Exercise 3.3.1

$\mathit{AK}=\left[\begin{array}{cccc}\hfill \mathit{Ab}\hfill & \hfill A^{2}b\hfill & \hfill \dots \hfill & \hfill A^{n}b\hfill \end{array}\right]$, and $\mathit{KB}=\left[\begin{array}{cccc}\hfill b\hfill & \hfill \mathit{Ab}\hfill & \hfill \dots \hfill & \hfill A^{n-1}b\hfill \end{array}\right]B=b{B}_{1,}+\mathit{Ab}{B}_{2,}+\cdots +A^{n-1}{B}_{n,}=\left[\begin{array}{ccccc}\hfill \mathit{Ab}\hfill & \hfill A^{2}b\hfill & \hfill \dots \hfill & \hfill A^{n-1}b\hfill & \hfill -b\hfill \end{array}\right]$

So $C=\mathit{AK}-\mathit{KB}=\left[\begin{array}{cccc}\hfill 0\hfill & \hfill 0\hfill & \hfill \dots \hfill & \hfill A^{n}b+b\hfill \end{array}\right]$