I am tasked to show that the set of affine transformations of ,
Where S is an invertible linear transformation and b is a vector, forms a group under the composition of maps.
It was easy to show that
Now I need to show that there is an identity element, and that all elements are invertible such that
Im a little unsure what to do at this point….Can anyone help me on this? Thanks!
Best Answer
To find the identity map, note that you are looking for a map such that both
$$e(f(x)) = f(x) = Sx+b$$
$$f(e(x)) = f(x) = Sx+b$$
This is obviously true for the map where $e(x)=Ix+0=x$.
To show that all elements are invertible means you are looking for a map where both
$$g(f(x)) = g(Sx+b) = S_2 (Sx+b)+b_2 = x$$
$$f(g(x)) = f(S_2x+b_2) = S(S_2x+b_2)+b = x$$
Doing some arithmetic, I think you'll be able to figure out what $S_2$ and $b_2$ end up being. Then because we defined $g(x)=S_2(x)+b_2$, it is trivially already in the group.