## Algebre lineaire by Florent.

By Florent.

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**Example text**

We have det A = ∑ sgn(π)a1π(1) · · · anπ(n) , π where the summation is over all permutations of 1, . . , n. Consider a term a1π(1) · · · anπ(n) , which is nonzero. Since π admits a cycle decomposition, such a term will correspond to some 2-cycles (i j) of π, which designate an edge joining i and j in G, as well as some cycles of higher order, which correspond to cycles in G. ) Thus, each nonzero term in the summation arises from an elementary subgraph of G with vertex set V (G). Suppose the term a1π(1) · · · anπ(n) corresponds to the spanning elementary subgraph H.

Then A is nonsingular if and only if T has a perfect matching. Proof. 8, det A = ∑(−1)n−c1 (H)−c(H) 2c(H) , where the summation is over all spanning elementary subgraphs H of T. If det A is nonzero then T has an elementary spanning subgraph. In the case of a tree, c(H) = 0 for any H. Thus, an elementary spanning subgraph consists exclusively of edges, which clearly must form a perfect matching. To prove the converse, first observe that if T has a perfect matching then it must be unique. This statement is easily proved by induction on the number of vertices.

There exists a lower triangular matrix T with non2 negative √ diagonal entries such that L(G) = T T . Then d1 = 11 = t11 , and hence t11 = √ d1 . Comparing the first column of both sides of L(G) = T T , we see that i1 = d1 ti1 , i = 1, 2, . . , n. Thus, the first diagonal entry of T T equals n 1 n ∑ ti12 = d1 ∑ i=1 i=1 2 i1 = 1 2 (d + d1 ) = d1 + 1. d1 1 The largest eigenvalue of T T exceeds or equals the largest diagonal entry of T T, and hence it exceeds or equals d1 + 1. The proof follows in view of the fact that the eigenvalues of L(G) = T T and T T are the same.