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Having the following Laplacian matrix :

0    0   0   0  0

0    0   0   0  0

-1  -1   2   0  0

0    0   0   0  0

0    0   0  -1  1

I got those eigenvalues and eigenvectors :

E = 0 : V : {1,0,0.5,0,0}

E = 0 : V : {0,1,0.5,0,0}

E = 0 : V : {0,0,0,1,1}

I know from the graph that vertices 1+2+3 are component connected and 4+5 are components connected by themself, how can i conclude this from the vectors above ?

user5132182
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  • You’ve either got too many rows or are missing a column in your matrix. Laplacian matrices are square. – amd Dec 22 '15 at 21:42
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    Possible duplicate of Connected components and eigenvectors – amd Dec 22 '15 at 21:44
  • @amd the col with the zeros is wrong, i edited my question, its 5,5 matrix, – user5132182 Dec 23 '15 at 07:21
  • Good. Still not quite there, though. A Laplacian matrix should also be symmetric. – amd Dec 23 '15 at 08:36
  • @amd I have a directed graph,, also what I really need to know is : 1) in wikipedia it says - one of the properties for undirected graph and its laplacian matrix is "The number of times 0 appears as an eigenvalue in the Laplacian is the number of connected components in the graph." , does this appeal if the graph directed ? 2) If I have that eigenvectors for any Laplacian matrix, how to understand which are the connected components vertices ? Thanks very much. – user5132182 Dec 24 '15 at 15:02
  • This still works with a directed graph, but you have to produce a proper Laplacian matrix for it. You need an undirected adjacency matrix for it, which will be symmetric. If you’ve got a directed incidence matrix $M$ instead, the Laplacian is $M^TM$. See http://math.stackexchange.com/a/1553633/265466 for details of how to determine the connected components from the basis of the kernel. – amd Dec 24 '15 at 18:02

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