<s>[INST] Characteristic Polynom of a Matrix for the complete Graph, I need to derive a formula for the characteristic polynom of a complete graph. Here are some example of the adjacency Matrices:
$$
K_2=
\begin{pmatrix}{}
 0 & 1 \\
 1 & 0 \\
\end{pmatrix}
$$
$$
\chi(K_2)=(-1 + x) (1 + x)
$$
$$
K_3=
\begin{pmatrix}
 0 & 1 & 1 \\
 1 & 0 & 1 \\
 1 & 1 & 0 \\
\end{pmatrix}
$$
$$
\chi(K_3)=-(-2 + x) (1 + x)^2
$$
$$
K_4=\left(
\begin{array}{cccc}
 0 & 1 & 1 & 1 \\
 1 & 0 & 1 & 1 \\
 1 & 1 & 0 & 1 \\
 1 & 1 & 1 & 0 \\
\end{array}
\right)$$
$$
\chi(K_4)=(-3 + x) (1 + x)^3
$$
$$
K_5=
\left(
\begin{array}{ccccc}
 0 & 1 & 1 & 1 & 1 \\
 1 & 0 & 1 & 1 & 1 \\
 1 & 1 & 0 & 1 & 1 \\
 1 & 1 & 1 & 0 & 1 \\
 1 & 1 & 1 & 1 & 0 \\
\end{array}
\right)$$
$$
\chi(K_5)=-(-4 + x) (1 + x)^4
$$
I see the pattern:
$$
\chi(K_n)=(-1)^{n}(-(n-1) + x) (1 + x)^{(n-1)}
$$
Although I have no idea how to proove it. Any ideas?
Application:
From the closed formula one can derive the number of closed walks in the graph
 [/INST] 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.
</s>