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The most appealing statement of the Bessis-Moussa-Villani conjecture is as follows:

Conjecture: For all Hermitian positive semidefinite $n\times n$ matrices $A$ and $B$, and all positive integer $m$, the polynomial function $$t \in \mathbb{R}\mapsto g(t) \equiv tr[(A + t B)^m] = \sum\limits_{ k=0}^m a_kt^k$$ has only nonnegative coefficients $a_k, k=1,\cdots,m$.

Most recent and past research concerns on the quantities $m$ and $n$. What about trying to prove the conjecture in the following way: first show that $a_m\ge0$, $a_0\ge0$, $a_{m-1}\ge0$, $a_1\ge0$, $a_{m-2}\ge0$, $a_2\ge0$. And then go on to show $a_3\ge 0$ (and so is $a_{m-3}\ge 0$)...

But it seems difficult to show $a_3\ge 0$. Can anyone share some idea on this particular coefficient?

UPDATED What is the largest term in $S_{2m,m}(AB)$ ? More precisely, consider the word in two positive definite letters $A^{k_1}BA^{k_2}B\cdots A^{k_m}B$ , where $(k1,k2,\cdots,k_m)$ is a pair of nonnegative integer solution of $k_1+k_2+\cdots+k_m=m$ . Is it true that $tr(A^{k_1}BA^{k_2}B\cdots A^{k_m}B)\le tr(A^mB^m)$ ?

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What do you mean by «Most recent and past research concerns on the quantities m and n»? – Mariano Suárez-Alvarez Mar 6 '10 at 4:05
@ M.S-Alvarez: for example, it is known $m = 6, n = 3$, the conjecture is true, and the case when $n < 3, with m\ge 1$ etc. – Sunni Mar 6 '10 at 4:29
He is probably referring to the fact that it has been confirmed for the case $m\le 13$ (or something close to that..) and for the case when n is large enough depending on m. I also don't think that the positivity of $a_3$ has been established in the general case, but supposedly the OP has some specific strategy in mind that might make this case simpler than the full conjecture? (I wouldn't mind seeing something like that) – Gjergji Zaimi Mar 6 '10 at 4:38
My personal view is that when people edit a question which other people have left comments, they should add material rather than replace it. Otherwise, it is hard to make sense of the comments. I have therefore taken the liberty of re-editing the original poster's last edit, so as to make this clear. – Yemon Choi Apr 6 '10 at 3:25
@Yemon Choi: Good advice. Thank you. – Sunni Apr 6 '10 at 13:38
up vote 2 down vote accepted

Look at

Here the fourth coefficient is shown to be non-negative which implies by Hillar's decent theorem also the third coefficient to be non-negative.

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There is a proof of the full BMV conjecture by Herbert Stahl on the preprint server that was added end of July 2011:

I have no opinions from colleagues about the proof so far.

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A year later, you did not forget to update this. ... Thanks, I will read it. – Sunni Aug 8 '11 at 13:44

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