= Solution
Write $n=\dim V$. If $Rv=\lambda v$ with $v\ne0$, then $0=R^av=\lambda^av$, hence $\lambda=0$. The <characteristic polynomial> splits over $\mathbb C$ and has degree $n$, so
$$
\boxed{\operatorname{spec}(R)=\{0\},\qquad\chi_R(t)=t^n.}
$$
There is indeed an <eigenvector> for the <eigenvalue> zero: a nonzero <vector> in the last nonzero <image> of $R$ lies in its <kernel of a linear map>. For the bound, choose $v$ with $R^{a-1}v\ne0$, using minimality. The vectors $v,Rv,\ldots,R^{a-1}v$ are <linearly independent>. Otherwise choose the least $j$ with nonzero coefficient in a relation and apply $R^{a-1-j}$; only its nonzero multiple of $R^{a-1}v$ survives, a contradiction. Therefore $\boxed{a\le n}$.
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