If the matrices $A$ and $B$ are similar, then are $AB$ and $BA$ similar?

Question

Let $A, B \in M_n(F)$. Suppose that $n \geq 2$. Prove or disprove: If $A$ and $B$ are similar, then $AB$ and $BA$ are similar.

I think that this statement is false. I tied to find A,B which are similar and not invertible which are $2 \times 2$ and are counterexample but I couldn't. Would appreciate help to find any counterexample $2 \times 2$ or $3 \times 3$.

Answer 1

This isn't true. Consider e.g. a field of characteristic $\ne2$ with $$ A=\pmatrix{1&1\\ 0&0}, \ B=\pmatrix{0&1\\ 0&1}, \ AB=\pmatrix{0&2\\ 0&0}\ne0=BA. $$

Answer 2

Indeed the answer is positive for $2\times2$ matrices. If one of both matrices is invertible, then $AB$ and $BA$ are obviously similar. If one of both is $0_2$, this is obvious again. There remains the case $$A=ab^T,\qquad B=uv^T,\qquad a,b,u,v\ne0.$$ (Since $A$ and $B$ are similar, we have $a\cdot b=u\cdot v$.) Then $$AB=(b\cdot u)av^T,\qquad BA=(v\cdot a)ub^T,$$ which are similar for the same reason, that is ${\rm Tr}AB={\rm Tr}BA$. Notice that we don't need at all that $A$ and $B$ be similar !