Let L:R^{n }-> R^{m} be linear transformation with matrix A. Show that ker(L) is orthogonal complement of row space of A.
Suppose $\displaystyle x\in Ker(A)$. Then $\displaystyle 0=(Ax,y)=(x,A^{T}y)$ means that $\displaystyle Ker(A)^{\bot}=Ran(A^{T})$. If we represent $\displaystyle A$ by its rows $\displaystyle w_{i}$, hence $\displaystyle A_{T}$ by its columns: $\displaystyle A^{T}=(w_{1}|w_{2}|...|w_{m})$, each row being an n-dimensional vector, and each y as $\displaystyle y=\sum_{i=1}^{i=n}{c_{i}e_{i}}$, then $\displaystyle A^{T}y=\sum_{i=1}^{i=n}{c_{i}A \cdot e_{i}=\sum_{i=1}^{i=n}{c_{i}w_{i}}}$, which means that $\displaystyle Ran(A^{T})=LinSpace({w_{i}})$.