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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.

This assertion is true. Apply the proved Characteristic polynomials of AB and BA identity to the two matrices $A$ and $BC$:It holds whether or not any of the three matrices is invertible.

The determinant isEquivalently it is the alternating multilinear function of the columns with value one on the identity matrix. To prove multiplicativity, write column $j$ of $AB$ as ${\sum }_k{b}_{kj}{A}_k$, where $A_k$ is column $k$ of $A$. Multilinear expansion gives a sum over choices $k_1,\dots ,k_n$; terms with repeated choices vanish by alternation. The remaining permutations contributeIf $C$ is invertible, $CB=C(BC)C^{-1}$, so multiplicativity makes $\det (CB{C}^{-1}-\lambda I)=\det (BC-\lambda I)$ and $p_{CB}=p_{BC}$. For arbitrary $A$, put $C=A+tI$. Its determinant is a nonzero polynomial in $t$, so it is invertible except for finitely many values. Thus for all other $t$,For each fixed $\lambda$ both sides are polynomials in $t$, and equality at infinitely many $t$ makes them identical. Setting $t=0$ proves the Characteristic polynomials of AB and BA identity $\boxed{p_{BA}=p_{AB}}$ without an invertibility assumption on $A$.

In the convention $p_A(\lambda )=\det (A-\lambda I)$, the coefficient of ${\lambda }^{n-1}$ is $(-1{)}^{n-1}\mathrm{tr}A$: in the determinant expansion such a term chooses one diagonal entry of $A$ and $n-1$ factors $-\lambda$; nonidentity permutations have at least two nonfixed positions and contribute lower degree. Therefore $\boxed{p_A=p_B⟹\mathrm{tr}A=\mathrm{tr}B}$.

The identity $(BA{)}^{j+1}=B(AB{)}^{j}A$ gives $m_{BA}(t)\mid t{m}_{AB}(t)$. If $AB$ is diagonalizable, its minimal polynomial has simple roots. Hence all nonzero Jordan blocks of $BA$ have size one and its zero blocks have size at most two. Squaring kills these nilpotent blocks, making both squared products diagonalizable. Their eigenvalue multisets agree by Characteristic polynomials of AB and BA, so the squared products are similar matrices.