**Question:** Given an arbitrary number of real matrices of the form $ A\_i=
\biggl(\begin{matrix}
C\_i+E\_i & B\_i \\
B\_i^T & D\_i-F\_i
\end{matrix} \biggr)
$, where $B\_i$ is an arbitrary $n\times n$ real matrix, $C\_i$ and $D\_i$ are $n\times n$ real anti-symmetric matrices, $E\_i$ and $F\_i$ are $n\times n$ real symmetric and positive semidefinite matrices, how to prove the following
$$
\det \biggl(I\_{2n} + \prod\_i e^{A\_i}\biggl)\ge 0 \,?
$$ 


**Background:** This is a stronger version of an earlier MO question [How to prove this determinant is positive?](https://mathoverflow.net/questions/204460/how-to-prove-this-determinant-is-positive) which was solved by GH from MO and Terry Tao. 
Their proof addressed the case of $E\_i=F\_i=0$, where $e^{A\_i}$ belongs to the split orthogonal group. 


The paper [arXiv:1601.01994v2](http://arxiv.org/abs/1601.01994v2) in fact contains a rigorous proof of the above statement using tools familiar to physicists, e.g. Majorana fermion and reflection positivity (cf Eq.(10)). It would be nice to see an alternate mathematical proof (perhaps of a similar nature to the split orthogonal group proof by [GH from MO](https://mathoverflow.net/a/204609/71225) and [Terry Tao](https://terrytao.wordpress.com/2015/05/03/the-standard-branch-of-the-matrix-logarithm/)). 


**Addendum:** The paper [arXiv:1601.01994v2](http://arxiv.org/abs/1601.01994v2) contains further results on complex matrices, which I do not know how to formulate in simple math language yet. 



How to prove this determinant is positive-II?
Let $q(x,y) = x^H J y$ for $x,y \in \mathbb{C}^{2n}$ where $J = diag(I\_n,-I\_n)$ and let $S = \{A \in M\_{2n}(\mathbb{R}) : q(Ax, Ax) \ge q(x,x) $ $\forall x \in \mathbb{C}^{2n}\}$.
Obviously $S$ is a semi group .
Furthermore the $e^{t A\_i}$ are in $S$ since $$\frac{d}{dt} q(e^{t A\_i} x,e^{t A\_i} x) = 2 (e^{t A\_i} x)^H diag (E\_i,F\_i) (e^{t A\_i} x) \ge 0$$ for all $x \in \mathbb{C}^{2n}$.


Now let $T : [0,1] \rightarrow S$ analytic where $det(I\_{2n}+T(0)) > 0$ and where $T(0)$ has no degenerate eigenvalues.


Let $E\_{\lambda}(t)$ be the generalized eigenspace of $T(t)$ to the eigenvalue $\lambda$ .
For $G \subset \mathbb{C}$ define $$E\_G(t) = \bigoplus\_{\lambda \in G}{E\_{\lambda}(t)}$$ .


Let $t\_0 \in [0,1]$ such that $det(I\_{2n}+T(t\_0)) = 0$ .


Now we want to show that $dim\,E\_{(-1,\infty)}(t)$ can only change by an even number near $t\_0$ . Therefore $det(I\_{2n}+T(t))$ can't change the sign.


Lemma 1 : Let $U \in S$, $x \in \mathbb{C}^{2n}$ with $0 = q(x,x) = q(Ux,Ux)$ . Then $q(x,y) = q(Ux,Uy)$ for all $y \in \mathbb{C}^{2n}$ .


Proof : We have $q(ax+y,ax+y) \le q(a Ux + Uy,a Ux + Uy)$ for all $a \in \mathbb{C}$ and therefore $0 \le 2 Re\,a (q(Uy,Ux) - q(y,x)) + q(Uy,Uy) - q(y,y)$ . But the right hand side can be made negative for appropriate a if $q(Uy,Ux) \neq q(y,x)$ .


Lemma 2 : Let $p$ a polynomial and $z \in \mathbb{C}$ . If $p(n) z^n$ is constant for all large enough $n \in \mathbb{N}$ then $p$ is constant and $z = 1$ or $p = 0$ or $z = 0$.


Proof left to the reader.


Lemma 3 : Let $U \in S$, $x$ a generalized eigenvector of $U$ to the eigenvalue $\lambda$, $y$ a generalized eigenvector of $U$ to the eigenvalue $\mu$ and $q(U^l x,U^l x) = 0$ for all $l \in \mathbb{N}\_0$. Then holds $\lambda \bar{\mu} = 1$ or $q(x,y) = 0$ .


Proof : By Lemma 1 we have $q(y,x) = q(U^l y, U^l x)$ for all $l \in \mathbb{N}$ . But $q(U^l y, U^l x)$ has the form $p(l) (\lambda \bar{\mu})^l$ for a polynomial $p$ for all large enough $l \in \mathbb{N}$ . From Lemma 2 then follows Lemma 3 . 


Lemma 4 : Let $U \in S$, $x$ a generalized eigenvector of $U$ to the eigenvalue $-1$ and $q(x,U^k x) = 0$ for all $k \in \mathbb{N}\_0$ . Then $q(U^k x,U^l x) = 0$ for all $k,l \in \mathbb{N}\_0$ .


Proof : Let $x\_k = (I\_{2n} + U)^k x$ and $m$ minimal such that $q(x\_j,x\_k) = 0$ for all $j,k \geq m$ . First we want to show that $q(x\_j,x\_k) = 0$ for $j \geq m$ and $k \geq 0$. If this is not the case then let $j \geq m$ and $k$ be maximal such that $q(x\_j,x\_k) \neq 0$ . Then $k > 0$ and by Lemma 1 $q(x\_j,x\_{k-1}) = q(U^l x\_j,U^l x\_{k-1}) = q(x\_j,x\_{k-1}) - l q(x\_j,x\_k)$ for all $l \in \mathbb{N}$ . Contradiction !
Now we get for $m > 1$ $q(x\_{m-2},x\_{m-2}) \leq q(U^l x\_{m-2},U^l x\_{m-2}) = l^2 q(x\_{m-1},x\_{m-1}) + O(l)$ and $q(x\_{m-2},x\_{m-2}) \geq q(U^{-l} x\_{m-2},U^{-l} x\_{m-2}) = l^2 q(x\_{m-1},x\_{m-1}) + O(l)$ . Contradiction to $q(x\_{m-1},x\_{m-1}) \neq 0$ ! Since $m = 1$ is impossible since $q(x\_0,x\_k) = 0$ for all $k \geq 0$ we are done .


Lemma 5 : The restriction of $q$ to $E\_{-1}(t\_0)$ is non degenerate .


Proof : Let $x \in E\_{-1}(t\_0)$ and $U = T(t\_0)$ . We want to show that there exists $y \in E\_{-1}(t\_0)$ such that $q(x,y) \neq 0$ . If there exists $n \in \mathbb{N}\_0$ such that $q(x,U^n x) \neq 0$ we are done. Otherwise ist follows from Lemma 4 and Lemma 3 that x is orthogonal w.r.t. q to all other generalized eigenspaces . But since q is non degenerate there exists $y \in E\_{-1}(t\_0)$ such that $q(x,y) \neq 0$ .


Lemma 6 : T(t) has degenerated eigenvalues only at isolated points.


Proof : The discriminant of T(t) is analytic in t and nonzero at t = 0 .


Now we can choose $\epsilon > 0$ and $r$ with $0 < r < 1$ such that for $\vert t-t\_0\vert < \epsilon$ holds :


i) $det(I\_{2n}+T(t)) \neq 0$ for $t \neq t\_0$ ,


ii) $E\_{\{z: \vert z+1 \vert \leq r\}}(t\_0) = E\_{-1}(t\_0)$ ,


iii) $\sigma (T(t)) \cap \{z: \vert z+1 \vert = r\} = \emptyset$ ,


iv) the signature of the restriction of $q$ to $V(t)$ is constant where $V(t) = E\_{\{z: \vert z+1 \vert \leq r\}}(t)$ ,


v) T(t) has no degenerate eigenvalues for $t \neq t\_0$ .


Let $D = \{z: \vert z+1 \vert \leq r\}$ .
For each eigenvalue $\lambda \in D$ of $T(t)$ with $\vert \lambda \vert = 1$ we can write $E\_{\lambda}(t) = E\_{\lambda}^+(t) \oplus E\_{\lambda}^-(t)$ such that the restriction of $q$ to $E\_{\lambda}^+(t)$ is positive semidefinite and the restriction of $q$ to $E\_{\lambda}^-(t)$ is negative definite and such that $E\_{\bar{\lambda}}^+(t) = \overline{E\_{\lambda}^+(t)}$ and $E\_{\bar{\lambda}}^-(t) = \overline{E\_{\lambda}^-(t)}$ .


Now we can write $V(t) = V^+(t) \oplus V^-(t)$ where
$$V^+(t) = E\_{D \cap \{z : \vert z \vert > 1\}}(t) \oplus \bigoplus\_{\lambda \in D , \vert \lambda \vert = 1} E\_{\lambda}^+(t)$$
and
$$V^-(t) = E\_{D \cap \{z : \vert z \vert < 1\}}(t) \oplus \bigoplus\_{\lambda \in D , \vert \lambda \vert = 1} E\_{\lambda}^-(t)$$ .


Now we want to show that for $\vert t-t\_0\vert < \epsilon$ and $t \neq t\_0$ the restriction of $q$ to $V^+(t)$ is positive semidefinite and the restriction to $V^-(t)$ is negative semidefinite :


For $x \in V^+(t)$ we get $$q(x,x) \geq \lim\_{m \rightarrow \infty} \frac{1}{m} \sum\_{l=1}^m q(U^{-l} x,U^{-l} x) = \sum\_{\lambda \in D, \vert \lambda \vert = 1} q(x\_{\lambda}^+,x\_{\lambda}^+) \geq 0$$ where $x\_{\lambda}^+$ is the component of x in $E\_{\lambda}^+(t)$ .
For $x \in V^-(t)$ we get $$q(x,x) \leq \lim\_{m \rightarrow \infty} \frac{1}{m} \sum\_{l=1}^m q(U^l x,U^l x) = \sum\_{\lambda \in D, \vert \lambda \vert = 1} q(x\_{\lambda}^-,x\_{\lambda}^-) \leq 0$$ where $x\_{\lambda}^-$ is the component of x in $E\_{\lambda}^-(t)$ .


Let $n\_+$ the number of positive eigenvalues of the restriction of q to $V(t)$ and $n\_-$ the number of negative eigenvalues.
We have shown that $n\_+ \geq dim\, V^+(t)$ and $n\_- \geq dim\, V^-(t)$ and therefore $n\_+ = dim\, V^+(t)$ .
Since $dim\, V^+(t) - dim\, E\_{(-1,-1-r)}$ is even $dim\, E\_{(-1,-1-r)}$ can only change by an even number.
And $dim\, E\_{[-1-r,-\infty)}$ can only change if a pair of complex conjugate eigenvalues gets real or vice versa and therefore also only by an even number.


So we have shown that $dim\,E\_{(-1,\infty)}(t)$ can only differ by an even number on different points in $\vert t-t\_0\vert < \epsilon, t\neq t\_0$ .


To finish the proof, choose $W$ such that $e^{t W} \in S$ for $t \in [0,1]$ and such that $e^W$ fullfills the requirements on $T(0)$ .
Then choose $$T(t) = e^{(1-t) W} \prod\_i e^{t A\_i}$$ .


