Real matrices (2 x 2)

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The 2 x 2 real matrices are the linear mappings of the Cartesian coordinate system into itself by the rule

$(x,y) \mapsto (x,y)\begin{pmatrix}a & c \\ b & d\end{pmatrix} = (ax + by, cx + dy).$

The set of all such real matrices is denoted by M(2,R). Two matrices p and q have a sum p + q given by matrix addition. The product matrix p q is formed from the dot product of the rows and columns of its factors through matrix multiplication. For

$q =\begin{pmatrix}a & c \\ b & d \end{pmatrix}\quad$ let $\quad q^{*} =\begin{pmatrix}d & -c \\ -b & a \end{pmatrix}$ .

Then q q * = (ad − bc) I, where I is the 2 x 2 identity matrix. The real number ad − bc is called the determinant of q. Evidently when ad − bc ≠ 0, q is an invertible matrix and then

$q^{-1} = q^*\ /(ad - bc)$ .

The collection of all such invertible matrices constitutes the general linear group GL(2,R). In terms of abstract algebra, the set of 2 by 2 real matrices and their associated addition and multiplication operators forms a ring, and GL(2,R) is its group of units. M(2,R) is also a four-dimensional vector space, so it is considered an associative algebra. It is ring-isomorphic to the coquaternions, but has a different profile.

[edit] Profile

Within M(2,R), the multiples by real numbers of the identity matrix I may be considered a real line. Since every matrix lies in a commutative subring of M(2,R) that includes this real line, the whole ring can be profiled by such subrings. Toward this end one needs matrices m such that m² ∈ { −I, 0, I } to form planes P_m = {x I + ym : x, y ∈ R}, which are in fact commutative subrings.

The square of the generic matrix is

$\begin{pmatrix}aa+bc & ac+cd \\ab+bd & bc+dd \end{pmatrix}$

which is diagonal when a + d = 0. Thus we assume d = −a when looking for m to form commutative subrings. When mm = −I, then bc = −1 − aa, an equation describing an hyperbolic paraboloid in the space of parameters (a, b, c). In this case P_m is isomorphic to the field of (ordinary) complex numbers. When mm = +I, bc = +1 − aa, giving a similar surface, but now P_m is isomorphic to the ring of split-complex numbers. The case mm = 0 arises when only one of b or c is non-zero, and the commutative subring P_m is then a copy of the dual number plane.

[edit] Equi-areal mapping

First transform one differential vector into another:

$\begin{align} (du, dv) & {} = (dx, dy) \begin{pmatrix}p & r\\ q & s \end{pmatrix} \\ & {} = (p\, dx + q\, dy ,\ r\, dx + s\, dy). \end{align}$

Areas are measured with density $dx \wedge dy$ , a differential 2-form which involves the use of exterior algebra. The transformed density is

$\begin{align} du \wedge dv & {} = 0 + ps\ dx \wedge dy + qr\ dy \wedge dx + 0 \\ & {} = (ps - qr)\ dx \wedge dy = (\det g)\ dx \wedge dy. \end{align}$

Thus the equi-areal mappings are identified with SL(2,R) = {g ∈ M(2,R) : det(g) = 1}, the special linear group. Given the profile above, every such g lies in a commutative subring P_m representing a type of complex plane according to the square of m. Since g g * = I, one of the following three alternatives occurs: