Complex-linear matrices I'm reading Tapp's introduction to matrix groups.The book introduced complex-linear matrices and it gives this characterization of complex-linear matrix:
Let $B\in M_{2n}(\mathbb R)$, let $J$ be the matrix 
$J = \left (\begin{array}{}
0 & -I \\
I  & 0
\end{array}\right )$,
where $I$ is the identity matrix ,  B is complex linear if and only if $BJ=JB$.
I understand why if $B$ is a complex-linear the $BJ=JB$ (by
using a commutative diagram) but i don't understand how i can prove 
the other side (if $BJ=JB$ then $B$ is complex-linear).
Is it enough to say that if $B$ commutes with $i$ ($J$ is "the copy of $i$" in $M_{2n}(\mathbb R)$) then $B$ is linear complex ?
Do you have any hints to give me for the proof? Thank you all.
{Note}: Definition of complex linear matrix: B is a complex linear iff  B is in the image of a map  $\rho_n: M_n(\mathbb C) \to M_{2n}(\mathbb R)$ . This map associates to each matrix $(X + iY)$ a matrix  $\left (\begin{array}{}
X & -Y \\
Y & X
\end{array}\right)$ , $X,Y \in  M_n(\mathbb R)$
 A: Expanding on David Hill's comment: suppose that
$$B = 
\begin{bmatrix} 
X & Y \\
Z & W
\end{bmatrix}.$$
Notice that $BJ = JB$ if and only if 
$$ 
\begin{bmatrix} 
Y & -X \\
W & -Z
\end{bmatrix}=
\begin{bmatrix} 
-Z & -W \\
X & Y
\end{bmatrix},$$
which holds if and only if $X=W$ and $Z=-Y$, i.e., if and only if 
$$B = 
\begin{bmatrix} 
X & Y \\
-Y & X
\end{bmatrix}.$$
Also, note that the definition of $\rho_n$ you state above is different than that stated in Tapp's book: 

In this case, 
$$J_n := 
\begin{bmatrix}
0 & -1 \\
1 & 0 \\
& & \ddots \\
& & & \\
& & & & 0 & -1 \\
& & & & 1 & 0
\end{bmatrix} \in M_{2n} (\mathbb{R}).
$$
A: More in line with Tapp's book, he notices (I use the same notations as Tapp) that it suffices to show that $F(iX)=iF(X)$, with $F=f_n^{-1}\circ R_{B}\circ f_{n}$. Also, $J_{2n}^{2}=-I$ so $BJ_{2n}=J_{2n}B$ implies $B=-J_{2n}BJ_{2n}$. Thus $F=f_n^{-1}\circ R_{-J_{2n}BJ_{2n}}\circ f_{n}=-f_n^{-1}\circ R_{J_{2n}} \circ R_B \circ R_{J_{2n}} \circ f_n$. Now, the point of $J_{2n}$ is that $ R_{J_{2n}} \circ f_n=f_n \circ R_{iI} $ so also, $f_n^{-1}\circ R_{J_{2n}} = R_{iI}\circ f_n^{-1} $. So, we get $F(iX)=-R_{iI}\circ f_n^{-1} \circ R_B\circ f_n \circ R_{iI} (iX)=-R_{iI}\circ f_n^{-1} \circ R_B\circ f_n (-X)=-R_{iI}\circ F(-X)$
Now, $F$ is $\mathbb{R}$-linear (as Tapp points out), so $F(iX)=R_{iI}(F(X))=iF(X)$
