On the existence of minimal polynomial of a square matrix How to prove that any annihilating polynomial $p(x)\in \Bbb C[x]$ of an $n \times n$ matrix $A$ with entries from $\Bbb C$ must contain all irreducible factors of the characteristic polynomial $\chi_A(x)$. In other words, why can't we annihilate a polynomial $q(x) \in \Bbb C[x]$ by plugging $A$ if $q$ lacks an irreducible factor from $\chi_A(x)$.
I feel that, we have to prove there exists a commen factor (a least commen factor later to be called as minimal polynomial $m_A(x)$) for each annihilating polynomial $p (x)\in \Bbb C[x]$.
Is it enough proving that $\chi_A(x)$ and $m_A(x)$ possess same set of irreducible factors to say any annihilating polynomial must contain all these irreducible factors? How it is?
For instance, $p(x)=x^2-1$ is an annihilating polynomial for the $2\times 2$ identity matrix $I_2$, but it's not the chara. polynomial while $p$ possess the commen factor $x-1$.
How to prove in general?
Thanks in advance.
 A: This simply stems from the definition of the minimal polynomial: it is the monic generator of the principal ideal of all annihilating polynomials.
A: $\chi_A(x) = g^{n_1}_{\lambda_1}(x) g^{n_2}_{\lambda_2}(x)....g^{n_L}_{\lambda_L}(x)$ where $g_{\lambda_i}(x)$ is the irreducible polynomial which has $\lambda_i$ as root. Since we are talking about complex numbers: $g_{\lambda_i}(x) = (x-\lambda_i)$.
Hence : $\chi_A(x) = (x-\lambda_1)^{n_1} (x-\lambda_2)^{n_2} .... (x-\lambda_L)^{n_L}$.
Now let $m(x)$ be minimal polynomial of $\chi_A(x)$. Hence $m(x) g(x) = \chi_A(x)$.
If wlog $(x-\lambda_1) \nmid m(x)$ and $m(x) = (x-\lambda_2)^{m_2} .... (x-\lambda_L)^{m_L}$ then let $v_{\lambda_1}$ be the eigen vector corresponding to eigenvalue $\lambda_1$. Then $m(A)v_{\lambda_1} = (\lambda_1-\lambda_2)^{m_2} .... (\lambda_1-\lambda_L)^{m_L} v_{\lambda_1} \neq 0$ a contradiction to $m(A) = 0$.
Hence we have $(x-\lambda_i) | m(x)$ for every $\lambda_i$ eigenvalue and $(x-\lambda_i) | m(x)$ iff $(x-\lambda_i) | \chi_A(x)$.
Since $I = \{f(x) : f \in \mathbb{C}[x], f(A) = 0\}$ has a polynomial of least degree, call it $p(x)$ then let $f(x) \in I$ then by division algorithm $f(x) = w(x)p(x)+r(x)$ with $deg(r) < deg(p)$ hence we have $r(A) = 0$ contradicting the minimality of degree of $p(x)$ and hence $r(x) = 0$ and hence $I$ generated by a single element $p(x)$, $I = <p(x)>$ i.e. $f(x) \in I$ iff $p(x) | f(x)$. Its clear that $m(x) = p(x)$ assuming both $m(x),p(x)$ are monic polynomials.
Let $q(A) = 0$. Hence $m(x) \ | \ q(x)$. Hence $(x-\lambda_i) \ | \ m(x) \
 | \ q(x)$ and hence $(x-\lambda_i) \ | \ q(x)$.
