Prove that $Tr((AB^2)A)=Tr(A^2B^2)$

Question

Prove that for every $n\times n$ matrices $A,B$: $$Tr((AB^2)A)=Tr(A^2B^2)$$

I need a solution that doesn't use expansion. One more question comes into my mind: given $A,B$ are square matrices. For which condition of $A,B$ we can conclude that $Tr(AB)=Tr(BA)$? Thanks in advance.

Answer 1

Note that for $A,B\in M_{n\times n}$ we have \begin{align*} \DeclareMathOperator{trace}{trace}\trace(AB) &= \sum_{k=1}^n [AB]_{kk} \\ &= \sum_{k=1}^n \sum_{j=1}^n[A]_{kj}[B]_{jk} \\ &= \sum_{j=1}^n \sum_{k=1}^n [B]_{jk}[A]_{kj} \\ &= \sum_{j=1}^n [BA]_{jj} \\ &= \trace(BA) \end{align*}

Answer 2

$${\rm Tr}(AB^2A) = {\rm Tr} \big((AB^2) A \big) = {\rm Tr} \big(A (AB^2) \big) = {\rm Tr}(A^2B^2)$$

using the property ${\rm Tr}(XY) = {\rm Tr}(YX)$.