 Solution

Write $n=\dim V$. If $Rv=\lambda v$ with $v\ne 0$, then $0=R^{a}v={\lambda }^{a}v$, hence $\lambda =0$. The characteristic polynomial splits over $\mathbb{C}$ and has degree $n$, soThere 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 $R^{a-1}v\ne 0$, using minimality. The vectors $v,Rv,\dots ,R^{a-1}v$ 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 $\boxed{a\le n}$.