A linear map $T: \mathbb{R^3 \to \mathbb{R^3}}$ has a two dimensional invariant subspace.

Question

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.

Answer 1

There is a simpler (and cleaner) solution: the adjoint has a real eigenvalue with associated eigenvector $v$. Now, the orthogonal complement of $span(v)$ is a $2$ dimensional invariant subspace of $T$.

Answer 2

If the minimal polynomial is of the form $(X-a)^3$, then there is some basis $B=\{e_1,e_2,e_3\}$ of $\mathbb{R}^3$ such that the matrix of $T$ with respect to $B$ is$$\begin{bmatrix}a&1&0\\0&a&1\\0&0&a\end{bmatrix}.$$So, consider the space spanned by $e_1$ and $e_2$.

Answer 3

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 $f\left(\ker(g)\right) \subseteq \ker(g)$.

Answer 4

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.