$A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

Question

Problem: The square real matrix $A$ is diagonalizable if the minimal and characteristic polynomial of $A$ are equal.

I think I've got this figured out if I can say that the characteristic polynomial has multiplicity 1 for all of its roots, but I don't know why that's true from what we're supposing here. Can anyone bridge that gap for me? Or is there an easier way to go about it that I'm not seeing?

Answer 1

It isn't true. The characteristic and minimal polynomial of $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ are both $x^2$.

Answer 2

As stated, this is not true. Take $A=\left(\begin{array}{cc}1&1\\ 0&1\end{array}\right)$. Then minimal and characteristic polynomial of $A$ are both $(x-1)^2$ (and thus equal), but $A$ is not diagonalizable.

A criterion for diagonalizability (over a suitable extension) is that the minimal polynomial is squarefree.