Nice — this is a very natural and useful way to see $\mathbb C$ as a real algebra of $2\times2$ matrices. I'll walk through the key facts, show why the map is an isomorphism, give formulas (inverse, determinant, eigenvalues, exponential), and point out the geometric picture.
The map and why it is an isomorphism
Define
$$
\varphi:\mathbb C\to M_2(\mathbb R),\qquad \varphi(a+bi)=M(a,b)=\begin{bmatrix} a & b\\ -b & a\end{bmatrix}.
$$
-
Additive and multiplicative homomorphism.
For $z=a+bi,\; w=c+di$,
$$
\varphi(z+w)=M(a+c,\;b+d)=M(a,b)+M(c,d),
$$
and a direct matrix multiplication gives
$$
M(a,b)M(c,d)=M(ac-bd,\;ad+bc)=\varphi((a+bi)(c+di)).
$$
So $\varphi$ preserves addition and multiplication.
-
Bijective.
The map is clearly linear over $\mathbb R$ and
$\varphi(a+bi)=0$ iff $a=b=0$, so it is injective; any matrix of the form above comes from exactly one complex number, so $\varphi$ is bijective onto the subset
$\{M(a,b):a,b\in\mathbb R\}$. Hence $\varphi$ is an isomorphism of real algebras.
Thus this subset of $2\times2$ real matrices is a copy (a real-algebra embedding) of $\mathbb C$.
Algebraic identities (useful formulas)
Write $I=\begin{bmatrix}1&0\\0&1\end{bmatrix}$ and $J=\begin{bmatrix}0&1\\-1&0\end{bmatrix}$. Then
$$
M(a,b)=aI+bJ,\qquad J^2=-I.
$$
Because $J^2=-I$, the algebra generated by $I$ and $J$ is isomorphic to $\mathbb C\simeq\{a+ b i\}$ via $i\mapsto J$.
Determinant and trace.
$$
\det M(a,b)=a^2+b^2=|a+bi|^2,\qquad \operatorname{tr} M(a,b)=2a=2\operatorname{Re}(a+bi).
$$
Inverse. If $a+bi\neq0$,
$$
M(a,b)^{-1}=\frac{1}{a^2+b^2}\,M(a,-b).
$$
This matches the complex formula $(a+bi)^{-1}=\dfrac{a-bi}{a^2+b^2}$.
Transpose and conjugation.
$$
M(a,b)^T=M(a,-b),
$$
so matrix transpose corresponds to complex conjugation $a+bi\mapsto a-bi$.
Orthogonality / rotations. If $a^2+b^2=1$ then $M(a,b)$ is orthogonal (indeed $M M^T=I$) and $\det=1$; these matrices represent rotations in the plane (they form a copy of the unit circle $S^1\subset\mathbb C$).
Eigenvalues and diagonalization
The characteristic polynomial is
$$
\lambda^2-2a\lambda+(a^2+b^2)=0,
$$
so the (complex) eigenvalues are
$$
\lambda=a\pm ib,
$$
i.e. the eigenvalues are exactly the complex number and its conjugate. Over $\mathbb R$ the matrix is not diagonalizable unless $b=0$, but over $\mathbb C$ it diagonalizes in the usual way.
Exponentials and analytic functions
Because $\varphi$ is an $\mathbb R$-algebra isomorphism, any real-coefficient power series evaluated at a complex number corresponds to evaluating the same power series at the matrix. In particular, for $z=a+ib$,
$$
\exp\big(M(a,b)\big)=M\big(\operatorname{Re}(e^z),\;\operatorname{Im}(e^z)\big)
= M(e^a\cos b,\; e^a\sin b).
$$
So matrix exponentiation corresponds exactly to the complex exponential.
Geometric interpretation
Think of $\mathbb R^2$ as the real plane. Multiplication by a complex number $z=re^{i\theta}$ is the linear map “scale by $r$” and “rotate by $\theta$”. Under $\varphi$, the matrix $M(a,b)$ implements that linear map on $\mathbb R^2$. Thus this embedding identifies $\mathbb C$ with the algebra of linear maps of $\mathbb R^2$ that are $\mathbb R$-linear and commute with the complex-structure operator $J$ (i.e. are of the form $aI+bJ$).
A concrete example
$(1+2i)(3+4i)=-5+10i$. Matrix side:
$$
M(1,2)M(3,4)=M(-5,10)=\begin{bmatrix}-5 & 10\\ -10 & -5\end{bmatrix}.
$$
Also $\det M(1,2)=1^2+2^2=5$, and
$$
M(1,2)^{-1}=\frac{1}{5}M(1,-2)=\frac{1}{5}\begin{bmatrix}1 & -2\\ 2 & 1\end{bmatrix}.
$$
Generalizations & remarks
- This exhibits $\mathbb C$ as a 2-dimensional real algebra inside $M_2(\mathbb R)$.
- The same idea generalizes: quaternions $\mathbb H$ can be represented inside $M_4(\mathbb R)$.
- Any real $2$-dimensional algebra containing an element whose square is $-1$ is (up to isomorphism) the complex numbers; here $J$ plays that role.
If you like, I can: (1) show a short proof that every matrix commuting with $J$ has the form $aI+bJ$; (2) draw the action of $M(a,b)$ on some sample vectors to illustrate the rotation/scale; or (3) show how this matrix representation relates to the usual complex-to-real 2×2 matrix $\begin{bmatrix}a&-b\\ b&a\end{bmatrix}$ (they are the same up to a simple change of coordinates). Which would you prefer?
