Let $F=\begin{pmatrix} 2 & -1 & 0\\ 1 & 0 & 0\\ -1 & 1 & 1\end{pmatrix}$, show that $W$ is an invariant subspace under $F$, with $W=span\{(1,1,1),(1,1,2)\}$. Does an invariant subspace $W'$ under $F$ exist such that $\mathbb{R}^3=W\oplus W'$?
The first part is done and I don't need help with that but rather with the second question, in my opinion it shouldn't have any problems to exist so I thought so but the book says that it can't exist such $W'$, why? It seems very reasonable to me
Let's suppose that there exists an invariant subspace $W'$ such that $\mathbb{R}^3=W \oplus W'$.
Since $W$ has dimension $2$, then $W'$ has dimension $1$ : there exists an eigenvector $x \in \mathbb{R}^3$ of $F$ such that $W'=\mathrm{span}(x)$ (let's recall that invariant subspaces of dimension $1$ are exactly the lines generated by eigenvectors).
But by a direct calculation, the only eigenvalue of $F$ is $1$, and the eigenspace associated is precisely $W$. So $x$ must belong both to $W$ and $W'$, which contradicts the fact that the sum $W\oplus W'$ is direct.