Exercise 3.5.12

A neat choice is $\left[\begin{array}{cc}1\hfill & 1\hfill \\ 0\hfill & 2\hfill \\ 1\hfill & 0\hfill \end{array}\right]\left[\begin{array}{ccc}1\hfill & 0\hfill & 1\hfill \\ 1\hfill & 2\hfill & 0\hfill \end{array}\right]=\left[\begin{array}{ccc}2\hfill & 2\hfill & 1\hfill \\ 2\hfill & 4\hfill & 0\hfill \\ 1\hfill & 0\hfill & 1\hfill \end{array}\right];r+(n-r)=n=3$ does not match $2+2=4$. Only $v=0$ is in both $N(A)$ and $C\left(A^T\right)$.