Verifying $A$ is complex linear if and only if $A$ commutes with $J$ I am trying to verify the statement that $A \in GL_{2n}(\mathbb{R})$ is complex linear if and only if $AJ_{2n} = J_{2n}A$, where $J_{2n}$ is the $2n \times 2n$ matrix whose diagonal blocks are all $\begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$. The definition of complex linear I am adopting is $A$ is complex linear if it is in the image of $d_n$, which takes each complex entry $a + bi$ to the $2\times2$ block $\begin{bmatrix} a & -b \\ b & a\end{bmatrix}$.
This statement is certainly true if $n =1$, when $A = 
\begin{bmatrix} 
a & c \\
b & d
\end{bmatrix}$ and $J_{2} = \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix}$.
So $AJ_2 = J_2A$ if and only if
$ 
\begin{bmatrix} 
c & -a \\
d & -b
\end{bmatrix}=
\begin{bmatrix} 
-b & -d \\
a & c
\end{bmatrix},$
which holds if and only if $a=d$ and $c=-b$, i.e., if and only if
$A = 
\begin{bmatrix} 
a & b \\
-b & a
\end{bmatrix}$, which is $d_n(a+bi)$.
But I don't know how to conveniently express $AJ_{2n}$ or $J_{2n}A$ when $n>1$ so I can do a direct comparison. Can someone help me with this or provide a smarter way to see this when $n>1$? Thanks in advance!
 A: Short answer: Use block matrix algebra.
Longer answer: By definition, $J_2=\begin{bmatrix}0&-1\\1&0\end{bmatrix}$ and $J_{2n}$ is an $n\times n$ block matrix
$$ J_{2n} = \begin{bmatrix}
J_2 & & & \\ & J_2 & & \\ & & \ddots & \\ & & & J_2
\end{bmatrix} $$
Then split $A$ up into an $n\times n$ block matrix with entries $A_{ij}$; that is, $A_{ij} = \begin{bmatrix}a_{2i-1,2j-1}&a_{2i-1,2j}\\a_{2i,2j-1}&a_{2i,2j}\end{bmatrix}$ where the $a_{k\ell}$ are the "actual" entries of $A$.
You should be able to take it from here: block matrices multiply like ordinary matrices (as long as you're careful about ordering the multiplications) and in the end you will just use the $n=1$ calculation a bunch of times.
A: You can make use of the complex numbers as the terminology suggests. Make $\def\R{\Bbb R}\R^{2n}$ into a vector space over $\def\C{\Bbb C}\C$ by extending scalar multiplication, which for real scalars remains as usual, by setting $\def\ii{\mathbf i}(a+b\ii)v=av+bJ_{2n}v$. The only axiom that requires a bit of attention to check is associativity of complex scalar multiplication: $(a+b\ii)(c+d\ii)v=(ac-bd+(ad+bc)\ii)v$, and it is satisfied because $(J_{2n})^2=-I_{2n}$. For this complex structure the space is of dimension$~n$, and has a basis$~\mathcal B$ obtained by grouping the standard basis of $\R^{2n}$ by two basis vectors at a time and retaining only the first of each pair. A map $f:\R^{2n}\to\R^{2n}$ is $\C$-linear if and only if it is both $\R$-linear and if $f(\ii v)=\ii f(v)$ for all $v\in\R^{2n}$, which means precisely that $f$ is given by multiplication by a real $2n\times2n$ matrix $A$ that satisfies $AJ_{2n}=J_{2n}A$. In this case $f$ has a complex $n\times n$ matrix $C$ with respect to the basis$~\mathcal B$, and if the coefficient of $C$ at position$~(i,j)$ is $a+b\ii$, then the coefficients of $A$ at the block of positions $(2i-1,2i)\times(2j-1,2j)$ (that is, the $2\times2$ block at position $(i,j)$ among those blocks) is $\bigl({a\atop b}~{-b\atop a}\bigr)$, as a simple computation shows. Conversely any complex matrix $C$ represents on the bases$~\mathcal B$ a $\C$-linear map$~f:\R^{2n}\to\R^{2n}$ given by left-multiplication by the real $2n\times2n$ matrix $A$ whose $2\times2$ blocks are so obtained from the entries of $C$. This proves the equivalence of the question.
A: I'll explain this more conceptually.
The real vector space $\mathbb R^{2n}$ can be turned into a complex vector space $\mathbb C^n$ by identifying $\begin{pmatrix} a_1 \\ b_1 \\ \vdots \\ a_n \\ b_n\end{pmatrix}$ with $\begin{pmatrix} a_1 + ib_1 \\ \vdots \\ a_n + ib_n\end{pmatrix}$.
The action of $\mathbb R$ on $\mathbb R^{2n}$ is clear. And if we know the action of $i$ on $\mathbb R^{2n}$, we would know how $\mathbb C$ acts on $\mathbb R^{2n}$ in full. By direct computation, we know $i$ acts on $\mathbb R^{2n}$ through the matrix $J_{2n}$.
Therefore a real linear transformation $A$ on $\mathbb R^{2n}$ is complex linear iff $AJ_{2n} = J_{2n}A$.
Now why this definition of complex linear is the same as yours? That's because if we have a complex linear map $B$ on $\mathbb C^n$, then its equivalent matrix representation on $\mathbb R^{2n}$ will turn each entry $a+bi$ of $B$ into a block $\begin{pmatrix} a & -b \\ b & a \end{pmatrix}$, as the matrix needs to turn $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$ (which corresponds to the complex entry $1+0i=1$) into $a+bi\sim\begin{pmatrix} a \\ b \end{pmatrix}$, and $\begin{pmatrix} 0 \\ 1\end{pmatrix}$ (which sorresponds to the complex entry $0+1i=i$ ) into $(a+bi)i=-b+ai\sim\begin{pmatrix} -b \\ a \end{pmatrix}$.
These ideas are used quite often in complex and symplectic geometries.
