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?


4 Answers 4



$A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \tag 1$


$AJ = JA; \tag 2$


$AJ = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -b & a \\ -d & c \end{bmatrix} \tag 3$


$JA = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} c & d \\ -a & -b \end{bmatrix}, \tag 4$

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 $a + bi$.


$I=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$ 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.


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}


$L : X \longmapsto Y$ $\mathbb{R}$-linear application between complex vectorial space is $\mathbb{C}$-linear iff commutes with $i$.

Infact $L$ it's $\mathbb{C}$-linear $\iff$ $iL = Li \iff J_{Y}L = LJ_{X}$


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