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](https://gist.github.com/wangleiphy/b7346ebe3cd877b1cb49) down here. If you run it for several times you will observe 


1. $\det(1+e^{A\_1}\ldots e^{A\_N})$ seems to be always larger than zero (the conjecture),
2. $M = \log(e^{A\_1}\ldots e^{A\_N})$ 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 $\det(1+e^M)=\prod\_l(1+e^
{\lambda\_l})\ge 0$,
3. 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](http://arxiv.org/abs/1506.05349). 



How to prove this determinant is positive?
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](https://www.jstage.jst.go.jp/article/kyushumfs/37/1/37_1_63/_pdf), 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](http://www.sciencedirect.com/science/article/pii/0021869380901349), 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](https://www.jstage.jst.go.jp/article/kyushumfs/37/1/37_1_63/_pdf) 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](https://terrytao.wordpress.com/2015/05/03/the-standard-branch-of-the-matrix-logarithm/) 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 $\det(A+A^{\*-1})>0$.


