4
$\begingroup$

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

$\endgroup$
0
10
$\begingroup$

Here is a topological approach.

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 $T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.