is identity of the space expanded by the vectors, namely $A\times A$ unit matrix. Of course it is diagonal.(2) for $B<A$ case, considering any vector $a$ and setting $b=Ma-a$, we find $Mb=0$. Namely any vector $a$ can be expressed as linear combination of invariant vector $c=Ma$ and null vector $b$ under action of $M$. So the whole linear space is composed of certain invariant vectors and null vectors under action of $M$. $M$ can be diagonalized in the representation with these vectors as basis. So we have: $$M(k,q_{0})=S^{-1}(k,q_{0})\overline{M}(k,q_{0})S(k,q_{0}),$$where $$\overline{M}(k,q_{0})=diag(1,1,\cdots ,1,0,0,\cdots ,0).$$Due to (\[eq:2\]), when $\psi _{st}(k,q_{0})$ is continuous, $M(k,q_{0})$ is also continuous. However $trM(k,q_{0})=tr\overline{M}(k,q_{0})=0,1,2,\cdot \cdot \cdot ,A$, which is discrete, so when $\psi _{st}$ is continuous, the value of $trM(k,q_{0})=A\psi_{00}(k,q_{0})$ is invariant for all $k$ and $q_{0}$. The trace of the projector is the zero order term of $\psi _{00}(k,q_{0})$ in Laurant expression of $e^{-ilk}$ and $e^{-2\pi iqA/l}$, so we have $$\begin{aligned}
trP &=&\int_{0}^{\frac{2\pi }{l}}dk\int_{0}^{\frac{l}{A}}dq\frac{A}{2\pi }%
\psi _{00}(k,q_{0}) \nonumber\\
&=&\int_{0}^{\frac{2\pi }{l}}dk\int_{0}^{\frac{l}{A}}dq\frac{1}{2\pi }%
trM(k,q_{0}) \nonumber\\
&=&\frac{1}{A}trM(k,q_{0}).\end{aligned}$$The projector is trivial for $trM(k,q_{0})=0,A$, indicating $P=0$ and $%
identity$. The nontrivial $trP=\frac{1}{A},\frac{2}{A},\cdots ,\frac{A-1}{A}$. In this paper, we only study the nontrivial projector with minimal trace($%
trM(k,q_{0})=1$). Thus

$$M(k,q_0)=s^{-1}(k,q_0) \left(
\begin{array}{cccc}
1 &  &  &  