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Let $C=left{M in M_{2n}(mathbb{R}): M^2 = -I_{2n}right}$. Why is $C = GL_{2n} (mathbb{R}) cdot J$ with $J=begin{bmatrix}0&-I_n\I_n&0end{bmatrix}$ and $GL_{2n} (mathbb{R} ) $ acting on $M_{2n} (mathbb{R})$ by left multiplication?

This is a question from an exercise sheet of mine and I don't know how to proceed at all. Firstly, I have put an $mathbb{C} $ - vector space structure on it by setting $(a+bi, v) = a+bMV$ and shown that the map $f(v) = Mv$ is linear with respect to this vector space. Moreover I know that a $mathbb{C}$- basis ${v_1, ldots, v_{2n}}$ corresponds to a real basis $lbrace Mv_1, ldots, Mv_{2n}, v_1, ldots , v_{2n} rbrace$. How do I proceed from this point on?

I don't think the question as set makes sense. As @Servaes points out you are asking for a proof that $C$ is the whole of $GL_{2n}(mathbb{C})$ which it is not: as I commented above $2Jnotin C$.

Here is what I think is true.

Suppose $Min M_{2n}(mathbb{R})$ and $M^2=-I$. Then there is a matrix $Pin GL_{2n}(mathbb{R})$ such that $P^{-1}M P=begin{pmatrix} O & -I_n\I_n&Oend{pmatrix}$.

This is easily got from the Rational Canonical Form. The minimal polynomial of $M$ is $X^2+1$, which is irreducible. So the RCF of $M$ is a sum of $2times 2$ blocks of the form $begin{pmatrix} 0 & -1\1 &0end{pmatrix}$. Re-arranging the basis gives what I have asserted.

[In other words the matrices must be over $mathbb{R}$ and the action of $GL$ on $J$ is conjugation not multiplication. If you insist on allowing $mathbb{C}$ I think you'll find that the matrix $begin{pmatrix} i & 0\0 &iend{pmatrix}$ is a counterexample to any adjustment you try to make to the question.]