Singular matrices possessing the triangle property

K. Kranthi Priya; K. C. Sivakumar

doi:10.1515/spma-2023-0112

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Singular matrices possessing the triangle property

K. Kranthi Priya and K. C. Sivakumar

Published/Copyright: January 29, 2024

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From the journal Special Matrices Volume 12 Issue 1

Abstract

It is known that the inverse of an invertible real square matrix satisfying the triangle property is a tridiagonal matrix. In this note, results that may be considered as analogues of this assertion are obtained for the Moore-Penrose inverse and the group inverse.

Keywords: triangle property; tridiagonal matrix; pentadiagonal matrix; group inverse; Moore-Penrose inverse

MSC 2010: 15A03; 15A09

1 Introduction

Let M m × n ( R ) denote the set of all matrices of order m × n with real entries. When m = n , we denote M m × n ( R ) by M n ( R ) . A matrix A = ( a i j ) ∈ M n ( R ) is said to be tridiagonal if a i j = 0 for ∣ i − j ∣ > 1 and pentadiagonal if ∣ i − j ∣ > 2 . Let A ∈ M n ( R ) be such that (a part of whose diagonal elements) a 22 , a 33 , … , a ( n − 1 ) ( n − 1 ) are nonzero. We say that A has the triangle property, if

a i j = a i k a k j a k k ,

for all i < k < j and i > k > j .

The motivation for this short note is an interesting and perhaps well-known characterization of inverses of matrices with the triangle property. We state it formally.

Theorem 1.1

[2, Theorem 1] Let A = ( a i j ) ∈ M n ( R ) be nonsingular and be such that the entries a 22 , a 33 , … , a ( n − 1 ) ( n − 1 ) are nonzero. Then A has triangle property iff A − 1 is tridiagonal.

In this note, we consider the question of whether analogous results hold for two of the most prominent classes of generalized inverses, viz., the Moore-Penrose inverse and the group inverse. First, we show that a verbatim statement for these two generalized inverses is false. We then identify a class of singular matrices with triangle property for which these generalized inverses turn out to be pentadiagonal.

It is pertinent to point to a recent work [3], where the question of when the Moore-Penrose inverse of a matrix is tridiagonal is studied. The results there are expressed in terms of the notion of semiseparability of matrices. The results of this note may be considered as complementary to those of [3].

2 Preliminaries

For a given A ∈ M n ( R ) , let R ( A ) , N ( A ) , and A T denote the range space, the null space, and the transpose of A , respectively. The Moore-Penrose inverse of A ∈ M m × n ( R ) is denoted by A † and is defined as the unique matrix X ∈ M n × m ( R ) that satisfies the equations A X A = A , X A X = X , ( A X ) T = A X , and ( X A ) T = X A . For a given A ∈ M n ( R ) , the equations A X A = A , X A X = X and A X = X A may or may not be satisfied by any matrix X ∈ M n ( R ) . However, when such a matrix exists, it is unique and called the group inverse of A and is denoted by A # . The group inverse A # , given A , exists iff rk ( A ) = rk ( A 2 ) , where “ rk ” stands for the rank of a matrix. Another characterization for its existence is that R ( A ) and N ( A ) are complementary subspaces. This in turn holds iff N ( A ) = N ( A 2 ) . Of course, when A is square and invertible, then A † = A # = A − 1 .

A tool that will be used here is the notion of the full-rank factorization. Consider a matrix A ∈ M n ( R ) with rk ( A ) = r > 0 . We say that A has a full-rank factorization A = F G , if there exist F ∈ M n × r ( R ) , G ∈ M r × n ( R ) such that rk ( F ) = rk ( G ) = r . It is well known that any nonzero matrix has a full-rank factorization. Let A = F G be a full-rank factorization of A . Then,

A † = G † F † = G T ( G G T ) − 1 ( F T F ) − 1 F T .

Furthermore, a necessary and sufficient condition for the existence of the group inverse A # is that G F is invertible. In such a case, we have

A # = F ( G F ) − 2 G .

For a comprehensive treatment on generalized inverses, their properties and applications, we recommend the treatise [1].

3 Main results

First, we show that verbatim versions of Theorem 1.1 are false for the Moore-Penrose and the group generalized inverses.

Example 3.1

A = 1 2 3 − 3 1 2 3 − 3 2 4 − 1 1 4 8 − 2 2 is a singular matrix endowed with the triangle property. It may be verified that

A † = 1 350 5 5 6 12 10 10 12 24 25 25 − 5 − 10 − 25 − 25 5 10 ,

and that (the group inverse exists and is given by)

A # = 1 63 1 2 − 9 9 1 2 − 9 9 − 6 − 12 − 9 9 − 12 − 24 − 18 18 ,

showing that neither the group inverse nor the Moore-Penrose inverse is tridiagonal.

Note from the definition that any principal submatrix of a matrix with triangle property inherits such a property. Let e n denote the vector with n coordinates, whose n th coordinate is 1.

In the next two results, we show in the presence of an assumption that the Moore-Penrose inverse and the group inverse (when it exists), of a singular matrix with triangle property, are pentadiagonal.

Theorem 3.2

Let A = ( a i j ) ∈ M n + 1 ( R ) be a singular matrix with triangle property such that the leading principal submatrix of A has rank n. Then A † is a pentadiagonal matrix.

Proof

We partition A as

A = B c d T a ( n + 1 ) ( n + 1 ) ,

where B ∈ R n × n is nonsingular and has the triangle property, and c , d ∈ R n . Set β n ≔ a n ( n + 1 ) a n n and α n ≔ a ( n + 1 ) n a n n . By the triangle property of A , we have

c = ( a 1 ( n + 1 ) , a 2 ( n + 1 ) , … , a n ( n + 1 ) ) T = a 1 n a n ( n + 1 ) a n n , a 2 n a n ( n + 1 ) a n n , … , a n n a n ( n + 1 ) a n n T = a n ( n + 1 ) a n n ( a 1 n , a 2 n , … , a n n ) T = β n B e n ,

where we have used the fact that the last column of the matrix B is ( a 1 n , a 2 n , … , a n n ) T . In a similar manner (by taking the transpose of A , for instance), we have

d = α n B T e n .

Since A singular and B is nonsingular, it follows that

a ( n + 1 ) ( n + 1 ) = d T B − 1 c = β n d T e n .

Define

F = B d T = I α n e n T B

and

G = I , B − 1 c = I , β n e n .

Then, A = F G is a full rank factorization. We then have

(3.1) F T F = B T I , α n e n I α n e n T B = B T ( I + α n 2 e n e n T ) B = B T D 1 B ,

where D 1 = I + α n 2 e n e n T is an invertible diagonal matrix. Since A = F G is a full rank factorization, F T F is invertible and one has

( F T F ) − 1 = B − 1 D 1 − 1 ( B T ) − 1 .

Now,

(3.2) F † = ( F T F ) − 1 F T = B − 1 D 1 − 1 ( B T ) − 1 B T I , α n e n = B − 1 D 1 − 1 I , α n e n .

Also,

(3.3) G G T = I , β n e n I β n e n T = I + β n 2 e n e n T .

Set D 2 ≔ I + β n 2 e n e n T . Then D 2 is also an invertible diagonal matrix. Thus,

(3.4) G † = G T ( G G T ) − 1 = I β n e n T D 2 − 1 .

Thus, the Moore-Penrose inverse of A is given by

(3.5) A † = G † F † = I β n e n T D 2 − 1 B − 1 D 1 − 1 I , α n e n = D 2 − 1 B − 1 D 1 − 1 α n D 2 − 1 B − 1 D 1 − 1 e n β n e n T D 2 − 1 B − 1 D 1 − 1 α n β n e n T D 2 − 1 B − 1 D 1 − 1 e n .

Now, B is a non-singular matrix with triangle property and so B − 1 is a tridiagonal matrix. Since D 1 − 1 , D 2 − 1 are diagonal matrices, it follows that the leading principal submatrix of A † , viz., D 2 − 1 B − 1 D 1 − 1 is also a tridiagonal matrix. The trailing principal submatrix is just a number. We observe that D 2 − 1 B − 1 D 1 − 1 e n ∈ R n . Since D 2 − 1 B − 1 D 1 − 1 is tridiagonal, the first n − 2 coordinates of the column vector α n D 2 − 1 B − 1 D 1 − 1 e n are zero. The property of being tridiagonal remains the same upon transposition and so an entirely similar argument shows that the first n − 2 coordinates of the row vector β n e n T D 2 − 1 B − 1 D 1 − 1 are zero.

Thus, A † is a pentadiagonal matrix.□

In what follows, we show that an analogous result holds for the group inverse after presenting a characterization for its existence.

Theorem 3.3

Let A ∈ M ( n + 1 ) ( R ) be a singular matrix such that the leading principal submatrix of A of order n × n is invertible. Suppose that A has the triangle property. Then A # exists iff a n n + a ( n + 1 ) ( n + 1 ) ≠ 0 . In this case, A # is a pentadiagonal matrix.

Proof

We begin with the full rank factorization of A , as in the proof of Theorem 3.2, so that F = I α n e n T B and G = I , β n e n . Then,

(3.6) G F = I , β n e n I α n e n T B = ( I + α n β n e n e n T ) B = D B ,

where D = I + α n β n e n e n T = diag ( 1 , 1 , … , 1 + α n β n ) . Since B is invertible, the matrix G F is invertible iff D is invertible, which in turn holds iff 1 + α n β n ≠ 0 .

By observing that the last coordinate of d is a ( n + 1 ) n , we have

a ( n + 1 ) ( n + 1 ) = β n d T e n = β n a ( n + 1 ) n = a n ( n + 1 ) a ( n + 1 ) n a n n ,

so that

a n n a ( n + 1 ) ( n + 1 ) = a n ( n + 1 ) a ( n + 1 ) n .

Thus, G F is invertible iff

0 ≠ 1 + α n β n = 1 + a ( n + 1 ) n a n ( n + 1 ) a n n 2 = 1 + a n n a ( n + 1 ) ( n + 1 ) a n n 2 = 1 + a ( n + 1 ) ( n + 1 ) a n n = 1 a n n ( a n n + a ( n + 1 ) ( n + 1 ) ) .

We may now conclude that A # exists iff a n n + a ( n + 1 ) ( n + 1 ) ≠ 0 . This proves the first part.

Next, assume that A # exists. Then, D is invertible and so we have

( G F ) − 1 = B − 1 D − 1 .

Note that D − 1 e n = 1 1 + α n β n e n and so B − 1 D − 1 e n = 1 1 + α n β n B − 1 e n . Then,

( G F ) − 1 G = B − 1 D − 1 ( I , β n e n ) = B − 1 D − 1 , β n 1 + α n β n B − 1 e n = ( B − 1 D − 1 , γ n B − 1 e n ) ,

where γ n ≔ β n 1 + α n β n .

Thus,

(3.7) A # = F ( G F ) − 1 ( G F ) − 1 G = ( F B − 1 ) D − 1 ( B − 1 D − 1 , γ n B − 1 e n ) = I α n e n T ( D − 1 B − 1 D − 1 , γ n D − 1 B − 1 e n ) = D − 1 B − 1 D − 1 γ n D − 1 B − 1 e n α n e n T D − 1 B − 1 D − 1 α n γ n e n T D − 1 B − 1 e n .

An entirely similar argument as in the proof of Theorem 3.2 applies here, as well. This completes the proof that A # is pentadiagonal.□

Next, we provide illustrative examples.

Example 3.4

Let B = 1 2 2 2 3 − 2 4 6 − 1 2 2 , c = − ( 3 , 6 , 2 ) T , and d = 1 2 ( − 1 , 2 , 2 ) T . Then B has the triangle property. If A = B c d T a 44 , then A = 1 2 2 2 3 − 6 − 2 4 6 − 12 − 1 2 2 − 4 − 1 2 2 − 4 has the triangle property and a 44 = d T B − 1 c , ensuring that A is singular. The Moore-Penrose inverse of A is given by

A † = 1 30 20 − 10 0 0 10 − 35 45 45 0 6 − 6 − 6 0 − 12 12 12 ,

whereas the group inverse (note that a 33 + a 44 ≠ 0 ) is

A # = 1 6 4 − 2 0 0 2 − 7 − 18 36 0 − 6 − 12 24 0 − 6 − 12 24 ,

both being pentadiagonal matrices.

The maximal rank condition of Theorem 3.2 or 3.3 is indispensable. This is evident from the details provided in Example 3.1.

Remark 3.5

The formula for the inverse of an invertible matrix satisfying the triangle property was given in [2]. In conjunction with that formula, we may compute the entries of the group inverse and the Moore-Penrose inverse of a singular matrix with the triangle property, using the - given in the proofs of Theorems 3.2 and 3.3, when the hypotheses apply.

Acknowledgement

The authors thank A.M. Encinas of the Polytechnic University of Barcelona, Spain for his suggestions and comments on the draft version.

Conflict of interest: The authors state no conflict of interest.
Data availability statement: There are no data associated with the study undertaken here.

References

[1] A. Ben-Israel and T. N. E. Greville, Generalized inverses - theory and applications, CMS Books in Mathematics vol. 15. Springer-Verlag, New York, 2003. Search in Google Scholar

[2] W. Barrett, A theorem on inverses of tridiagonal matrices, Linear Algebra Appl. 27 (1979), 211–217. 10.1016/0024-3795(79)90043-0Search in Google Scholar

[3] M. I. Bueno and S. Furtado, Singular matrices whose Moore-Penrose inverse is tridiagonal, Appl. Math. Comput. 459 (2023), 128154. 10.1016/j.amc.2023.128154Search in Google Scholar

Received: 2023-11-11

Accepted: 2023-12-08

Published Online: 2024-01-29

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

Articles in the same Issue

https://doi.org/10.1515/spma-2023-0112

Keywords for this article

triangle property; tridiagonal matrix; pentadiagonal matrix; group inverse; Moore-Penrose inverse

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