How to Prove $\{Ax + b \mid Fx = g \}$ is Affine?

Question

In order to prove $\{Ax + b \mid Fx = g \}$ is affine, is the following sufficient for the proof?

$\{A\left(\theta x_1+\left(1-\theta\right)x_2\right) + b \mid F\left(\theta x_1+\left(1-\theta\right)x_2\right) \stackrel{?}{=} g \}$

So,

$F\left(\theta x_1+\left(1-\theta\right)x_2\right)=\theta Fx_1+Fx_2-\theta Fx_2 = \theta g+g -\theta g=g$

Do we need to show the same thing for $A\left(\theta x_1+\left(1-\theta\right)x_2\right) + b$? In general, is applying the affine definition (which is sometimes hard) the only way to prove a set is affine?

Answer 1

Let's call $S=\{Ax+b\mid Fx=g\}$

Your title is misleading, what you are seemingly trying to do, is to prove that $S$ is convex.

In this case let's take two points $(X,Y)\in S^2$ and $t\in[0,1]$

$\begin{array}{ll} Z &= tX+(1-t)Y\\ &= t(Ax+b)+(1-t)(Ay+b)\\ &= A(tx+(1-t)y)+b(t+(1-t))\\ &= Az+b\end{array}\qquad$ with $z=tx+(1-t)y$

Then $Fz=tFx+(1-t)Fy=tg+(1-t)g=g\qquad$ so $\quad Z\in S$.

Thus $S$ is convex.

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Anyway, the proof for an affine space derived from $S$ should go like this:

Let's define $\vec{S}=\{Au\mid u\in\ker F\}$

Since $A,F$ are linear applications, then it is immediate that $\vec S$ is a vector space.

• $\vec 0\leftrightarrow u=0$
• $\vec U+\vec V\leftrightarrow Au+Av=A(u+v)$ and $F(u+v)=Fu+Fv=0+0=0$
• $\alpha\vec U\leftrightarrow \alpha Au=A(\alpha u)$ and $F(\alpha u)=\alpha Fu=\alpha.0=0$
• and so on...

Now we can verify that $S\times\vec S$ is affine

• $X+\vec 0=Ax+b+A0=Ax+b=X$
• $(X+\vec U)+\vec V=Ax+B+Au+Av=Ax+b+A(u+v)=X+(\vec U+\vec V)$
• $X-Y=Ax+b-Ay-b=A(x-y)=\vec U\in\vec S$ since $F(x-y)=Fx-Fy=g-g=0$