Brouwer Fixed Point Theorem and Matrix Eigenvalues

Proposition

A square matrix with all positive entries has a positive eigenvalue.

Proof.

Consider the standard simplex:

$$\begin{array}{}\text{(1)}& {\mathrm{\Delta }}^{n-1}:=\left\{\sum _{i=1}^n{\theta }_i{e}_i\right|{\theta }_i\ge 0,\text{and}\sum _{i=1}^n{\theta }_i=1\},\end{array}$$

which consists of points in ${\mathrm{\Delta }}^{n-1}$$\Delta^{n-1}$ with ${\ell }^1$$\ell^1$-norm equal to 1.

For a matrix $A:{\mathbb{R}}^n\to {\mathbb{R}}^n$$A: \mathbb{R}^n \to \mathbb{R}^n$, define the function:

$$\begin{array}{}\text{(2)}& f(x)=\frac{Ax}{\parallel Ax{\parallel }_1},\end{array}$$

which maps points in ${\mathrm{\Delta }}^{n-1}$$\Delta^{n-1}$ back to ${\mathrm{\Delta }}^{n-1}$$\Delta^{n-1}$ (since the entries of $A$$A$ are all positive) and is a continuous function.

By the Brouwer Fixed Point Theorem:

Every continuous function from a nonempty convex compact subset $K$$K$ of a Euclidean space to $K$$K$ itself has a fixed point.

We know that $f(x)$$f(x)$ has at least one fixed point, i.e., there exists $x\in {\mathrm{\Delta }}^{n-1}$$x \in \Delta^{n-1}$ such that:

$$\begin{array}{}\text{(3)}& Ax=\parallel Ax{\parallel }_{1}x.\end{array}$$