Let $f$ be a linear transformation of $\mathbb R^n\to \mathbb R^n$ that has rank $r$.
Prove the existence of a degree $r+1$ polynomial that annihilates $f$
I have a proof : consider $g$ which is $f$ restricted to $Im(f)$. Then $\chi_g=\sum_0^ra_k X^k$ which has degree $r$ annihilates g.
Thus, $\forall x\in \mathbb R^n, 0= \sum_0^r a_kg^k(f(x)) = \sum_0^r a_kf^{k+1}(x)$.
Therefore $ \sum_0^r a_kX^{k+1}$ is what we were looking for.
Let $(e_1,\ldots,e_{n-r})$ a basis of $\ker f$ and we complete it on $(e_1,\ldots,e_n)$ a basis of $\Bbb R^n$. Let $V=\operatorname{span}(e_{n-r+1},\ldots, e_n)$ and the endomorphism $g$ the restriction of $f$ on $V$: $g\colon V\rightarrow \operatorname{im}(f)$ so the characteristic polynomial $\chi_g$ of $g$ has the degree $r=\dim V$ and it annihilates $g$ hence the polynomial $$P(x)=x\chi_g(x)$$ has the degree $r+1$ and it annihilates $f$. In fact, let $x=x_1+x_2$ where $x_1\in\ker f$ and $x_2\in V$ then (notice that $f$ and $\chi_g(f)$ commute) $$f(\chi_g(f)(x))=\chi_g(f)(f(x))=\chi_g(f)(f(x_2))=f(\chi_g(f)(x_2))=f(0)=0$$