Why is a graph's Laplacian matrix positive semidefinite? Can anyone provide an intuitive explanation and a proof?
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Rodrigo de Azevedo
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user1559897
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3It's symmetric and diagonally dominant. – Adam Hughes Oct 30 '14 at 01:26
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2See also the Gershgorin circle theorem – Ben Grossmann Oct 30 '14 at 01:36
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Two nice proofs that the Laplacian matrix of a graph is PSD: https://people.orie.cornell.edu/dpw/orie6334/lecture7.pdf – JStrahl Jan 07 '22 at 09:20
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If $B$ is the incidence matrix of an orientation of $G$, then $L=BB^T$. So
$$ x^TLx = \|Bx\|^2 \ge 0$$
for all $x$.
The matrix has rows indexed by the vertices, columns by edges and the $ij$-entry is 1 if the $i$-th vertex is the head of the $j$-th edge, $-1$ if its the tail and 0 otherwise.
Rodrigo de Azevedo
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Chris Godsil
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