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$A(t)$ is a continuous mapping from $\mathbb R$ to $GL(\mathbb R^{2n})$ such that $A^2(t)=-id$ for all $t$. Is there a $\epsilon>0$ and a continuous mapping $B(t)$ from $(-\epsilon,\epsilon)$ to $GL(\mathbb R^{2n})$, such that

${B^{ - 1}}(t)A(t)B(t) = \left( {\begin{array}{*{20}{c}} {\begin{array}{*{20}{c}} 0&{ - 1}\\ 1&0 \end{array}}& \cdots &0&0\\ \vdots & \ddots & \vdots & \vdots \\ 0& \cdots &0&{ - 1}\\ 0& \cdots &1&0 \end{array}} \right)$ for all $t\in (-\epsilon,\epsilon)$

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Yes. I'll call the matrix of standard complex structure $I$. First, make $A=A(0)= I$ by a change of coordinates. That is, build such a basis $\beta$ that in it $A$ is $I$. Start with any $v$, take $b_1=v$, then add $b_2=Av$ (these are independent as $A$ has not real eigenvectors). Then take any not in the span of $b_1$ and $b_2$. Take $b_3=w$ and $b_4=Aw$ (still independent - suppose opposite and apply $A$ to the resulting equation, derive contradiction). Pick $u$ not in span of $b_1, \ldots b_4$. Proceed until you get a basis $\beta=\{b_1 \ldots b_{2n}\}$. Now, the claim is that for all t small enough, $\beta(t)=\{b_1, A(t)b_1, b_3, A(t)b_3, \ldots b_{2n-1}, A(t)b_{2n-1}\}$ is a basis in which $A(t)$ is $I$. Proof: Being independent is open condition, so for small $t$, we have that $\beta(t)$ is indeed a basis. The rest is obvious.

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