find minimal polynomial of matrix product $AB$ where $AB=BA$ Let $x^2-1$ and $x^2+1$ be minimal polynomials of $A,B\in M_n(\mathbb{R})$, respectively.
If $AB=BA$, find the minimal polynomial of $AB$.
 A: Notice that $A^2 = I$ and $B^2 = -I$. Therefore $(AB)^2 = A^2 B^2 = -I$, so the minimal polynom of $AB$ divides $x^2+1$. Because $x^2+1$ is monic and irreducible over $\mathbb{R}$ this already is the minimal polynomial.
A: Note that $X^2-1=(X-1)(X+1)$ is split over $\Bbb R$ with simple roots, so that $A$ is diagonalisable with set of eigenvalues $\{1,-1\}$, while $X^2+1$ is irreducible so that $B$ is not diagonalisable over$~\Bbb R$. Since $A$ and $B$ commute, each of the two eigenspaces for $A$ is a $B$-stable subspace. The minimal polynomial of the restriction of $B$ to such an eigenspace is a divisor of $X^2+1$, and since the eigenspaces have dimension${}>0$ it is not $1$, so the minimal polynomial of each restriction of$~B$ must be $X^2+1$. This implies that the complexification of those restrictions of$~B$ are both diagonalisable with set of eigenvalues $\def\ii{\mathbf i}\{\ii,-\ii\}$. Then over $\Bbb C$, the product $AB$ is diagonalisable with eigenvalues $\{1,-1\}*\{\ii,-\ii\}$ (the set of products), which set equals $\{\ii,-\ii\}$. Then the minimal polynomial of $AB$ is $(X-\ii)(X+\ii)=X^2+1$.
