Trace of a power of a matrix product

Question

Suppose I have two 2x2 matrices $A$ and $B$. What can I say about $Tr(A^k B^k)$ versus $Tr((AB)^k)$?

I know that if there is some cyclic permutation that takes $A\cdot A\cdots A B\cdot B\cdots B$ to $A\cdot B\cdots A\cdot B$ then they have the same trace. So, is there an easy way to see that there is such a cyclic permutation?

Otherwise is there some other way to compare these two traces?

Answer 1

In fact, it is an occasionally useful property that for Hermitian matrices $A,B$, $$ \DeclareMathOperator{\tr}{trace} \tr(ABAB) \leq \tr(A^2B^2) $$ Where equality holds if and only if $A$ and $B$ commute. The proof can be summarized as follows:

• $AB - BA$ is skew-symmetric
• The square of a skew-symmetric matrix is negative-definite (i.e. Hermitian with negative eigenvalues)
• $0 \geq \tr(AB - BA)^2 = 2\tr(AABB) - 2\tr(ABAB)$

As you might infer, there is no cyclic permutation from $ABAB$ to $AABB$.

Answer 2

Hint: Use the identity $Tr(AB)=Tr(BA)$

Answer 3

You can say very little in general. Take $A$ to be strictly lower triangular of order $n\times n$, with all entries non-negative. Take $B=A^T$. If $k>n$ then $A^k=B^k=0$ but $\mathbb{tr}((AA^T)^k)>0$ (and will increase as $k$ increases.