How to show every linear transformation  :→ has an invariant subspace of dimension 1 or 2 Suppose that $V$ is a finite dimensional vector space over $\mathbb{R}$. Prove that every $T \in \mathcal{L}(V)$ has an invariant subspace of dimension one or two.
I know that for a subspace $W$ to be $T$-invariant, $T(x) \in W$ for all $x \in W$.  I know that there must be some invariant subspaces of dimension $1$ or $2$ if $V$ is finite dimensional, because they would be contained in $V$ which has a finite dimension of $n$.  But, I'm having trouble putting this into proof form.  Also, all of this could be wrong -- I have trouble with proofs. I would love some help to just get me started at least, or an outline?  Thank you ^-^

We have to assume that $\dim V>0$ (to avoid a trivial counterexample). Then the characteristic polynomial $\chi_T(x)=\det(xI-T)$ of $T$, which is a polynomial of degree $n$, has a root in $\Bbb C$, perhaps even in $\Bbb R$.
If $T$ has a real eigenvalue $\lambda$ and $v$ is an eigenvector accordingly, then $v\Bbb R$ is a one-dimensional invariant subspace.
If $T$ has a (non-real) complex eigenvalue $\lambda$and $v$ is an eigenvector accordingly, then $\overline v$ is eigenvector with eigenvalue $\overline\lambda$ and in $V\otimes \Bbb C$, $v\Bbb C$ and $\overline v\Bbb C$ are a complex-one-dimensional invariant subspaces. Note that $u:=v+\overline v$ and $w:=i(v-\overline v)$ are real vectors (because $\overline u=u$ and $\overline w=w$). As $Tu=\lambda v+\overline\lambda \overline v=\frac{\lambda+\overline \lambda}2u+\frac{\lambda-\overline \lambda}{2i}w$ and we find a similar expression for $Tw$, we see that $u\Bbb R+w\Bbb R$ is a two-dimensional invariant subspace. 

Alternatively, write $\chi_T(x)=f_1(x)\cdots f_k(x)$ as a product of irreducible (over $\Bbb R$) factors. As $\chi_T(T)$ is singular, one of the $f_j(T)$ must be singular, i.e., for such $j$, there exists $v\ne0$ with $f_j(T)(v)0$. We know that a real irreducible polynomial can only be of the form $x+a$ or $x^2+bx+c$. In the first case, $T(v)=-av$, and in the second, $T(T(v))=-Tv+v$. We conclude that the space spanned by $v$ in the first case or by $v$ and $Tv$ in the second case is $T$-invariant.
