I have a map: $F_{A} (X) :GL\left(2n,\mathbb{R}\right) \longrightarrow\mathbb{\mathfrak{M}_{\mathit{2n\times2n}}\left(\mathbb{R}\right)}$ such as \begin{eqnarray} & F_{A}(X) & =XAX^{-1}-A \end{eqnarray}
where $X\in GL(2n,\mathbb{R})$ and $A=\left(\begin{array}{cc} 0 & -I_{n}\\ I_{n} & 0 \end{array}\right)$
And I need to find the rank of this map
I tried to develop in for $n=1$ just to see what's going. What I obtained $$F_{A}(X)= \frac{1}{\det X} \left(\begin{array}{cc} X_{1} & X_{2}\\ X_{3} & X_{4} \end{array}\right) \left(\begin{array}{cc} 0 & -I_{n}\\ I_{n} & 0 \end{array}\right)\left(\begin{array}{cc} -X_{4} & +X_{2}\\ +X_{3} & -X_{1} \end{array}\right)-\left(\begin{array}{cc} 0 & -I_{n}\\ I_{n} & 0 \end{array}\right)$$
$$F_{A}(X)=\frac{1}{\det X}\left(\begin{array}{cc} \left(-X_{1}X_{3}-X_{2}X_{4}\right) & \left(X_{1}X_{1}+X_{2}X_{2}\right)\\ \left(-X_{3}X_{3}-X_{4}X_{4}\right) & \left(X_{3}X_{1}+X_{4}X_{2}\right) \end{array}\right)-\left(\begin{array}{cc} 0 & -I_{n}\\ I_{n} & 0 \end{array}\right)$$ but I'm quite stuck. I could go on with brute force and resolve the $n=1$ case, but I'm pretty sure that since it seems a really natural application there should be a known formalism or even a known formula where this kind of things are evident. Can You give me an hint?
Let us compute $dF(X)$ using directional derivatives: for $M\in\mathbb R^{2n\times 2n}$ we have $$ dF(X)M=\lim_{t\to0}\frac{F(X+tM)-F(X)}{t}=\lim_{t\to0}\frac{(X+tM)A(X+tM)^{-1}-XAX^{-1}}{t}. $$ Now for $|t|$ small we can use the expression $$ (X+tM)^{-1}=X^{-1}-tX^{-1}MX^{-1}+t^2P, $$ so that $$ (X+tM)A(X+tM)^{-1}-XAX^{-1}=t\big(MAX^{-1}-XAX^{-1}MX^{-1}\big)+t^2Q, $$ and we conclude the limit above is $$ dF(X)M=MAX^{-1}-XAX^{-1}MX^{-1}. $$ Awful as it is, we can use that expression to find out the rank. Indeed,we first look at the kernel of $dF(X)$. It consists of all matrices such that $$ MAX^{-1}=XAX^{-1}MX^{-1} $$ Multiplying by $X^{-1}$ on the left and $X$ on the right, this is equivalent to $$ (X^{-1}M)A=A(X^{-1}M). $$ We conclude that $M\mapsto N=X^{-1}M$ gives an isomorphism from the kernel onto the space $E$ of matrices $N$ commuting with $A$. All in all we obtain $$ \text{rank }dF(X)=4n^2-\dim\ker df(X)=4n^2-\dim E. $$ Finally, after some computations one sees that $E$ consists of all matrices of the form $$ N=\left(\begin{array}{cc} G&H\\-H&G \end{array}\right), $$ and consequently $E$ has dimension $2n^2$. Hence in the end: $$ \text{rank }dF(X)=2n^2. $$