statement, we have $$\begin{aligned}
 {\mathrm{proj}\big[ U \big]}\circ\mathcal{T}(\pmb{u}) 
&= \sum_{i=1}^n u_i\cdot{\mathrm{proj}\big[ U \big]}\circ\mathcal{T}(\pmb{b}_i) 
= \sum_{i=1}^n u_i\cdot\mathcal{T}\circ{\mathrm{proj}\big[ U \big]}(\pmb{b}_i) \\
&= \mathcal{T}\circ{\mathrm{proj}\big[ U \big]}\left(\sum_{i=1}^n u_i\pmb{b}_i\right)
= \mathcal{T}\left(\pmb{u}\right). \end{aligned}$$

In accordance with Lemma \[ProjOperatorCommutative\], $({\mathrm{Aut} \hspace{0.4mm} V_{\lambda}})_v = \{ \sigma \in {\mathrm{Aut} \hspace{0.4mm} V_{\lambda}} : \sigma \hspace{0.5mm} {\scriptstyle \circ} \hspace{0.5mm} {\mathrm{proj}\big[ V_{\lambda} \big]}(\pmb{e}_v) = {\mathrm{proj}\big[ V_{\lambda} \big]}(\pmb{e}_v) \}$. Hence, one can easily obtain a partition of $[n]$ relevant to $\Pi^*_v(\lambda)$, which is induced by coordinates of the vector ${\mathrm{proj}\big[ V_{\lambda} \big]}(\pmb{e}_v)$, [*i.e.,*]{} two vertices belong to one cell of the partition if the coordinates corresponding to them are the same. As a matter of fact, we can work out a refined partition more close to $\Pi^*_v(\lambda)$ in virtue of a geometric feature of ${\mathrm{Aut} \hspace{0.4mm} V_{\lambda}}$ — region.

Let $X$ be a subspace of $\operatorname{\mathbb{R}}^n$. As we shall see below, a region of $X$ can be defined in two ways — outside or inside. Let us first define region outside. Suppose $\pmb{b}_1,\ldots,\pmb{b}_m$ is a group of vectors in $\operatorname{\mathbb{R}}^n$ such that ${\mathrm{span} \hspace{0.4mm} \{ \pmb{b}_1,\ldots,\pmb{b}_m \}} = \operatorname{\mathbb{R}}^n$ and $\| \pmb{b}_i \| > 0$ ($i=1,\ldots,m$) where ${\mathrm{span} \hspace{0.4mm} \{ \pmb{b}_i : i \in [m] \}}$ stands for the subspace spanned by $\pmb{b}_1,\ldots,\pmb{b}_m$. Clearly, 