Binary Moore-Penrose Inverses of Set Inclusion Incidence Matrices

This note is a supplement to some recent work of R.B. Bapat on Moore-Penrose inverses of set inclusion matrices. Among other things Bapat constructs these inverses (in case of existence) forH(s, k) mod p, p an arbitrary prime, 0 ≤ s ≤ k ≤ v − s. Here we restrict ourselves to p = 2. We give conditions for s, k which are easy to state and which ensure that the Moore-Penrose inverse of H(s, k) mod 2 equals its transpose. E.g., H(s, v − s) mod 2 has this property. Furthermore KerH(s, v − s) mod 2 is nonzero if 0 < 2s < v ≤ 3s and then there is a decomposition KerH(s, v − s) ≡ ∑ 0≤j≤s−1 2 | ( v−2s ) ImH(v − s, v − j) mod 2. Also, refinements of this decomposition are given. 1. Let F be a field. If A is a m× n-matrix with entries in F, then a n×mmatrix G (with entries in F) is called a generalized inverse (g-inverse) of A if AGA = A. A Moore-Penrose inverse of A, denoted by A, is a n×m-matrix G satisfying the equations AGA = A, GAG = G, (AG) = AG, (GA) = GA. Note that if A is square and invertible, G = A−1 is a Moore-Penrose inverse of A. Thus A (if it exists) can be thought of as a substitute for the inverse of A, even if A is non-square. • Any real or complex matrix admits a unique Moore-Penrose inverse. In general, the following theorem gives a necessary and sufficient condition for a matrix to admit a Moore-Penrose inverse over an arbitrary field. Theorem (R.B. Bapat, K.P.S. Bhaskara, K. Manjunatha, [2]) A m × n-matrix of rank r over an arbitrary field admits a Moore-Penrose inverse if and only if the sum of the squares of the r × r minors of A is nonzero. Moore-Penrose inverses were introduced by Penrose for complex matrices with a slight modification replacing in the above four defining equations the transpose of a matrix by its complex-conjugate transpose. However, the proof that Moore-Penrose inverses are uniquely determined (if they exist) carries over to the definiton given above. For the convenience of the reader we reproduce the proof. So let X, Y two Moore-Penrose inverses of A. Then X = XAX = X(AX) = XXA = XX (AY A ) = X(AX) (AY ) = X(AXA)Y = XAY = = (XA) (Y AY ) = (XA) (Y A)Y = (AXA )Y Y = = AY Y = (Y A)Y = Y. Note that all four defining equations have been used in the proof. Now we recall the definition of set-inclusion incidence matrices. In fact there are two families of incidence matrices. Let s, k, v be in N0 and s, k ≤ v. Let Hv(s, k) = H(s, k) denote the integer (0, 1)-matrix with rows and columns indexed respectively by the s-subsets and k-subsets of a fixed v-set with ijentry equal to one if and only if the i-th s-subset is contained in the j-th k-subset.