Is my proof correct? linear transformation over $\Bbb R$ has invariant subspace $\dim(U)=1$ or $2$

Question

$V$ is a Vector space over $\Bbb R$, and $\dim(V)=n$. A linear transformation $T$ from $V$ to $V$. Then, T has an invariant subspace $U$ such that $\dim(U)=1$ or $2$

I read in many books, which contain a proof, there is a complex root $\lambda$, such that $p(\lambda)=0$, so there exist $X_1, X_2\in\Bbb R^n$ such that

$${\bf T}(X_1+iX_2)=\lambda(X_1+iX_2)$$

I try to construct a proof by myself: But I'm not sure

Consider characteristic polynomia $p(x)$. If $p(x)$ has a real eigenvalue, then $\bf T$ has an eigenvector.

In not, $p(x)$ has no real root. $\forall \alpha\in V, \alpha\ne0$, then $\alpha, T(\alpha) $ are linearly independent.

Let

$$p(x)=(x^2+a_kx+b_k).... (x^2+a_1x+b_1)$$

$x^2+a_ix+b_i(i=1,2,\dotsc,k)$ all has no real root.

According to Hamilton-Cayley theorem,

$$(T^2+a_1T+b_1E).... (T^2+a_kT+b_kE)=0$$

$\forall \alpha, \alpha\ne0 $, we have

$$(T^2+a_kT+b_kE).... (T^2+a_1T+b_1E)\alpha=0$$

there exists $j(0\leq j\lt k)$ such that

$$(T^2+a_jT+b_jE)\dotsb(T^2+a_1T+b_1E)\alpha\ne0\quad (T^2+a_{j+1}T+b_{j+1}E)\dotsb(T^2+a_1T+b_1E)\alpha=0$$

Denote $\beta =(T^2+a_jT+b_jE)... (T^2+a_1T+b_1E)\alpha $, then

$$ (T^2+a_{j+1}T+b_{j+1}E)\beta=0$$

or

$$ T^2\beta=-a_{j+1}T\beta-b_{j+1}\beta$$

As $\beta, T\beta$ are linearly independent, so $\beta,T\beta$ span a invariant subspace of dimension $2$

Is the above proof correct? Thank you!

Answer 1

Long comment: For any polynomial $p$, the kernel $\ker p(T)$ is an invariant subspace. For most polynomials it will be trivial, i.e., zero-dimensional.

Let $p$ be a polynomial so that $\ker p(T)$ is non-trivial and there is a decomposition $p=q_1q_2$ in mutually prime factors. Then there are, by the Bezout identity, polynomials $u_1,u_2$ with $$ 1=u_1q_1+u_2q_2 $$ and $u_1(T)q_1(T)$ is a projector on the kernel of $q_2(T)$ and vice versa. And at least one of $\ker q_1(T)$ and $\ker q_2(T)$ is non-trivial.

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This decomposition allows you to split the polynomial until there is only a power $q^m$ of an irreducible (over the given field) polynomial $q$ left. At this point you have to examine the chain of $\ker q(T)^m$, $\ker q(T)^{m-1}$, ... , $\ker q(T)$ and find the largest power such that there is a drop in the dimensions. From that you can then construct a minimal invariant subspace.