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The algebraic connectivity of a graph G is the second-smallest eigenvalue of the Laplacian matrix of G. This eigenvalue is greater than 0 if and only if G is a connected graph. The magnitude of this value reflects how well connected the overall graph is.

for an example, "adding self-loops" does not change laplacian eigenvalues (specially algebraic connectivity) of graph. Because, laplacian(G)= D-A is invariant with respect to adding self-loops.

My question is:

Does anyone has studied effect of different operations (such as edge contraction) on spectrum of laplacian? do you know good references?

Remark: the exact definition of the algebraic connectivity depends on the type of Laplacian used. For this question I prefer to use Fan Chung definition in SPECTRAL GRAPH THEORY. In this book Fan Chung has uesed a rescaled version of the Laplacian, eliminating the dependence on the number of vertices.

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It would help if you provide some motivation and background. Please see How to ask a good question? and the site's FAQ. –  Kaveh Mar 11 '11 at 21:54
    
I'm also interested in the edge contraction case. I've spend some time previously trying to find references about the relation between eigenvalues and minor operations, with no success. –  Hsien-Chih Chang 張顯之 Mar 16 '11 at 1:33
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To me, the motivation seems fairly clear. –  Suresh Venkat Sep 24 '11 at 20:12
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I second Suresh, knowing how various operations influence the Laplacian is interesting in itself and this problems shows up in various contexts. –  Marcin Kotowski Sep 24 '11 at 20:50

1 Answer 1

up vote 5 down vote accepted

Intuitively operations that preserve connectivity will not decrease the eigenvalues. For example, adding edges to the graph does not decrease the connectivity.

In general, if H is a subgraph of a graph G, by interlacing we know that the i-th largest Laplacian eigenvalue of H is no larger than the i-th largest Laplacian eigenvalue of G. A proof can be found in Proposition 3.2.1 of the book "Spectra of graphs" by Brouwer and Haemers. Note that the definition of Laplacian used in the book is not normalized; it has node degrees on the diagonal and -1 (or 0 if there is no edge) in the off-diagonal entries.

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Thanks Chang. Your answer is really useful for me. But if we use the definition of Laplacian that is not normalized, then many of comparisons seems to be meaningless. For example we have Algebraic Connectivity(K10)=10 and Algebraic Connectivity(K20)=20. however, both graphs are fully connected simple graphs. But if we use the normalized Laplacian, then NormalizedAlgebraicConnectivity(K10) = NormalizedAlgebraicConnectivity(K20)=1 and therefore comparison of normalized version seems to be more rational and natural. –  b.a Mar 16 '11 at 2:03
    
@behnam: I agree with you. But after the normalization, some of the nondecreasing properties may differ. (Say one can ensure a strict decreasing on the largest Laplacian when deleting edges for the unnormalized ones, but not for the normalized ones.) –  Hsien-Chih Chang 張顯之 Mar 16 '11 at 17:16
    
@Hsien-Chih Chang "Intuitively operations that preserve connectivity will not decrease the eigenvalues. For example, adding edges to the graph does not decrease the connectivity." Are you sure? Do you have a proof? Is the following a counter example? Start with a path graph and add edges to form the Lollipop graph. The cover time gets worse; it goes from $\Theta(n^2)$ to $\Theta(n^3)$. What is happening to the eigenvalues in this example? –  Tyson Williams Sep 25 '11 at 13:36

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