Directions of zeros and poles for transfer matrices (revised)

Before reading the valuable book Control System Synthesis: A Factorization Approach, written by Prof. M. Vidyasagar, I have no idea of the essences of directions of zeros for transfer matrices.

In verifying the right (left) coprimeness of two transfer matrices, the right (left) direction of the transfer matrices should be used correctly. Suppose $G = NM^{-1}$ is a right factorization of a tansfer matrix $G$, we need to use right direction to check if the factorization has zero-pole cancellation or not. Ditto for the left factorization and left direction. Precisely, if  $N$ and $M$ have no common zeros with the same right direction, then $N$ and $M$ are right coprime. Otherwise, there exist $z_0 \in \mathbb{C}$ and nonzero vector $v$ such that
\begin{align*}
\begin{bmatrix}
N(z_0)\\
M(z_0)
\end{bmatrix}v = 0.
\end{align*} Then the elements in the vector $\left[\begin{smallmatrix}
N(s)\\
M(s)
\end{smallmatrix}\right]v$ have a common factor $(s-z_0)$. That is, $\left[\begin{smallmatrix}
N(s)\\
M(s)
\end{smallmatrix}\right]v = \left[\begin{smallmatrix}
n_1(s)\\
m_1(s)
\end{smallmatrix}\right](s-z_0)$ for some polynomials $n_1(s)$ and $m_1(s)$. Now let $V = \left[v ~~ V_1\right]$ be an invertible matrix augmented by $v$ and some $V_1$. Then
\begin{align*}
\begin{bmatrix}
N(s)\\
M(s)
\end{bmatrix}V =
\begin{bmatrix}
N(s)v & N(s)V_1\\
M(s)v & M(s)V_1
\end{bmatrix} =
\begin{bmatrix}
n_1(s) & N(s)V_1\\
m_1(s) & M(s)V_1
\end{bmatrix}\begin{bmatrix}
s-z_0 & 0\\
 0 & I
\end{bmatrix}
\end{align*} or, equivalently, \begin{align*}
\begin{bmatrix}
N(s)\\
M(s)
\end{bmatrix} =
\begin{bmatrix}
N_1(s)\\
M_1(s)
\end{bmatrix}\left(\begin{bmatrix}
s-z_0 & 0\\
 0 & I
\end{bmatrix}V^{-1}\right)
\end{align*} where $N_1(s) = [n_1(s)~~N(s)V_1]$, $M_1(s) = [m_1(s)~~M(s)V_1]$. As $\det\left(\left[\begin{smallmatrix}
s-z_0 & 0\\
 0 & I
\end{smallmatrix}\right]V^{-1}\right) = (s-z_0)\det(V^{-1})$ is not a nonzero constant, $N(s),M(s)$ are not right coprime, and vice versa.

However, to my mind it is meaningless to discuss the left common zero direction in right factorization. On the one hand, the dimensions of the left zero directions of $N$ and $M$ may not be the same. On the other hand, even though, in the square case, there exists a $z_0$ and a nonzero vector $w$ such that $w^*N(z_0)=0,w^*M(z_0)=0$, $N(s)$ and $M(s)$ are probably right coprime. For an example, suppose $G(s)$ has a following right factorization:
\begin{align*}
  N(s) = \begin{bmatrix}
           s+1 & 0 \\
           0 & s+2
         \end{bmatrix},~~M(s) = \begin{bmatrix}
           0 & s+1 \\
           s+3 & 0
         \end{bmatrix}.
\end{align*} Then it is easy to show that $z_0 = -1$ and $w = \begin{bmatrix}
            1 \\ 0
          \end{bmatrix}$ would satisfy the left zero direction condition, while the right coprimeness of $N(s)$ and $M(s)$ can be proved. That is, $NM^{-1}$ is exactly a right coprime factorization of $G(s)$. In this case, we cannot say $M^{-1}N$ is NOT a left coprime factorization of $G(s)$ as $M^{-1}N \ne G$ indeed.

In conclusion, $N$ and $M$ are right (left) coprime if and only if they have no common zero with the same right (left) directions. 

On the other hand, the left direction of the factor $M(s)$ is itself meaningful. For a zero, say $z_0$, of $M(s)$, a vector $\bar{w}_0$ is an input pole vector of $M(s)$ if $\bar{w}_0^*M(z_0) = 0$. Now let a (minimal) state-space representation of $G(z)$ given by $(A,B,C,D)$. Then it proves that \begin{align*}
  M(s) = \left[ \begin{array}{c|c}
           A-BF & -B \\ \hline
           F & I
         \end{array}\right].
\end{align*} Since $z_0$ is an eigenvalue of $A$, there exists a nonzero vector ${w}_0$ such that ${w}_0^*A = z_0 {w}_0^*$. It shows that $\bar{w}_0 = B^*{w}_0$. 

This can be seen from the fact that \begin{align*} \begin{bmatrix} {w}_0^* & \bar{w}_0^* \end{bmatrix} = \left[ \begin{array}{cc} z_0I - A + BF & -B \\ F & I \end{array} \right] = 0. \end{align*}
Moreover, the dyadic expansion of $A$ derives \begin{align*} G(s) = \sum_{i=1}^n \displaystyle \frac{ C v_i w_i^* B }{s-p_i} + D = \sum_{i=1}^n \displaystyle \frac{ \big(C v_i \big) \big(B^* w_i\big)^* }{s-p_i} + D \end{align*} where $v_i$'s and $w_i$'s are the right and left eigenvectors of $A$ corresponding to the eigenvalues $p_i$'s.
The direction vector $C v_i$ is formally called the output (pole) vector of the system.