Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$A(t)$ 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 $t\in (-\epsilon,\epsilon)$

share|cite|improve this question
up vote 2 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.