By L. Mirsky

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This quantity includes the lawsuits of the 7th Workshop in Lie thought and Its purposes, which used to be held November 27-December 1, 2009 on the Universidad Nacional de Cordoba, in Cordoba, Argentina. The workshop was once preceded via a different occasion, oEncuentro de teoria de Lieo, held November 23-26, 2009, in honour of the 60th birthday of Jorge A.

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We pose the following problem, which was first stated in [30]. 8 Characterise all nonnegative and irreducible matrices A for which Q# = (r(A)I − A)# is an M-matrix. In Chapter 6 we will provide one answer to this question using mean first passage times for Markov chains. However, without appealing to mean first passage times, the authors of this book have studied several classes of matrices A in Φn,n for which the associated M -matrix Q = r(A)I − A has a group inverse which is an M-matrix. For example, if x and y are positive n-vectors such that y t x = 1, we find readily that for the rank one matrix A = xy t , we have (I − A)# = I − xy t , which is certainly an M-matrix.

It follows that we can find an orthonormal basis of eigenvectors of N (B), say u1 , . . , un−r , and further n−r that B # B = BB # = I − j=1 uj u∗j . 14), so that necessarily B# = B†. 16 Group Inverses of M-Matrices and their Applications Thus, in particular, we see that if B is Hermitian, then B # = B † . The following simple example further illustrates the distinction between B # and B † . 1 Suppose that we have vectors u, v ∈ Rn such that ut v = 0, and consider the rank 1 matrix B = uv t . Then 0 is a semisimple eigenvalue of B of multiplicity n − 1.

Let X = diag(x), Y = diag(y), and consider the following two matrices: G = Y Q# X and B = Y QX. Observe that since G is formed from Q# by multiplication by positive diagonal matrices, it suffices to show that G is in the class P0n . Note also that B1 = 0 and 1t B = 0t , from which we find that B is both row and column diagonally dominant. 1]), we find that B + B t is a positive semidefinite matrix; in particular we find that for any vector v ∈ Rn , v t Bv ≥ 0. We claim that for any vector u ∈ Rn , there is a vector v such that u Gt u = v t Bv.