Sum of the eigenvalues of adjacency matrix Let $G$ be a simple undirected graph with $n$ vertices, and let $A_G$ be the corresponding adjacency matrix.  Let $\kappa_1, \dots , \kappa_n$ be the eigenvalues of the adjacency matrix $A_G$.  I have read that
$$
\kappa_1 + \dots + \kappa_n = 0,
$$
and I have checked this by hand for $n \leq 3$ along with a few "random" graphs.  It is true in all the cases I have checked.
It seems that we want to write $A_G = U^{T} D U$, where $D$ is a diagonal matrix with trace $\text{tr}(D) = 0$, but I have been unable to supply a proof.  How would one prove this?

If there are no self loops, diagonal entries of a adjacency matrix are all zeros which implies $trace(A_G)=0$. Also, it is a symmetric matrix. Now use the connection between the trace of a symmetric matrix and sum of the eigenvalues (they are equal). To prove this, since $A_G$ is symmetric, $A_G=U^{-1}DU$ for some unitary matrix $U$. Now, note that trace has circularity property, i.e. $trace(ABC)=trace(BCA)$. So 
\begin{align}
0=trace(A_G)=trace(U^{-1}DU)=trace(DUU^{-1})=trace(D)
\end{align}
and $trace(D)$ is the sum of eigen values. 
