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

By Jason J. Molitierno

''Preface at the floor, matrix concept and graph thought are possible very diverse branches of arithmetic. even if, those branches of arithmetic engage because it is frequently handy to symbolize a graph as a matrix. Adjacency, Laplacian, and occurrence matrices are prevalent 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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Ii) There exists a vector x >> 0 such that Ax = 0. (iii) Each principal submatrix of A other than A itself is a nonsingular M-matrix. Proof: Since A is a singular M-matrix, A = sI − B for some nonnegative matrix B and s = ρ(B). If A is irreducible, then so is B. 20, ρ(B) is a simple eigenvalue of B. Hence zero is a simple eigenvalue of A, proving (i). 20 there exists an eigenvector x >> 0 of B such that Bx = ρ(B)x. Thus Ax = (sI − B)x = sx − ρ(B)x = 0 since s = ρ(B). This proves (ii). To prove (iii), let Aˆ be a principal submatrix of A where Aˆ = A.

For all x ∈ In addition (ii) λn = max xT Ax = max xT Ax xT x xT x=1 (iii) λ1 = min xT Ax = xT x x=0 and x=0 min xT Ax. xT x=1 Proof: Since A is symmetric, there exists a unitary matrix U ∈ Mn such that A = U DU T where D = diag(λ1 , . . , λn ). For any vector x ∈ n , we have n xT Ax = xT U DU T x = (U T x)T D(U T x) = λi |(U T x)i |2 . i=1 Since each term |(U T x)i |2 is nonnegative, it follows that n n |(U T x)i |2 ≤ xT Ax = λ1 i=1 n λi |(U T x)i |2 ≤ λn i=1 |(U T x)i |2 . i=1 Since U is unitary, it follows that n n |(U T x)i |2 = i=1 n |xT U U T x| = i=1 |xi |2 = xT x.

Therefore (I + A)n−1 is not positive. Suppose now that A is irreducible, then so is I +A. Let Y be the set of all nonnegative nonzero vectors in n with at least one entry of zero. Since I +A is irreducible, it follows by matrix-vector multiplication that (I + A)y will have fewer zero entries than y for each vector y ∈ Y . Hence (I +A)n−1 y is positive for all vectors y ∈ Y . The only way that this can hold for every vector y ∈ Y is for (I +A)n−1 to be positive. 19 If A ∈ Mn , A is nonnegative, and Ak is positive for some k ≥ 1, then ρ(A) is a simple eigenvalue of A.

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