Does the non-symmetric irreducible matrix with positive eigenvalues is positive definite?

Question

If $A$ is irreducible and all of the eigenvalues of a matrix $A$ are positive, then can we conclude $x^T A x>0$ for all $x≠0∈ℝ^n$?

Answer 1

No, we can't. Take the matrix $$\begin{pmatrix}1&2\\\epsilon&1\end{pmatrix}$$ (with sufficiently small positive $\epsilon$) as a counterexample. The symmetric matrix which defines the same quadratic form is $$\begin{pmatrix}1&1+\epsilon/2\\1+\epsilon/2&1\end{pmatrix},$$ so it isn't positive definite. On the other hand if $\epsilon$ is small, the eigenvalues of the original matrix are close to one (and positive).