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Suppose, $K_n$ is a complete simple graph with each edge label $k$. Then its adjacency matrix $A(K_n)$ has all the entries zero along the diagonal, and each non-diagonal entries are $k$.

What are the eigenvalues or, the spectrum of $K_n$? Is there any known formula for this?

gete
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  • Well, you know that since the sum of each row is $(n-1)k$, then it is an eigenvalue. Maybe there's more, that's just the first thing that comes in mind. – GSofer Sep 20 '20 at 15:49
  • @GSofer Still, I couldn't figure out as how is this question duplicate? It seems like finding all $n$ Eigen values remained a difficult problem for me. Well do you mean that all the Eigenvalues will be $(n-1)k$ in this case? Could you please elaborate? – gete Sep 20 '20 at 16:34
  • this is definitely an interesting problem, off the top of my head I'm not sure what are the remaining eigenvalues, this was just the first thing that came to mind – GSofer Sep 20 '20 at 17:53

1 Answers1

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If $A$ is your matrix, then $A+kI$ is the matrix where all entries are $k$'s, which has rank $1$. Therefore $-k$ is an eigenvalue of your matrix with multiplicity $n-1$.

(In general, $\lambda$ is an eigenvalue with multiplicity $m$ if the rank of $A - \lambda I$ is $n-m$.)

This gives us all but one of the eigenvalues. But we know that the trace of $A$ is $0$, so the eigenvalues have to add up to $0$. If $n-1$ of them are $-k$, then the remaining one should be $(n-1)k$.

Misha Lavrov
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