Given a vector space $V$, we will observe the linear map $T:V\to V$ such that $$T_{A,v}(x) := Ax+v$$ for $A:V\to V$ a reversible linear map, and $v \in V$. We will call this set $Aff(V)$. Prove that $Aff(V)$ is a group under function composition.
So what I'm having problem with is I am not sure that my answer is very correct. I am starting with:
- The identity element is supposed to be $$1_{Aff} := T_{I, 0}(x) := x$$ if we will check for $f \in Aff(v)$: $$1_{Aff} \circ f = Ax + v = f \circ 1_{Aff}$$
- The inverse element: for each $f\in Aff(V)$ we will take as $f^{-1} := T_{A^{-1}, -A^{-1}(v)}(x)$ so that $$f^{-1} \circ f(x) = f^{-1}(Ax+v) = A^{-1}(Ax + v) - A^{-1}v =\ldots= x,$$ $$f \circ f^{-1}(x) = f(A^{-1}x - A^{-1}v) = A(A^{-1}x - A^{-1}v) + v =\ldots= x.$$ Hence $$ f^{-1}\circ f=f\circ f^{-1}=1_{Aff}$$
- Associativity: for each $f,h,g \in Aff(v)$ $$(h \circ g)(x) = h(A_gx + v_g) = A_h(A_gx + v_g) + v_h$$ $$(f \circ h)(x) = f(A_hx + v_h) = A_f(A_hx + v_h) + v_f$$ $$f \circ (h \circ g) = A_f(A_h(A_gx + v_g) + v_h) + v_f = (f \circ h)(A_gx + v_g) = (f \circ h) \circ g$$
Does this look correct? I'm not so sure about the last part. I would like any feedback you have :)