Irreducible factors of characteristic polynomial and minimal polynomial are same? I've seen in my linear algebra textbook that one can prove that the irreducible factors of a characteristic polynomial and minimal polynomial are the same using Primary Decomposition Theorem, but I have no idea how this happens.
So the fact that irreducible factors of a minimal polynomial are irreducible factors of a characteristic polynomial is trivial due to Cayley-Hamilton Theorem, however the converse does not seem so easy. I think I found a solution using field extensions and eigen values, but I want to try to prove this using Primary Decompostion Theorem.
I've found Showing that minimal polynomial has the same irreducible factors as characteristic polynomial this question, it says,
Let the minimal polynomial of a linear transfomation $T:V \to V$ over field F be  $m = {f_1}^{m_1} ... {f_n}^{m_n}$. Now restrict $T$ to $ker({f_i}^{m_i})$, and use the Cayley-Hamilton Theorem, and the fact that $f_i$ is irreducible to prove the statement.
How ever I don't see how this happens, if I let $W_i = ker({f_i}^{m_i})$, I can show that the minimal polynomial of  $T|_{W_i}$ is ${f_i}^{m_i}$ but using Cayley-Hamilton only gives that the characteristic polynomial of $T|_{W_i}$is a multiple of ${f_i}^{m_i}$, and I don't think this really helps.
I'm thinking that I have to use some unique properties of a characteristic polynomial other than the fact that it's an annihilator, since not every annihilators have the same irreuducible factors as minimal polynomials. Can you help me with this?
+) Okay so to be exact, there's a Lemma on my Textbook:
Let $V$ be a vector space over field $F$, and suppose $T$ is a linear transformation from vector space $V$ to $V$, and let $f_1(t), ...f_k(t)$ be mutually relatively prime monic polynomials over field $F$. if $f_1(t)f_2(t)...f_k(t)$ is an annihilator of $T$, $V$ can be decomposed into direct sums of $kerf_i(T)$.
and there's a exercise that says to prove the following using the Lemma above:
Let $V$ be a vector space over field $F$, and suppose $T$ is a linear transformation from $V$ to $V$, then the 'set of monic irreducible divisor over field F' is the same for the characteristic polynomial of $T$ and the minimal polynomial of $T$.
I have no knowledge about modules, and I know characteristic polynomial as $det(tI-T)$ and minimal polynomial as the minimal annihilator of T. 
This isn't my homework or anything, I just got stuck while studying with this textbook on myself :(( Any help?
 A: You did the hardest step.
Assume that the characteristic polynomial of $T_{W_i}$ is $f_i^{t_i}g_i$ where $g_i$ is prime to $f_i$. 
Then $W_i=\ker(f_i^{t_i}(T))\oplus \ker(g_i(T))$. It is easy to see (for example by writing down a representative matrix in a basis adapted to this direct sum), that the minimal polynomial of $T_{W_i}$ is the lcm of the minimal polynomial of $T_{\ker(f_i^{t_i}(T))}$ and of the minimal polynomial of $T_{\ker(g_i(T))}$.
Thie first one is a power of $f_i$. The second  divides $g_i$, so is prime to $f_i$. Hence their lcm is the product of the two.
We then get that the minimal polynomial $T_{W_i}$ is NOT a power of $f_i$, unless $g_i=1$, and this concludes your proof.
To finish, I would like to give you an alternative approach to prove the theorem you are interested in, which only use polynomials.
Let $u:E\to E$ be an endomorphism, where $E$ has dimension $n$, with minimal polynomial $\mu_u$ and characteristic polynomial $\chi_u$. Then $\mu_u\mid \chi_u\mid \mu_u^n$.
Caylet-Hamilton implies $\mu_u\mid\chi_u$.
Now notice that for all $m\geq 1,$ we have 
$$ X^m-Y^m =(X-Y)(X^{m-1}+X^{m-2}Y+\cdots+XY^{m-2}+Y^{m-1})\in  K(X)[Y].$$
If $d$ is the degree of  $\mu_u,$ we deduce easily that there exist polynomials $P_0,\ldots,P_{d-1}\in K[X]$ such that
$$\mu_u(X)-\mu_u(Y)=(X-Y)(P_0+P_1Y+\cdots +P_{d-1}Y^{d-1})\in K(X)[Y].$$
Let $M$ be the representative matrix of $u$ in a fixed basis of $E.$
Then$$\mu_M(M)=0\in M_n(K)\subset M_n\big(K(X)\big).$$
Considering $M$ as an element of  $M_n\big(K(X)\big)$, this implies
$$\mu_u(X)I_n=(XI_n-M)Q\in M_n\big(K(X)\big),$$
où $Q=(P_0I_n+P_1M+\cdots +P_{d-1}M^{d-1})\in M_n\big(K(X)\big),$
since $\mu_M(M)=0$. Taking determinants , we get $$\mu_u^n=\chi_u \det(Q) \in K(X).$$ Notice now that $Q\in M_n(K[X]).$ Thus $\det(Q)\in K[X],$ hence the result.
Now the divisibilty relations $\mu_u\mid\chi_u\mid \mu_u^n$ easily implies that $\mu_u$ and $\chi_u$ have the same irreducible factors. 
