# How to show every linear transformation 𝑇 :𝑉→𝑉 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.