Talk:Kirchhoff's theorem

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I copied the original material from spanning tree (mathematics) and added a different formulation of Kirchhoff's theorem using the cofactor of the admittance matrix. Then I wrote a new article admittance matrix and deleted most of the explaination on this page in terms of discrete Laplace operator (which was just a definition of the admittance matrix). Now the article is quite terse, perhaps someone else can add some material to illustrate the theorem. Although I do not think it is a good idea to duplicate the definition of the admittance matrix here. MathMartin 17:32, 30 Jan 2005 (UTC)

hmm?

Fair enough we an make a spanning tree, a loop free mechanism to determine the nodes, we can also determine the number of spanning trees in the graph (logically if there are k Vertices in the graph there is always a way to connect all k via some loop free path) however i doubt how it helps to say that path 102 is the same as path 201

Cayleys formula

Seeing that Cayley's formula follows from Kirchhoff's theorem as a special case is easy: every vector with 1 in one place, -1 in another place, and 0 elsewhere is an

eigenvector of the Laplacian matrix of the complete graph, with the corresponding eigenvalue being n. These vectors together span a space of dimension n-1, so there

are no other non-zero eigenvalues.

I don't think this is fully correct. First it should be mentioned that we're talking about $M_{11}$ , not $Q$ . $M_{11}$ is $n-1\times n-1$ and has $n-2$ eigenvectors of the form $(\dots ,0,1,0,\dots ,0,-1,0,\dots )$ to the eigenvalue $n$ and one eigenvector $(1,1,\dots ,1)$ to the eigenvalue 1. Multiplying those eigenvalues gives obviously $n^{(}n-2)$ . —Preceding unsigned comment added by 134.169.77.186 (talk) 13:33, 14 November 2008 (UTC)[reply]

Illustration

In case anyone is interested, here are illustrations that could make the counting of spanning trees more intuitive:

=>

In this pecular case there are 11 spanning trees covering the original graph.

I hope that helps, --MathsPoetry (talk) 23:59, 16 March 2012 (UTC)[reply]

Error in introduction

The determinant of the Laplacian matrix is 0, because the constant vector 1,1,1...,1 is in the matrix's kernel. So this determinant is not the number of spanning trees. — Preceding unsigned comment added by Vincent Semeria (talk • contribs) 19:52, 9 May 2020 (UTC)[reply]

I agree. I added "minor" in the first sentence, which I think fixes the problem. (Ehm, I was very close to marking it as a "minor edit".) --St.nerol (talk) 14:14, 24 January 2022 (UTC)[reply]