Similar matrices and minimal polynomial I guess that I'm missing here something...
How to prove that, if two matrices are similar, then their minimal polynomials are the same one. 
Thanks!
 A: Minimal polynomials are defined for linear operators $\phi:V\to V$ on a finite dimensional vector space. The minimal polynomial of a square matrix is that of a linear operator it represents on some basis. Two square matrices are similar if and only if they represent the same linear operator $\phi:V\to V$, each in a basis of $V$. End of proof.
Since julien questioned this point, let me elucidate the second sentence. I did not say that the minimal polynomial of a square matrix is defined as that of a linear operator it represents on some basis. But it is easily seen to be equal to it. Let $\def\B{\mathcal B}\B$ be an ordered basis of $V$, with $n$ elements. The map $\def\Mat{\operatorname{Mat}}\def\End{\operatorname{End}}\Mat_\B:\End(V)\to M_{n,n}(k)$ that sends a linear operator to its matrix with respect to$~\B$ is an isomorphism of $K$-vector spaces and of rings (since that is how matrix multiplication is defined). Let $\phi\in\End(V)$ and $A=\Mat_\B(\phi)\in M_{n,n}(k)$. Let $f_1:K[X]\to\End(V)$ be the substitution morphism of $\phi$ for $X$ (sending $P(X)\mapsto P(\phi)$ if you will, though I abhor of this notation) and let $f_2:K[X]\to M_{n,n}(k)$ similarly be substitution of $A$ for $X$. By definition the minimal polynomial of $\phi$ is the monic polynomial of minimal degree in $\ker f_1$, and the minimal polynomial of $A$ is the monic polynomial of minimal degree in $\ker f_2$. Now the diagram commutes, $f_2=\Mat_\B\circ f_1$, since both sides are the same on $K$ and on the element $X$, by the universal property of polynomial rings. But since $\Mat_\B$ is an isomorphism, $\ker(f_1)=\ker(f_2)$, so the minimal polynomials are equal. "Look mom, no conjugations!". All this may seem a lot of learned language, but it just amounts to saying that matrices do what they are supposed to do, allow concretely determining properties of linear maps.
A final remark: the above is based on the fact that minimal polynomials can be defined without having to introduce matrices. While this is easy to see, it is not something insignificant. By comparison, I don't think there is any simple way to define the characteristic polynomial without introducing matrices.
