Is the maximum of the eigenvalues of any symmetric positive?

Question

Let $A$ be a symmetric matrix having dimension $n \times n \;, \; \; n\geq 2$. If one wants to pick the maximum of its eigenvalues, will the value be positive? Suppose A was an adjacency matrix, should one expect the same results?

Answer 1

If one wants to pick the maximum of its eigenvalues, will the value be positive?

Of course not. Consider

$$ \begin{pmatrix} -1 & 0\\ 0 & -2 \end{pmatrix} $$

For an adjacency matrix, it's true that the largest eigenvalue is non-negative, but that's not trivial; see eg here or here or here.

Update: as A.G. notes in a comment, for a symmetric adjacency matrix (more in general: if all eigenvalues are real) it's actually trivial that the largest eigenvalue is non-negative: because, elsewhere the trace would be negative, which is imposible. By the same observation, restricting to symmetric adjacency matrices distinct from zero, then we can assert that the largest eigenvalue is strictly positive.

Answer 2

If $A$ is simetric and positive you have $(Ax,x)>0$ and if $x$ is an eigenvector you have $(\lambda x,x)>0$. So $\lambda >0$.