Let $T : \mathbb R^3 \to \mathbb R^3$ be the linear application have for matrix
$$\begin{pmatrix} 2 \ 0 \ 0 \\ 1 \ 2 \ 0 \\ 0 \ 0 \ 3 \\ \end{pmatrix}$$
Let's say $W = \ker(T-2Id)$. How do I prove that no $T$-invariant $U$ subspace exists so that $\mathbb R^3 = W \oplus U$?
Let's denote $\{e_1,e_2,e_3\}$ the canonical basis and $S=T - 2 Id$. We have $\ker S = \mathbb R e_2 = W$.
Suppose that $U$ is a linear subspace $U \subseteq \mathbb R^3$ with $\mathbb R e_2 \oplus U = \mathbb R^3$. We can write $e_3 = \lambda e_2 + u$ with $u \in U$. Then
$$e_3 = S.e_3 = \lambda S.e_2 + S.u= S.u.$$
As $U$ is supposed to be $T$-invariant, it is also $S$-invariant and $e_3$ belongs to $U$.
Let's now focus on $e_1$. We can write $e_1 = \alpha e_2 + a$ where $a \in U$. Applying $S$ on both sides of this equality we get
$$e_2 = S.e_1 = \alpha S.e_2 + S.a = S.a$$
and therefore $e_2 \in U$ in contradiction with $\mathbb R e_2 \oplus U = \mathbb R^3$.