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Put $C=A+yI$. For every $y$ except the finitely many roots of $\det (A+yI)$, $C$ is invertible and $CB$ is similar to $BC$, since $C^{-1}(CB)C=BC$. HenceBoth sides are polynomials in $y$. Agreement for infinitely many $y$ makes them identical, so evaluating at $y=0$ proves Characteristic polynomials of AB and BA:Their minimal polynomials need not agree. Forwe have $AB=B\ne 0$, $B^2=0$, and $BA=0$. The two minimal polynomials are respectively $t^2$ and $t$, although both characteristic polynomials are $t^2$.

For any $p(t)={\sum }_j{p}_j{t}^j$, the identity $(BA{)}^{j+1}=B(AB{)}^{j}A$ givesTaking $p=m_{AB}$ makes the right-hand side zero. Every annihilating polynomial is divisible by the minimal polynomial, soIf $AB$ is diagonalizable, $p$ has only simple roots. Therefore each nonzero eigenvalue of $BA$ has only size-one Jordan blocks, and the zero-eigenvalue blocks have size at most two. Squaring kills the latter nilpotent blocks, so $(BA{)}^2$ is diagonalizable, as is $(AB{)}^2$.

By their common characteristic polynomial, $AB$ and $BA$ have the same eigenvalues counted with algebraic multiplicity. Squaring produces the same multiset of squared eigenvalues, including combined multiplicities if two eigenvalues have the same square. The two diagonalizable squares therefore have the same diagonal normal form up to permutation. This proves similarity of squared matrix products with one diagonalizable product:It does not imply that $AB$ and $BA$ themselves are conjugate: the possible size-two nilpotent blocks at zero are exactly what squaring removes.