Can anybody help me with this problem. If $A$ and $B$ are Hermitian Matrices, then $\text{tr}(ABAB) \geq 0$.
It is easy to show that the trace is a real number, but I cannot prove that it must be non-negative.
2026-03-25 06:03:24.1774418604
If A and B are Hermitian Matrices, $\text{tr}(ABAB) \geq 0$.
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You need further conditions. Take $$ A = \pmatrix{1 & 0 \\ 0 & -1}, B=\pmatrix{0 & 1\\ 1 & 0}.$$ Then $AB$ is a rotation by 90 degrees and $(AB)^2=-I$ has trace -2.