= Characteristic polynomials of AB and BA
{c}
{title2=$\chi_{AB}=\chi_{BA}$}
For square <matrices> $A,B$ of the same size over any <field>,
$$
\det(tI-AB)=\det(tI-BA).
$$
If $A$ is <invertible matrix>[invertible], $BA=A^{-1}(AB)A$, so the products are <similar matrices>. To cover singular $A$, introduce an <indeterminate> $s$ and work over the <rational function field> $F(s)$. The <matrix> $A-sI$ is <invertible matrix>[invertible] there, hence $(A-sI)B$ and $B(A-sI)$ have the same <characteristic polynomial>. Both sides are <polynomials> in $s,t$, so specialize $s=0$. This proof also works over a <finite field>; choosing infinitely many scalars in the original <field> would not.
Equal <characteristic polynomials> do not force the products to be similar. With $A=\operatorname{diag}(1,0)$ and $B=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$, $AB=B\ne0$ but $BA=0$. Both have <characteristic polynomial> $t^2$, while their <minimal polynomials> are $t^2$ and $t$.
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