# Applications of combinatorial matrix theory to Laplacian by Jason J. Molitierno

By Jason J. Molitierno

''Preface at the floor, matrix conception and graph thought are possible very diverse branches of arithmetic. despite the fact that, those branches of arithmetic have interaction because it is frequently handy to symbolize a graph as a matrix. Adjacency, Laplacian, and prevalence matrices are widely used to symbolize graphs. In 1973, Fiedler released his first paper on Laplacian matrices of graphs and confirmed what number houses of Read more...

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Handbook of Algebra Vol III

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1 0 1  1 1  1 1   0 0 1       0 0 0 1    1 1 0 0 0 1  , =  2  ... ... ... ... ... ...    1 0 0 0 1       1 0 0 1  1 1 0 ✐ ✐ ✐ ✐ ✐ ✐ “molitierno˙01” — 2011/12/13 — 10:46 — ✐ ✐ 30 Applications of Combinatorial Matrix Theory to Laplacian Matrices of Graphs and C3 = 1 n−1 (J − I). Then (a) C1 , C2 , and C3 are symmetric doubly stochastic matrices that commute with each other. (b) µ(C1 ) = 12 , µ(C2 ) = 1, and µ(C3 ) = 1. (c) The eigenvalues 1 = γ1 ≥ γ2 ≥ .

Molitierno˙01” — 2011/12/13 — 10:46 — ✐ Matrix Theory Preliminaries ✐ 37 Hence 0 is also an eigenvalue of A# also with corresponding eigenvector x. 1) that λx = Ax = AA# Ax = AA# λx. 1) again, we obtain x = AA# x = A# Ax = λA# x. Thus A# x = λ1 x. Therefore the nonzero eigenvalues of A# are the reciprocals of the nonzero eigenvalues of A. Moreover, if x is an eigenvector of A corresponding to λ, then x is also an eigenvector of A# corresponding to 1/λ. Exercises: 1. Find the Moore-Penrose inverse and group inverse of   2 1 1   A =  4 0 3  8 2 5 2.

K+1 , . . , λn , where ρ = ρ(A) is an eigenvalue of A with algebraic multiplicity k ≥ 1; the eigenvalues λi are all such that |λi | < ρ(A) for all k + 1 ≤ i ≤ n (by part (iii)). 6 we have  L = lim m→∞ 1 A ρ(A) m     = U    ∗ ...     = U lim  m→∞      m 1 1 λk+1 ρ 0 ... λn ρ      UT      1 ∗ ...     T U    1 0 0 ... 0 where the diagonal entry 1 is repeated k times in the last two expressions, and the diagonal entry 0 is repeated n − k times. 7), we conclude that k > 1 is impossible, thus proving (iv).