#math/linear-algebra For any linear transformation $A : V \to W$ we can associate two subspaces, namely, its kernel, or null space $ \operatorname{Ker} A = \operatorname{Null} A := \{\mathbf{v} \in V : A\mathbf{v} = \mathbf{0}\} \subset V, $ and its range $ \operatorname{range} A = \{\mathbf{w} \in W : \mathbf{w} = A\mathbf{v} \text{ for some } \mathbf{v} \in V\} \subset W. $ However if we also consider $\text{range }A^{\top}$ and $\text{Ker }A^{\top}$, all four make up the `fundamental subspaces` of matrix $A.$ For a linear transformation (matrix) $A$, its rank, rank $A$ is given by the dimension of the range of $A$. $ \text{rank }A:=\text{dim range } A. $ The pivot columns of the original matrix $A$ give us a basis of $\text{range }A$. Think $\text{dim }Tv_{1},\ldots,Tv_{n}$. The pivot rows of echelon form $A_{e}$ give us a basis in the row space give us a basis in the row space. The non-zero rows of $A_{e}$ transposed into columns thus give us $\text{range }A^{\top}$. To find a basis in the null space $\text{Ker }A$, we must solve the homogenous equation $Ax=0$. Row reduce $A$ and solve $Ax=0$. Then write $x$ as a linear combination of the free variables. You must do the same for $A^{\top}$, no shortcut.