Thus, the concept of irreducibility should be reformulated. Notice that $[P_{0}+P_{1}+P_{2},Q_{3}]=0$, and the total invariant space is the complete, leading to $P_{0}+P_{1}+P_{2}=\mathbbm{1}$. The idempotency requires $P_{0}^{2}=P_{0}$, $P_{1}^{2}=P_{1}$ and $P_{2}^{2}=P_{2}$. From these constraints, the invariant subspace projectors can be derived as $$P_{0}=\frac{1}{3}\left[ 
\begin{array}{ccc}
\mathbbm{1} & U_{1} & U_{1}U_{2} \\ 
U_{1}^{\dagger } & \mathbbm{1} & U_{2} \\ 
U_{2}^{\dagger }U_{1}^{\dagger } & U_{2}^{\dagger } & \mathbbm{1}\end{array}\right] ,$$where $U_{1}$ and $U_{2}$ are unitary block matrices with the same dimensions. Since $A^{[i]}P_{j}=P_{j}A^{[i]}P_{j}$, we can obtain $$\begin{aligned}
A^{[i]} &=&\left[ 
\begin{array}{ccc}
a_{0}^{[i]} & 0 & 0 \\ 
0 & U_{1}^{\dagger }a_{0}^{[i]}U_{1} & 0 \\ 
0 & 0 & U_{2}^{\dagger }U_{1}^{\dagger }a_{0}^{[i]}U_{1}U_{2}\end{array}\right] ,|i|=0;  \notag  \label{Z3nontrivial} \\
\text{ }A^{[i]} &=&\left[ 
\begin{array}{ccc}
0 & a_{0}^{[i]} & 0 \\ 
0 & 0 & U_{1}^{\dagger }a_{0}^{[i]}U_{2} \\ 
U_{2}^{\dagger }U_{1}^{\dagger }a_{0}^{[i]}U_{1}^{\dagger } & 0 & 0\end{array}\right] ,|i|=1;  \notag \\
\text{ }A^{[i]} &=&\left[ 
\begin{array}{ccc}
0 & 0 & a_{0}^{[i]} \\ 
U_{1}^{\dagger }a_{0}^{[i]}U_{2}^{\dagger }U_{1}^{\dagger } & 0 & 0 \\ 
0 & U_{2}^{\dagger }U_{1}^{\dagger }a_{0}^{[i]]}U_{2}^{\dagger } & 0\end{array}\right] ,|i|=2.\text{ }  \notag \\
&&\end{aligned}$$The gauge transformation $G=\mathbbm{1}\oplus U_{1}\oplus \left(
U_{1}U_{2}\right) $ can be used to rewrite them in the standard forms. After substituting $a^{[i]}$ for $a_{0}^{[i]}$ if $|i|=0$, $a_{0}^{i}U_{1}^{\dagger }$ if $|i|=1$, and $a_{0}^{i}U_{2}^{\dagger }U_{1}^{\dagger }$ 