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 $$T(t) = e^{(1-t) W} \prod_i e^{t A_i}$$ .

1more comment