$Q_{3}$ is the $\mathbb{Z}_{3}$ charge matrix, we can further derive $$A^{[i]}P_{1}=P_{1}A^{[i]}P_{1},A^{[i]}P_{2}=P_{2}A^{[i]}P_{2},  \notag$$where $P_{1}=Q_{3}P_{0}Q_{3}^{-1}$ and $P_{2}=Q_{3}^{-1}P_{0}Q_{3}$ are also invariant subspace projectors. Since $P_{0}$ is already associated with an irreducible invariant space, $P_{0}$, $P_{1}$ and $P_{2}$ must be either the same or mutually orthogonal, otherwise it will contradict with the fact that $P_{0}$ is already associated with an irreducible invariant space. So there are two different situations, we discuss them separately.

1\. $\left[ P_{0},Q_{3}\right] =0$

It can be simply determined that $P_{0}=P_{1}=P_{2}$, and $P_{0}$ contains all three charge sectors, indicating that $A^{[i]}$ in $P_{0}$ irreducible invariant subspace preserve the $\mathbb{Z}_{3}$ charge symmetry. Thus all $A^{[i]}$ in this invariant subspace will have the initial structures shown in Eq. (\[block\]) and span a trivial simple $\mathbb{Z}_{3}$ graded algebra.

2\. $\left[ P_{0},Q_{3}\right] \neq 0$

In this situation, $P_{0}$, $P_{1}$ and $P_{2}$ are mutually orthogonal projectors, the corresponding invariant subspaces do not contain all $\mathbb{Z}_{3}$ charge sectors. For the parafermionic MPS, the $\mathbb{Z}_{3}$ charge symmetry can never be broken and the MPS can not be reduced, since the invariant spaces do not contain all charge sectors. The reduced matrices also break the $\mathbb{Z}_{3}$ graded structures of local matrices and no longer span a $\mathbb{Z}_{3}$ graded algebra. 