Solution Sets of Linear Systems and Kernel Theorem

The solution set $S$ of $Ax=b$ is a subspace $⟺b=0$

• With $S$ being a subspace, then its kernel is $Ker(T_A)$

Proof

1. Suppose that the solution set of $Ax=b$ is a subspace
2. We know that it has solutions in the form $x_p+Ker(T_A)$
3. Equivalently, $x=x_p+x_n$, if the solution is a subspace, then $\overrightarrow{0}$ is a solution. Thus, we get $A\overrightarrow{0}=b⟹\overrightarrow{0}=b$
4. Suppose that $b=\overrightarrow{0}$, then we get the solution set in the form $x_p+Ker(T_A)$
5. Suppose $b=0$, then we get the solution set $x_p+Ket(T_A)$
6. $=\overrightarrow{0}+Ker(T_A)$ by picking $x_p=0$ to be the solution
7. $=Ket(T_A)$
8. Therefore, the solution set is a subspace