Some new characterizations of the normality for group invertible matrices

1.   Introduction and Preliminaries

In this paper, the matrices that we consider are all $n\times n$ complex matrices, and we denote the set of all $n\times n$ matrices over the complex number field $\mathbb{C}$ by $\mathbb{C}^{n\times n}$. Let $A\in \mathbb{C}^{n\times n}$; then there always exists a matrix $B\in \mathbb{C}^{n\times n}$ such that

where $A^H$ denotes the conjugate transpose matrix of $A$. The matrix $B$ is called the Moore$-$Penrose inverse matrix of $A$, which is uniquely determined and is denoted by $A^{\dagger}$^[1,2,3].

The matrix $A$ is said to be group invertible if there is a matrix $B\in \mathbb{C}^{n\times n}$ such that

The matrix $B$ is called the group inverse matrix of $A$. If it exists, then it is unique^[4] and is always denoted by $A^{\#}$.

We use $G_n(\mathbb{C})$ to denote the set of all group invertible matrices in $\mathbb{C}^{n\times n}$. It is well known that $A\in G_n(\mathbb{C})$ if and only if ${\rm{rank}}(A) = {\rm{rank}}(A^2).$ Hence, for any $A\in \mathbb{C}^{n\times n}$, $AA^H,$ $A^HA$, and $A+A^H$ are all group invertible matrices.

$A \in G_n(\mathbb{C})$ is called an $EP$ matrix if $A^{\#} = A^{\dagger}$. It is known that $A$ is $EP$ if and only if $AA^{\dagger} = A^{\dagger}A$^[2]. For the study of $EP$ matrices, we can also refer to ^[5]. And $A$ is called a normal matrix if $A^HA = AA^H$. In ^[6], it gives some counterexamples on normal matrices. In recent years, the normality of matrices was wildly studied. For example, in [7, Theorem 5.1], it is shown that $A\in G_n(\mathbb{C})$ is normal if and only if the equation

is consistent and the general solution is given by

In [8, Theorem 3.1], it is proved that $A\in G_n(\mathbb{C})$ is normal if and only if the equation

has at least one solution in $\{A, A^\#, A^\dagger, \ A^H, (A^\dagger)^H, (A^\#)^H\}$. For the studies of normality in a ring with involution, the readers can refer to ^[5,9,10,11].

Let $A, B$ and $C\in \mathbb{C}^{n\times n}$. Recall that $A$ is said to be a projection if $A^2 = A = A^H$. Clearly, $A$ is a projection if and only if $A = AA^H$ or $A = A^HA$always ^[12]. $B$ and $C$ are called left (right) $A$-equality if $AB = AC \ (BA = CA)$. $A$ is called left (right) $B$-idempotent if $A^2 = BA \ (A^2 = AB).$ Clearly, $A$ is left and right $A$-idempotent. And $A$ is an idempotent matrix if and only if $A$ is left $2A$-$E_n$-idempotent.

In ^[13], using the projections, many characterizations of $EP$ elements were founded. Utilizing the generalized inverse of matrices to construct the inverse matrices of related matrices, the authors provide some characterizations of normal matrices in ^[7]. Inspired by these, we use the projection, the generalized inverse, one-sided $X$-equality, and $X$-idempotency of matrices to investigate the normality of the group invertible matrices. It seems to be the first time that the projection, one-sided $X$-equality, and one-sided $X$-idempotency are used to characterize the normality of matrices.

This paper is organized as follows: in terms of projections, many new and interesting characterizations of normal matrices are obtained in Section 2. In Section 3, we characterize normal matrices by constructing the inverse of the product of some matrices. In Sections 4 and 5, utilizing one-sided $A$-equalitivity and one-sided $X$-idempotency, we give some properties and characterizations of normal matrices. We conclude in Section 6.

2.   Using projections to characterize normal matrices

We begin with the following lemma, which appears in [8, Lemma 2.6].

Lemma 2.1. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $(A^{\dagger})^HA^\#A^H = A^{\dagger}$.

Lemma 2.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is a projection if and only if $A^H = AA^{\dagger}$.

Proof. Necessity. If $A$ is a projection, then $A^2 = A = A^H$ and it follows that $A = A^2 = A^HA$ and $AA^{\dagger} = A^HAA^{\dagger} = A^H$.

Sufficiency. Suppose that $A^H = AA^{\dagger}$. Then $A = (AA^{\dagger})^H = AA^{\dagger} = A^H$ and so $A^2 = (AA^{\dagger})^2 = AA^{\dagger} = A$. Hence $A$ is a projection. □

From Lemma 2.1, one knows that if $A$ is normal, then $(A^{\dagger})^HA^\#A^HA = A^{\dagger}A$ is a projection. Using Lemma 2.2, we can obtain

This implies $((A^{\dagger})^HA^\#A^HA)^{\dagger} = A^\#(A^{\dagger})^HAA^HA^HA(A^\#)^HA^{\dagger}$, if $A$ is normal. The following lemma points out exactly what $((A^{\dagger})^HA^\#A^HA)^{\dagger}$ is for any $A \in G_n(\mathbb{C})$.

Lemma 2.3. Let $A \in G_n(\mathbb{C})$. Then

$(1)$ $((A^{\dagger})^HA^\#A^HA)^{\dagger} = A^{\dagger}(A^\#)^HA^{\dagger}A^2A^H$.

$(2)$ $((A^{\dagger})^HA^\#A^H)^{\dagger} = (A^{\dagger})^HAA^H$.

$(3)$ $((A^\#)^HA^\#A^H)^{\dagger} = (A^{\dagger})^HA^2A^{\dagger}A^HA^{\dagger}A$.

$(4)$ $((A^\#)^HA^\#A^H)^\# = (A^\#)^HA^{\dagger}A^3A^{\dagger}A^H$.

Proof. It is routine, we omit the proof.□

Theorem 2.1. Let $A \in G_n(\mathbb{C})$. Then the following statements are equivalent:

$(1)$ $A$ is normal;

$(2)$ $(A^{\dagger})^HAA^H = A$;

$(3)$ $(A^\#)^HAA^HA^{\dagger}$ is a projection;

$(4)$ $(A^{\dagger})^HAA^HA^\#$ is a projection.

Proof. (1)$\Rightarrow$(2). Since $A$ is normal, by Lemma 2.1, $(A^{\dagger})^HA^\#A^H = A^{\dagger}$. It follows from Lemma 2.3 that

(2)$\Rightarrow$(3). If $A = (A^{\dagger})^HAA^H$, then

Hence by [5, Theorem 1.2.1], $A$ is EP and

is a projection.

(3)$\Rightarrow$(4). Under the assumption, one obtains

Hence $(A^{\dagger})^HAA^HA^\#$ is a projection.

(4)$\Rightarrow$(1). Assuming that $(A^{\dagger})^HAA^HA^\#$ is a projection, then

and

Multiplying (2.2) on the right by $AA^{\dagger}$, one has

Multiplying (2.3) on the left by $(A^{\dagger})^HA^\#A^H$, one yields

Hence, by [5, Theorem 1.2.1], $A$ is EP, which induces $A^\#A = AA^{\dagger}.$ Now multiplying (2.1) on the left by $(A^{\dagger})^HA^\#A^H$, one obtains

and

It follows from Lemma 2.3 that

By Lemma 2.1, $A$ is normal.□

Example 2.1. Let

then $A$ is normal and $A^\dagger = A^\# = \left(\begin{array}{ccc}\frac{1}{2} & \frac{1}{2} & 0\\\frac{1}{2} & \frac{1}{2} & 0\\0 & 0 & -i\end{array}\right)$. Hence

and

is a projection.

Theorem 2.2. Let $A \in G_n(\mathbb{C})$. Then the following statements are equivalent:

$(1)$ $A$ is normal;

$(2)$ $(A^{\dagger})^HA^\#A^HA$ is a projection;

$(3)$ $A^{\dagger}(A^{\dagger})^HAA^H$ is a projection.

Proof. (1)$\Rightarrow$(2). Assume that $A$ is normal. Then $(A^{\dagger})^HA^\#A^H = A^{\dagger}$ by Lemma 2.1. It follows that $(A^{\dagger})^HA^\#A^HA = A^{\dagger}A$ is a projection.

(2)$\Rightarrow$(3). The projectivity of $(A^{\dagger})^HA^\#A^HA$ implies

Multiplying the equality on the left by $(A^{\dagger})^HAA^H$, one gets

This gives

Hence $A$ is EP [5, Theorem 1.2.1], which induces

Applying the involution, one has

and

Therefore,

is a projection.

(3)$\Rightarrow$(1). Using the projectivity of $A^{\dagger}(A^{\dagger})^HAA^H$, one obtains

Multiplying the equality on the right by $(A^{\dagger})^HA^\#A^H$, one gets

and

Hence $A$ is normal by [5, Theorem 1.3.2].□

Example 2.2. Let

then $A$ is not normal; $A$ is group invertible with $A^\# = A$ and

Hence both

and

are not projections.

From Lemma 2.2, we know that $A$ is a projection matrix if and only if $A^H$ is a projection matrix. Hence, we can obtain the following corollary.

Corollary 2.1. Let $A \in G_n(\mathbb{C})$. Then the following statements are equivalent:

$(1)$ $A$ is normal;

$(2)$ $A^HA(A^\#)^HA^{\dagger}$ is a projection;

$(3)$ $AA^HA^{\dagger}(A^{\dagger})^H$ is a projection.

3.   The representation form of the inverses of matrices and normal matrices

According to [5, Lemmas 1.3.2 and 1.3.3], $A$ is normal if and only if $A$ is EP and $A^{\dagger}A^H = A^HA^{\dagger}$. Using the representation of the Moore-Penrose inverse of the product of generalized inverse elements, we can get the following conclusion.

Theorem 3.1. Let $A \in G_n(\mathbb{C})$. Then $A$ is a normal matrix if and only if

Proof. Necessity. Suppose that $A$ is normal. Then $A$ is EP and $A^{\dagger}A^H = A^HA^{\dagger}$. By Lemma 2.3,

Sufficiency. Using the hypothesis and Lemma 2.3, one yields

Multiplying the equality on the left by $A^{\dagger}A^\#A^H$, one obtains

It follows that

Hence, $A$ is EP, which infers

Thus, $A$ is normal.□

Corollary 3.1. Let $A \in G_n(\mathbb{C})$. Then $A$ is a normal matrix if and only if

Proof. Necessity. Assume that $A$ is normal. Then $A$ is EP and $A^{\dagger}A^H = A^HA^{\dagger}$. By Lemma 2.1, $(A^{\dagger})^HA^\#A^H = A^{\dagger}$. Then

Hence, by Theorem 2.4,

Sufficiency. From the assumption and Lemma 2.3, one yields

By [14, Lemma 2.11], one gets

Multiplying the equality on the right by $(A^{\dagger})^HA^\#A^\#$, one has

So $A^{\dagger} = A^{\dagger}A^{\dagger}A$, this implies $A$ is EP. It follows from Lemma 2.3 that

By Theorem 2.4, $A$ is normal.□

It is well known that for a group invertible matrix $A$, $A+E_n-AA^\#$ is invertible and

In [7, Theorem 4.2], it is proved that a group invertible matrix $A$ is normal if and only if

In the following, we shall characterize normal matrices by using $(AA^\#)^H$ to construct invertible matrices. Noting that $((A^\#)^HA^\#A^H)((A^\#)^HA^\#A^H)^\# = AA^\#$ by Lemma 2.3, we can get

Theorem 3.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $(A^\#)^HA^\#A^H+E_n-(AA^\#)^H$ is invertible with

Corollary 3.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if

Proof. Necessity. Since $A$ is normal, $(AA^\#)^H = AA^\# = AA^{\dagger}$. By Theorem 3.3, we are done.

Sufficiency. Using the assumption, one has

This gives

By Theorem 3.4, $A$ is normal.□

Corollary 3.3. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if

Proof. Necessity. It follows from Corollary 3.4 and the fact that $AA^{\dagger} = A^{\dagger}A.$

Sufficiency. Under the assumption, one obtains

This induces

Hence $A$ is EP, which infers $A^{\dagger}A = AA^{\dagger}$. By Corollary 3.4, $A$ is normal.□

In [8, Theorem 4.3], it is shown that a group invertible matrix $A$ is normal if and only if

for some $X\in\{A^{\dagger}, A^H, (A^\#)^H\}$. This inspired us to consider the Moore$-$Penrose inverse; we have the following theorem.

Theorem 3.3. Let $A \in G_n(\mathbb{C})$. Then $A$ is a normal matrix if and only if

for some $X\in\{A, A^\#, A^{\dagger}, A^H, (A^\#)^H, (A^{\dagger})^H\}$.

Proof. Necessity. It is an immediate corollary of Theorem 3.1.

Sufficiency. (1) If $X = A$, then $((A^\#)^HA^\#A)^{\dagger} = A^{\dagger}A^2A^HA^{\dagger}$.

Noting that $((A^\#)^HA^\#A)^{\dagger} = A^{\dagger}A^2A^{\dagger}A^HA^{\dagger}A.$ Then one obtains

Multiplying the equality on the left by $(A^{\dagger})^HA^\#$, one yields

and so

By [14, Corollary 2.10], $A^{\dagger} = A^{\dagger}A^{\dagger}A.$ Hence $A$ is $EP$, which gives

and

Hence $A$ is normal.

(2) If $X = A^\#$, then

Since $((A^\#)^HA^\#A^\#)^{\dagger} = A^{\dagger}A^4A^{\dagger}A^HA^{\dagger}A$, one obtains

Multiplying the equality on the left by $(A^{\dagger})^H(A^\#)^3$, one gets

By the proof of (1), one obtains that $A$ is normal.

(3) If $X = A^{\dagger}$, then

Multiplying the equality on the left by $A^{\dagger}A^\#$, one has

By the proof of (1), $A$ is normal.

(4) If $X = A^H$, then $A$ is normal by Theorem 3.1.

(5) If $X = (A^{\dagger})^H$, then

Multiplying the equality on the left by $A^{\dagger}(A^{\dagger})^H$, one obtains

By (1), $A$ is normal.

(6) If $X = (A^\#)^H$, then

Multiplying the equality on the left by $A^{\dagger}AA^\#(A^\#)^H$, one has

Hence $A$ is normal by (1).□

4.   Using one-sided equalities to characterize normal matrices

From Lemma 2.2, one knows that $A\in \mathbb{C}^{n\times n}$ is a projection if and only if $A^{\dagger}A$ and $(A^{\dagger})^H$ are right $A$-equality.

From Lemma 2.1 and Theorem 2.4, we have

Theorem 4.1. Let $A \in G_n(\mathbb{C})$. Then the following statements are equivalent:

(1) $A$ is normal;

(2) $(A^{\dagger})^HA^\#, \ A^{\dagger}(A^{\dagger})^H$ are right $A^H-$equality;

(3) $A^{\dagger}(A^{\dagger})^HAA^H, \ A^{\dagger}A$ are left $A$-equality.

Lemma 4.1. Let $B$ and $C$ be right $A$-equality. Then $B^2$ and $BC$ are right $A$-equality.

Proof. It is clear.□

Lemma 4.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $(A^{\dagger})^HA^\# = A^{\dagger}(A^{\dagger})^H$.

Proof. Necessity. Suppose that $A$ is normal. Then $(A^{\dagger})^HA^\#A^H = A^{\dagger}$ by Lemma 2.1, it follows that $(A^{\dagger})^HA^\# = (A^{\dagger})^HA^\#A^{\dagger}A = (A^{\dagger})^HA^\#A^H(A^{\dagger})^H = A^{\dagger}(A^{\dagger})^H$.

Sufficiency. Applying the condition $"(A^{\dagger})^HA^\# = A^{\dagger}(A^{\dagger})^H"$, one obtains

By Lemma 2.1, $A$ is normal.□

Theorem 4.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if both $((A^{\dagger})^HA^\#)^2$ and $A^{\dagger}(A^{\dagger})^H(A^{\dagger})^HA^\#$ are right $A^H$-equality.

Proof. Necessity. Under the assumption, $(A^{\dagger})^HA^\#, \ A^{\dagger}(A^{\dagger})^H$ are right $A^H$-equality by Theorem 4.1, and at once, $((A^{\dagger})^HA^\#)^2, (A^{\dagger})^HA^\#A^{\dagger}(A^{\dagger})^H$ are right $A^H$-equality by Lemma 4.2. Hence $((A^{\dagger})^HA^\#)^2, \ A^{\dagger}(A^{\dagger})^H(A^{\dagger})^HA^\#$ are right $A^H$-equality by Lemma 4.3.

Sufficiency. From the hypothesis, one gets

Multiplying (4.1) on the right by $(A^{\dagger})^HAA^HA$, one obtains

and

By Lemma 4.3, $A$ is normal.□

Observing the formula (4.1), we have the following result.

Corollary 4.1. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if both $(A^{\dagger})^HA^\#$ and $A^{\dagger}(A^{\dagger})^H$ are right $(A^{\dagger})^HA^\#A^H$-equality.

Noting that

Then Corollary 4.5 induces

Corollary 4.2. Let $A \in G_n(\mathbb{C})$. Then the following statements are equivalent:

$(1)$ $A$ is normal;

$(2)$ $(A^{\dagger})^HA^\#A^+$ and $A^{\dagger}(A^{\dagger})^HA^+$ are right $A(A^{\dagger})^HA^\#A^H$-equality;

$(3)$ $(A^{\dagger})^HA^\#A$ and $A^{\dagger}(A^{\dagger})^HA$ are right $A^\#(A^{\dagger})^HA^\#A^H$-equality;

$(4)$ $(A^{\dagger})^HA^\#A^\#$ and $A^{\dagger}(A^{\dagger})^HA^\#$ are right $A(A^{\dagger})^HA^\#A^H$-equality.

5.   Using one-sided idempotents to characterize normal matrices

Theorem 5.1. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $(A^{\dagger})^HAA^H$ is right $A$-idempotent.

Proof. Necessity. Since $A$ is normal, $(A^{\dagger})^HAA^H = A$ by Theorem 2.4. Hence $(A^{\dagger})^HAA^H$ is right $A$-idempotent.

Sufficiency. Applying the assumption, one yields

Multiplying the equality on the left by $A^{\dagger}AA^\#A^H$, one has

i.e., $A^{\dagger}A^2A^H = A^HA$. It follows that

Hence, $A$ is normal by Theorem 2.4.□

Theorem 5.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $(A^\#)^HAA^H$ is left $A$-idempotent.

Proof. Necessity. Assume that $A$ is normal. Then $A^{\dagger} = A^\#$ and, by Theorem 2.4, $(A^{\dagger})^HAA^H = A$. It follows that $(A^\#)^HAA^H = A$. Hence $(A^\#)^HAA^H$ is left $A$-idempotent.

Sufficiency. Applying the condition, one obtains

Multiplying the equality on the right by $(A^{\dagger})^HA^{\dagger}A^HA^{\dagger}A$, one obtains

and

Hence $A$ is EP, which leads to $A = (A^\#)^HAA^HA^{\dagger}A = (A^{\dagger})^HAA^H$. By Theorem 2.4, $A$ is normal.□

Corollary 5.1. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $A-(A^\#)^HAA^H$ is right $A$-idempotent.

Proof. It follows from Theorem 5.2 and the fact: $C$ is left $D$-idempotent if and only if $D-C$ is right $D$-idempotent. □

It is easy to see that $C$ is right $D$-idempotent if and only if $C, \ D$ are left $C$-equality. Hence Theorem 5.1 induces the following results.

Corollary 5.2. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if both $(A^{\dagger})^HAA^H$ and $A$ are left $(A^{\dagger})^HAA^H$-equality.

Corollary 5.3. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if both $(A^{\dagger})^HAA^H+E_n-AA^{\dagger}$ and $A+E_n-AA^{\dagger}$ are left $(A^{\dagger})^HAA^H$-equality.

Theorem 5.3. Let $A \in G_n(\mathbb{C})$. Then $A$ is normal if and only if $(AA^\#)^HA$ is left $(A^{\dagger})^HAA^H$-idempotent.

Proof. Necessity. Assume that $A$ is normal. Then $(A^{\dagger})^HAA^H = A$ by Theorem 2.4 and

By Theorem 5.2, one obtains $(AA^\#)^HA$ is left $(A^{\dagger})^HAA^H$-idempotent.

Sufficiency. Applying the condition, one obtains

Multiplying the equality on the right by $A^{\dagger}A^{\dagger}$, one obtains

Hence, $A$ is EP, which leads to

Therefore, $A$ is normal by [5, Theorem 1.3.2].□

6.   Conclusions

In this paper, we have given many new characterizations of the normality for the group invertible matrices. These characterizations concern the projection, the Moore-Penrose inverse, one-sided $X$-equality and $X$-idempotency of matrices. To our knowledge, it is the first time that the projection, one-sided $X$-equality, and one-sided $X$-idempotency are used to characterize the normality of matrices. We shall consider the similar characterizations for the normality in $C^*$-algebra or a ring with involution. Moreover, we shall investigate some interesting applications of our results, for instance in some fields such as degenerate polynomial and stochastic equations^[15,16].

Author contributions

Zhirong Guo: Writing-original draft, review and editing; Qianglian Huang: Supervision, Writing-review and editing, Funding acquisition. All authors have read and approved the final version of the manuscript for publication.

Use of Generative-AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This research is supported by the National Natural Science Foundation of China (12471133, 11771378).

Conflict of interest

The authors declare that there are no conflicts of interest.