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Let be two Hermitian matrices. When and commute, we have the identity
When and do not commute, the situation is more complicated; we have the Baker-Campbell-Hausdorff formula
where the infinite product here is explicit but very messy. On the other hand, taking determinants we still have the identity
Recently I learned (from Emmanuel Candes, who in turn learned it from David Gross) that there is another very nice relationship between and , namely the Golden-Thompson inequality
The remarkable thing about this inequality is that no commutativity hypotheses whatsoever on the matrices are required. Note that the right-hand side can be rearranged using the cyclic property of trace as ; the expression inside the trace is positive definite so the right-hand side is positive. (On the other hand, there is no reason why expressions such as need to be positive or even real, so the obvious extension of the Golden-Thompson inequality to three or more Hermitian matrices fails.) I am told that this inequality is quite useful in statistical mechanics, although I do not know the details of this.
To get a sense of how delicate the Golden-Thompson inequality is, let us expand both sides to fourth order in . The left-hand side expands as
while the right-hand side expands as
Using the cyclic property of trace , one can verify that all terms up to third order agree. Turning to the fourth order terms, one sees after expanding out and using the cyclic property of trace as much as possible, we see that the fourth order terms almost agree, but the left-hand side contains a term whose counterpart on the right-hand side is . The difference between the two can be factorised (again using the cyclic property of trace) as . Since is skew-Hermitian, is positive definite, and so we have proven the Golden-Thompson inequality to fourth order. (One could also have used the Cauchy-Schwarz inequality for the Frobenius norm to establish this; see below.)
Intuitively, the Golden-Thompson inequality is asserting that interactions between a pair of non-commuting Hermitian matrices are strongest when cross-interactions are kept to a minimum, so that all the factors lie on one side of a product and all the factors lie on the other. Indeed, this theme will be running through the proof of this inequality, to which we now turn.