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I am trying to prove or disprove the following trace inequality for positive-semidefinite Hermitian matrices $A_1$, $A_2$, $A_3$:

$$ Eq. (1) \qquad |Tr( A_1 A_2 A_3 )|^2 \le Tr(A_1A_2) Tr(A_2A_3) Tr(A_1A_3). $$

One can use the scale invariance of Eq. (1), $A \to x A$ for $x >0$, to restrict to psd Hermitian matrices of trace 1.

My numerical search for counterexamples produced no such.

I consider the following generalisation for psd Hermitian matrices of trace 1:

$$ Eq.(2) \qquad \left|Tr(A_1A_2\ldots A_k)\right|^{k-1} \le \prod_{j=1}^k \left|Tr\left(\prod_{l\ne j}A_l\right)\right| $$ where on the r.h.s. the product of the matrices in each trace is ordered one (by the natural order as on the l.h.s. with a missing matrix of index $j$).

Note that for k = 2 we have true Eq. (2) $Tr(A_1A_2) \le 1 = Tr(A_1)Tr(A_2)$.

Any idea?

P.S. The inequality in Eq. (2) (if true of course) can be ascribed a physical meaning of multipartite fidelity bound. It appears, for instance, in the partial distinguishability theory of identical particles.

valery shchesnovich
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  • One can use tensor product and permutation operators on tensor product of spaces to rewrite both sides of Eqs. (1) and (2) as a single trace. This way does not seem to help. – valery shchesnovich May 10 '17 at 22:20

1 Answers1

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Notice that for the Frobenius norm and any matrices $A,B,C$ we have $$ |tr(ABC)| \le \|A\| \|B\|\|C\|, $$ by Cauchy-Schwarz and sub-multiplicativity.

Now, for hermitian $A,B,C$ we have $$ |tr(AABBCC)| = |tr((AB)(BC)(CA))| \le \|AB\| \|BC\| \|CA\|$$ and $$ \|AB\|^2 = tr(ABBA) = tr(AABB). $$

user251257
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  • Thanks, that seems to answer the question. – valery shchesnovich May 11 '17 at 02:50
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    Thanks again! I did not notice this simple trick. The general case, Eq. (2), is actually false. I found a counterexample for N = 4. The analysis shows that, for instance, for N = 4 the only possible (and actually true) bound is by the traces of pairs, $$|Tr(ABCD)|^2≤|Tr(AB)Tr(BC)Tr(CD)Tr(DA)|.$$ – valery shchesnovich May 11 '17 at 13:43