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eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2020; 60(1): 71-72

Published online March 31, 2020

### Quasi-reversibility of the Ring of 2×2 Matrices over an Arbitrary Field

Dariush Heidari∗, Bijan Davvaz

Faculty of science, Mahallat Institute of Higher Education, Mahallat, Iran
e-mail : dheidari82@gmail.com
Department of Mathematics, Yazd University, Yazd, Iran
e-mail : davvaz@yazd.ac.ir

Received: September 14, 2019; Accepted: January 29, 2020

### Abstract

A ring R is quasi-reversible if 0 ≠ abI(R) for a, bR implies baI(R), where I(R) is the set of all idempotents in R. In this short paper, we prove that the ring of 2×2 matrices over an arbitrary field is quasi-reversible, which is an answer to the question given by Da Woon Jung et al. in [Bull. Korean Math. Soc., 56(4) (2019) 993–1006].

Keywords: quasi-reversible ring, matrix ring.

### 1. Introduction

Let R be a ring. Use I(R) to denote the set of all idempotents in R and I(R) = I(R){0}. Let Matn(R) be n×n matrix ring over R. Following Da Woon Jung et. al. [1] a ring R is quasi-reversible provided that if abI(R) for a, bR, then baI(R).

### Theorem 1.1

([1, Theorem 1.8]) Mat2(ℤ2) is quasi-reversible.

### 2. Main Result

Now, we propose an answer to Question 1 stated in Da Woon Jung et. al. [1].

### Question 2.1

Let K be a field. Is Mat2(K) quasi-reversible?

### Theorem 2.2

Let K be a filed. Then Mat2(K) is quasi-reversible.

Proof

Let K be a field and A,BMat2(K) such that ABI(Mat2(K)). If A is invertible, then

$BA=A-1(AB)A=A-1(ABAB)A=BABA.$

Hence BAI(Mat2(K)). Similarly, if B is invertible, then BA is idempotent. It remains to consider the case that A and B are non-invertible. To this end we prove the following claims:

Claim 1

If M = (mij) is a singular 2 × 2 matrix, then M2 = tr(M)M where, tr(M) = m11 + m22.

Proof of Claim 1

Since |M| = 0, it follows that m11m22 = m12m21. Consequently, we have

$M2=(m112+m12m21m11m12+m12m22m11m21+m21m12m12m21+m222)=(m112+m11m22m12(m11+m22)m21(m11+m22)m11m22+m222)=(m11+m22)M=tr(M)M.$

### Claim 2

If M = (mij) is a non-zero singular and idempotent 2 × 2 matrix, then tr(M) = 1.

Proof of Claim 2

By Claim 1, we have 0 ≠ M = M2 = tr(M)M thus (tr(M) – 1)M = 0 and hence tr(M) = 1.

Now, Let A and B be two non-invertible matrices such that ABI(Mat2(K)). Then, Claim 2 concludes tr(AB) = 1. Thus, by Claim 1 we deduce that

$(BA)2=tr(BA)BA=tr(AB)BA=BA.$

Therefore, BA is idempotent and the proof is completed.

### References

1. DW. Jung, CI. Lee, Y. Lee, S. Park, SJ. Ryu, HJ. Sung, and SJ. Yun. On reversibility related to idempotents. Bull Korean Math Soc., 56(4)(2019), 993-1006.