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

Published online March 31, 2020

Copyright © Kyungpook Mathematical Journal.

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

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.

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.

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.

  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.