Null Space of Reduced Echelon Form

Theorem
Let $\mathbf A$ be a matrix such that:


 * $\mathbf A \mathbf x = \mathbf 0$

is a homogeneous system of linear equations.

The null space of $\mathbf A$ is the same as that of the null space of the reduced row echelon form of $\mathbf A$:


 * $\operatorname{N} \left({\mathbf A}\right) = \operatorname{N}\left({\operatorname{rref}\left({\mathbf A}\right)}\right)$

Proof
By the definition of null space,


 * $\mathbf x \in \operatorname{N}\left({\mathbf A}\right) \iff \mathbf A \mathbf x = \mathbf 0$

From the corollary to Row Equivalent Matrix Has Same Solutions,


 * $\mathbf A \mathbf x = \mathbf 0 \iff \operatorname{rref}\left({\mathbf A}\right)\mathbf x = \mathbf 0$

Hence the result, by the definition of set equality.