This question was asked in my linear algebra quiz and I was unable to solve it.
Let A , B be n cross n matrices. Which one of the following equals $trace ( A^2 B^2) $ ?
(i) ${(trace (AB))}^2 $
(ii) $trace ( AB^2 A) $
(iii) $trace ({(AB)}^2) $
(iv) trace (BABA)
I thought of using that Trace is commutative but by that all four options come out to be correct. So, Can you please tell me what mistake I am making in understanding this result regarding trace and what is the right way to do the question?
Many thanks!!
$\newcommand{tr}[1]{\mbox{tr}(#1)}$The trace is not commutative. But it is cyclic.
What does that mean? Well, we know that $\tr{AB} = \tr{BA}$ for all $A,B$. But then if you take three matrices $X,Y,Z$, this doesn't give that $\tr{XYZ} = \tr{XZY}$ for example, because you cannot actually use the commutativity here : if you use it you just get $\tr{YZ} =\tr{ZY}$ but this does not imply that $\tr{XYZ} = \tr{XZY}$. You can easily find a counterexample.
However, the trace is cyclic : if you take a matrix at the end, and put it in the front, then the trace does not change. Why? Because you can apply the result for two matrices, by using the block of all matrices except the last as $A$ and the last one as $B$. So for example, by taking $XY = A$ and $Z=B$, we do get : $$ \tr{XYZ} = \tr{ZXY} = \tr{YZX} $$
so we can always take the last one, and put it as the first one and the trace would not change.
In this example, we have $A^2B^2 = AABB$. So, by applying the result, we get : $$ \tr{AABB} = \tr{BAAB} = \tr{BBAA} = \tr{ABBA} $$
which is the same as : $$ \tr{A^2B^2} = \tr{BA^2B} = \tr{B^2A^2} = \tr{AB^2A} $$
Further, any other combination of $A$ and $B$ need not have the same trace as any of these matrices.