For a given matrix, say $A$, of dimension $n$, in $\mathbb{F}^{n\times n}$, the characteristic polynomial of $A$ is of the form $$c_A = p_1^{d_1}*...*p_k^{d_k},$$ and the minimal polynomial of the form $$c_A = p_1^{e_1}*...*p_k^{e_k},$$ where $p_i$s are prime factors in $\mathbb{F}[x]$.
By Primary Decomposition theorem, we know that $$V = \oplus_i^k V_i\quad where \quad V_i = Ker(p_i^{e_i}(A))$$
Then, by Cyclic decomposition theorem, we can decompose each $V_i$ as $$V_i = \{0\} \oplus [ \oplus_j^{r_i} Z(\alpha_{i,j},A) ]$$ for some $\alpha_{i,j} \in V_i$.
Question:
My question is that for each $V_i$ how to find the vectors $\alpha_{i,j}\in V_i$, $j=1,...,r_i$.
My thoughts:
We have already proved that for the vector $\alpha_{i,j} \in V_i$, the vectors $\{\alpha_{i,j}, A(\alpha_{i,j}),..., A^{k_{i,j}} (\alpha_{i,j})\}$ form a basis for $Z(\alpha_{i,j}, A)$, they cannot be zero, and they have to be linearly independent.Moreover, since the minimal polynomial of $A|_{V_i}$ is $p_i^{e_i}$, (and by Cyclic decomposition theorem) the $A$-annihilator of $\alpha_{i,1}$ is $p_i^{e_i}$, $$p_i^{e_i-1} (A) (\alpha_{i,1}) \not =0,$$ so, for example, would any vector $v_1 in V_i$ s.t $p_i^{e_i-1} (A) (\alpha_{i,1}) \not =0$ work as $\alpha_{i,1}:= v_1$ ?
Even we could find $\alpha_{i,1}$, how to find $\alpha_{i,2}$ and so on, which satisfies the necessary conditions ??