Nondiagonalizable Matrix and Polynomials

Question

I got the following problem:

If $A$ is a nondiagonalizable square matrix of order $n$ over field $\mathbb{F}$ then there exists a polynomial $P$ of degree $n-1$ over $\mathbb{F}$ such that $(P(A))^2=0$
Over $\mathbb{F}=\mathbb{C}$ ?
Over $\mathbb{F}=\mathbb{R}$ ?

I tried to solve it but didn't manage to proceed that much. Thanks...

Answer 1

For $\mathbb{C},$ $A$ nondiagonalizable means that the minimal polynomial of $A$ is of smaller degree than the characteristic polynomial, so at most $n-1,$ so you don't need to square - $P(A)=0,$ where $P$ is the minimal polynomial.