suppose that $G$ is connected graph and for every eigenvalue of its adjacency matrix we have $\lambda \geq -1$. prove that $G$ is complete graph.
I think that the easiest way is to show that we have just two distinct eigenvalues,it forces the graph to have diameter 1!so it is complete!
I don't know how to show it.any hint or Idea or theorem that make it possible will be great .thanks a lot.
You probably mean adjacency matrix.
For $|G| = 2$, $G$ is connected iff $G$ is complete so it is true.
For $|G| = 3$, the only case needed to check is the one where the graph is a path, but this has an eigenvalue $-\sqrt{2}$.
Now, suppose it is true for $|G| < n$ ($n \ge 4$).
Now, we prove for $|G| = n \ge 4$. Let $A$ be the adjacency matrix of $G$.
$A$ has every eigenvalue $\lambda \ge -1$ iff $A + I$ is positive semidefinite.
By Sylvester's Criterion, this means that $B + I$ is also positive semidefinite, where $B$ is the matrix obtained by deleting the first row and first column of $A$, i.e. $B$ also has every eigenvalue $\ge -1$. By induction, if the graph corresponding to $B$ is connected the it is complete.
So, we've proven that if the subgraph induced by some $n-1$ vertices of $G$ is connected then it is complete. Now, there exists one vertex which does'nt disconnnect the graph (consider a leaf of the spanning tree), so we know there exists a $(n-1)$-clique. Let $V = \{v_1,\cdots,v_n\}$, with $\{v_2,\cdots, v_n\}$ spanning a regular graph, and assume $(v_1,v_2) \in E$. We only need to show that the $1$st vertex is connected to all of them.
Now, the graph induced by deleting $v_3$ is connected, so it must be complete, hence $v_1$ is connected to $v_4,\cdots, v_n$. In particular, this implies that the graph induced by deleting $v_2$ is connected, so it must also be complete, hence $v_1$ is connected to $v_3$, completing the proof.