Showing that minimal polynomial has the same irreducible factors as characteristic polynomial

If you believe in invariant factors, this is easily seen as follows. The minimal polynomial is the last of the invariant factors (arranged so that each divides the next) and the characteristic polynomial is their product. It follows that the minimal polynomial divides the characteristic one (Cayley-Hamilton) and that the characteristic polynomial divides the minimal polynomial to the power the number of invariant factors. With each one dividing a power of the other, they must have the same irreducible factors.

Or if you don't (want to) believe in invariant factors, you can see the non-C-H direction by induction on the dimension, as follows. Leaving the $0$-dimensional base case as exercise, choose any nonzero vector$~v$, let $W$ be the cyclic submodule it generates and let $P\in k[X]$ be the minimal degree monic polynomial such that $P[T](v)=0$. Then $P$ is the characteristic polynomial of the restriction to$~W$ of$~T$ (a well known fact about cyclic modules; it also follows from C-H and dimension consideration). Then the full characteristic polynomial$~\chi$ of$~T$ is the product of $P$ and the characteristic polynomial$~\chi'$ of the linear operator $T_{/W}$ that $T$ induces in the quotient module $V/W$ (this follows by considering the block triangular matrix of$~T$ on a basis of$~V$ that extends one of$~W$). Now let $f$ be an irreducible factor of$~\chi$. If $f$ divides$~P$ we are done, since the minimal polynomial$~\mu$ of$~T$ is a multiple of$~P$ (since $\mu(T)(v)=0$). So suppose the contrary, then $f$ must divide$~\chi'$. By the inductive hypothesis $f$ divides the minimal polynomial$~\mu'$ of $T_{/W}$, which is the minimal degree monic polynomial such that $\mu'[T](V)\subseteq W$. But $\mu$ satisfies $\mu[T](V)=\{0\}\subseteq W$, so $\mu$ is a multiple of $\mu'$, and we conclude $f\mid\mu'\mid\mu$. QED

There is another proof using the properties of the determinant.

Claim: $\chi$ divides $m^n$ in $k[X]$.

To see this, fix a basis of $V$ and consider the matrix $M$ of $T$ in said basis. We see $\mathcal{M}_n(k),\mathcal{M}_n(k[X])$ as a subrings of $\mathcal{M}_n(k(X))$ given the natural embeddings $k\rightarrow k[X] \rightarrow k(X)$. In this context $\chi = \det(X.I_n-M)$ is a regular determinant of $X.I_n-M\in \mathcal{M}_n(k(X))$.

Write $m=\sum \limits_{k=0}^d a_kX^k$. We have: $m(X.I_n)=m(X.I_n) - m(M)=\sum \limits_{k=0}^d a_k(X^k.I_n-M^k)= \sum \limits_{k=1}^d a_k(X^k.I_n-M^k)=(X.I_n-M)\sum \limits_{k=1}^da_k\sum \limits_{p=0}^{k-1}X^p.M^{k-1-p}$.

Let $B:=\sum \limits_{k=1}^da_k\sum \limits_{p=0}^{k-1}X^p.M^{k-1-p}$. Since $B\in \mathcal{M}_n(k[X])$ where $k[X]$ is a ring, we have $\det(B) \in k[X]$.

We then compute the determinant of those matrices: $m^n=m^n.\det(I_n)=\det(m(X.I_n))=\det((X.I_n-M)B)=\det(X.I_n-M)\det(B)=\chi\det(B)$, hence the result.

One can reason using only some field theory, rather than linear algebra, if one prefers. I will assume the Cayley-Hamilton theorem, and the fact that all eigenvalues are roots of the minimal polynomial$~\mu$, which holds because an eigenvector for$~\lambda$ is killed by $P[T]$ (if and) only if $P$ is a multiple of $X-\lambda$.

First observe that the result is true when $\chi$ splits over $k$ (and hence so does$~\mu$, which divides$~\chi$ by the Cayley-Hamilton theorem), since the factors $X-\lambda$ of either polynomial are precisely those with $\lambda$ an eigenvalue. This in particular takes care of the case where $k$ is algebraically closed; the remainder deals with the case where $\chi$ does not split into linear factors in $k[X]$.

If $K/k$ is a field extension, one can extend scalars to obtain $\def\ext{\otimes_kK}T\ext: V\ext\to V\ext$. Since one gets identical matrices$~A$ for $T$ and for $T\ext$, using some $k$-basis of $V$ and the corresponding $K$-basis of $V\ext$, they have the same characteristic polynomials. They have the same minimal polynomials as well, since the minimal polynomial can be found from the unique solution of the linear system $x_0I+X_1A+\cdots+x_{d-1}A^{d-1}=A^d$ (one equation for each matrix coefficient) for the smallest $d$ for which this system has a solution at all; solving a linear system with coefficients in$~k$ over a larger field$~K$ will not change the existence of a solution, or the solution in case it is unique.

So over a sufficiently large field$~k$ (a splitting field of$~\chi$ will do), the polynomials $\mu$ and $\chi$ split, and give rise to the same set of linear factors $X-\lambda$ (though some may occur with larger multiplicity for$~\chi$ than for$~\mu$). In order to conclude that they also have the same set of irreducible factors in $k[X]$, I propose two arguments. One is that the presence of an irreducible factor$~f$ in $k[X]$ can be read off from the presence of any one of the factors into which $f$ splits in $K[X]$, because no two (monic) irreducible polynomials over$~k$ share a root in$~K$: if $\alpha\in K$ is a root of a monic irreducible $f\in k[X]$, then the minimal polynomial of$~\alpha$ over$~k$ is $f$ (this is actually the minimal polynomial of multiplication by$~\alpha$ viewed as $k$-linear map $K\to K$). The other argument is that, having the same set of (linear) factors, $\chi$ divides in $K[X]$ some power of $\mu$. But this division is valid in $k[X]$ as well (where both $\mu$ and $\chi$ live), which shows that the irreducible factors of$~\chi$ occur among those of$~\mu$ (and C-H gives the converse).