 Solution

For every nonnegative integer $k$,Consequently, if $m_{AB}(AB)=0$, thenBy the divisibility characterization of the minimal polynomial,Interchanging $A$ and $B$ also gives $m_{AB}(t)\mid t,m_{BA}(t)$. Therefore the Minimal polynomials of AB and BA can differ only in the exponent of the factor $t$, and that exponent differs by at most one.

If $AB$ is diagonalizable, its minimal polynomial has no repeated roots. The preceding divisibility says that every nonzero Jordan block of $BA$ has size one and every zero-eigenvalue block has size at most two. Squaring $BA$ preserves the one-dimensional nonzero blocks and sends every size-at-most-two nilpotent block to zero. Thus the Jordan normal form of $(BA{)}^2$ is diagonal, so