Why is a feedback loop not represented by the least order transfer function?
Not really an answer, but let's work it out with $H(z) = h$ and $C(z) = \frac{K}{z-\alpha}$.:

$$
\begin{array}
\ L(z) &amp;=&amp; \frac{H(z)C(z)}{1+H(Z)C(z)}\\
&amp;=&amp; \frac{h \frac{K}{z-\alpha}}{ 1 + h \frac{K}{z-\alpha}}\\
&amp;=&amp; \frac{h K}{ z - \alpha + h K}\\
&amp;=&amp; \frac{h K}{ z - \alpha + h K}\ \ \frac{z - \beta}{ z - \beta}
\end{array}
$$

I added the last line to show what I mean by a pole-zero cancellation.  $\beta$ can be any value; if the pole and the zero at $\beta$ cancel out, the overall value of $\frac{z - \beta}{ z - \beta}$ is 1.  

The trick will be to figure out why matlab inserts the extra $\frac{z - \alpha}{ z - \alpha}$ in there in the first plcae,