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Theorems and Lemmas on Spanning Trees in Graph Theory

Explore key theorems such as Kirchhoff's and Binet-Cauchy Lemmas related to spanning trees in undirected graphs, adjacency matrices, and incidence matrices in graph theory. Understand the computational complexity and applications.

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Theorems and Lemmas on Spanning Trees in Graph Theory

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  1. [Section 1.1] Spanning Trees • Recall the definitions of: • graphs, the vertex set V={0,1,2,…,n-1}, the edge set E • n = |V|, m = |E| • adjacency matrix • spanning tree

  2. [Section 1.1] Spanning Trees Thm 1.1 (Kirchhoff): Let A be the adjacency matrix of an undirected graph and D be the diagonal matrix with the degrees of vertices on its diagonal. Then, for every i2 {0,1,2,…,n-1}: # spanning trees of G = det (D-A)ii Where (D-A)ii is an (n-1)x(n-1) submatrix obtained from D-A by removing the i-th row and the i-th column.

  3. [Section 1.1] Spanning Trees Thm (Kirchhoff): Let A be the adjacency matrix of an undirected graph and D be the diagonal matrix with the degrees of vertices on its diagonal. Then, for every i2 {0,1,2,…,n-1}: # spanning trees of G = det (D-A)ii Where (D-A)ii is an (n-1)x(n-1) submatrix obtained from D-A by removing the i-th row and the i-th column. Note: determinant can be computed in time O(n3)

  4. [Section 1.1] Spanning Trees Lemma (Binet-Cauchy): If A is an (rxm) matrix and B is an (mxr) matrix, then: det(A.B) = Sµ [m], |S|=r det(A*S).det(BS*) Where A*S is the (rxr) matrix obtained from A by keeping only the columns in S, and BS* is the (rxr) matrix obtained from B by keeping only the rows in S.

  5. [Section 1.1] Spanning Trees Def: Incidence matrix of a directed graph H: an (nxm) matrix N=(ºve) such that: +1 if e starts at v ºve = -1 if e ends at v 0 o\w

  6. [Section 1.1] Spanning Trees Fact 1.6: rank N = n - # weakly connected components of H

  7. [Section 1.1] Spanning Trees Fact 1.6: rank N = n - # weakly connected components of H

  8. [Section 1.1] Spanning Trees Fact 1.7: If B is a square matrix with entries in {+1,-1,0} and every column has at most one +1 and at most one -1. Then, det(B) is +1, -1, or 0.

  9. [Section 1.1] Spanning Trees Thm 1.1 (Kirchhoff): # spanning trees = det (D-A)ii Proof:

  10. [Section 1.1] Spanning Trees Thm 1.1 (Kirchhoff): # spanning trees = det (D-A)ii Proof:

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