vector $\pmb{v}$ in $\operatorname{\mathbb{R}}^n$. It is easy to check that a permutation on $[n]$ is an isometry on $\operatorname{\mathbb{R}}^n$.

\[ProjOperatorCommutative\] Let $\mathcal{T}$ be an isometry on $\mathbb{R}^n $, and let $U$ be a subspace of $\mathbb{R}^n $. Then the following statements are equivalent.

1.  $U$ is $\mathcal{T}$-invariant.

2.  $\mathcal{T}\circ{\mathrm{proj}\big[ U \big]} ={\mathrm{proj}\big[ U \big]}\circ\mathcal{T}$, where ${\mathrm{proj}\big[ U \big]}$ is the orthogonal projection onto the subspace $U$.

3.  There exists a basis $\pmb{b}_1,\ldots,\pmb{b}_n$ of $\operatorname{\mathbb{R}}^n$ so that $\mathcal{T}\circ{\mathrm{proj}\big[ U \big]}(\pmb{b}_i) ={\mathrm{proj}\big[ U \big]}\circ\mathcal{T}(\pmb{b}_i)$, $i=1,\ldots,n$.

We first verify that [*i)*]{}$\Rightarrow$[*ii)*]{}. Let $\pmb{v}$ be a vector of $\operatorname{\mathbb{R}}^n$. Then there exist uniquely $\pmb{u}\in U$ and $\pmb{u}'\in U^{\bot}$ so that $\pmb{v}=\pmb{u}+\pmb{u}'$. Consequently, $\mathcal{T}\circ{\mathrm{proj}\big[ U \big]}(\pmb{v})=\mathcal{T}\circ{\mathrm{proj}\big[ U \big]}(\pmb{u}+\pmb{u}')=\mathcal{T}(\pmb{u})={\mathrm{proj}\big[ U \big]}(\mathcal{T}(\pmb{u})+\mathcal{T}(\pmb{u}'))={\mathrm{proj}\big[ U \big]}\circ\mathcal{T}(\pmb{v})$ since $\mathcal{T}$ is an isometry and $U$ is an $\mathcal{T}$-invariant subspace.

Clearly, the 2nd statement can imply the 3rd one. So now we turn to the last part and show that the 3rd statement implies the 1st one.

Let us first recall a fact that $$\mathcal{T}U=U \mbox{ if and only if }\mathcal{T}(\pmb{u}) = {\mathrm{proj}\big[ U \big]}\circ\mathcal{T}(\pmb{u}), ~\forall \pmb{u}\in U.$$ Since $\pmb{b}_1,\ldots,\pmb{b}_n $ is a basis of $\mathbb{R}^n$, for any vector $\pmb{u}\in U$, $\pmb{u}=\sum_{i=1}^n u_i\pmb{b}_i$ where $u_i\in\operatorname{\mathbb{R}}$ and $i=1,\ldots,n$. In accordance with the 3rd 