Fdtxjr

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

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
$$
pmatrixe_1 cr e_2cdot (a+bi) ;=; pmatrixa & b cr -b & apmatrixe_1cr e_2
$$
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 $$rbeginpmatrixcos theta & -sin theta \ sin theta & cos theta endpmatrix$$

Since $r e^itheta=rcos theta + ir sin theta = a +ib$, we have $$a +ib = beginpmatrixa & -b \ b & a endpmatrix$$

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 $phicolon MtomathbbC$ by
$$
beginpmatrix alpha & beta \ -beta & alphaendpmatrixmapsto alpha+ibeta.
$$
Note that this function has inverse $phi^-1$ defined by $alpha+ibetamapstobeginpmatrix alpha & beta \ -beta & alphaendpmatrix$. 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
$$
A=beginpmatrix alpha & beta \ -beta & alphaendpmatrix,qquad
B=beginpmatrix gamma & delta \ -delta & gammaendpmatrix.
$$
Then
$$
phi(A+B)=phibeginpmatrix alpha+gamma & beta+delta \ -beta-delta & alpha+deltaendpmatrix=(alpha+gamma)+i(beta+delta)=(alpha+ibeta)+(gamma+idelta)=phi(A)+phi(B).
$$
Also,
$$
phi(AB)=phibeginpmatrix alphagamma-betadelta & alphadelta+betagamma \ -betagamma-alpha-delta & -betadelta+alphagammaendpmatrix=(alphagamma-betadelta)+i(alphadelta+betagamma)=(alpha+ibeta)(gamma+idelta)=phi(A)phi(B).
$$
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 $mathbbC$ 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 $,[1,,it i,]^T:$

$$rm (a+b,it i,) left[ beginarrayc 1 \ it i endarray right]
,=, left[beginarrayrrm a+b,it i\rm -b+a,it i endarray right]
,=, left[beginarrayrrrm a &rm b\rm -b &rm a endarray right]
left[beginarrayc 1 \ it i endarray right]$$

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 $I=beginbmatrix1&0\0&1endbmatrix$ and $J=beginbmatrix0&-1\1&0endbmatrix$ 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, $aI+bJ=beginbmatrixa&-b\b&aendbmatrix$ 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 $mathcalQ$ is the group consisting of elements $pm1, pm hati, pm hatj, pm hatk$ equipped with multiplication that satisfies the rules according to the diagram

$$hati rightarrow hatj rightarrow hatk.$$

Now what is more interesting is that you can let $mathcalQ$ become a four dimensional real vector space with basis $1,hati,hatj,hatk$ equipped with an $BbbR$ - bilinear multiplication map that satisfies the rules above. You can also define the norm of a quaternion $a + bhati + chatj + dhatk$ as

$$||a + bhati + chatj + dhatk|| = a^2 + b^2 + c^2 + d^2.$$

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

$$textrmSU(2) = lefthspace3mm a,b,c,d in BbbR, hspace3mm a^2 + b^2 + c^2 + d^2 = 1 right.$$

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 $a + bhati + chatj + dhatk$ to the matrix $$left(beginarraycc a + bi & -c + di \ c + di & a-bi endarrayright).$$

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 $textrmSU(2)$ as boring old matrices that you say "meh" to you now have a geometric understanding of what $textrmSU(2)$. 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 $BbbR^4$ has now basis elements consisting of $1,hati,hatj,hatk$, you have given $BbbR^4$ a multiplication structure and it becomes not just an $BbbR$ - module but a module over itself.