# Talk:Kirchhoff's theorem

## Untitled

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 ${\displaystyle M_{11}}$, not ${\displaystyle Q}$. ${\displaystyle M_{11}}$ is ${\displaystyle n-1\times n-1}$ and has ${\displaystyle n-2}$ eigenvectors of the form ${\displaystyle (\dots ,0,1,0,\dots ,0,-1,0,\dots )}$ to the eigenvalue ${\displaystyle n}$ and one eigenvector ${\displaystyle (1,1,\dots ,1)}$ to the eigenvalue 1. Multiplying those eigenvalues gives obviously ${\displaystyle n^{(}n-2)}$. —Preceding unsigned comment added by 134.169.77.186 (talk) 13:33, 14 November 2008 (UTC)

## 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)