Let $A$ denote the adjacency matrix of a graph $G$ with $n$ vertices and $e$ edges. Let $\lambda_1\ge \lambda_2\ldots \ge \lambda_n$ be the eigen values of $A$ .
Show that :
$\sum_{i=1}^n \lambda_i^2=2e$
My try: Applying induction on $n$ . Checking for $n=2$; $\sum_{i=1}^2 \lambda_i^2=(\lambda_1+\lambda_2)^2-2\lambda_1\lambda_2$
Since trace of adjacency matrix is zero so $(\lambda_1+\lambda_2)=0$ and $\lambda_1\lambda_2=-e$ since sum of all $2\times 2$ principal minors is equal to $-e$
How to proceed after this ?Please help.
Using the definition of an adjacency matrix $A$, it follows that $[A^2]_{ii}$ is equal to the number of edges connected to $i$, i.e. the degree of $i$: $d_i$. Now recall (or easily rederive) that $\sum_i d_i=2e$. Finally use the fact that the eigenvalues of $A^2$ are $\lambda_i^2$ along with the fact that their sum is equal to the trace of $A^2$.