Dtjytyk

Look at the arithmetic operations and their actions. With + and *, these matrices form a field. And we have the isomorphism
$$a + ib mapsto left[matrix{a&-bcr b &a}right].$$

As to the "where did it come from?", rather than verifying that it does work: this is a special case of a "rational representation" of a bigger collection of "numbers" as matrices with entries in a smaller collection. ("Fields" or "rings", properly, but it's not clear what our context here is.)

That is, the collection of complex numbers is a two-dimensional real vector space, and multiplication by $a+bi$ is a real-linear map of $mathbb C$ to itself, so, with respect to any $mathbb R$-basis of $mathbb C$, there'll be a corresponding matrix. For example, with $mathbb R$-basis $e_1=1,,e_2=i$,
$$
(a+bi)cdot e_1 = a+bi = ae_1+be_2
hskip40pt
(a+bi)cdot e_2 = (a+bi)i = -b+ai = -be_1+ae_2
$$
So
$$
pmatrix{e_1 cr e_2}cdot (a+bi) ;=; pmatrix{a & b cr -b & a}pmatrix{e_1cr e_2}
$$
Oop, I guessed wrong, and got the $b$ and $-b$ interchanged. Maybe using $e_2=-i$ instead will work... :)

But this is the way one finds such representations.

What I think of a complex number is a scaling and 2D rotation operation, where the absolute value $r$ is scaling factor, and the phase $theta$ is the rotation angle.

The same operation can be described by scalar multiplication of a rotation matrix as $$rbegin{pmatrix}cos theta & -sin theta \ sin theta & cos theta end{pmatrix}$$

Since $r e^{itheta}=rcos theta + ir sin theta = a +ib$, we have $$a +ib = begin{pmatrix}a & -b \ b & a end{pmatrix}$$

I had something written up on this lying around. The $-$ sign is off, but it's more or less the same, I hope it helps.

Let $M$ denote the set of such matrices. Define a function $phicolon Mtomathbb{C}$ by
$$
begin{pmatrix} alpha & beta \ -beta & alphaend{pmatrix}mapsto alpha+ibeta.
$$
Note that this function has inverse $phi^{-1}$ defined by $alpha+ibetamapstobegin{pmatrix} alpha & beta \ -beta & alphaend{pmatrix}$. This function is well defined, since $alpha+ibeta=gamma+idelta$ if and only if $alpha=gamma$ and $beta=delta$, and thus it is never the case that a complex number can be written in two distinct ways with different real part and different imaginary part. So $phi$ is invertible.

Now let
$$
A=begin{pmatrix} alpha & beta \ -beta & alphaend{pmatrix},qquad
B=begin{pmatrix} gamma & delta \ -delta & gammaend{pmatrix}.
$$
Then
$$
phi(A+B)=phibegin{pmatrix} alpha+gamma & beta+delta \ -beta-delta & alpha+deltaend{pmatrix}=(alpha+gamma)+i(beta+delta)=(alpha+ibeta)+(gamma+idelta)=phi(A)+phi(B).
$$
Also,
$$
phi(AB)=phibegin{pmatrix} alphagamma-betadelta & alphadelta+betagamma \ -betagamma-alpha-delta & -betadelta+alphagammaend{pmatrix}=(alphagamma-betadelta)+i(alphadelta+betagamma)=(alpha+ibeta)(gamma+idelta)=phi(A)phi(B).
$$
So $phi$ respects addition and multiplication. Lastly, $phi(I_2)=1$, so $phi$ also respects the multiplicative identity. Hence $phi$ is a field isomorphism, so $M$ and $mathbb{C}$ are isomorphic as fields.

The matrix rep of $rm:alpha = a+b,{it i}:$ is simply the matrix representation of the $:Bbb R$-linear map $rm:xto alpha, x:$ viewing $,Bbb Ccong Bbb R^2$ as vector space over $,Bbb R.,$ Computing the coefficients of $,alpha,$ wrt to the basis $,[1,,{it i},]^T:$

$$rm (a+b,{it i},) left[ begin{array}{c} 1 \ {it i} end{array} right]
,=, left[begin{array}{r}rm a+b,{it i}\rm -b+a,{it i} end{array} right]
,=, left[begin{array}{rr}rm a &rm b\rm -b &rm a end{array} right]
left[begin{array}{c} 1 \ {it i} end{array} right]$$

As above, any ring may be viewed as a ring of linear maps on its additive group (the so-called left-regular representation). Informally, simply view each element of the ring as a $1!times! 1$ matrix, with the usual matrix operations. This is a ring-theoretic analog of the Cayley representation of a group via permutations on its underlying set, by viewing each $,alpha,$ as a permutation $rm,xtoalpha,x.$

When, as above, the ring has the further structure of an $rm,n$-dimensional vector space over a field, then, wrt a basis of the vector space, the linear maps $rm:xto alpha, x:$ are representable as $rm,n!times!n,$ matrices; e.g. any algebraic field extension of degree $rm,n.,$ Above is the special case $rm n=2.$

The matrices $I=begin{bmatrix}1&0\0&1end{bmatrix}$ and $J=begin{bmatrix}0&-1\1&0end{bmatrix}$ commute (everything commutes with $I$), and $J^2=-I$. Everything else follows from the standard properties (associativity, commutativity, distributivity, etc.) that matrix operations have.

Thus, $aI+bJ=begin{bmatrix}a&-b\b&aend{bmatrix}$ behaves exactly like $a+bi$ under addition, multiplication, etc.

Since you put the tag quaternions, let me say a bit more about performing identifications like that:

Recall the quaternions $mathcal{Q}$ is the group consisting of elements ${pm1, pm hat{i}, pm hat{j}, pm hat{k}}$ equipped with multiplication that satisfies the rules according to the diagram

$$hat{i} rightarrow hat{j} rightarrow hat{k}.$$

Now what is more interesting is that you can let $mathcal{Q}$ become a four dimensional real vector space with basis ${1,hat{i},hat{j},hat{k}}$ equipped with an $Bbb{R}$ - bilinear multiplication map that satisfies the rules above. You can also define the norm of a quaternion $a + bhat{i} + chat{j} + dhat{k}$ as

$$||a + bhat{i} + chat{j} + dhat{k}|| = a^2 + b^2 + c^2 + d^2.$$

Now if you consider $mathcal{Q}^{times}$, the set of all unit quaternions you can identify $mathcal{Q}^{times}$ with $textrm{SU}(2)$ as a group and as a topological space. How do we do this identification? Well it's not very hard. Recall that

$$textrm{SU}(2) = left{ left(begin{array}{cc} a + bi & -c + di \ c + di & a-bi end{array}right) |hspace{3mm} a,b,c,d in Bbb{R}, hspace{3mm} a^2 + b^2 + c^2 + d^2 = 1 right}.$$

So you now make an ansatz (german for educated guess) that the identification we are going to make is via the map $f$ that sends a quaternion $a + bhat{i} + chat{j} + dhat{k}$ to the matrix $$left(begin{array}{cc} a + bi & -c + di \ c + di & a-bi end{array}right).$$

It is easy to see that $f$ is a well-defined group isomorphism by an algebra bash and it is also clear that $f$ is a homeomorphism. In summary, the point I wish to make is that these identifications give us a useful way to interpret things. For example, instead of interpreting $textrm{SU}(2)$ as boring old matrices that you say "meh" to you now have a geometric understanding of what $textrm{SU}(2)$. You can think about each matrix as being a point on the sphere $S^3$ in 4-space! How rad is that?

On the other hand when you say $Bbb{R}^4$ has now basis elements consisting of ${1,hat{i},hat{j},hat{k}}$, you have given $Bbb{R}^4$ a multiplication structure and it becomes not just an $Bbb{R}$ - module but a module over itself.