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I need to derive a formula for the characteristic polynomial of a complete graph. Here are some example of the adjacency matrices:

$$ K_2 = \begin{pmatrix}{} 0 & 1 \\ 1 & 0 \end{pmatrix} , \qquad \chi(K_2)=(-1 + x) (1 + x) $$

$$ K_3 = \begin{pmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{pmatrix} , \qquad \chi(K_3)=-(-2 + x) (1 + x)^2 $$

$$ K_4 = \begin{pmatrix} 0 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{pmatrix}, \qquad \chi(K_4)=(-3 + x) (1 + x)^3 $$

$$ K_5 = \begin{pmatrix} 0 & 1 & 1 & 1 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 1 & 1 \\ 1 & 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 1 & 0 \end{pmatrix}, \qquad \chi(K_5)=-(-4 + x) (1 + x)^4 $$

I see the pattern

$$ \chi(K_n)=(-1)^{n}(-(n-1) + x) (1 + x)^{(n-1)} $$

but I have no idea how to prove it. Any ideas?

Application: from the closed formula one can derive the number of closed walks in the graph.

Rodrigo de Azevedo
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thetha
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1 Answers1

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If you add the identity then you get a rank-one matrix whose characteristic polynomial is $$\chi(K_n + I) = (-1)^n x^{n-1} (x - n).$$ The factor $(x-n)$ coming from the lone eigenvalue $n$ with eigenvector $(1,1,...1).$ When passing from $K_n+I$ to $K_n$ you substitute $x \mapsto x+1$ and get the formula you want.

user420261
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