Characterizations of EP，normal and Hermitian elements in rings using generalized inverses

(1School of Economics and Management, Southeast University, Nanjing 211189, China)(2School of Mathematics, Southeast University, Nanjing 211189, China)

Abstract：The properties and some equivalent characterizations of equal projection (EP), normal and Hermitian elements in a ring are studied by the generalized inverse theory. Some equivalent conditions that an element is EP under the existence of core inverses are proposed. Leta∈R

, then a is EP if and only if aa

a#=a#aa

. At the same time, the equivalent characterizations of a regular element to be EP are discussed. Let a∈R, then there exist b∈R such that a=aba and a is EP if and only if a∈R

, a

ba. Similarly, some equivalent conditions that an element is normal under the existence of core inverses are proposed. Let a∈R

, then a is normal if and only if a*a

a*. Also, some equivalent conditions of normal and Hermitian elements in rings with involution involving powers of their group and Moore-Penrose inverses are presented. Let a∈R

∩R#, n∈N, then a is normal if and only if a*a

(a#)n=a#a*(a

)n. The results generalize the conclusions of Mosi

et al.

Key words：equal projection (EP) elements; normal elements; Hermitian elements; core inverse; Moore-Penrose inverse; group inverse

Let R be an associative ring with unity. An involution * is an anti-automorphism of degree 2 in R, i.e., (a*)*=a, (a+b)*=a*+b* and (ab)*=b*a* for all a,b∈R. Let R be a ring with involution and a∈R, we say that x∈R is the Moore-Penrose inverse (MP-inverse for short) of a, if the following equations hold:

Under this condition, x is unique if it exists, which is denoted by a

. The set of all MP-invertible elements of R is denoted by R

. If a satisfies the equation axa=a, then, we state that a is regular. An element a∈R has a group inverse if there exists x∈R such that

Then x is called a group inverse of a and it is unique when it exists, denoted by a#. We use R# to represent the set of all group invertible elements of R.

In Ref.[1], Baksalary and Trenkler introduced the core inverse of complex matrices. Later, Raki

et al.[2] generalized the core inverse of complex matrices to the case of elements in rings, and they used five equations to characterize the core inverse of elements, i.e., let a∈R, if there exist x∈R satisfying

then x is the core inverse of a. In this case, x is unique if it exists, denoted by a

.The set of all core invertible elements of R is denoted by R

. They also showed that each core invertible element is group invertible. It is clear to see that if a∈R

∩R#, then, we have a∈R

, and the converse does not hold. Recently, Xu et al.[3] proved that a

=x if and only if

An element a∈R satisfying a*a=aa* is called normal. If a∈R such that a*=a, then, a is called Hermitian (resp. symmetric). Let n∈N, an element a∈R satisfying a*an=ana* is called generalized normal. Any a∈R satisfying an=(a*)n is called generalized Hermitian. An element a∈R is called EP (equal projection) if a∈R

∩R# and a

=a#. The set of all EP elements of R is denoted by REP.

In Refs.[4-8], the authors focused on EP matrices or EP linear operators on Banach or Hilbert spaces. Later, Mosi

et al.[9] presented some characterizations, for example, elements of rings are EP by using their group inverses and MP-inverses, which extended the results in Refs.[4-7]. They showed that if a∈R

∩R#, then, a is EP if and only if aa

a#=a#aa

. Motivated by them, we consider similar conditions for elements in rings in terms of their core inverses. For example, we prove that if a∈R

, then, a is EP if and only if aa

a#=a#aa

. Moreover, Mosi

et al.[10] also considered the necessary and sufficient conditions for regular elements to be EP. They showed that a is EP with a=aba if and only if a∈R

∩R#, a

ba. Inspired by them, we investigate similar cases for elements in rings by their core inverses. For example, we present that a is EP with a=aba if and only if a∈R

, a

ba.

For a long time, normal and Hermitian matrices, as well as normal and Hermitian linear operators on Banach or Hilbert spaces have attracted much attention[4-7,11-12]. In Ref.[13], Mosi

et al. gave several equivalent conditions for elements in rings to be normal by using their group inverses and MP-inverses, which generalized the results in Refs.[4-7]. Mosi

et al. proved that if a∈R

, then, a is normal if and only if a∈R#, a*a#=a#a*. Inspired by them, we consider the corresponding conditions for elements in rings by their core inverses. For example, we show that if a∈R

, then, a is normal if and only if a*a

a*. Later, in Ref.[14], Mosi

et al. also presented some equivalent characterizations for elements in rings to be generalized normal (resp. generalized Hermitian) involving powers of their group inverses and MP-inverses. Motivated by Refs.[13-14], we investigate some new characterizations for elements in rings to be normal (resp. Hermitian) by powers of their group inverses and MP-inverses. We prove that if a∈R

∩R#, n∈N, then, a is normal if and only if a*a

(a#)n=a#a*(a

)n.

Lemma 1[15] Let a∈R#, b∈R such that ab=ba. Then, a#b=ba#.

Lemma 2[9] a∈REP if and only if a∈R

and aa

Lemma 3[16] Let a∈R

and b∈R. If ab=ba and a*b=ba*, then, a

b=ba

Lemma 4[2] Let a∈R. The following assertions are equivalent: 1) a∈REP; 2) a∈R

, a#=a

Lemma 5[17] Let a∈R. Then the following statements are equivalent: 1) a∈REP; 2) a∈R# and Ra⊆Ra*; 3) a∈R# and Ra*⊆Ra.

In the following theorem, we use core inverses to characterize EP elements.

Theorem 1 Let a∈R

. Then, a∈REP if and only if one of the following equivalent conditions holds:

1) aa

a#=a#aa

; 2) aa#a

=a#a

a; 3) aa#a

aa#; 4) a#a

=(a#)2; 5) a

aa#=a

; 6) a#a

a=a

Proof If a∈REP, by Lemma 4 it is easy to verify that conditions 1) to 6) hold. Conversely, to conclude that a∈REP, we show that one of the conditions of Lemma 4 or Lemma 5 holds.

1) The equality aa

a#=a#aa

gives aa

a=aaa

by Lemma 1. Then, a=aa

a=aaa

=a(a

)*a*, which follows Ra⊆Ra*. By Lemma 5, we have a∈REP.

2) Assume that aa#a

=a#a

a. Then, Ra*=Ra

=Raa#a

=Ra#a

a⊆Ra, which implies a∈REP by Lemma 5.

3) Since aa#a

aa#, we have aa

=(aaa#)a

=a(a

aa#)=aa#. Thus, a

=a#. Then, a∈REP by Lemma 4.

4) Using the equality a#a

=(a#)2, we obtain a

=aa#a(a

)2=aa#a

=a(a#)2=a#. Then, a∈REP by Lemma 4.

5) The assumption a

aa#=a

gives Ra*=Ra

=Ra

aa#⊆Ra. Thus, a∈REP by Lemma 5.

6) It is similar to 5).

Now, we consider the equivalent conditions of an EP element a which satisfies a=aba, where b∈R.

Theorem 2 Let a∈R. Then, the following conditions are equivalent:

Proof 1)⟹2). Assume that a∈REP, then, a

=a# by Lemma 4. It follows that

2)⟹3). Suppose that a*=a*ba, we have

3)⟹1). The assumption a

ba gives

=aa

ba=aa

baa#a=aa

a#a=aa

a(a#)2a=a#a

Therefore, we obtain a

=a#. Then, a∈REP by the Lemma 4. Hence, it follows that a=aba.

Next, we investigate some results such that elements in rings are normal when their core inverses exist. Now, we begin with an auxiliary lemma.

Lemma 6 Let a∈R

with aa*=a*a, then, a∈REP.

Thus, we have a

=a#, which implies a∈REP by Lemma 4.

In the following, motivated by Ref.[13], some characterizations of normal elements in rings are presented under the condition that a∈R

Theorem 3 Let a∈R

. Then, a is normal if and only if one of the following equivalent conditions holds:

1) a*a

a*; 2) a

a*a=a*; 3) aa*a=a*aa; 4) aa*a#=a*; 5) a*a

a#=a#a*a

; 6) aa*a

=a*a

Proof As aa*=a*a and Lemma 6, we have a∈REP. Thus, it is easy to verify 1) to 7). Conversely, we show that a is normal.

1) As a*a

a*, we have

a)*= a*(a

)*=a*aa

)*=

a*(a

a2)a

)*=a

a*a2a

)*=

a*a2a

)*=a

aa*a

a2a

)*=

aa*aa

)*=a

aa*(a

)*=a

a(a

a)*

Then, we can obtain a

. Thus, a(a

)2=a

, a

a2=a, which implies a

a=a

aaa

=aa

. By Lemma 2, we have a#=a

. Thus, a*a

a* yields a*a#=a#a*. Then, by Lemma 1 we have a*a=aa*.

2) From a

a*a=a*, we have a

=a#. Thus, a

a*a=a*, it follows that a*a=aa*.

3) The equality aa*a=a*aa gives a#a=aa

. So, we obtain a

=a#. Thus, aa*a=a*aa implies that aa*=aa*aa

=aa*aa#=a*aaa#=a*a.

4) We can easily obtain that 3) holds.

5) Using a*a

a#=a#a*a

, we have aa#a*=a*. It follows that a#a=aa

. Thus, a

=a#. Hence, a*a

a#=a#a*a

yields a*(a#)2=a#a*a#. So, a*a#=a#a*. By Lemma 1, we have a*a=aa*.

6) By Lemma 1, we have 5) holds.

7) Since a*a#a

a*a#, we have a

=a#. Then, a*a#a

a*a# implies aa*=a*a. Therefore, a is normal.

In Refs.[13-14], the authors showed that the equivalent conditions such that a∈R is a normal element, are closely related to the powers of the group inverse and MP-inverse of a. Motivated by Ref.[13], in the following two theorems we present some new characterizations for elements in rings to be normal (resp. Hermitian) involving powers of their group inverses and MP-inverses.

Theorem 4 Let a∈R

, n∈N. Then a is normal if and only if a∈R# and one of the following conditions holds:

1) a*a

(a#)n=a#a*(a

)n; 2) a*a#(a

)n=a

a*(a#)n; 3) (a

)na*a#=(a#)na

a*; 4) (a

)na#a*=(a#)na*a

; 5) a

(a#)na*=(a#)na*a

; 6) a

a*(a#)n=a*(a#)na

Proof If a is a normal element, then a

=a# by Lemma 3. We also have a*a#=a#a* by Lemma 1. Thus, it is easy to verify that conditions 1) to 6) hold. Conversely, we prove a is normal.

1) Suppose that a*a

(a#)n=a#a*(a

)n, we have a

a=a#a. So, a

=a#. Therefore, a*a

(a#)n=a#a*(a

)n yields a*(a#)n+1=a#a*(a#)n. Thus, aa*=a*a.

2) The equation a*a#(a

)n=a

a*(a#)n implies aa

=aa#. So, we obtain a

=a#. Thus, a*a#(a

)n=a

a*(a#)n becomes a*(a#)n+1=a#a*(a#)n, which follows aa*=a*a.

3) Assume that (a

)na*a#=(a#)na

a*, we have aa

=aa#. So, a

=a#.

Therefore, (a

)na*a#=(a#)na

a* gives (a#)na*a#=(a#)n+1a*. Thus, aa*=a*a.

4) Using the assumption (a

)na#a*=(a#)na*a

, we can obtain a

a=a#a. Then, we have a

=a#. Hence, (a

)na#a*=(a#)na*a

implies (a#)n+1a*=(a#)na*a#. So, aa*=a*a.

5) The equality a

(a#)na*=(a#)na*a

gives a#a=a

a. Then, we obtain a

=a#.

Therefore, a

(a#)na*=(a#)na*a

yields (a#)n+1a*=(a#)na*a#. So, aa*=a*a.

6) Since a

a*(a#)n=a*(a#)na

, we obtain aa

=aa#. Then, we obtain a

=a#. Also, a

a*(a#)n=a*(a#)na

implies a*a#=a#a*. Therefore, we have a*a=aa* by Lemma 1.

Corollary 1[13] Let a∈R

. Then, a is normal if and only if a∈R# and one of the following conditions holds:

1) a*a

a#=a#a*a

; 2) a*a#a

a*a#; 3) a

a*a#=a#a

a*; 4) a

a#a*=a#a*a

Next, we consider the cases of Hermitian elements in rings.

Theorem 5 Let a∈R

and n∈N. Then a is Hermitian if and only if a∈R# and one of the following conditions holds:

1) a(a#)n=a*(a

)n; 2) a*(a

)n+1=(a#)n; 3) a*(a#)n+1=(a#)n; 4) a*a

(a#)n=(a#)n; 5) (a#)na*a#=(a

)n; 6) a#a*(a#)n=(a

)n; 7) (a

)na=(a#)na*; 8) ana*a

=an; 9) (a*)n+1+1a#=(a*)n.

Proof Since Hermitian elements are normal, we have aa

a by Lemma 3. Using a*=a, we can obtain a*a

a*. Thus, it is easy to verify that conditions 1) to 9) hold. Conversely, we show that a is Hermitian.

1) Suppose that a(a#)n=a*(a

)n, we have aa

=aa#. Thus, a

=a#. It follows that a(a#)n=a*(a

)n yields a(a#)n=a*(a#)n, which implies a=a*.

2) The equality a*(a

)n+1=(a#)n gives aa

=aa#. So, we obtain a

=a#. Therefore, a*(a

)n+1=(a#)n implies a*(a#)n+1=(a#)n. Thus, we have a*=a.

3) Using the assumption a*(a#)n+1=(a#)n, we have a

a=a#a. Then, a

=a#. Therefore, by a*(a#)n+1=(a#)n, we have a*=a.

4) If a*a

(a#)n=(a#)n, then, a

a=a#a. So, we obtain a

=a#. It follows that

a*a

(a#)n=(a#)n yields a*(a#)n+1=(a#)n. Thus, a*=a.

5) From the equality (a#)na*a#=(a

)n, we have aa

=aa#. Thus, we obtain a

=a#. It follows that (a#)na*a#=(a

)n turns into (a#)na*a#=(a#)n. So, a*=a.

6) By a#a*(a#)n=(a

)n, we obtain a#a=a

a. Then, a

=a#. Therefore, a#a*(a#)n=(a

)n gives a#a*(a#)n=(a#)n. Thus, a*=a.

7) Let (a

)na=(a#)na*, we obtain a

a=a#a. Then, we deduce that a

=a#. Therefore, (a

)na=(a#)na*yields (a#)na=(a#)na*. Thus, a=an(a#)na=an(a#)na*=a*.

8) The assumption ana*a

=an implies aa

=a#a. Then, a

=a#. Hence, ana*a

=an implies ana*a#=an. Thus, a*=a.

9) Using the equality (a*)n+1a#=(a*)n, we have aa

=aa#. Then, a

=a#. From (a*)n+1a#=(a*)n, it follows that (a*)n+1=(a*)na. Thus, we can obtain a=a*.

Corollary 2[13] Let a∈R

. Then, a is Hermitian if and only if a∈R# and one of the following conditions holds:

1) aa#=a*a

; 2) a*(a

)2=a#; 3) a*(a#)2=a#; 4) a*a

a#=a#; 5) a#a*a#=a

; 6) a

a=a#a*; 7) aa*a

=a; 8) (a*)2a#=a*.

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环上EP元、正规元和对称元的广义逆刻画

(1东南大学经济管理学院, 南京 211189)(2东南大学数学学院, 南京 211189)

摘要：结合广义逆理论研究了环中平等投影(EP)元、正规元和对称元的性质和一些等价刻画.给出了在核逆存在的情况下元素为EP元的一些等价条件.设a∈R

，那么a是EP元当且仅当aa

a#=a#aa

.同时，讨论了正则元是EP元的等价刻画.设a∈R，那么存在b∈R,使得a=aba且a是EP元当且仅当a∈R

，a

ba.同样地，给出了在核逆存在的情况下元素为正规元的一些等价条件.设a∈R

，那么a是正规元当且仅当a*a

a*.而且在群逆和Moore-Penrose逆存在的情况下给出了元素为正规元和对称元的一些涉及次数的等价条件.设a∈R

∩R#，且存在n∈N，那么a是正规元当且仅当a*a

(a#)n=a#a*(a

)n.结果推广了Mosi

等人的结论.

Foundation item：s：The National Natural Science Foundation of China (No.11371089), the Specialized Research Fund for the Doctoral Program of Higher Education (No.20120092110020), the Natural Science Foundation of Jiangsu Province (No.BK20141327).

Citation：：Ma Yilin, Chen Jianlong, Han Ruizhu. Characterizations of EP, normal and Hermitian elements in rings using generalized inverses[J].Journal of Southeast University (English Edition),2017,33(2):249-252.

Biographies：Ma Yilin (1992─), male, graduate; Chen Jianlong (corresponding author), male, doctor, professor, jlchen@seu.edu.cn.

Characterizations of EP, normal and Hermitian elementsin rings using generalized inverses

环上EP元、正规元和对称元的广义逆刻画