show that matrices sharing a minimal polynom, also share the Jordan form

Question

Let $A$ and $B$ be square matrices of the same dimension. Show that they have the same minimal polynom if and only if they have the same Jordan form.

(btw, is it possible to prove with the charastaristic polynom instead?)

thanks!

Answer 1

Both statements are wrong - the Jordan form is a full invariant of similarity and contains more information than just the minimal polynomial or the characteristic polynomial. For a counter-example, consider the matrices

$$ A = \left( \begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} \right), \,\,\, B = \left( \begin{matrix} 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 \end{matrix} \right). $$