Why matrices commuting with $\small\begin{bmatrix} 0&1\\-1&0\end{bmatrix}$ represent complex numbers? I am trying to understand which $2$ by $2$ real matrices represent complex numbers in following way.
Let $J=\begin{bmatrix} 0&1\\-1&0\end{bmatrix}$ and $A=\begin{bmatrix} a&b\\c&d\end{bmatrix}$ be any real matrix.
If $A$ represents a complex matrix (by standard embedding of complex field into matrix ring) then $A$ should commute with the matrix $J$, which image of complex number $i$.
Q. I want to understand why the matrices commuting with $J$ are precisely the matrices representing complex numbers?
 A: $I=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$ and $J=\begin{bmatrix} 0&1\\-1&0\end{bmatrix}$ form the basis for modelling complex numbers as real-valued matrices. Not all $2\times 2$ matrices fit into this model, only those of the form
$C=\begin{bmatrix} a&b\\-b&a\end{bmatrix}$
A real number $a$ is modelled as $A=aI$, and an imaginary number $b$ is modelled as $B=bJ$. The complex number $c=a+ib$ is modelled as $C=aI+bJ$, which is the matrix above.
A: Let's consider $\varphi:\mathbb C\rightarrow M_2(\mathbb C)$, $\varphi(a+ib)=\pmatrix{a & b \\ -b & a}$ the standard embedding of $\mathbb C$ into the matrix ring. 
Consider $Z(J)=\{A\in M_2(\mathbb C)\ | \ JA=AJ\}$ the set of the matrix commuting with $J$.
Your question is equivalent to show that $Z(J) = \varphi(\mathbb C)$.
And this is true because:
\begin{gather}
A=\pmatrix{a &b \\ c & d}\in Z(J) \Longleftrightarrow \pmatrix{a &b \\ c & d}\pmatrix{0 & 1 \\ -1 & 0} = \pmatrix{0 &1 \\ -1 & 0}\pmatrix{a &b \\ c & d} \Longleftrightarrow\\
\begin{cases}
-b = c\\
a=d
\end{cases}\Longleftrightarrow A=\pmatrix{a &b \\ -b & a}\in \varphi(\mathbb C)
\end{gather}
A: $L : X \longmapsto Y$ $\mathbb{R}$-linear application between complex vectorial space is $\mathbb{C}$-linear iff commutes with $i$.
Infact $L$ it's $\mathbb{C}$-linear $\iff$ $iL = Li \iff J_{Y}L = LJ_{X}$
A: Let
$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \tag 1$
with
$AJ = JA; \tag 2$
writing
$AJ = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -b & a \\ -d & c \end{bmatrix} \tag 3$
and
$JA = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ -a & -b \end{bmatrix}, \tag 4$
we see in light of (2) that 
$c = -b, \tag 5$
$d = a; \tag 6$
thus $A$ takes the form
$A = \begin{bmatrix} a & b \\ -b & a\end{bmatrix}; \tag 7$
we note that may write
$A = aI + bJ, \tag 8$
which evidently commutes with $J$; thus every matrix satisfying (2) is of the form (8).  And under the correspondence
$i \longleftrightarrow J, \tag 9$
$A$ corresponds to the complex number $a + bi$.
