I am going through the book "Linear System Theory and Design" by Chi-Tsong Chen, where they talk about the controllable canonical form and how it is a realization of a transfer matrix.
In the book, we have the following passage:
What is this shifting property that they are talking about? The closest thing I can find is this relationship between Companion Form Matrices and Linear Recurrence Relations, but not sure how that applies here.
There's nothing fancy here, but elementary matrix multiplication and \$A\$ and \$B\$ being in a particular companion form (which is given in 4.34).
$$A=\left( \begin{array}{ccccc} \alpha _1 I_p & \alpha _2 I_p& \cdots & \alpha _{r-1} I_p & \alpha _r I_p \\ I_p & 0 & \cdots & 0 & 0 \\ 0 & I_p & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & I_p & 0 \\ \end{array} \right) \ \ \ B=\left( \begin{array}{c} I_p \\ 0 \\ 0 \\ \vdots \\ 0 \\ \end{array} \right)$$
Now use (4.35), (4.37) and the above form of \$A\$ and \$B\$ to get the following.
$$s\left( \begin{array}{c} Z_1 \\ Z_2 \\ Z_3 \\ \vdots \\ Z_r \\ \end{array} \right)=\left( \begin{array}{ccccc} \alpha _1 I_p & \alpha _2 I_p& \cdots & \alpha _{r-1} I_p & \alpha _r I_p \\ I_p & 0 & \cdots & 0 & 0 \\ 0 & I_p & \cdots & 0 & 0 \\ \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & \cdots & I_p & 0 \\ \end{array} \right)\left( \begin{array}{c} Z_1 \\ Z_2 \\ Z_3 \\ \vdots \\ Z_r \\ \end{array} \right)+\left( \begin{array}{c} I_p \\ 0 \\ 0 \\ \vdots \\ 0 \\ \end{array} \right)$$
And perform block matrix multiplication. Row blocks 2 to the end will give \$s Z_2=Z_1\$, \$s Z_3 =Z_2\$, etc.
This pattern is because row block 2 onward have the identity matrix in the lower off diagonal. This will shift the elements of \$Z\$ one down (https://en.wikipedia.org/wiki/Shift_matrix). And conveniently, all the corresponding elements of \$B\$ are \$0\$.
