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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)?

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  • $\begingroup$ Do you assume that $A$ is diagonalizable? (Since otherwise, not all if the eigenvectors $v_1, ..., v_N$ exist.) $\endgroup$ Mar 27 at 23:39
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    $\begingroup$ I think I didn't formulate it well: I don't need the eigenvectors to form a basis, and actually even don't need them to be distinct (the top one and a combination of bottom ones are used to construct a Lyapunov function for some stochastic process, so I don't need all of them). So, rather, it should be "let $v_1,\ldots,v_N$" be some corresponding eigenvectors; then they can be chosen in such a way that (2) holds. $\endgroup$ Mar 28 at 9:24
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(1) No. Counterexample: the symmetric $3 \times 3$ matrix $$ M(a,b) = \left[ \begin{array}{ccc} 0 & a & b \cr a & 0 & b \cr b & b & 0 \end{array} \right] $$ with $0 < b < a$ has $\lambda_3 = -a$ with 1-dimensional eigenspace generated by $(1,-1,0)$.

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  • $\begingroup$ Thanks! Still hoping for (2) then :) $\endgroup$ Mar 23 at 21:22
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Partial answer: For the special case of self-adjoint matrices, the answer to (2) is yes. Funnily enough, this has nothing to do with the non-negativity of the matrix:

Proposition. Let $A \not= 0$ be a self-adjoint complex $N \times N$ matrix with all diagonal entries equal to $0$, let $\lambda_1 \ge \dots \ge \lambda_N \in \mathbb{R}$ be its eigenvalues, and let $v_1, \dots, v_n \in \mathbb{C}^N$ be an orthonormal basis of eigenvectors. Assume that every entry of $v_1$ is non-zero.

Then for each $j \in \{1,\dots,N\}$ there exists an eigenvector for a negative eigenvalue whose $j$-the component is non-zero.

Proof. Fix $j$ and write the $j$-th canonical unit vector $e_j$ as $$ e_j = \sum_{k=0}^N \alpha_k v_k, $$ where $\alpha_k = \langle v_k, e_j \rangle$ for each $k$ (here, I use the "physical" convention that the inner product be linear in the second component). We have $$ 0 = \langle e_j, A e_j \rangle = \sum_{k,\,\ell=1}^N \overline{\alpha_k} \alpha_\ell \langle v_k, Av_\ell \rangle = \sum_{k=1}^N \lvert \alpha_k \rvert^2 \lambda_k. $$ Since $\lambda_1 > 0$ and $\alpha_1 \not= 0$ by assumption, it follows that there exists $k_0$ such that $\lambda_{k_0} < 0$ and $\alpha_{k_0} \not= 0$. $\square$

Remark. The assumption that every entry of $v_1$ be non-zero can be weakened; without this assumption, the following is still true:

Fix $j$. If there exists an eigenvector for a positive eigenvalue whose $j$-th component is non-zero, then there also exists an eigenvector for a negative eigenvalue whose $j$-th component is non-zero. The proof is the same.

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  • $\begingroup$ Oh, that's a nice proof! (Doesn't quite solve my issue, but thanks anyway!) $\endgroup$ Mar 28 at 9:46

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