The core inverse and constrained matrix approximation problem

Hongxing Wang; Xiaoyan Zhang

doi:10.1515/math-2020-0178

Open Access Published by De Gruyter Open Access July 6, 2020

The core inverse and constrained matrix approximation problem

Hongxing Wang and Xiaoyan Zhang

From the journal Open Mathematics

https://doi.org/10.1515/math-2020-0178

Abstract

In this article, we study the constrained matrix approximation problem in the Frobenius norm by using the core inverse:

| | M x − b | | F = min subject to x ∈ ℛ ( M ) ,

where M ∈ ℂ n CM . We get the unique solution to the problem, provide two Cramer’s rules for the unique solution and establish two new expressions for the core inverse.

Keywords: core inverse; Cramer’s rule; constrained matrix approximation problem

MSC 2010: 15A24; 15A29; 15A57

1 Introduction

Let M ∗ , ℛ ( M ) and N ( M ) stand for the conjugate transpose, range space and null space of M ∈ ℂ m × n , respectively. The symbol M ( i → b ) denotes a matrix from M by replacing the ith column of M by b ∈ ℂ n . The symbol e i denotes the ith column of I n in which 1 ≤ i ≤ n . The Moore-Penrose inverse of M is the unique matrix X ∈ ℂ n × m satisfying the relations: M X M = M , X M X = X , ( M X ) ∗ = M X and ( X M ) ∗ = X M and is denoted by X = M † [1,2,3].

Let M ∈ ℂ n × n be singular. The smallest positive integer k for which rk ( M k + 1 ) = rk ( M k ) is called the index of M and is denoted by Ind ( M ) . The index of a non-singular matrix is 0 and the index of a null matrix is 1. Furthermore,

(1.1) ℂ n CM = { M | Ind ( M ) ≤ 1 , M ∈ ℂ n × n } .

Let M ∈ ℂ n × n with Ind ( M ) = k . A matrix X is the Drazin inverse of M if M X M k = M k , X M X = X and M X = X M . We write X = M D for the Drazin inverse of M. In particular, when M ∈ ℂ n CM , the matrix X is the group inverse of M and is denoted by X = M # [1,2,3].

The core inverse of M ∈ ℂ n CM is defined as the unique matrix X ∈ ℂ n × n satisfying the equations: M X M = M , M X 2 = X and ( M X ) ∗ = M X and is denoted by X = M ⊛ [4,5]. It is noteworthy that the core inverse is a “least squares” inverse [6,7]. Moreover, it is proved that M ⊛ = M # M M † [4].

Recently, the relevant conclusions of the core inverse are very rich. In [7,8,9,10], generalizations of core inverse are introduced, for example, the core-EP inverse and the weak group inverse. In [11,12,13,14,15], their algebraic properties and calculating methods are studied. In [16,17], the studying of them is extended to some new fields, for example, ring and operator. Moreover, those inverses are used to study partial orders in [4,5,10,18,19].

Consider the following equation:

(1.2) M x = b .

Let M ∈ ℂ n × n with Ind ( M ) = k and b ∈ ℛ ( M k ) . Campbell and Meyer [20] show that x = M D b is the unique solution of (1.2) with respect to x ∈ ℛ ( M k ) . Wei [21] gets the minimal P-norm solution of (1.2), where P is nonsingular, P − 1 M P is the Jordan canonical form of M and ∥ x ∥ p = ∥ P − 1 x ∥ 2 . Furthermore, let M ∈ ℂ m × n . Wei [22] considered the unique solution of

W M W x = b subject to x ∈ ℛ ( ( M W ) k 1 ) ,

where W ∈ ℂ n × m , k 1 = Ind ( M W ) , k 2 = Ind ( W M ) and b ∈ ℛ ( ( W M ) k 2 ) . More results of (1.2) under some certain conditions can be found in [3,21,23,24,25,26,27].

It is well known that b ∈ ℛ ( M ) if and only if (1.2) is solvable. Let b ∈ ℛ ( M ) and the index of M is 1, then x = M # b is the unique solution with x ∈ ℛ ( M ) [20]. It follows from M ⊛ = M # M M † that M # b = M ⊛ b [28]. Furthermore, the unique solution x = M ⊛ b is given by Cramer’s rule [28, Theorem 3.3].

When b ∉ ℛ ( M ) , (1.2) is unsolvable, yet, it has least-squares solutions. Motivated by the aforementioned works, it is natural to consider the least-squares solutions of (1.2) under the certain condition x ∈ ℛ ( M ) , i.e.,

(1.3) ∥ M x − b ∥ F = min subject to x ∈ ℛ ( M ) ,

where M ∈ ℂ n CM , rk ( M ) = r < n and b ∈ ℂ n .

2 Preliminaries

Lemma 2.1

[1] Let M ∈ ℂ n × n be idempotent. Then, M = P ℛ ( M ) , N ( M ) with ℛ M ⊕ N M = ℂ n . In contrast, if F ⊕ G = ℂ n , then there exists an idempotent P F , G such that ℛ ( P F , G ) = F and N ( P F , G ) = G .

Furthermore, I n − P F , G = P G , F .

Lemma 2.2

[3] Let M ∈ ℂ n × n . Then, Ind ( M ) = k if and only if

(2.1) ℛ ( M k ) ⊕ N ( M k ) = ℂ n .

Lemma 2.3

[3] Let M X M = M and X M X = X . Then,

X M = P ℛ ( X ) , N ( M ) and M X = P ℛ ( M ) , N ( X ) .

Lemma 2.4

[3] Let F ⊕ G = ℂ n . Then,

P F , G M = M ⇔ ℛ ( M ) ⊆ F ;
M P F , G = M ⇔ N ( M ) ⊇ G .

Lemma 2.5

[14] Let M ∈ ℂ n CM with rk ( M ) = r . Then, there exists a unitary matrix V such that

(2.2) M = V T S 0 0 V ∗ ,

where T ∈ ℂ r × r is nonsingular. Furthermore,

(2.3) M ⊛ = V T − 1 0 0 0 V ∗ .

3 Main results

3.1 Solution of (1.3)

Theorem 3.1

Let M ∈ ℂ n CM and b ∈ ℂ n . Then,

(3.1) x = M ⊛ b

is the unique solution of ( 1.3 ).

Proof

From x ∈ ℛ ( M ) , it follows that there exists y ∈ ℂ n for which x = M y . Then, x is the solution of (1.3) if and only if y is the solution of

(3.2) ∥ M 2 y − b ∥ F = min .

Let the decomposition of M be as in (2.2). Denote

(3.3) V ⁎ y = y 1 y 2 , V ⁎ b = b 1 b 2 and M ⊛ b = V T − 1 b 1 0 ,

where y 1 , b 1 and T − 1 b 1 ∈ ℂ rk ( M ) . It follows that

∥ M x − b ∥ F 2 = T 2 y 1 + T S y 2 − b 1 − b 2 F 2 = ∥ T 2 y 1 + T S y 2 − b 1 ∥ F 2 + ∥ b 2 ∥ F 2 .

Since T is invertible, we have min y 1 , y 2 ∥ T 2 y 1 + T S y 2 − b 1 ∥ F 2 = 0 , that is, ∥ M 2 y − b ∥ F = min = ∥ b 2 ∥ F , in which y 2 ∈ ℂ n − rk ( M ) is arbitrary, and y 1 = T − 2 b 1 − T − 1 S y 2 . It follows that

x = M y = V T S 0 0 V ⁎ y = V T y 1 + S y 2 0 = V T − 1 b 1 0 = M ⊛ b ,

that is, (3.1) is the unique solution of (1.3).□

3.2 Determinantal formulas

When M ∈ ℂ n × n is nonsingular, it is well known that the solution of (1.2) is unique and x = M − 1 b . Let x = ( x 1 , x 2 , … , x n ) T . Then,

(3.4) x i = det ( M ( i → b ) ) det ( M ) , i = 1 , 2 , … , n ,

is called Cramer’s rule for solving (1.2). In [29], Ben-Israel gets a Cramer’s rule for obtaining the least-norm solution of the consistent linear system (1.2),

x i = det M ( i → b ) U V ∗ ( i → 0 ) 0 det M U V ∗ 0 , i = 1 , 2 , … , n ,

where U and V are of full column rank, ℛ ( U ) = N ( M ∗ ) and ℛ ( V ) = N ( M ) . In [26], Wang gives a Cramer’s rule for the unique solution x ∈ ℛ ( M k ) of (1.2), where b ∈ ℛ ( M k ) and Ind ( M ) = k . In [30], Ji proposes two new condensed Cramer’s rules for the unique solution x ∈ ℛ ( M k ) of (1.2), where b ∈ ℛ ( M k ) and Ind ( M ) = k . More details of Cramer’s rules for finding restricted solutions of (1.2) can be found in [1,3,31,32,33,34,35,36]. In Theorems 3.4 and 3.6, we will give two Cramer’s rules for the unique solution of (1.3).

First of all, we give the following two lemmas to prepare for a Cramer’s rule for core inverse in Theorem 3.4.

Lemma 3.2

Let M ∈ ℂ n CM with rk ( M ) = r , and let L ∈ ℂ n × ( n − r ) with rk ( L ) = n − r and ℛ ( L ) = N ( M ∗ ) . Then,

(3.5) M ⊛ M + ( I n − M ⊛ M ) L ( L ∗ L ) − 1 L ∗ = I n .

Proof

Let M be as in (2.2), applying Lemma 2.2, we see that

(3.6) ℛ ( M ) ⊕ N ( M ) = ℂ n .

Denote M 1 = I n − M ⊛ M and M 2 = L ( L ∗ L ) − 1 L ∗ .

Applying Lemmas 2.1, 2.3 and M ⊛ M = M # M , we have

(3.7) M ⊛ M = P ℛ ( M ) , N ( M ) ,

(3.8) M 1 = I − M ⊛ M = P N ( M ) , ℛ ( M ) .

Since ( L ( L ∗ L ) − 1 ) L ∗ ( L ( L ∗ L ) − 1 ) = L ( L ∗ L ) − 1 and L ∗ ( L ( L ∗ L ) − 1 ) L ∗ = L ∗ , applying Lemma 2.3, we obtain

(3.9) M 2 = P ℛ ( L ) , N ( L ⁎ ) = P N ( M ⁎ ) , ℛ ( M ) .

Since ℛ ( L ) = N ( M ∗ ) , we obtain M 2 M 1 = M 2 and

(3.10) M 1 M 2 = P N ( M ) , ℛ ( M ) .

Therefore, applying Lemma 2.1, (3.7) and (3.10), we gain

M ⊛ M + M 1 M 2 = P ℛ ( M ) , N ( M ) + P N ( M ) , ℛ ( M ) = I n ,

i.e., (3.5).□

In [28, Theorems 3.2 and 3.3], let M ∈ ℂ n C M , b ∈ ℂ n and b ∈ ℛ ( M ) , and let M b and M c be of the full column ranks with N ( M ∗ ) = ℛ ( M b ) and N ( M c ∗ ) = ℛ ( M ) . Then,

M M b M c ∗ 0

is invertible and the unique solution x = M ⊛ b of (1.2) satisfying

x i = det M i → b M b M c ∗ i → 0 0 / det M M b M c ∗ 0 ,

where i = 1 , 2 , … , n . In Lemma 3.3 and Theorem 3.4, we give the unique least-squares solution of (1.3) in a similar way.

Lemma 3.3

Let M and L be as in Lemma 3.2. Then,

(3.11) G = M L L ∗ 0

is invertible and

(3.12) G − 1 = M ⊛ ( I n − M ⊛ M ) L ( L ∗ L ) − 1 ( L ∗ L ) − 1 L ∗ 0 .

Proof

Since ℛ ( L ) = N ( M ∗ ) and M ⊛ = M # M M † , we have M ⊛ L = M # M M † L = 0 and ( L ∗ L ) − 1 L ∗ M = 0 . Furthermore, applying (3.5), we have

M ⊛ ( I n − M ⊛ M ) L ( L ∗ L ) − 1 ( L ∗ L ) − 1 L ∗ 0 M L L ∗ 0 = M ⊛ M + ( I n − M ⊛ M ) L ( L ∗ L ) − 1 L ∗ M ⊛ L ( L ∗ L ) − 1 L ∗ M ( L ∗ L ) − 1 L ∗ L = I 2 n − r ,

that is, G is invertible and G − 1 is of the form (3.12).□

Based on Lemmas 3.2 and 3.3, we get a Cramer’s rule for the unique solution of (1.3).

Theorem 3.4

Let M and b be as in Lemma 3.2, and let L be as in Lemma 3.2. Then, (1.3) has the unique solution x = ( x 1 , x 2 , … , x n ) T satisfying

(3.13) x i = det M ( i → b ) L L ∗ ( i → 0 ) 0 / det M L L ∗ 0 ,

where i = 1 , 2 , … , n .

Proof

Since G is invertible, applying Lemma 3.3, we get the unique solution x ˆ = G − 1 b ˆ of G x ˆ = b ˆ , in which x ˆ ∗ = [ x ∗ y ∗ ] ∗ and b ˆ ∗ = b ∗ 0 ∗ . It follows from (3.12) that

x y = M ⊛ ( I n − M ⊛ M ) L ( L ∗ L ) − 1 ( L ∗ L ) − 1 L ∗ 0 b 0 = M ⊛ b ( L ∗ L ) − 1 L ∗ b .

Applying (3.4) we obtain (3.13).□

In the following theorem, we give a characterization of the core inverse and prepare for a Cramer’s rule for the core inverse in Theorem 3.6.

Theorem 3.5

Let M and L be as in Lemma 3.2. Then,

(3.14) M ⊛ = ( M M ∗ M + L L ∗ ) − 1 M M ∗ .

Proof

Since ℛ ( L ) = N ( M ∗ ) , M ∈ ℂ n CM and ℛ ( M ) ⊥ = N ( M ∗ ) , we obtain

( L L ∗ ) ( L L ∗ ) † = P N ( M ∗ ) , ℛ ( M ) , ( M M ∗ M ) ( M M ∗ M ) ⊛ = P ℛ ( M ) , N ( M ∗ )

and

( M M ⁎ M + L L ⁎ ) ( ( M M ⁎ M ) ⊛ + ( L L ∗ ) † − ( M M ∗ M ) ⊛ M M ∗ M ( L L ∗ ) † ) = ( M M ∗ M ) ( M M ∗ M ) ⊛ + ( L L ∗ ) ( L L ∗ ) † = P ℛ ( M ) , N ( M ∗ ) + P N ( M ⁎ ) , ℛ ( M ) = I n .

Therefore, M M ∗ M + L L ∗ is invertible.

Since ( L L ∗ ) † M M ∗ = 0 and ( M M ∗ M ) ⊛ M M ∗ = M ⊛ , we have

( M M ∗ M + L L ∗ ) − 1 M M ∗ = ( M M ∗ M ) ⊛ M M ∗ + ( L L ∗ ) † M M ∗ = M ⊛ .

It follows that we get (3.14).□

Theorem 3.6

Let M and L be as in Lemma 3.2. Then, (1.3) has the unique solution x = ( x 1 , x 2 , … , x n ) T satisfying

(3.15) x j = det ( M M ∗ M + L L ∗ ) ( j → M M ∗ b ) det ( M M ∗ M + L L ∗ ) ,

where j = 1 , 2 , … , n .

Proof

Applying Theorems 3.5 to 3.1, we have

x = ( M M ∗ M + L L ∗ ) − 1 M M ∗ b ,

that is,

( M M ∗ M + L L ∗ ) x = M M ∗ b .

It follows from (3.4) that we get (3.15).□

In [30], Ji obtains the condensed determinantal expressions of M † and M D . By using Theorem 3.5, we get a condensed determinantal expression of M ⊛ .

Theorem 3.7

Let M and L be defined as in (3.11). Then, the core inverse M ⊛ is given by:

(3.16) M i , j ⊛ = det ( M M ∗ M + L L ∗ ) ( i → ( M M ∗ ) e j ) det ( M M ∗ M + L L ∗ ) ,

where 1 ≤ i , j ≤ n .

Proof

Since M M ∗ M + L L ∗ is invertible, we consider

( M M ∗ M + L L ∗ ) x = ( M M ∗ ) e j

and get the solution

e i T x = det ( M M ∗ M + L L ∗ ) ( i → ( M M ∗ ) e j ) det ( M M ∗ M + L L ∗ ) ,

in which i , j = 1 , … , n .

It follows from (3.14) and M i , j ⊛ = e i T M ⊛ e j that we get (3.16).□

3.3 Examples

In the following examples, we show that our results are effective.

Example 3.1

Let M = 1 2 2 0 0 0 0 0 0 , L = 0 0 1 0 0 1 and b = 1 1 1 . It is easy to check that ℛ ( L ) = N ( M ∗ ) . By applying Lemma 3.3, we have M ⊛ = 1 0 0 0 0 0 0 0 0 . Then, ( I n − M ⊛ M ) L ( L ∗ L ) − 1 = − 2 − 2 1 0 0 1 , ( L ∗ L ) − 1 L ∗ = 0 1 0 0 0 1 , G = 1 2 2 0 0 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 , det ( G ) = 1 and G − 1 = 1 0 0 − 2 − 2 0 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 1 0 0 .

By applying Theorem 3.1, we get the solution of (1.3) is x = M ⊛ b = 1 0 0 .

For det 1 0 0 − 2 − 2 1 0 0 1 0 1 0 0 0 1 0 1 0 0 0 0 0 1 0 0 = 1 and det 1 1 2 0 0 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 0 0 1 0 0 = det 1 2 1 0 0 0 0 1 1 0 0 0 1 0 1 0 1 0 0 0 0 0 0 0 0 = 0 , by applying Theorem 3.4, we get x 1 = 1 1 , x 2 = 0 1 and x 3 = 0 1 . Therefore, the solution of (1.3) is x = 1 0 0 .

For det ( M M ∗ M + L L ∗ ) = det 9 18 18 0 1 0 0 0 1 = 9 , M M ∗ b = 9 0 0 ,

det ( M M ∗ M + L L ∗ ) ( 1 → M M ∗ b ) = det 9 18 18 0 1 0 0 0 1 = 9 ,

det ( M M ∗ M + L L ∗ ) ( 2 → M M ∗ b ) = 0 , det ( M M ∗ M + L L ∗ ) ( 3 → M M ∗ b ) = 0 , and by applying Theorem 3.6, we get x 1 = 9 9 , x 2 = 0 9 and x 3 = 0 9 . Therefore, the solution of (1.3) is x = 1 0 0 .

Example 3.2

Let

M = 1 / 4 1 / 8 1 / 8 1 / 4 1 / 8 1 / 8 1 / 4 0 1 / 4 .

Then,

M # = 1 3 − 2 1 3 − 2 1 − 5 6 , M # = 4 / 3 4 / 3 0 8 / 3 8 / 3 − 4 − 4 / 3 − 4 / 3 4 , M ⊛ = 2 2 − 2 2 2 − 2 − 2 − 2 6

and

L = 1 − 1 0

with rk ( L ) = n − r and ℛ ( L ) = N ( M ∗ ) . It is easy to check that

M M ∗ M + L L ∗ = 137 / 128 − 125 / 128 3 / 64 − 119 / 128 131 / 128 3 / 64 5 / 64 3 / 128 7 / 128 .

By applying Theorem 3.5, we get

( M M ∗ M + L L ∗ ) − 1 M M ∗ = 2 2 − 2 2 2 − 2 − 2 − 2 6 = M ⊛ .

For

det ( M M ∗ M + L L ∗ ) = 3 / 4096 , det ( M M ∗ M + L L ∗ ) ( 1 → ( M M ∗ ) e 1 ) = 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 1 → ( M M ∗ ) e 2 ) = 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 1 → ( M M ∗ ) e 3 ) = − 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 2 → ( M M ∗ ) e 1 ) = 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 2 → ( M M ∗ ) e 2 ) = 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 2 → ( M M ∗ ) e 3 ) = − 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 3 → ( M M ∗ ) e 1 ) = − 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 3 → ( M M ∗ ) e 2 ) = − 3 / 2048 , det ( M M ∗ M + L L ∗ ) ( 3 → ( M M ∗ ) e 3 ) = 9 / 2048 ,

by applying Theorem 3.7, we get

M 11 = 2 , M 12 = 2 , M 13 = − 2 , M 21 = 2 , M 22 = 2 , M 23 = − 2 , M 31 = − 2 , M 32 = − 2 , M 33 = 6 ,

that is,

M ⊛ = 2 2 − 2 2 2 − 2 − 2 − 2 6 .

Acknowledgments

Hongxing Wang was supported partially by the Guangxi Natural Science Foundation (grant number 2018GXNSFAA138181), the Special Fund for Science and Technological Bases and Talents of Guangxi (grant number GUIKE AD19245148), the Xiangsihu Young Scholars Innovative Research Team of Guangxi University for Nationalities (grant number 2019RSCXSHQN03) and the Special Fund for Bagui Scholars of Guangxi (grant number 2016A17). Xiaoyan Zhang was supported partially by the National Natural Science Foundation of China (grant number 11361009) and High Level Innovation Teams and Distinguished Scholars in Guangxi Universities (grant number GUIJIAOREN201642HAO).

Conflict of interest: The authors report no potential conflict of interest.

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Received: 2019-08-30

Revised: 2020-03-07

Accepted: 2020-04-28

Published Online: 2020-07-06

This work is licensed under the Creative Commons Attribution 4.0 International License.

The core inverse and constrained matrix approximation problem

Abstract

1 Introduction

2 Preliminaries

Lemma 2.1

Lemma 2.2

Lemma 2.3

Lemma 2.4

Lemma 2.5

3 Main results

3.1 Solution of (1.3)

Theorem 3.1

Proof

3.2 Determinantal formulas

Lemma 3.2

Proof

Lemma 3.3

Proof

Theorem 3.4

Proof

Theorem 3.5

Proof

Theorem 3.6

Proof

Theorem 3.7

Proof

3.3 Examples

Example 3.1

Example 3.2

Acknowledgments

References

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