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suppose $F$ is a forest,prove that the determinant of adjacency matrix of this forest is $-1$ or $0$ or $1$ .

I focused on trees of this forest, say that if I know eigenvalues of trees,I will multiply all together hoping they will be 1 ,-1 or 0.but I don't know anything about them.

in my searching I found useful theorem that say : $T$ is a tree with $n$ vertex,if we have complete matching $det(A)=(-1)^{\frac{n}{2}}$ otherwise it is 0.

1.can you give me some result about eigenvalues of trees.

2.I don't have the proof of theorem I mentioned,can you give me reference to provide it.

3.if I have determinant of trees,if I multiply them, do I have the determinant of forest?

thank you very much.

kpax
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  • the 3 is true, since you can renominate the nodes so that your adjacency matrix is block diagonal, with each block corresponding to a tree. – Exodd Sep 27 '14 at 15:34
  • if I do that the result is similar matrix,and nothing change about determinant,am I right? – kpax Sep 27 '14 at 15:37
  • yes, since it is a base change. (It is also orthogonal, but that's not important here) – Exodd Sep 27 '14 at 15:42
  • Here (http://www.cs.elte.hu/~lovasz/scans/eigentrees.pdf) an article on the eigenvalues/eigenvectors of trees, where this theorem is also proved – Exodd Sep 27 '14 at 15:50

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