I found this statement in a book : "$GL(n, \Bbb C)$ is the group of $\Bbb C$-linear automorphisms of $(R^{2n} , J)$, i.e. $\{A| AJ= JA\}$. Where $J$ is any complex structure" but I don't understand it Some one give me a hint to prove it please?
2026-04-03 22:59:04.1775257144
description of $Gl(n,\Bbb C)$
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Let $J^2 =-I_{2n}$.
Then its minimal polynomial is $x^2+1$ which has no real root, so $J$ has no eigenvector.
Choose a basis of $\Bbb R^{2n}$ in the following way:
Let $e_1$ be arbitrary nonzero vector, and $f_1:=Je_1$. Then let $e_2$ be arbitrary not in the span of $e_1,f_1$, and let $f_2:=Je_2$. And so on..
Note that $Jf_k=-e_k$, so $\mathrm{span}(e_k, f_k)$ is a $J$-invariant subspace.
Of course, $f_k$ will play the role of $i\cdot e_k$, so that, as a vector space over $\Bbb C$, it has $e_1,\dots,e_n$ as a basis.
Now, $AJ=JA$ means that $A$ is determined by its values on $e_1,\dots, e_n$:
Indeed, if $AJ=JA$, then $Af_k=AJe_k=JAe_k$.
Conversely, let $Ae_k$ be arbitrarily given, and define $Af_k:=JAe_k$. Verify that then $AJ=JA$ follows.