We are given non-oriented graph without loops. Task is to prove that adjacency matrix of that graph has negative eigenvalue.
I put some effort into drawing a proof here , but it seems that I'm missing some links between statements. So any pointers would be appreciated.
According to eigenvalue definition, $det(A - \lambda \cdot I) = 0$ should hold.
Taking in account given description of graph, matrix would be somewhat like:
$ A =\left( \begin{array}_
0 & a_{1,2} & ... & a_{1,n} \\
a_{2,1} & 0 & ... & a_{2,n} \\
... & ... &... & ... \\
a_{n,1} & a_{n,2} & ... & 0 \\ \end{array} \right)$
where $a_{i,j} > 0$.
Also it might be important that since it's an adjacency matrix, it's symmetric, hence diagonalizable.
Actually, it is not true without further assumptions. If the graph has no edges, the only eigenvalue will be 0. However, except for that case --
Hint: As you have noticed, the matrix can be diagonalized. What is the trace of a diagonal matrix? What is the trace of the adjacency matrix?