Eigenvalues of Laplacian are non-negative

tarnat · Mar 17, 2014

Hi, I need to learn the following proof and I'm having trouble getting my head round it. Any help would be appreciated.

Show that if vector x in R^n with components x=(x1,x2,...,xn), then
x.Lx=0.5 sum(Aij(xi-xj)^2)
where A is the graphs adjacency matrix, L is laplacian.
Then use this result to prove that the eigen values of L are non-zero for all 1<j<n.

Thanks.

MarkFL · Mar 17, 2014

I have deleted the duplicate of this thread posted in the Discrete Mathematics subforum.

We ask that a question be posted only once and in the appropriate subforum. This eliminates the possibility of duplication of effort on the part of our helpers, whose time is valuable. :D

Eigenvalues of Laplacian are non-negative

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Eigenvalues of Laplacian are non-negative

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