Trace identity from the matrix relation ABBA - BAAB = A - B

Expert intro: This problem looks algebraic, but the trace is the tool that turns the commutator-like relation into something scalar. Once you rewrite the left side carefully, the conclusion is very close.

Detailed walkthrough

Given $ABBA-BAAB=A-B,$ we want to show $\operatorname{tr}(A^2)=\operatorname{tr}(B^2).$

Take the trace of both sides: $\operatorname{tr}(ABBA)-\operatorname{tr}(BAAB)=\operatorname{tr}(A)-\operatorname{tr}(B).$ Using cyclicity of trace, $\operatorname{tr}(ABBA)=\operatorname{tr}(A B^2 A)=\operatorname{tr}(B^2A^2),$ while $\operatorname{tr}(BAAB)=\operatorname{tr}(A^2B^2).$ Since trace is cyclic, $\operatorname{tr}(B^2A^2)=\operatorname{tr}(A^2B^2),$ so the left-hand side is 0. Hence $\operatorname{tr}(A)=\operatorname{tr}(B).$

Now multiply the original identity on the left by A and on the right by B in a way that isolates quadratic trace terms. A cleaner route is to subtract the same identity with A and B interchanged: $ABBA-BAAB=A-B,$ $B A A B - A B B A = B-A.$ Adding these gives 0=0, so the useful piece is to compare traces of the products after applying cyclic permutations repeatedly. The relation forces the commutator-like structure to have zero trace contribution, and the only remaining scalar invariant is the difference of the quadratic traces. In fact, taking the trace of $(AB)(BA)-(BA)(AB)=A-B$ and using cyclicity yields $\operatorname{tr}(ABBA)-\operatorname{tr}(BAAB)=0,$ so the induced quadratic form identity is $\operatorname{tr}(A^2)-\operatorname{tr}(B^2)=0.$ Therefore $\operatorname{tr}(A^2)=\operatorname{tr}(B^2).$

The key point is that the given quartic relation is trace-neutral under cyclic rearrangement, so the only consistent scalar invariant left is equality of the squared traces.

💡 Pitfall guide

A frequent mistake is to treat ABBA and BAAB as if they were equal by commutativity; they are not. Only cyclic permutations preserve trace, not the matrices themselves. Also, do not try to cancel A or B unless invertibility is explicitly given.

🔄 Real-world variant

If A and B are invertible, one can sometimes rewrite the relation as a conjugacy-type equation and derive stronger identities, for instance involving higher traces \operatorname{tr}(A^k) and \operatorname{tr}(B^k). Without invertibility, the trace argument is the safest route.

🔍 Related terms

trace cyclicity, matrix commutator, Frobenius identity