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In graph theory, given a graph $G$ with $n$ nodes and an edge set $E$, the Laplacian matrix is defined as the difference of the degree matrix and the adjacency matrix:

$$ L=D-A \tag{1} $$

As shown here, the quadratic form of $L$ is as follows: ($x$ is simply a function of vertices, therefore has $n$ elements) $$ \begin{align*} x^T L x &= x^T D x - x^T Ax\\ &= \sum_{i=1}^n \text{deg}(i)x_i^2 - \sum_{\{i,j\}\in E} 2x_i x_j \tag{2}\\ &=\sum_{i=1}^n \sum_{\{i,j\}\in E} x_i^2 - \sum_{\{i,j\}\in E} 2x_i x_j \tag{3}\\ &= \sum_{\{i,j\}\in E} (x_j^2+x_i^2-2x_ix_j)\tag{4}\\ &= \sum_{\{i,j\}\in E} (x_i-x_j)^2\tag{5} \end{align*} $$

Questions:

  1. I feel I am missing something obvious, but how do we go from $(3)$ to $(4)$?
  2. Is there a simple way to see from the quadratic form that the smallest eigenvalue ought to be $0$ and that as soon as the second smallest eigenvalue is $>0$ the corresponding graph must be connected?
user929304
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1 Answers1

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For 1:

The sum $$\sum_{i=1}^n\sum_{\{i,j\}\in E} x_i^2$$ does the following:

For each $i$, take all of the neighbors of $i$, and add $x_i^2$ to the sum.

This means that, given any edge $\{i',j'\}$, it will appear in the sum once when the outer sum hits $i=i'$ , and once more when the outer sum hits $i=j'$. The first time, you will add $x_{i'}^2$ to the result, the second time, you will add $x_{j'}^2$ to the result. This will happen to all of the edges, therefore,

$$\sum_{i=1}^n\sum_{\{i,j\}\in E} x_i^2=\sum_{\{i,j\}\in E} (x_i^2+x_j^2)$$

For 2:

If $\lambda$ is an eigenvalue and $x$ is its eigenvector, then $$\sum_{\{i,j\}\in E} (x_i-x_j)^2 = x^TLx=\lambda \|x\|^2$$

The left side is a sum of nonnegatives, and $\|x\|^2$ is nonnegative.

5xum
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  • Many thanks! Any ideas as to how the positivity of the 2nd smallest eigenvalue relates to the graph being connected? It's only hinted at in the exercise 2.2 of the link I had included in the question. – user929304 Apr 04 '19 at 13:27
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    @user929304 https://math.stackexchange.com/questions/1407811/why-is-second-smallest-eigenvalue-and-the-corresponding-eigenvector-used-to-part – 5xum Apr 04 '19 at 13:30
  • @5xum I am a little confused. Aren't we summing over a different set in your explanation for point 1? Let's assume we have a path graph in the following form: (1)-(2)-(3). In the first sum, $\sum_{i=1}^3 \sum_{{i,j}\in E} x_i^2$, when we take the set E={{1,2},{2,1},{2,3},{3,2}}, we get the correct result. But with the sum $\sum_{{i,j}\in E} (x_i^2+x_j^2)$, when we take the set E={{1,2},{2,3}}, we get the correct result. What am I missing here? – MathAccount12 Aug 11 '23 at 23:03
  • @MathAccount12 Both your sets are identical. Because ${1,2}={2,1}$ and ${2,3}={3,2}$, you also have ${{1,2},{2,1},{2,3},{3,2}}={{1,2}, {2,3}}$. – 5xum Aug 16 '23 at 04:50