ∈ T {1, 4}. Lemma 3 [17, Exercise 1.6] Let A ∈ R n×m , B ∈ R m×n , C ∈ R n×n . (1) I + AB is left invertible if and only if I + B A is left invertible. Moreover, C(I + AB) = I if and only if (I − BC A)(I + B A) = I . (2) I + AB is right invertible if and only if I + B A is right invertible. Moreover, (I + AB)C = I if and only if (I + B A)(I − BC A) = I . (3) I + AB is invertible if and only if I + B A is invertible. Moreover, (I + B A)−1 = I − B(I + AB)−1 A. Lemma 4 [6, Theorem 2.3] Let T ∈ R n×n . D exists if and only if T D exists and T k {1, 3}  = ∅, where k ≥ i(T ). (1) T  D In this case, T  = T D T k (T k )(1,3) for any (T k )(1,3) ∈ T k {1, 3}. D exists and T k {1, 4}  = ∅, where k ≥ i(T ). (2) T D exists if and only if T k (1,4) In this case, T T k T D for any (T k )(1,4) ∈ T k {1, 4}. D = (T ) D (3) T  and T exist if and only if T D and (T k )† exist, where k ≥ i(T ). D D k † k D. In this case, T  = T D T k (T k )† and T D = (T ) T T

Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with T k = P AQ for some positive integer k. Suppose P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s . The following result gives a characterization for T to have pseudo core inverse, under the assumption that A{1, 4}  = ∅. Proposition 1 Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with A{1, 4}  = ∅ and T k = P AQ for some positive integer k. Suppose P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s . Let A(1,4) ∈ A{1, 4}. Then the following are equivalent: D (1) T  exists with i(T ) ≤ k. (1,4) (2) A AQT P A + I − A(1,4) A and (P A)∗ P A + I − A(1,4) A are invertible.

In this case, D = P A[A(1,4) AQT P A + I − A(1,4) A]−1 Q P A[(P A)∗ P A + I − A(1,4) A]−1 (P A)∗ . T

Y. Gao, J. Chen

Proof

Let A(1,4) ∈ A{1, 4}, by Lemma 1, T D exists if and only if A(1,4) AQT P A + I − A(1,4) A is invertible,

in which case, T D = P A[A(1,4) AQT P A + I − A(1,4) A]−1 Q. Then by Lemma 2, T k {1, 3}  = ∅ if and only if (P A)∗ P A + I − A(1,4) A is invertible, in which case, Q  A(1,4) A[(P A)∗ P A + I − A(1,4) A]−1 (P A)∗ ∈ T k {1, 3}. D From Lemma 4, T  exists if and only if T D exists and T k {1, 3}  = ∅, where k ≥ i(T ). Thus, (1) is equivalent to (2). Moreover, for any (T k )(1,3) ∈ T k {1, 3}, D = T D T k (T k )(1,3) T

= P A[A(1,4) AQT P A + I − A(1,4) A]−1 Q P AQ Q  A(1,4) A ×[(P A)∗ P A + I − A(1,4) A]−1 (P A)∗ = P A[A(1,4) AQT P A + I − A(1,4) A]−1 Q P A[(P A)∗ P A + I − A(1,4) A]−1 (P A)∗ . 

Dually, we have the following result, which characterizes the dual pseudo core invertibility of T , under the assumption that A{1, 3}  = ∅. Proposition 2 Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with A{1, 3}  = ∅ and T k = P AQ for some positive integer k. Suppose P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s . Let A(1,3) ∈ A{1, 3}. Then the following are equivalent: (1) T D exists with i(T ) ≤ k. (2) AQT P A A(1,3) + I − A A(1,3) and AQ(AQ)∗ + I − A A(1,3) are invertible. In this case, ∗ ∗ (1,3) −1 T ] AQ P[AQT P A A(1,3) + I − A A(1,3) ]−1 AQ. D = (AQ) [AQ(AQ) + I − A A

If k = 1 in Propositions 1 and 2, then we get a characterization for the existence of the core inverse (resp. dual core inverse) of a product P AQ. Corollary 1 Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with T = P AQ. Suppose P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s . (1) If A{1, 4}  = ∅. Let A(1,4) ∈ A{1, 4}. Then T # exists if and only if both A(1,4) AQ P A + I − A(1,4) A and (P A)∗ P A + I − A(1,4) A are invertible. In this case, T # = P A[A(1,4) AQ P A + I − A(1,4) A]−1 [(P A)∗ P A + I − A(1,4) A]−1 (P A)∗ . (2) If A{1, 3}  = ∅. Let A(1,3) ∈ A{1, 3}. Then T# exists if and only if both AQ P A A(1,3) + I − A A(1,3) and AQ(AQ)∗ + I − A A(1,3) are invertible. In this case, T# = (AQ)∗ [AQ(AQ)∗ + I − A A(1,3) ]−1 [AQ P A A(1,3) + I − A A(1,3) ]−1 AQ. (3) If A† exists. Then T # and T# exist if and only if A† AQ P A + I − A† A, (P A)∗ P A + I − A† A and AQ(AQ)∗ + I − A A† are invertible. In this case, T # = P A[A† AQ P A + I − A† A]−1 [(P A)∗ P A + I − A† A]−1 (P A)∗ and T# = (AQ)∗ [AQ(AQ)∗ + I − A A† ]−1 [AQ P A A† + I − A A† ]−1 AQ. In what follows, we consider the existence criteria and formulae of the pseudo core inverse of T , under the assumption that A is regular, generalizing Theorem 3.1 in [16]. We begin with an auxiliary lemma.

The pseudo core inverse of a lower triangular matrix

Lemma 5 [18] Let A ∈ R m×s . If there exists X ∈ R m×s such that X A∗ A = A, then X ∗ ∈ A{1, 3}. Proposition 3 Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with A{1}  = ∅ and T k = P AQ for some positive integer k. Let A− ∈ A{1} and let U = A− AQT P A + I − A− A, V = AQ(P AQ)∗ P A A− + I − A A− . Then the following are equivalent: (1) T D exists with i(T ) ≤ k, A = P  P A = AQ Q  and Z  AQ(P AQ)∗ P AQ = P AQ for some P  ∈ R m×n , Q  ∈ R n×s , Z  ∈ R n×m . (2) U is invertible and V is left invertible. (3) X AQT P A = A = AQT P AY , Z AQ(P AQ)∗ P A = A for some X ∈ R m×m , Y ∈ R s×s , Z ∈ R m×m . D exists with In this case, T 

∗

D T = P AU −1 Q P AQ Q ∗ A∗ Z ∗ P ∗ = P AU −1 Q P AQ Q ∗ A∗ Z  .

Proof By Lemma 1, T D exists with i(T ) ≤ k, and A = P  P A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s if and only if U is invertible. Next, we prove that U is invertible if and only if X AQT P A = A = AQT P AY for some X ∈ R m×m , Y ∈ R s×s . In fact, U = A− AQT P A + I − A− A = I + A− (AQT P A − A). By Lemma 3, U is invertible if and only if U  = AQT P A A− + I − A A− is left invertible and U = A− AQT P A + I − A− A is right invertible. Assume that U is invertible. Then U  is left invertible, thus there exists X such that XU  = X (AQT P A A− + I − A A− ) = I. Hence, X AQT P A = A. From the fact that U is right invertible, it follows that there exists Y such that U Y = (A− AQT P A + I − A− A)Y = I . We thus have AQT P AY = A. Conversely, If X AQT P A = A = AQT P AY for some X ∈ R m×m , Y ∈ R s×s , then (X A A− + I − A A− )(AQT P A A− + I − A A− ) = I and (A− AQT P A + I − A− A)(A− AY + I − A− A) = I, namely, U  is left invertible and U is right invertible. Hence U is invertible. In the following, we prove that V is left invertible if and only if there exists Z such that Z AQ(P AQ)∗ P A = A if and only if there exists Z  such that Z  AQ(P AQ)∗ P AQ = P AQ. If V is left invertible, then there exists Z such that Z V = Z [AQ(P AQ)∗ P A A− + I − A A− ] = I. Thus, Z AQ(P AQ)∗ P A = A. Conversely if Z AQ(P AQ)∗ P A = A, then (Z A A− + I − A A− )V = I , namely, V is left invertible. If Z AQ(P AQ)∗ P A = A, then P Z AQ(P AQ)∗ P AQ = P AQ. Conversely if  Z AQ(P AQ)∗ P AQ = P AQ, then P  Z  AQ(P AQ)∗ P A = A because of P  P A = A = AQ Q  . From the above, we derive the equivalences of (1)–(3). In this case, T D = P AU −1 Q

Y. Gao, J. Chen (1,3)

follows from Lemma 1, and T k T k = P AQ Q ∗ A∗ Z ∗ P ∗ = P AQ Q ∗ A∗ Z  ∗ follows from Lemma 5. D Hence by applying Lemma 4, T  exists with D = T DT kT k T

(1,3)

∗

= P AU −1 Q P AQ Q ∗ A∗ Z ∗ P ∗ = P AU −1 Q P AQ Q ∗ A∗ Z  . 

If k = 1 in Proposition 3, then we recover Theorem 3.1 of [16]. Next, we consider the case that A has a {1, 3}-inverse, generalizing Theorem 3.3 in [16]. It turns to be a useful tool to characterize the pseudo core invertibility of a lower triangular matrix. Proposition 4 Let P ∈ R n×m , A ∈ R m×s , Q ∈ R s×n , T ∈ R n×n with A{1, 3}  = ∅ and T k = P AQ for some positive integer k. Let A(1,3) ∈ A{1, 3}. Suppose P(I − A A(1,3) ) = I − A A(1,3) , P  P A = A = AQ Q  for some P  ∈ R m×n , Q  ∈ R n×s and let U = AQT P A A(1,3) + I − A A(1,3) , V = P ∗ P A A(1,3) + I − A A(1,3) . Then the following are equivalent: (1) (2) (3) (4)

D exists with i(T ) ≤ k. T U is invertible and T k {1, 3}  = ∅. U is invertible and V is left invertible. U is invertible and V is right invertible.

D In this case, T  = PU −1 AQ P A A(1,3) V −1 P ∗ . D exists if and only if T D exists and T k {1, 3}  = ∅. Proof By Lemma 4, T   Since P P A = A = AQ Q  , by Lemma 1, T D exists with i(T ) ≤ k if and only if A(1,3) AQT P A + I − A(1,3) A is invertible, which is equivalent to U is invertible by Lemma 3. In this case, T D = PU −1 AQ. We thus get (1) ⇔ (2). By [14, Theorem 2.10], we obtain that (P A){1, 3}  = ∅ if and only if V is left invertible if and only if V is right invertible, in which case, A(1,3) V −1 P ∗ ∈ (P A){1, 3}. And notice that when A = AQ Q  , if (P AQ){1, 3}  = ∅, then Q(P AQ)(1,3) ∈ (P A){1, 3}; conversely if (P A){1, 3}  = ∅, then Q  (P A)(1,3) ∈ (P AQ){1, 3}. Thus, T k {1, 3} = (P AQ){1, 3}  = ∅ if and only if (P A){1, 3}  = ∅ whenever A = AQ Q  . Hence, the equivalences of (2)–(4) follow. In this case, D T = T D T k (T k )(1,3) = PU −1 AQ P AQ Q  A(1,3) V −1 P ∗

= PU −1 AQ P A A(1,3) V −1 P ∗ . 

If k = 1 in Proposition 4, letting Q = I , then we recover Theorem 3.3 of [16].

3 The pseudo core inverse of a lower triangular matrix Many authors have derived a number of results on various generalized inverses of lower triangular matrices. We refer readers to [16,19–22]. D In this section, some applications of Propositions 1 and 4 are indicated. R  and R D denote the sets of all pseudo core invertible and dual pseudo core invertible elements in R, respectively.

The pseudo core inverse of a lower triangular matrix

  a0 It is known that if a (resp. d) is Drazin invertible, then T = ∈ R 2×2 is Drazin bd invertible if and only if d (resp. a) is Drazin invertible [19, Theorem 1]. One may guess whether it is true for the pseudo core inverse.   a0 Example 1 (1) Both a and d are pseudo core invertible may not imply that T = ∈ bd   10 ∈ C2×2 with transpose as R 2×2 is pseudo core invertible. For example, let T = i 0 involution, then both 1 and 0 are pseudo core invertible, but for any positive integer k, T k = T has no {1, 3}-inverse.   Hence T is not pseudo core invertible. a0 (2) a (resp. d) and T = ∈ R 2×2 are pseudo core invertible may not imply that bd d (resp. a) is pseudo core invertible. For example, ⎤ ⎡ 1000       ⎢0 1 0 0 ⎥ 10 00 10 4×4 with transpose ⎥ (i) let a = ,b = ,d = and T = ⎢ ⎣0 0 1 0 ⎦ ∈ C 01 10 i 0 10i 0 as involution, then T 2 = T and ⎤ ⎡ 0 0 −i 1 ⎢ 0 1 0 0⎥ D ⎥ = T # = T # T T (1,3) = T T (1,3) = ⎢ T ⎣−i 0 2 i ⎦ , 1 0 i 0 but d is not pseudo core invertible.

⎤ ⎡ 1000     ⎢ i 0 0 0⎥ 10 00 00 4×4 with transpose ⎥ (ii) let a = ,b= ,d= and T = ⎢ ⎣0 0 1 0 ⎦ ∈ C i 0 10 01 1000 ⎤ ⎡ 1 i 01 ⎢ i −1 0 i ⎥ D ⎥ =⎢ as involution, then T  ⎣0 0 1 0⎦, but a is not pseudo core invertible. 1 i 01 



 However, it is true for a special lower triangular matrix T =

 00 ∈ R 2×2 . bd

  00 D D exists if and only if d  exists. Moreover, Theorem 1 Let T = ∈ R 2×2 . Then T  bd   0 0 D = T D . 0 d D D   Proof Suppose  that d exists with i(d) = k, then it is easy to check that T exists with 0 0 D = T D , as well as 0 d ⎧ if d is invertible ⎨ 1, D if (1 − d  d)d k−1 b = 0 . i(T ) = k, ⎩ D  k + 1, if (1 − d d)d k−1 b  = 0

Y. Gao, J. Chen D Conversely, if T  exists, according to the definition of the pseudo core inverse, then we  00 D D  for some x ∈ R. And hence, d  = x. 

must have T = 0x

In what follows, we give some existence criteria  and formulae for the pseudo core inverse a0 of a general lower triangular matrix T = ∈ R 2×2 , making use of Propositions 1 and bd 4 respectively. Firstly, we have the following result by applying Proposition 1.   a0 D k Theorem 2 Let T = ∈ R 2×2 such that d ∈ R  ∩ R D with i(d) = k, a {1, 4}  = ∅, bd

D k ) ]c[1 − (a k )(1,4) a k ] with w{1, 4}  = ∅. Then the c = i+ j=k−1 d i ba j , w = [1 − d k (d  following are equivalent: D exists with i(T ) ≤ k. (1) T  (2) ξ and η are invertible, where

ξ = (a k )(1,4) a 2k+1 + w (1,4) wa k+1 + 1 − (a k )(1,4) a k − w (1,4) w, η = (a k )∗ a k + c∗ [1 − d k (d k )† ]c + 1 − (a k )(1,4) a k − w (1,4) w.

In this case,

D T

  μν = , where δ ω

μ = a k ξ −1 a k η−1 (a k )∗ , ν = a k ξ −1 a k η−1 c∗ [1 − d k (d k )† ], δ = [1 − d k (d k )† ]cξ −1 a k η−1 (a k )∗ − d D γ ξ −1 a k η−1 (a k )∗ + d D (d k )† ca k η−1 (a k )∗ + d D [1 − d k (d k )† ]cη−1 (a k )∗ , ω = [1 − d k (d k )† ]cξ −1 a k η−1 c∗ [1 − d k (d k )† ] − d D γ ξ −1 a k η−1 c∗ [1 − d k (d k )† ] + d D (d k )† ca k η−1 c∗ [1 − d k (d k )† ] + d D [1 − d k (d k )† ]cη−1 c∗ [1 − d k (d k )† ] D + d ,

γ = (d k )† ca k+1 + (d k )† d k ba k + (d k )† d k+1 [1 − d k (d k )† ]c. D D and (d k )† exist. Proof Applying Lemma 4, d ∈ R  ∩ R D with i(d) = k if and only if d D k k † k k  k+1 k+1 k k k † k In this case, d (d ) = d (d ) , (d ) = (d ) D d D d = (d ) d . a0 Since T = , then bd

Tk =

    k  k   1 0 ak 0 1 0 a0 a 0 = = bd [1 − d k (d k )† ]c(a k )(1,4) 1 w d k (d k )† c 1 c dk

:= P AQ.

The pseudo core inverse of a lower triangular matrix

  k (1,4) (1,4) w [1 − d k (d k )† ] (a ) ∈ A{1, 4}. Then we derive 0 (d k )†  k (1,4) (1,4)   (a ) w [1 − d k (d k )† ] a k 0 (1,4) A A= w dk 0 (d k )†  k (1,4) k  (a ) a + w (1,4) w 0 = , 0 (d k )† d k  k (1,4) 2k+1  (a ) a + w (1,4) wa k+1 0 A(1,4) AQT P A + I − A(1,4) A = γ (d k )† d 2k+1   1 − (a k )(1,4) a k − w (1,4) w 0 + 0 1 − (d k )† d k   ξ 0 , = γ (d k )† d 2k+1 + 1 − (d k )† d k   k ∗ k 0 (a ) a + c∗ [1 − d k (d k )† ]c ∗ (1,4) (P A) P A + I − A A= 0 (d k )∗ d k   0 1 − (a k )(1,4) a k − w (1,4) w + 0 1 − (d k )† d k   η 0 = . 0 (d k )∗ d k + 1 − (d k )† d k

Observe that

k k = d k+1 + 1 − Since (d k+1 )# exists, then d k+1 + 1 − (d k )† d k = d k+1 +1 − (d D ) d k+1 k+1 (d ) is invertible (see [9, Corollary 1]). From Lemma 3, it follows that (d k )† d 2k+1 + D d 1 − (d k )† d k is invertible.

Since (d k )† exists, (d k )∗ d k + 1 − (d k )† d k is invertible (see [13, Theorem 1]). D According to Proposition 1, T  exists with i(T ) ≤ k if and only if both A(1,4) AQT P A + (1,4) ∗ I−A A and (P A) P A + I − A(1,4) A are invertible, which equivalent to both ξ and η are invertible. Moreover, D T = P A[A(1,4) AQT P A + I − A(1,4) A]−1 Q P A

× [(P A)∗ P A + I − A(1,4) A]−1 (P A)∗    ξ −1 0 0 ak = k+1 γ ξ −1 d k+1 + 1 − (d k )† d k [1 − d k (d k )† ]c d k −d D D    −1   η 0 0 ak × 0 (d k )† (d k )†∗ + 1 − (d k )† d k (d k )† ca k + [1 − d k (d k )† ]c d k   k ∗ ∗ (a ) c [1 − d k (d k )† ] × 0 (d k )∗   μν = . δ ω 

With an application of Proposition 4, we obtain the following result.   a0 D Theorem 3 Let T = ∈ R 2×2 such that a ∈ R  with i(a) = k, d k {1, 3}  = ∅, bd

D k k ) a ] with w{1, 3}  = ∅. Then the c = i+ j=k−1 d i ba j , w = [1 − d k (d k )1,3 ]c[1 − (a  following are equivalent:

Y. Gao, J. Chen D (1) T  exists with i(T ) ≤ k. (2) α and β are invertible, where D k ∗ D k α = 1 + [c(a  ) ] [1 − d k (d k )(1,3) − ww (1,3) ]c(a  ) ,

β = d k+1 ww (1,3) + d 2k+1 (d k )(1,3) + 1 − ww (1,3) − d k (d k )(1,3) . D = In this case, T 

 t1 t2  t3 t4 , where

D t1 = a  α −1 , D D k ∗ α −1 [c(a  ) ] [1 − d k (d k )(1,3) − ww (1,3) ], t2 = a  D k −1 D ) α − β −1 d k ba  α −1 t3 = β −1 d k [1 − d k (d k )(1,3) − ww (1,3) ]c(a  D k+1 −1 D k+1 −1 − β −1 d k+1 [1 − d k (d k )(1,3) ]c(a  ) α + [1 − d k (d k )(1,3) ]c(a  ) α

t4 = β −1 d 2k (d k )(1,3) + β −1 d k ww (1,3) + β −1 d k [1 − d k (d k )(1,3) − ww (1,3) ] D k −1 D k ∗ ) α [c(a  ) ] [1 − d k (d k )(1,3) − ww (1,3) ] − β −1 d k+1 × c(a  D k+1 −1 D k ∗ × [1 − d k (d k )(1,3) ]c(a  ) α [c(a  ) ] [1 − d k (d k )(1,3) − ww (1,3) ] D D k ∗ − β −1 d k ba  α −1 [c(a  ) ] [1 − d k (d k )(1,3) − ww (1,3) ] D k+1 −1 D k ∗ ) α [c(a  ) ] [1 − d k (d k )(1,3) − ww (1,3) ]. + [1 − d k (d k )(1,3) ]c(a 

Proof Since T =

Tk =

  a0 , then bd

   k  k    1 0 1 0 ak 0 a0 a 0 = = D k bd ) 1 w d k (d k )(1,3) c 1 [1 − d k (d k )1,3 ]c(a  c dk

:= P AQ. It is easy to check that

   D k k ) a ]w (1,3) (a D )k [1 − (a  ∈ A{1, 3}, and 0 (d k )(1,3)

P[I − A A(1,3) ] =

  D 1 − aa  0 = I − A A(1,3) , 0 1 − ww (1,3) − d k (d k )(1,3)

V = P ∗ P A A(1,3) + I − A A(1,3)   D k ∗ D k D k ∗ [c(a  ) ] [1 − d k (d k )(1,3) ]c(a  ) + 1 [c(a  ) ] ww (1,3) = D k [1 − d k (d k )(1,3) ]c(a  ) 1       D k ∗  (1,3) 1 0 α 0 1 [c(a ) ] ww = , D 0 1 [1 − d k (d k )(1,3) ]c(a  )k 1 0 1   k+2  D a a D + 1 − aa  0 , U = AQT P A A(1,3) + I − A A(1,3) = θ β D D D k where θ = d k (d k )(1,3) ca 2 a  + d k baa  + d k+1 [1 − d k (d k )(1,3) ]c(a  ) . D  By applying Proposition 4, T exists if and only if U and V are invertible. Since a D D exists with i(a) = k, then (a k+1 )# exists. Thus, a k+1 + 1 − aa  is invertible by [9, Corollary D D D 1]. Again, applying Lemma 3, a k+2 a  + 1 − aa  is invertible. Hence T  exists if and only if α and β are invertible. Moreover,

The pseudo core inverse of a lower triangular matrix

T

D 

= PU

−1

AQ P A A

(1,3)

V

−1

  t1 t2 . P = t3 t4 ∗



Fact 1 [6, Remark 1.3] Let a ∈ R. Then a is pseudo core invertible if and only if a ∗ is dual D ∗  pseudo core invertible. In this case, (a ∗ ) ) . D = (a D ∗ ∗  Evidently, the above expression can be reworded as b D = ((b ) ) . By applying Fact 1 to Propositions 3, 4 and Theorems 1, 2, 3, we can get the corresponding results concerning the dual pseudo core inverse. Acknowledgements The authors are highly grateful to the responsible editor and the anonymous referees for their valuable and helpful comments and suggestions, which led to great improvements of the paper. The first author is grateful to China Scholarship Council for supporting her further study in University of Minho, Portugal. Compliance with ethical standards Funding This research is supported by the National Natural Science Foundation of China (no. 11771076), the Scientific Innovation Research of College Graduates in Jiangsu Province (no. KYZZ16_0112), the Natural Science Foundation of Jiangsu Province (no. BK20141327).

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