The minimal and characteristic polynomials have exactly the same monic irreducible factors

Given: A finite-dimensional endomorphism $T$ and a monic irreducible polynomial $q\in F[x]$.

The minimal polynomial divides the characteristic polynomial (The minimal polynomial divides the characteristic polynomial, ${\mu }_T\mid {\chi }_T$).

For monic irreducible $q$, the quotient $K=F[x]\mathrm{/}(q)$ is a field extension containing $a=x+(q)$ with $q(a)=0$ ($F[x]\mathrm{/}(p)$ for monic irreducible $p$ is a field extension containing the root $x+(p)$ with unique reduced representatives).

Extending scalars from $F$ to $K$ leaves the minimal polynomial unchanged (For a matrix over a field, extending the scalar field does not change its minimal polynomial).

A scalar is an eigenvalue exactly when it is a root of the characteristic polynomial (For every finite-dimensional space, ${\sigma }_F(T)$ is exactly the set of roots in $F$ of ${\chi }_T$).

If $a$ is algebraic over $F$, the kernel of evaluation at $a$ is generated by its unique monic irreducible minimal polynomial, and $f(a)=0$ exactly when that polynomial divides $f$ (The evaluation kernel and the unique monic irreducible minimal polynomial of an algebraic element).

Polynomial evaluation at an endomorphism is $p(T)={\sum }_{k\ge 0}{a}_k{T}^k$ for $p(x)={\sum }_{k\ge 0}{a}_k{x}^k$ (Polynomial evaluation at an endomorphism: $p(T)={\sum }_k{a}_k{T}^k$).

An endomorphism and its matrix in any ordered basis have the same minimal polynomial; in particular, this polynomial is invariant under changing the basis (The minimal polynomial is unchanged by choosing a matrix representation or replacing a matrix by a similar one).

In any ordered basis, the characteristic polynomial of an endomorphism is the characteristic polynomial of its representing matrix, independently of the chosen basis (The basis-independent characteristic polynomial ${\chi }_T$ of an endomorphism of a finite-dimensional space, including ${\chi }_T=1$ in dimension zero).