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Feb 18, 2020 at 13:01 comment added N math @Tony Huynh Thanks for your proof.
Feb 18, 2020 at 12:50 vote accept N math
Feb 15, 2020 at 12:32 comment added Tony Huynh @BrendanMcKay Yes, I swept that part under the rug. Thanks for your comment!
Feb 15, 2020 at 12:17 history edited Tony Huynh CC BY-SA 4.0
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Feb 15, 2020 at 10:34 comment added Brendan McKay A tiny bit is missing in order to go to the complement. Namely, a regular graph cannot be cospectral to an irregular graph. This is an old result of Sachs: a graph is regular iff the sum of the squares of the eigenvalues equal $n$ times the largest eigenvalue.
Feb 15, 2020 at 8:09 comment added Tony Huynh You're welcome. I also added a proof of Proposition 6 here. If you are satisfied with the answer, you can click on the green check mark to show that it has been answered.
Feb 15, 2020 at 8:07 history edited Tony Huynh CC BY-SA 4.0
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Feb 14, 2020 at 12:54 comment added N math Thanks for your answer.
Feb 14, 2020 at 10:57 comment added Tony Huynh Yes, that's correct.
Feb 14, 2020 at 9:54 comment added N math Great. Thanks for your answer. You mean since these graphs are regular, we can take $\overline{A}$, instead of A, the adjacency matrix, it's right?
Feb 14, 2020 at 6:08 history answered Tony Huynh CC BY-SA 4.0