Exercise 4.1.3

(a) One way is to use these two columns directly: $A=\left[\begin{array}{ccc}\hfill 1& \hfill 2& \hfill -3\\ \hfill 2& \hfill -3& \hfill 1\\ \hfill -3& \hfill 5& \hfill -2\end{array}\right]$

(b) Impossible because $N(A)$ and $C\left(A^T\right)$ are orthogonal subspaces : $\left[\begin{array}{c}\hfill 2\\ \hfill -3\\ \hfill 5\end{array}\right]$ is not orthogonal to $\left[\begin{array}{c}1\hfill \\ 1\hfill \\ 1\hfill \end{array}\right]$

(c) $\left[\begin{array}{c}1\hfill \\ 1\hfill \\ 1\hfill \end{array}\right]$ and $\left[\begin{array}{c}1\hfill \\ 0\hfill \\ 0\hfill \end{array}\right]$ in $C(A)$ and $N\left(A^T\right)$ is impossible: not perpendicular

(d) Rows orthogonal to columns makes $A$ times $A=$ zero matrix $\rho$. An example is $A=$ $\left[\begin{array}{cc}1\hfill & -1\hfill \\ 1\hfill & -1\hfill \end{array}\right]$

(e) $(1,1,1)$ in the nullspace (columns add to the zero vector) and also $(1,1,1)$ is in the row space: no such matrix.