If $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factors I'm trying to prove that if a linear operator $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factors. This is what I have so far:
Let $f$ be diagonalisable. So there exists a basis relative to which $f$ has a diagonal matrix, say $D$. So the characteristic polynomial of $f$ is given by $p_f(x)=(x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_s),$ where $\lambda_i$ are the diagonal entries of $D$. 
I know that the minimal polynomial must divide the characteristic polynomial and have the same linear factors. Without loss of generality let the first $i$ linear factors be distinct. So I claim that the minimal polynomial $m_f(x)=\pm (x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_i).$ 
However, how can I now verify that $m_f(D)=0$ ?  
 A: Denote $\lambda_1,\ldots,\lambda_s$ all distinct eigenvalues of $A$ and consider
$$
\pi_A(\lambda)=\prod_{k=1}^s(\lambda-\lambda_k),
$$
then if $A=SDS^{-1}$ we have
$$
\pi_A(A)=\prod_{k=1}^s(A-\lambda_kI)=S\prod_{k=1}^s(D-\lambda_kI)S^{-1}=S\cdot 0\cdot S^{-1}=0
$$
since for every diagonal position there is always some $D-\lambda_kI$ in the product that has zero there. It means that this particular $\pi_A$ with simple roots annihilates $A$, hence, the minimal polynomilal must have simple roots too as it divides $\pi_A$.
P.S. Actually $\pi_A$ is the minimal polynomial, so you claim is, in fact, if and only if.
A: This is very basic, and you do not need to use the characteristic polynomial or the fact that the minimal polynomial divides it at all. You just need to realise that "diagonalisable" means that the sum of the eigenspaces fills the whole space, so a linear operator is zero if (and obviously only if) it is zero on each of the eigenspaces.
Now on the eigenspace for an eigenvalue$~\lambda$, our $f$ acts by scalar multiplication by$~\lambda$. It easily follows that on this eigenspace any polynomial $P[f]$ acts by scalar multiplication by$~P[\lambda]$ (just check that $f^k$ acts by multiplication by $\lambda^k$, and then combine the monomials of the polynomial $P$ linearly). So by the above, $P[f]=0$ iff $P[\lambda]=0$ for every eigenvalue$~\lambda$. The minimal monic polynomial$~P$ with that property is the product of (just) one factor $X-\lambda$ for each distinct eigenvalue$~\lambda$ of$~f$; there are distinct linear factors.
A: Without loss of generality, let the diagonal form $D$ have all the $i$ distinct eigenvalues in the top rows.  For every eigenvalue $\lambda$, one of the top $i$ rows of the matrix $(D - \lambda I)$ will consist of zero entries.  Hence, the determinant of the top left $i$-by-$i$ submatrix of $(D - I \lambda)$ is zero.  But, this determinant is $m_{f}(\lambda)$.
A: Question: "I'm trying to prove that if a linear operator f is diagonalisable then its minimal polynomial is the product of distinct linear factors."
Remark: If we define an $n \times n$-matrix $A \in M(n\times n,k):=R$ with coefficients in a field $k$ to be diagonalizable iff there is an invertible matrix $P\in R$ with $P^{-1}AP =D$ with $D$ a diagonal matrix, it follows the zero matrix $D(0)$ (the matrix with zeros as coefficients) is a diagonalizable matrix. For any invertible matrix $P$ it follows
$$D(0)=P^{-1}D(0)P.$$
The matrix $D(0)$ has all eigenvalues equal to zero.  By definition: The minimal polynomial $f(t)$ of $D(0)$ is the unique monic polynomial generating the ideal $I$ with
$$I:=\{f(t) \in k[t]: f(D(0))=0 \}$$
and $I=(t)$ has $f(t):=t$ as unique monic generator. And this polynomial
has one linear factor. The characteristic polynomial of $D(0)$ is $t^n$, and this polynomial has repeated roots. Hence the zero matrix is not a counterexample to the above proposition.
