TITLE: Minimal polynomial of diagonalizable matrix
QUESTION [9 upvotes]: Prove that a matrix $A$ over $\mathbb{C}$ is diagonalizable if and only if its minimal polynomial's roots are all of algebraic multiplicity one.

REPLY [5 votes]: Evaluating a polynomial$~P$ at a matrix commutes with change of basis: $C^{-1}P[A]C=P[C^{-1}AC]$ for any invertible matrix $C$. This means that the polynomials annihilating $A$ are the same as those annihilating $C^{-1}\!AC$, and in particular their minimal polynomials are the same. Now if $A$ is diagonalisable then there is some diagonal $D=C^{-1}\!AC$, and evaluation $P$ at $D$ is a diagonal matrix with diagonal entries obtained by evaluating $P$ at the corresponding diagonal entry (eigenvalue) of$~D$. It follows that in this case the minimal polynomial of$~D$, and of$~A$, is a product of distinct factors $(X-a_i)$, one for each distinct eigenvalue$~a_i$; in particular it has simple roots.
Conversely if any product $P$ of distinct factors $(X-a_i)$ has $P[A]=0$ (in particular if the minimal polynomial of$~A$ has this form) then because all these factors are pairwise relatively prime the kernel decomposition theorem says
$$
  V=\ker(0)=\ker(P[A])=
  \ker(A-a_1I)\oplus\ker(A-a_2I)\oplus\cdots\oplus\ker(A-a_kI),
$$
and the nonzero factors on the right are the eigenspaces of $A$, showing that $A$ is diagonalisable.