TITLE: A linear map $T: \mathbb{R^3 \to \mathbb{R^3}}$ has a two dimensional invariant subspace.
QUESTION [6 upvotes]: Let $T: \mathbb{R^3 \to \mathbb{R^3}}$ be an $\mathbb{R}$-linear map. Then I want to show that $T$ has a $2$ dimensional invariant subspace of $\mathbb{R^3}.$

I considered all possible minimal polynomial of $T$ and applying canonical forms I found some obvious $2$ dimensional invariant subspaces.

I stuck when the minimal polynomial is of the form $(X-a)^3$ for some real number $a.$ 

In this situation since the minimal polynomial and the characteristic polynomial coincides $T$ has a cyclic vector. But I can't complete it further. I need some help. Thanks.

REPLY [2 votes]: The characteristic polynomial of the transpose map $f^T$ is cubic with real coefficients. Hence it has a real root $\alpha$, and there is a real eigenvector $v=(a,b,c)$ such that $f(v)=\alpha v$. Then the 2-dim space given by the equation
$$ax+by+cz=0$$ is invariant under $f$. That is because
$$\langle v, f(w)\rangle=\langle f^T(v),w\rangle=\alpha\langle v, w\rangle=0$$
for any vector $w$ in the subspace.