Highly Influential

Let M= ( A B C 0 ) be a complex square matrix where A is square. When BCB = 0, rank(BC) = rank(B) and the group inverse of ( BAB 0 CB 0 ) exists, the group inverse of M exists if and only if rank(BC + A ( BAB ) π BA) = rank(B). In this case, a representation of M in terms of the group inverse and Moore-Penrose inverse of its subblocks is given. Let A be a… (More)

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