This is exercise 1.1.24 in "Graph Theory" by Bondy and Murty:
$(a)$ No eigenvalue of a Graph $G$ has absolute value greater than $\Delta$
$(b)$ If $G$ is a connected graph and $\Delta$ is an eigenvalue of $G$, then $G$ is regular
$(c)$ If $G$ is a connected graph and $-\Delta$ is an eigenvalue of $G$, then $G$ is regular and bipartite
I think I have managed to proof $(a)$, but would like somebody to confirm: (Note that $\Delta$ is the maximum degree and I write $d(x_i)$ for the degree of the vertex $x_i$.
Let $\lambda$ be some eigenvalue of $A$ with eigenvector $v \neq 0$. Choose $j$ such that $| v_j |$ is maximal among all $| v_i |$.
Evaluating $Av = \lambda v$ at $j$ we get $\sum\limits_{i \neq j}a_{ij}v_i=\lambda v_j$ and thus $$| \lambda | = \frac{| \sum\limits_{i \neq j}a_{ij}v_i |}{| v_j |} \leq \frac{ \sum\limits_{i \neq j}a_{ij}|v_i |}{| v_j |} \leq \frac{ \sum\limits_{i \neq j}a_{ij}|v_j |}{| v_j |} = \sum\limits_{i \neq j}a_{ij} = d(x_j) \leq \Delta$$
I am not sure how to tackle $(b)$ and $(c)$. I do know that to show regularity it is necessary and sufficient that $\Delta$ is an eigenvalue of $A$ with eigenvector $(1,\ldots,1)$, but I don't see how $G$ being connected helps me... Any hint or partial solution for either $(b)$ or $(c)$ would be much appreciated.