By Geir T. Helleloid

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Xn ))), we have clearly defined addition and multiplication operations. But what is the situation with regard to inverses and composition? The answers are that f (x) has an inverse if and only if f (0) = 0 and the composition f (g(x)) is well-defined if g(0) = 0. In the case of composition, this is because computing the coefficient of xn in f (g(x)) requires a finite number of steps if g(0) = 0, but requires infinitely many if g(0) = 0 (and, say, f (x) is not a polynomial). It is time for the fundamental result in this lecture (which specializes to the Exponential Formula).

This is called the Smith normal form of L and the di are the invariant factors. 23. Let d1 , . . , dk be the invariant factors of Ls (G). Then CF (G) ∼ = Z/d1 Z × · · · × Z/dk Z. Example. Let G = Kn+1 , the complete graph on n + 1 vertices, with one vertex designated as the sink. Then n −ut −u (n + 1)I − J Ls (G) = (n + 1)I − J = , where J is the n × n matrix of all 1’s, I is the (n − 1) × (n − 1) identity matrix, J is the (n − 1) × (n − 1) matrix of all 1’s, and u is the (n − 1) × 1 matrix of all 1’s.

2. Given an oriented spanning tree with root v, we can construct Eulerian tours. v∈V outdeg(v) − 1)! To prove (1), just observe that 1. T has n − 1 edges. 2. T does not have two arcs going out of the same vertex. 3. T does not have an arc going out of v. 4. T does not have a cycle. To prove (2), given T , we construct an Eulerian tour by starting at e and continue to choose any edge possible except we don’t choose f ∈ T unless we have to. The set of last exits of the tour coincide with the set of edges of T .