Given matrices
$$A_i= \biggl(\begin{matrix}
0 & B_i \\
B_i^T & 0
\end{matrix} \biggr)$$
where $B_i$ are real matrices and $i=1,2,\ldots,N$, how to prove the following?
$$\det \big( I + e^{A_1}e^{A_2}\ldots e^{A_N} \big) \ge 0$$
This seems to be true numerically.
Update1: As was shown in below, the above inequality is related to another conjecture $\det(1+e^M)\ge 0$, given a $2n\times 2n$ real matrix $M$ that fulfills $\eta M \eta =-M^T$ and $\eta=diag(1_n, -1_n)$. The answers of Christian and Will, although inspiring, did not really disprove this conjecture as I understood.
Update2: Thank you all for the fruitful discussion. I attached my Python script down here. If you run it for several times you will observe
-
$\det(1+e^{A_1}\ldots e^{A_N})$ seems to be always larger than zero (the conjecture),
-
$M = \log(e^{A_1}\ldots e^{A_N})$ are sometimes indeed pure real and it fulfills the condition mentioned in update1. In this case the eigenvalues of $M$ are either pure real or in complex conjugate pairs. Thus it is easy to show $\det(1+e^M)=\prod_l(1+e^
{\lambda_l})\ge 0$, -
However, sometimes the matrix $M = \log(e^{A_1}\ldots e^{A_2})$ can be complex and they are indeed in the form written down by Suvrit. In this case, it seems that the eigenvalues of $M$ will contain two sets of complex values: $\pm a+i\pi$ and $\pm b + i\pi$. Therefore, $\det(1+e^{M})\ge 0$ still holds because $(1-e^a)(1-e^{-a})(1-e^{b})(1-e^{-b})\ge 0$.
Update3: Thank you GH from MO, Terry and all others. I am glad this was finally solved. One more question: how should I cite this result in a future academic publication ?
Update4: Please see the publication out of this question at arXiv:1506.05349.
Best Answer
Here are some ideas how to decide the conjecture. (EDIT: In fact these ideas lead to a proof of the conjecture as Terry Tao explained in two comments below.)
As Christian Remling and Will Sawin showed, the conjecture is equivalent to $\det(I+T)\geq 0$ for any $T\in\mathrm{SO}^0(n,n)$.
We can assume that $-1$ is not an eigenvalue of $T$. Up to conjugacy, $T$ is a sum of indecomposable blocks as in Theorem 1 of Nishikawa's 1983 paper, and then $\det(I+T)$ is the product of the determinants of the corresponding blocks of $I+T$. Hence, by the idea of jjcale, we can forget about the blocks that are of exponential type. By page 83 in Djoković's 1980 paper, the remaining blocks are of type $\Gamma_m(\lambda,\lambda^{-1})$ with $\lambda<0$ and $\lambda\neq -1$, which in turn are described on page 77 of the same paper. Such a block contributes $(1+\lambda)^{2m+2}/\lambda^{m+1}$ to $\det(I+T)$, hence we can forget about the blocks where $m$ is odd.
To summarize, we can assume that $T$ is composed of $(2m+2)\times(2m+2)$ blocks of type $\Gamma_m(\lambda,\lambda^{-1})$ with $\lambda<0$ and $\lambda\neq -1$ and $m$ even. The conjecture is true if and only if the number of such blocks is always even. For this, the explicit description of $\mathrm{SO}^0(n,n)$ on page 64 of Nishikawa's 1983 paper might be useful (see also page 68 how to use this criterion for $m=1$). Based on this, I verified by hand that one cannot have a single block for $m=2$, which also shows that the smallest possible counterexample to the conjecture is of size $10\times 10$ (i.e. $n\geq 5$).
Added 1. Terry Tao realized and kindly added that in the remaining case we are done. Read his comments below. To summarize and streamline his ideas, we have in this case \begin{align*}\det(I_{2n}+T) &=\det(I_n+A)\det(I_n+A^{*-1})\\ &=\det(A)\det(I_n+A^{-1})\det(I_n+A^{*-1})\\ &=\det(A+A^{*-1})\frac{\det(I_n+A^{-1})^2}{\det(I_n+A^{-1}A^{*-1})}, \end{align*} where $(A+A^{*-1})/2$ can be described as the restriction of $T$ to a totally positive subspace followed by the orthogonal projection to this subspace. Now we have $\det(A+A^{*-1})>0$ by $T\in\mathrm{SO}^0(n,n)$, while the fraction on the right is clearly positive, hence we conclude $\det(I_{2n}+T)>0$.
Added 2. Terry Tao wrote a great blog entry on this topic.
Added 3. Let me add a variation on Terry's original argument. Djoković defines $\mathrm{SO}(n,n)$ via $J:=\begin{pmatrix} 0 & I_n \\ I_n & 0 \end{pmatrix}$, while Nishikawa defines it via $K:=\begin{pmatrix} I_n & 0 \\ 0 & -I_n\end{pmatrix}$. These two matrices are connected via $J=M^*KM$, where $M:=\frac{1}{\sqrt{2}}\begin{pmatrix} I_n & I_n\\ -I_n & I_n\end{pmatrix}$, hence any matrix $T$ in Djoković's $\mathrm{SO}(n,n)$ corresponds to $MTM^*$ in Nishikawa's $\mathrm{SO}(n,n)$. We need to examine the case of $T = \begin{pmatrix} A & 0 \\ 0 & A^{*-1} \end{pmatrix}$, which corresponds to $MTM^*=\frac{1}{2}\begin{pmatrix} A+A^{*-1} & -A+A^{*-1} \\ -A+A^{*-1} & A+A^{*-1} \end{pmatrix}$. This lies in Nishikawa's $\mathrm{SO}^0(n,n)$, whence $\det(A+A^{*-1})>0$.