Prove that if $S$ is subspace then there must exist $Ax=0, x \in S$ I would like to show the first direction of the following statement.
$S$ is a subspace if and only if there exists a matrix $A$ s.t. $S = \{x \in \mathbb{R}^n: Ax =0 \}$.
Definition: $S \subset \mathbb{R}^n$ where $S\neq \emptyset$  is called a subspace if $x \in S$ implies $\lambda x \in S, \forall \lambda \in \mathbb{R}$ and $x,y \in S$ implies $x+y \in S$. In addition,
$``\Leftarrow"$If there exists a matrix $A$ s.t. $S = \{x \in \mathbb{R}^n: Ax =0 \}$, then we can select $x, y \in \mathbb{R}^n$ as the solutions for $Ax =0$. It is clear to show that:
$A(x+y) = Ax + Ay = 0$ and $A(\lambda x) = \lambda (Ax) = 0$. Hence, $S$ is a subspace.
How can I show the other direction? If I assume that $S$ is a subspace, how can I show that there must exists a matrix $A$?

Hint.
Suppose $S$ is a subspace of $\mathbb{R}^n$. Then the exists a basis $\beta$ of $S$. Extend $\beta$ to be a basis $\alpha$ of $\mathbb{R}^n$.
Define $A$ to be a linear transformation on $\mathbb{R}^n$ such that $Av=0$ for every $v\in\beta$ and $Av=v$ for every $v\in\alpha\setminus\beta$. Extend $A$ linearly.
