0
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ No condition, this holds for all matrices. $\endgroup$
    – Winther
    Feb 9, 2015 at 5:14
  • $\begingroup$ Why not use expansion? I think that's the best way to prove it within elementary linear algebra. $\endgroup$
    – Vim
    Feb 9, 2015 at 5:15
  • 3
    $\begingroup$ @Arashium: That's not true. $\endgroup$
    – anomaly
    Feb 9, 2015 at 5:16
  • $\begingroup$ If expansion means writing down the trace in terms of the explicit entries of $A$ and $B$, that's a terrible way of solving the problem. The relation $\operatorname{tr}(XY) = \operatorname{tr}(YX)$ holds for arbitrary $X, Y$ (square of the same dimension), and that's enough to prove the statement in your question. $\endgroup$
    – anomaly
    Feb 9, 2015 at 5:18

2 Answers 2

5
$\begingroup$

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*}

$\endgroup$
3
  • $\begingroup$ @Arashium true. This answer is brilliant but I think the OP wants some novel one without expansion :P $\endgroup$
    – Vim
    Feb 9, 2015 at 5:20
  • $\begingroup$ @Arashium absolutely not. Merely that the trace of $AB$ is the same as the trace of $BA$. $\endgroup$ Feb 9, 2015 at 5:20
  • $\begingroup$ Thanks for reminding me that $tr(AB)=tr(BA)$. How silly I can forget such a property! $\endgroup$ Feb 9, 2015 at 5:24
1
$\begingroup$

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

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .