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How can I prove that is a graph is vertex transitive, then a eigenvalue is 1?

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How do you know this is true? – Igor Rivin Feb 17 2012 at 3:46
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 It's not true, so "with difficulty" is the answer to your question. – gordon-royle Feb 17 2012 at 3:50
One way to prove that $k$ is an eigenvalue is exhibit an eigenvector. A square (4 cycle) labelled 1 x 1 x with x=-1 shows that -2 is an eigenvalue. – Aaron Meyerowitz Feb 17 2012 at 5:57

closed as too localized by Will Jagy, Mark Sapir, Yemon Choi, Aaron Meyerowitz, Alain Valette Feb 17 2012 at 8:21

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