Proving existence of $T$-invariant subspace Let $T:\mathbb{R}^{3}\rightarrow \mathbb{R}^{3}$ be a linear transformation. I'm trying to prove that there exists a T-invariant subspace $W\subset \mathbb{R}^3$ so that $\dim W=2$.
How can I prove it? Any advice?

Just look at its Jordan nomal form.
If there're three Jordan blocks of dimension 1,  the result is clear.
If you have one Jordan block of dimension 1 and one of dimension 2, then then the result is obvious, too.
Thus, we need to study the case where you matrix is a Jordan block of dimension 3, i.e.
$$\begin{pmatrix}\lambda&1&0\\0&\lambda&1\\0&0&\lambda\end{pmatrix}.$$Note, that in this case $\lambda$ is real. Clearly, the subspace $span\{e_1,e_2\}$ is invariant, which finalises the proof.
An edit to reflect the questions in comments
Obviously, we talk about the real Jordan normal form as described in this wiki article.
Let's take a look at possible eigenvalue structure our matrix can have.
First of all, it has at least one real eigenvalue (because the dimension is odd). This leaves us with two possible choices: either other eigenvalues are real, or they are complex conjugate. If they are complex conjugate (let's call them $a\pm ib$), then there exist two eigenvectors in $\mathbb C^3$, noted by  $v_\pm$ respectively. It's easy to show that, in fact $\bar v_- =v_+$ (i.e. they differ only by the sign at their imaginary part). Let's look what happens when we apply $A$ to $\Re v_+$ and to $\Im v_+$:
$$(a+ib)v_+=Av_+=A\Re v_+ +iA\Im v_+ = (a+ib)( \Re v_+ +i \Im v_+)=a\Re v_+ -b \Im v_+ +i(a\Im v_+ +b\Re v_+ ),$$
hence
$$A\Re v_+ =a\Re v_+ -b \Im v_+,\quad A\Im v_+= a\Im v_+ +b\Re v_+,$$
so $span \{ \Im v_+, \Re v_+\} $ is an invariant subspace.
In the case where two other eigenvalues are real, too, we can safely apply the usual Jordan normal form (upper-triangular) and deduce the necessary result.
