# Given a matrix with non-negative real entries, can you algebraically prove that it has a non-negative eigenvalue?

I am looking specifically for an algebraic proof, but if you can offer both algebraic and topological proofs, I will appreciate it even more.

Name the associated linear function $T$, $T:\mathbb{R}^n\rightarrow\mathbb{R}^n$. If $T$ has a $0$-eigenvectors, then we are done. Let's consider the other cases.
Since all entries of $T$ are positive, we can define a map $T^*$, $T^*:S_{\geq0}^n\rightarrow S_{\geq0}^n$ by $T(x)= \frac{T(x)}{\|T(x)\|}$. Here, $S_{\geq0}^n$ denotes the intersection of the unit sphere with the "positive quarter" of $\mathbb{R}^n$. Now, because $S_{\geq0}^n$ is homeomorphic to $D^{n-1}$, due to Brouwer fixed-point theorem, there exists a point $x_0$ such that $T^*(x_0) = x_0$. Equivalently, $x_0 = \frac{T(x_0)}{\|T(x_0)\|}$ or $T(x_0) = \|T(x)\| x_0$. So, $x_0$ is an eigenvector of $T$ and $\|T(x)\|$ is a positive eigenvalue of $T$.