# 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?

• Did you notice that $J^2=-I$? May 9, 2020 at 18:11

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 $$a + bi$$.

$$I=\begin{bmatrix} 1&0\\0&1\end{bmatrix}$$ 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 $$\iff$$ $$iL = Li \iff J_{Y}L = LJ_{X}$$