# Eigenvector with non-negative coordinates if matrix has non-negative entries

Suppose $A$ is a $n\times n$ matrix with non-negative real entries. Why is there an eigenvector with non-negative coordinates?

Before, I have proven the Brouwer fixed-point theorem, i.e. that every continuous function $f:\mathbb{D}^m\to\mathbb{D}^m$ has a fixed-point. Maybe this helps here.

Unfortunately I don't have any ideas, so I can't say anything to my solutions.

• Consider the simplex $S = \{ x \in \mathbb{R}^n : 0 \leqslant x_k \text{ for all }k, \text{ and } x_1 + \dotsc + x_n = 1\}$. Use $A$ to define a continuous $f\colon S \to S$. – Daniel Fischer May 14 '17 at 20:10
• I think math.stackexchange.com/a/268899/161825 may interest you. – Jonas Dahlbæk May 14 '17 at 21:51
• Alright but how can Brouwer's fixed-point theorem be applied here? This is valid for continuous functions $\mathbb{D}^m\to\mathbb{D}^m$. I can prove that it is also valid for balls $B_R(x)$ rather than for $\mathbb{D}^m=B_1(0)$. But what now? – user369147 May 15 '17 at 7:29
• $\mathbb{D}^{n-1}$ is homeomorphic to the simplex $\{ x \in \mathbb{R}^n : 0 \leqslant x_k \text{ for all }k, \text{ and } x_1 + \dotsc + x_n = 1\}$ – JJR May 15 '17 at 17:43
• Why that? Was is an homoemorphism? – user369147 May 15 '17 at 21:04

As the elements of your matrix are non-negative, then A is also a linear transformation on the subspace of elements with non-negative entries (that is, your transformation maps non-negative vectors to the subspace of non-negative vectors). Next, considering the transformation on that subspace, observe that it will also be a positive semi-definite operator (that is, any scalar product $(Ax, x) \ge 0$). And according to the fundamental theorem of algebra, operator A will have eigenvalues, which will also be non-negative (because any eigenvalue $\lambda$ can be written in a form of $\lambda = (Ax,x)$, where $x$ is a normalized eigenvector). In case they are real, the corresponding eigenvectors will also have non-negative real entries.