“…Applying Theorem 6 with B = SS (2) M,N and recalling that S (2) M,N SS (2) M,N = S (2) M,N is valid, we get B 2 = B. Again, since S (2) M,N is a {2}-generalized inverse of S we have rank(SS (2) M,N C) = rank(S (2) M,N C). Analogously, rank(SS (2) M,N ) = rank(S (2) M,N ).…”