Prove that an affine transformation maps an affine subspace on an affine subspace

Question

I am trying to prove that if $f:A_1 \rightarrow A_2$ is an affine transformation and $a+F$ is an affine subspace of $A_1$,then $f(a+F)$ is an affine subspace of $A_2$ and $f(a+F)=f(a)+f'(F)$, where $f'$ is the linear part of $f$. I know I need to use the fact that $f(a+\vec{u})=f(a)+f'(\vec{u}) \forall a \in A_1, \forall \vec{u} \in F$ but I don't know really how to do it formally because I have always proved that $a+F$ is an affine subspace if and only if $F=\{\vec{ab}|a,b \in a+F\}$ is a vector subspace but this seems not operative in this case. Thanks.

Answer 1

To clarify a bit things

Let $X$ and $Y$ vectors spaces. Let $a_1\in X$ and $a_2\in Y$. You have that $A_1=\{a_1+v\mid v\in V\}$ for some sub vector space $V$ of $X$ and $A_2=\{a_2+w\mid w\in W\}$ for some vector space $W$ of $Y$. Then $f:A_1\to A_2$ is defined by $$f(a_1+v)=a_2+\bar f(v),$$ for all $a_1+v\in A_1$, where $\bar f:V\to W$ is a linear map. Let $a_1+F$ be a subspace of $A_1$, i.e. $$a_1+F=\{a_1+u\mid u\in F\},$$ where $F$ is a subspace of $V$.

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Hint to prove your statement

It remains to prove that $f(a_1+F)=\{a_2+\bar f(u)\mid u\in F\}$ is an affine subspace of $A_2$, which is nothing else than proving that $\bar f(F)$ is a vector space.

Answer 2

$f(a+F)=\{f(a+u):u\in F\}=\{f(a)+f'(u):u \in F\}=f(a)+\{f'(u):u \in F\}=f(a)+f'(F)$

$F$ is a vector subspace and $f'$ is linear. Then $f'(F)$ is a vector subspace. And $f(a)+f'(F)$ is an affine subspace of $A_2$.

So $f(a+F)$ is an affine subspace of $A_2$, equal to $f(a)+f'(F)$ .