Let $A$ be an $N\times N$ nonnegative matrix with all diagonal entries equal to zero and such that there is $n_0$ such that all entries of $A^{n_0}$ are strictly positive. Let $\lambda_1,\ldots, \lambda_N$ be its eigenvalues ordered in the decreasing order with respect to their real parts, and $v_1,\ldots, v_N$ be the corresponding (left) eigenvectors. Perron and Frobenius tell us that $\lambda_1$ is a strictly positive real number and therefore (since the sum of eigenvalues must be zero) there must also be eigenvalues with strictly negative real part; let $\lambda_{k_0},\ldots, \lambda_N$ be those.

Questions:

(1) is it true that the "smallest" (with respect to the real part of the corresponding eigenvalue) eigenvector $v_N$ can be chosen in such a way that all of its entries are nonzero?

(2) if the above doesn't hold, is it at least true that for any $j\in \{1,\ldots,N\}$ we can find $m\geq k_0$ such that $v_m$ has nonzero $j$th component (that is, the set of eigenvectors corresponding to eigenvalues with negative real part cannot have a common all-zero entry index)?