Single transformation matrix of $A \circ B$ and $B \circ A$ with certain conditions

Question

Let $A$ is 2x1 translation matrix and $B$ is 2x2 matrix of reflection or rotation matrix (reflection, rotation, etc.).

Suppose I want to find the mapping of a $y=mx+c$ line and the mapping is done by two steps. I know about how to differentiate between $A \circ B$ and $B \circ A$. Thus $A \circ B=A \cdot B$ and $B \circ A=B \cdot A$.

Usually if the both matrices have 2x2 matrix, I could find the single transformation matrix by multiply both matrices, inverse the multiplication, and substitute the values to original line. But not with 2x1 and 2x2 matrices.

1. I'd like to know how do you make single transformation matrix of $A \circ B$ and $B \circ A$.

2. I'd like also to know how do you make single transformation of $A \circ B$ and $B \circ A$ with $A$ is 2x2 translation matrix and $B$ is matrix of reflection about $y=mx$ with $M_{y=mx}=\frac{1}{1+m^{2}} \begin{pmatrix} 1-m^{2} & 2m \\ 2m & m^{2}-1 \end{pmatrix}$ with $m$ as gradient of $y=mx$ line.

Detail help is appreciated! Thanks.

Answer 1

I think the difficulty here is that translation by a fixed vector isn't a linear transformation. One way to see this is that a sum of translates isn't the translate of the sum.

Thus not going to be able to cook up a matrix to get the job done.

Answer 2

Translation is not a linear map, but can still be represented by a matrix by adding one new coordinate and setting it equal to$~1$. In your example you are dealing with a plane, so you have two coordinates $x,y$ say; you add a third coordinate $z$ and force it to be equal to$~1$. In other words you identify your (affine) plane with $P=\{\,(x,y,z)\in\Bbb R^3\mid z=1\,\}$. You are only interested in linear transformations of $\Bbb R^3$ that globally stabilise the plane $P$, which means that the last row of their matrices must be $(0~~0~~1)$. Now you can see that you can realise your translations respectively (rotation, reflection) maps that fix a chosen origin $O=(0,0,1)\in P$ by matrices of the form $$ A=\pmatrix{0&0&a_1\\0&0&a_2\\0&0&1} \text{ respectively } B=\pmatrix{b_{1,1}&b_{1,2}&0\\b_{2,1}&b_{2,2}&0\\0&0&1}, $$ whose action restricted to $P$ has the desired effect. Now composition of maps is given simply by matrix multiplication. Note that $A\cdot B$ and $B\cdot A$ differ in general, as expected.