Determine the Jordan form (over $\mathbb{C}$) of nxn matrices $M$ satisfying $M^2=-I$, where $I$ is the identity matrix.
The minimal polynomial divides $x^2+1$, so I would like to conclude that the minimal polynomial is either $x+i$ or $x-i$, but is it possible $x-i,x+i$ are not the minimal polynomial? I see that $x^2+1=0$ implies one of $x\pm i=0$ because $\mathbb{C}[x]$ is an integral domain, but matrices with multiplication don't form an integral domain. So isn't it possible that $(M+iI)(M-iI)=0$ while neither $M\pm iI=0$? One can easily construct 2x2 nonzero matrices whose product is zero.
You can't conclude that the minimal polynomial is either $x+i$ or $x-i$. For instance, your matrix could be $\begin{pmatrix}i & 0\\ 0 & -i\end{pmatrix}$. Indeed, any monic polynomial can be the minimal polynomial of a matrix; a general way to do this is to take the companion matrix.
So if you want to find the Jordan forms by thinking about the minimal polynomial, you will have three different cases: the minimal polynomial is $x-i$, the minimal polynomial is $x+i$, and the minimal polynomial is $x^2+1$. (Actually, there is a fourth case: the minimal polynomial is $1$. This case is only possible for $n=0$, though.) But the easiest way to solve this problem is not to think about minimal polynomials at all. Instead, just suppose you have a matrix in Jordan form such that $M^2=-I$. Then each Jordan block of $M$ also satisfies $M^2=-I$. When does a Jordan block square to $-I$?