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**Group Inverse of 2 × 2 Block Matrices over Minkowski Space M** ()

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*Advances in Linear Algebra & Matrix Theory*,

**6**, 75-87. doi: 10.4236/alamt.2016.63009.

1. Introduction

Let F be a skew field and be the set of all matrices over F. For, the matrix is said to be the group inverse of A, if

.

and is denoted by, and is unique by [1] .

In [12] the existence of anti-reflexive with respect to the generalized reflection anti- symmetric matrix and solution of the matrix equation in Minkowski space is given. In [13] necessary and sufficient condition for the existence of Re-nnd solution has been established of the matrix equation where and. In [14] partitioned matrix in Minkowski space was

taken of the form to yield a formula for the inverse of

in terms of the Schur complement of.

In this paper and denote the conjugate transpose and Minkowski adjoint of a matrix P respectively. denotes the identity matrix of order. Minkowski Space is an indefinite inner product space in which the metric matrix associated with the indefinite inner product is denoted by G and is defined as

satisfying and.

G is called the Minkowski metric matrix. In case, indexed as, G is called the Minkowski metric tensor and is defined as [12] . For any, the Minkowski adjoint of P denoted by is defined as where is the usual Hermitian adjoint and G the Minkowski metric matrix of order n. We establish the necessary and sufficient condition for the existence

and the representation of the group inverse of a block matrix or

in Minkowski space, where. We also give a sufficient condition for to be similar to.

2. Lemmas

Lemma 1. Let. If

,

then there are unitary matrices such that

where and.

Proof. Since there are two unitary matrices such that

where

.

Now

and

From we have

and from we get

So,

Lemma 2. Let

.

Then the group inverse of M exists in if and only if the group inverse of

exists in and. If the group inverse of exists in M,

then

Proof. Since, suppose group inverse of exists in and. Now

.

But because exists. There-

fore exists in.

Conversely, suppose the group inverse of M exists in, then it satisfies the following conditions: 1) 2) and 3). Also

.

Let then,

1)

2)

3)

Lemma 3. Let, and. Then the

group inverse of M exists in if and only if the group inverse of exists in and. If the group inverse of M exists in, then,

Proof. The proof is same as Lemma 2.

Lemma 4. Let. If

then the following conclusions hold:

1)

2)

3)

4)

5)

Proof. Suppose, then by Lemma 1 we have

where. Then

Since we have that is invertible. By using Lemma 2 and 3 we get

Then, 1)

Similarly we can prove 2) - 5).

3. Main Results

Theorem 1. Let where, then

1) The group inverse of M exists in if and only if

.

2) If the group inverse of M exists in, then, where

Proof. 1) Given. Suppose then,

. We know that

so,.

Therefore the group inverse of M exists. Now we show that the condition is ne- cessary,

.

Since the group inverse of M exists in if and only if, we have

Also

Then and. Therefore,

.

From

and

,

we have

Since

and

,

we get

.

Thus

.

Then there exists a matrix such that. Then

.

So, we get

.

2) Let, we will prove that the matrix X satisfies the conditions of

the group inverse in. Firstly we compute

Applying Lemma 4 1), 2) and 5) we have

Now

□

Theorem 2. Let in, where,

Then,

1) the group inverse of M exists in if and only if

.

2) if the group inverse of M exists in, then, where

Proof. 1) Given. Suppose then,

.

We know that

so,

.

Therefore the group inverse of M exists in. Now we show that the condition is necessary,

Since the group inverse of M exists in if and only if. We know

Also

Then and Therefore

From

and

we have

Since

and

,

we get

.

Thus

Then there exist a matrix such that Thus

So, we get.

2) Proof is same as Theorem 1 2).

Theorem 3. Let if

.

Then and are similar.

Proof. Suppose, then by using Lemma 1, there are unitary matrices such that

,

where. Hence

So and are similar.

Conflicts of Interest

The authors declare no conflicts of interest.

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