Kernel of a "degenerate" diagonally dominant matrix

Question

I was trying to prove the existence of a fixed point and I am stuck with a technical problem I cannot overcome. I give a detailed explanation in case someone kindly knows or has some idea on how to solve this:

Assume we are given a matrix $Q = (q_{ij}) \in M_n(\mathbb{R})$ satisfying the following:

• For any $i \neq j$, $q_{ij} \geq 0$.
• For any $i \in [n]$, $\; 0 > q_{ii} = - \sum_{j \neq i} q_{ij}.$ In other words, $(1, \cdots , 1) \cdot Q = \vec{0},$ and the diagonal terms cannot be zero.

Since the matrix $Q$ has $(1, \cdots , 1)$ as left eigenvector with eigenvalue $0$, it has a right eigenspace associated to this eigenvalue $0$, with dimension at least $1$. Differently explained, I can consider the nontrivial vectorial space $$A = \{ v \in \mathbb{R}^n \; | \; Q \cdot v = \vec{0} \}.$$ My question is: Does there exist a nonzero $v \in A$ such that each of its components is nonnegative (maybe even positive?) ?

However, this is what I have not been able to do, and why ask you for help. I guess the result is true, at least so it is for cases $n = 2,3$.

Thanks a lot for your advise, any suggestion is welcome!

Answer 1

First, let's consider the case when all $q_{ij}>0$ if $i\ne j$. Take the matrix $P=\alpha I+Q$ where $\alpha+q_{ii}>0$ for all $i$. The matrix $P$ has all its elements positive. This means that we can apply the Perron-Frobenius theorem to it. This means that there is a unique positive eigenvector $(v_1,\dots,v_n)^T$ which corresponds to the maximum eigenvalue. Also, if we apply this theorem to $P^T$ we'll get the same conclusion (unique positive eigenvector, and it corresponds to the maximum eigenvalue), but since we already know that $(1,\dots,1)$ is such the eigenvector (and $P^T(1,\dots,1)^T=(\alpha I+Q^T)(1,\dots,1)^T=\alpha(1,\dots,1)^T$, we can conclude that the maximum eigenvalue is $\alpha$.

To summarize: for $Q$ which meets the problem conditions and have strictly positive off-diagonal elements there is a unique positive eigenvector which corresponds to the zero eigenvalue.

If the matrix $P=\alpha I+Q$ were irreducible, you could apply the Perron-Frobenius theorem to it as well, but if it isn't you can't prove the same strong statement (uniqueness, positivity).

Though if you take a sequence of matrices $Q_m$ with elements $q^m_{ij}=q_{ij}+1/m$, for $i\ne j$, and $q^m_{ii}=q_{ii}-(n-1)/m$, then you'll find that for $Q_m$ there exists a positive eigenvector $v_m$ such that $|v_m|=1$ and $Q_mv_m=0$. Since the unity sphere is compact, we can find a converging sequence $v_{m_k}\to v$ (here $m_k\to\infty$ as $k\to\infty$, so $Q_{m_k}\to Q$). Therefore, $Qv=\lim_{k\to\infty}Q_{m_k}v_{m_k}\equiv0$. Also, the vector $v$ is non-negative.