A question on minimal polynomials (with relation to some basis) Let $\mathscr{A}$ be a linear transformation on $n$-dimensional vector space $V$. $0\neq\alpha\in V$ (which a vector field on an arbitrary field $\mathbb{F}\supset \mathbb{Q}$).
Then it is easy to see that there exists a unique polynomial $m_\alpha(\lambda)$ such that $m_\alpha(\mathscr{A})(\alpha)=0$, and for any polynomial $f(\lambda)$ satisfying $f(\mathscr{A})(\alpha)=0$, then $m_\alpha(\lambda)\mid f(\lambda)$.
Now, let $\xi_1,\cdots,\xi_n$ be a basis of $V$, what is the relationship between $m_{\xi_1}(\lambda),\cdots,m_{\xi_n}(\lambda)$ and the minimal polynomial $m_\mathscr{A}(\lambda)$ of $\mathscr{A}$.
I suspect it is 
$$m_{\xi_1}(\lambda)\cdots m_{\xi_n}(\lambda)=m_\mathscr{A}(\lambda)$$
or the least common multiplier of
$$m_{\xi_1}(\lambda),\cdots,m_{\xi_n}(\lambda)$$ is $$m_\mathscr{A}(\lambda)$$
But at this moment I have no idea to show. What I could see is only that $m_{\xi_i}(\lambda)\mid m_{\mathscr{A}}(\lambda)$.

Let $f(x)=LCM (m_{e_1}(x),..,m_{e_n}(x)).$ We have $f=m_A$. PROOF: For each $i=1...n$ there exists a polynomial $g_i$ such that $g(x)m_{e_i}(x)=f(x)$, which implies $f(A)(e_i)=g(A)m_{e_i}(A)(e_i)=0.$ Since $e_1,...,e_n$ is a basis and $f(A)(e_i)=0$ for each $i,$ we have $f(A)=0.$ Now  for each $i=1,..,n$ let $$h_i=\gcd (m_A,m_{e_i}).$$ There exist polynomials $p_i(x)$ and $q_i(x)$ such that $$h_i(x)=p_i(x)m_A(x)+q_i(x)m_{e_i}(x).$$ Then $h_i(A)(e_i)=0$, and the minimality of $m_{e_i}$ implies that $$m_{e_i}(x)|h_i(x),$$  while by the def'n of $h_i$ we have $$h_i(x)|m_A(x).$$ Therefore each $m_{e_i}$ is a divisor of $m_A.$ In other words $m_A$ is a common multiple of $ (m_{e_1},..,m_{e_n}),$ so the def'n of $f$ implies that $$f|m_A.$$ But $f(A)=0$ so the minimality of $m_A$ implies that $$m_A|f.$$   So to within a non-zero constant multiple we have $m_A=f.$...QED...... For an extreme example where $m_A\ne \prod_1^nm_{e_i}$ take $n>1$ and for $i=1,..,n$ let  $e_{n+i}=e_i$  and  $A(e_i)=e_{i+1}$. Then $m_A(x)=x^n-1$ while $\prod_1^nm_{e_i}(x)=(x^n-1)^n.$
