Kernel is orthogonal complement

Suppose $x\in Ker(A)$. Then $0=(Ax,y)=(x,A^{T}y)$ means that $Ker(A)^{\bot}=Ran(A^{T})$. If we represent $A$ by its rows $w_{i}$, hence $A_{T}$ by its columns: $A^{T}=(w_{1}|w_{2}|...|w_{m})$, each row being an n-dimensional vector, and each y as $y=\sum_{i=1}^{i=n}{c_{i}e_{i}}$, then $A^{T}y=\sum_{i=1}^{i=n}{c_{i}A \cdot e_{i}=\sum_{i=1}^{i=n}{c_{i}w_{i}}}$, which means that $Ran(A^{T})=LinSpace({w_{i}})$.