I hope the question is not a duplicate. Let me know if that's the case.
I found the following exercise in a linear algebra exercise book:
Exercise : Let $f : \mathbb{R}^3 \longmapsto \mathbb{R}^3$ a linear application. Prove that exists a vector subspace $\pi$ of $\mathbb{R}^3$ of dimension $2$ such that $f(\pi) \subseteq \pi$.
The latter is relatively simple using Jordan canonical form. I know for sure an other awesome proof involving linear algebra. To you the pleasure of discovering it.
I was wondering whether there were other proofs involving linear algebra or other advanced theory, so I'm asking if you know other proofs of the problem.
Edit: To answer Martin in the comments: No it doesn't. I do know two proofs of the facts. What I'm asking for are alternative proofs just for the sake of the pleasure of discovering new proofs of probably one of the most inspiring exercises of linear algebra I've ever done. The question was inspired from a friend of mine, who originally wanted to solve the problem analytcally, or through theorems such as Borsuk-Ulam for example.
Consider the dual linear map $$ f^t : \text{Hom}(\mathbb{R}^3,\mathbb R) \longrightarrow \text{Hom}(\mathbb R^3, \mathbb R), \qquad g\mapsto f^t(g) = g\circ f $$ This map has an eingenvector $g\in \text{Hom}(\mathbb R^3, \mathbb R)$ different from $0$. The kernel of $g$ is a two dimensional subspace of $\mathbb R^3$ and it is easy to show that $f\left(\ker(g)\right) \subseteq \ker(g)$.