# Lower eigenvectors of nonnegative matrices with zero trace

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)?

• Do you assume that $A$ is diagonalizable? (Since otherwise, not all if the eigenvectors $v_1, ..., v_N$ exist.) Mar 27 at 23:39
• 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. Mar 28 at 9:24

(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 $$(1,-1,0)$$.

• Thanks! Still hoping for (2) then :) Mar 23 at 21:22

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.

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