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How can somebody apply PCA on a set of graphs? Is it possible to define a meaningful graph kernel for my problem, and then follow the typical procedure on the derived matrix of pairwise distances (kernel PCA) ?

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How can somebody apply PCA on a set of graphs?

It is perfectly legal to apply PCA on a set of graphs. After all, one graph is just a point in a space, and given a distance measure (kernel) in that space you can apply PCA as usual.

Is it possible to define a meaningful graph kernel for my problem, and then follow the typical procedure on the derived matrix of pairwise distances?

That's absolutely fine -- once you have a matrix of pairwise distances you're back to good old PCA.

If you're looking for some interesting kernels I would recommend you to search online for "graph distance measures" or "graph similarity measures," and you will find a lot of them. One particularly easy example is the Hamming distance, which is basically the number of edge additions or deletions needed to transform one graph into another.

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  • $\begingroup$ Thanks for your answer. Commenting on your point about seeing the graph as a point in a space - well I can think of that one such mapping would be using the adjacency matrix rows as coordinates in a high dimensional space. But this becomes a bit problematic when it comes to graphs of different sizes. And particularly when there is no mapping among the nodes of two graphs (modulo the permutations of the nodes) - do you have in mind a more natural instantiation of a graph as a point in space? $\endgroup$ – Dionysis M May 24 '17 at 7:11
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    $\begingroup$ @DionysisM Unfortunately I don't think thinking about graphs in terms of coordinates will take you very far. It's important to know the difference between a Euclidean space (where you have lengths and distances and you can actually plot points in it) and a metric space (where you only have distances between points and nothing else). $R^n$ is a Euclidean space while graphs live in a metric space, so there is no way you can map graphs onto $R^n$ without extra assumptions. I'm afraid you'll have to get familiar with the abstract mathematical notions of space and metric. $\endgroup$ – Pedro Mediano May 24 '17 at 8:41

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