This is not a duplicate of How to prove: Every affine set can be expressed as the solution set of a system of linear equations.
I have already read the provided answer, but I don't understand the logic of the answer. Here are the definitions and theorems I have proved.
Definition. A set $C \subseteq \mathbb{R}^n$ is affine if $$x_1, x_2 \in C \text{ and }\theta \in \mathbb{R} \implies \theta x_1+(1-\theta)x_2\in C\text{.}$$ Theorem. If $C$ is affine and $x_1, \dots, x_k \in C$, $\sum_{i=1}^{k}\theta_i = 1$, then $\sum_{i=1}^{k}\theta_i x_i \in C$.
Theorem. Let $C$ be affine and $x_0 \in C$. Then $$V = C - x_0 = \{x-x_0:x \in C\}$$ is a subspace of $\mathbb{R}^n$.
Theorem. The set $C = \{x : Ax = b\}$ is affine, with $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$.
Theorem. With $C = \{x : Ax = b\}$, the set $V$ (defined above) is the nullspace of $A$.
What I want to prove: Every affine set can be expressed as the solution set of a system of linear equations.
What I think this means: If $C \subseteq \mathbb{R}^n$ is affine, then $C = \{x: Ax = b\}$ for some $A \in \mathbb{R}^{m \times n}$ and $b \in \mathbb{R}^m$. This means that we need to show that $C \subseteq \{x: Ax = b\}$ and $\{x: Ax = b\} \subseteq C$.
How does this imply the answer in the link above?
Fix an element $x_0 \in C$.
Claim: The set $K = \{x-x_0: x \in C\}$ is a linear subspace of $\Bbb R^n$.
Claim: There exists a linear map $T$ whose kernel is precisely $K$.
Claim: $C$ is the solution to the linear system $Tx = Tx_0$ on $x$.
Please note that I'm very rusty on linear transformations and the associated terminology.
Here is an outline.
Assume $C\ne\emptyset$ (otherwise it is trivial $A=0$, $b=1$). We are to show that there exist $A$, $b$ such that $$ Ax=b \iff x\in C. $$ Step 1: Take any $x_0\in C$ (exists as $C\ne\emptyset$) and consider $$ C_0=\{z\colon z=x-x_0,\ x\in C \}. $$ Claim 1 says that it is an easy exercise that $C$ affine iff $C_0$ linear (subspace). To work with a subspace is easier as now instead of searching for $A$, $b$ that $Ax=b$ we can search for simply $A$ such that $$ A(x-x_0)=0\iff x-x_0\in C_0. $$ If we manage to find such $A$ then we are done (just choose $b=Ax_0$).
Step 2: How to find such $A$? Every subspace $C_0$ in ${\mathbb R}^n$ is complementable, so there exists a subspace $C_0^\bot$ (orthogonal complement to $C_0$) such that $C_0\,\bot\, C_0^\bot$ and $C_0+C_0^\bot={\mathbb R}^n$. It means that $$ z\in C_0\iff z\bot a,\ \forall a\in C_0^\bot. $$ In words, vectors in $C_0$ are those and exactly those that are orthogonal to all vectors in $C_0^\bot$.
Now let $\{a_1,\ldots,a_k\}$ be a basis of $C_0^\bot$. Then $$ z\bot a,\ \forall a\in C_0^\bot \iff z\bot a_i,\ i=1,\ldots,k,\iff a_i^Tz=0,\ \forall i=1,\ldots,k. $$ If we define a matrix $A$ with rows being $a_i^T$ then the last orthogonality condition is precisely $$ Az=0. $$ Step 3: Now recall that $z\in C_0$ iff $z=x-x_0$, $x\in C$, that is we have found $A$ such that $$ A(x-x_0)=0\iff x-x_0\in C_0 \iff x\in x_0+C_0=C. $$ To finish, set $Ax_0=b$ and conclude that $$ Ax=b\iff x\in C. $$