Let $L:V\to V$ be a linear operator with the same characteristic and minimal polynomial $p_L(x)=m_L(x)$.
i.Show that no primary component of $L$ can be split into a direct sum of nonzero invarians subspaces.(I was thinking of getting it from Primary Decomposition Thrm.)
ii. Deduce from(i) that every component of $L$ is cyclic.
If you know the structure theorem of finitely generated modules over $K[X]$ (which is a PID), then you get from $p_L=m_L$ that the module is cyclic as explained in this answer, or more directly: if there were more than one cyclic factor in a decomposition into cyclic modules, with the order (minimal polynomial) of each factor dividing the next, then the order of the last factor is $m_L$, while $p_L$ is the product of all the order, so $m_L\neq p_L$, a contradiction. Having a cyclic vector (generator) for the whole module, projection on the primary components gives cyclic vectors of those components.
A more direct explanation is given in this answer. If a primary component$~C$ has minimal polynomial $P^k$ with $P$ an irreducible polynomial, and the fact that $m_L=p_L$ implies that $\dim_K(C)=\deg(P^k)$. For every vector$~v$ of $~C$ the minimal monic polynomial$~Q$ with $Q[L].v=0$ divides $P^k$, so is a power of$~P$, and given the dimension condition, $v$ is a cyclic vector for the primary component if and only if $Q=P^k$. But $P^{k-1}$ does not annihilate$~C$, so such a cyclic vector exist (in fact most of the vectors are).
There remains the initial question (for which I had no use) of why $C$ cannot decompose into a nontrivial direct sum of submodules. If that were the case, the minimal polynomial $P^k$ of$~C$ would be the least common multiple of the minimal polynomials of the summands, which necessarily has lower degree than the dimension (over$~K$) of$~C$. But then also $m_L$ will have lower degree than $p_L$, a contradiction.
