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  .
In  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  necessary and sufficient condition for the existence of Re-nnd solution has been established of the matrix equation where and. In  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
G is called the Minkowski metric matrix. In case, indexed as, G is called the Minkowski metric tensor and is defined as  . 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.
Lemma 1. Let. If
then there are unitary matrices such that
Proof. Since there are two unitary matrices such that
From we have
and from we get
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,
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
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:
Proof. Suppose, then by Lemma 1 we have
Since we have that is invertible. By using Lemma 2 and 3 we get
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
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
Then and. Therefore,
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
Theorem 2. Let in, where,
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
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
Then and Therefore
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
So and are similar.
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