Let $\mathbf{x}$ be a random Gaussian vector in $\mathbb{R}^n$, i.e. $\mathbf{x}\sim\mathcal{N}(\mathbf{0},\mathbf{I}\_n)$. Then for any fixed unit vector $\mathbf{u}$, one has $\mathbf{u}\mathbf{u}^\top \mathbf{x}$ is independent of $(\mathbf{I}\_n-\mathbf{u}\mathbf{u}^\top)\mathbf{x}$ and trivially their sum is $\mathbf{x}$ itself.


My question is as following.
Let $\mathbf{y}$ be a random vector sampled from a uniform coordinate distribution, that is $\mathbf{y}$ is distributed uniformly across $\{\sqrt{n}\mathbf{e}\_1, \sqrt{n}\mathbf{e}\_2,\cdots,\sqrt{n} \mathbf{e}\_n\}$, where $\mathbf{e}\_i$ is the $i$-th standard basis vector. 


Can we similarly decompose $\mathbf{y}$ into two parts such that those two parts are independent of each other?



Independent decomposition of coordinate distribution
$\newcommand{\al}{\alpha}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\tf}{\tilde f}
\newcommand{\tg}{\tilde g}$


Let $Y:=\mathbf{y}/\sqrt n$ and $e\_i:=\mathbf e\_i$, so that $Y$ is uniformly distributed on the set $\{e\_1,\dots,e\_n\}$. Let also $I$ denote the identity operator on $\R^n$. 


It appears reasonable to interpret the OP's question as follows: Do there exist linear operators $A$ and $B$ on $\R^n$ such that $B=I-A$ and the random vectors $AY$ and $BY$ are (stochasticaly) independent and each of them is not almost surely constant? Let us show that the answer to this question is **no**. 


Indeed, let $f\_1,\dots, f\_k$ denote all the distinct vectors in the set $\{Ae\_1,\dots,Ae\_n\}$, and then let 
\begin{equation}
 I\_r:=\{p\in[n]\colon Ae\_p=f\_r\} \tag{1}
\end{equation} 
for $r\in[k]$,  

where $[n]:=\{1,\dots,n\}$. Note that $f\_1,\dots, f\_k$ are all the distinct values of the random vector $AY$, and 
$$\PP(AY=f\_r)=|I\_r|/n$$ 
for $r\in[k]$, where $|\cdot|$ denotes the cardinality. 
Similarly, let $g\_1,\dots, g\_\ell$ denote all the distinct vectors in the set $\{Be\_1,\dots,Be\_n\}$, and then let 
\begin{equation}
 J\_s:=\{p\in[n]\colon Be\_p=g\_s\} \tag{2}
\end{equation}
for $s\in[\ell]$, so that 
$$\PP(BY=g\_s)=|J\_s|/n$$ 
for $s\in[\ell]$. (At this point, let us pretend to forget the condition $B=I-A$, which implies $\ell=k$ and, without loss of generality, $J\_r=I\_r$ for all $r\in[k]=[\ell]$.)  

We also have 
$$\PP(AY=f\_r,BY=g\_s)=|I\_r\cap J\_s|/n$$ 
for $r\in[k]$ and $s\in[\ell]$. 


So, the independence of $AY$ and $BY$ means that for all $r\in[k]$ and $s\in[\ell]$ 
\begin{equation\*}
 |I\_r\cap J\_s|=|I\_r|\,|J\_s|/n. \tag{2.5}
\end{equation\*}
Note that the sets $I\_r$ and $J\_s$ are nonempty, and so, by (2.5), $|I\_r\cap J\_s|$ is a natural number (that is, a strictly positive integer). 


We have the following number-theoretic proposition: 



> 
> **Proposition 1.** Suppose that, for all $r\in[k]$ and $s\in[\ell]$, 
>  \begin{equation\*}
> n\_{rs}=a\_rb\_s, 
> \end{equation\*}
>  where $n\_{rs}$ is a natural number and $a\_r$ and $b\_s$ real numbers. Then there exist natural numbers $u\_1,\dots,u\_k,v\_1,\dots,v\_\ell$ such that 
>  \begin{equation\*}
> n\_{rs}=u\_rv\_s 
> \end{equation\*}
>  for all $r\in[k]$ and $s\in[\ell]$. 
> 
> 
> 


This will be proved a bit later in this answer. At this point, let us just use Proposition 1 in conjunction with (2.5), to see that 
\begin{equation\*}
|I\_r\cap J\_s|=u\_rv\_s 
\end{equation\*}
for some natural numbers $u\_1,\dots,u\_k,v\_1,\dots,v\_\ell$ and 
all $r\in[k]$ and $s\in[\ell]$. 
Let 
\begin{equation\*}
 u:=\sum\_1^k u\_r,\quad v:=\sum\_1^\ell v\_s. 
\end{equation\*}
Note also that $(I\_r\cap J\_s\colon r\in[k], s\in[\ell])$ is a partition of the set $[n]$. 
So, there is a $u\times v$ matrix $B$ consisting of $u\_r\times v\_s$ blocks $B\_{rs}$ such that 
for all $r\in[k]$, $s\in[\ell]$, and $p\in[n]$ we have 
\begin{equation\*}
 p\in I\_r\cap J\_s\iff p\Vvdash B\_{rs}, 
\end{equation\*}
where $\Vvdash$ means "is an entry of". 
Note also that $uv=n$ and the set of all entries of the matrix $B$ is precisely the set $[n]$. 
Moreover, we have the one-to-one correspondence  

\begin{equation\*}
 [u]\times[v]\ni(i,j)\leftrightarrow p=b\_{i,j}\in[n]
\end{equation\*}
between the sets $[u]\times[v]$ and $[n]$, where $b\_{i,j}$ is the $ij$-entry of the matrix $B$. 
Accordingly, for $(i,j)\in[u]\times[v]$ let 
\begin{equation\*}
 e\_{ij}:=e\_p\quad\text{if }p=b\_{i,j}; \tag{3}
\end{equation\*}
so, we get a double-index re-enumeration $e\_{ij}$ of $e\_p$. 


Next, note that any row index $i\in[u]$ of the matrix $B$ uniquely determines the corresponding block row index $r\_i$ so that for any $j\in[v]$ there is a (unique) $s\in[\ell]$ such that $b\_{ij}\Vvdash B\_{r\_i s}$. Similarly, any column index $j\in[v]$ of the matrix $B$ uniquely determines the corresponding block column index $s\_j$. 
Therefore, in view of (1), (2), and (3), 
\begin{equation\*}
 Ae\_{ij}=\tf\_i:=f\_{r\_i},\quad Be\_{ij}=\tg\_j:=g\_{s\_j}
\end{equation\*}
for all $i\in[u]$ and $j\in[v]$. 


If $u=1$, then $k=1$, so that the set $\{Ae\_1,\dots,Ae\_n\}$ is a singleton set, and hence the random vector $AY$ is actually a (nonrandom) constant. Similarly, if $v=1$, then the random vector $BY$ is a constant. 


Excluding these two trivial cases, suppose now that $u,v\ge2$, and also recall at this point that $B=I-A$. We have $e\_{ij}=Ae\_{ij}+Be\_{ij}=\tf\_i+\tg\_j$ for $i,j$ in $\{1,2\}$ (say). Hence, 
\begin{equation\*}
 0=(\tf\_1+\tg\_1)+(\tf\_2+\tg\_2)-(\tf\_1+\tg\_2)-(\tf\_2+\tg\_1)
 =e\_{11}+e\_{22}-e\_{12}-e\_{21},
\end{equation\*}
which is a contradiction. 


Thus, to complete the answer, it remains to prove Proposition 1. For each pair $(r,s)\in[k]\times[\ell]$, consider the prime factor decomposition 
\begin{equation\*}
 n\_{rs}=\prod\_{\al=1}^\infty p\_\al^{N\_{rs\al}}, 
\end{equation\*}
where $p\_\al$ is the $\al$th prime number and the $N\_{rs\al}$'s are some nonnegative integers (only finitely many of then nonzero). The condition $n\_{rs}=a\_rb\_s$ implies that 
\begin{equation\*}
 N\_{11\al}+N\_{rs\al}=N\_{r1\al}+N\_{1s\al}
\end{equation\*}
for all $r,s$. So, Proposition 1 reduces to 



> 
> **Lemma 1.** Suppose that $N\_{rs}$'s are nonnegative integers such that 
>  \begin{equation\*}
> N\_{11}+N\_{rs}=N\_{r1}+N\_{1s}
> \end{equation\*}
>  for all $(r,s)\in[k]\times[\ell]$. Then there exist nonnegative integers $m\_1,\dots,m\_k,n\_1,\dots,n\_\ell$ such that 
>  \begin{equation\*}
>  N\_{rs}=m\_r+n\_s
> \end{equation\*}
>  for all $(r,s)\in[k]\times[\ell]$. 
> 
> 
> 


It remains to prove Lemma 1. This can be done by the following explicit construction of $m\_1,\dots,m\_k,n\_1,\dots,n\_\ell$: for all $(r,s)\in[k]\times[\ell]$, 
\begin{equation\*}
 m\_1:=\max\_1^k(N\_{11}-N\_{r1})\ge N\_{11}-N\_{11}=0, 
\end{equation\*}
\begin{equation\*}
 r\ge2\implies m\_r:=N\_{r1}-N\_{11}+m\_1\ge N\_{r1}-N\_{11}+N\_{11}-N\_{r1}=0, 
\end{equation\*}
\begin{align\*}
 n\_s:=N\_{1s}-m\_1
 &=N\_{1s}+\min\_{r=1}^k(N\_{r1}-N\_{11}) \\ 
 &=\min\_{r=1}^k(N\_{1s}+N\_{r1}-N\_{11})
 =\min\_{r=1}^k N\_{rs}\ge0. 
\end{align\*}
So, $m\_1,\dots,m\_k,n\_1,\dots,n\_\ell$ are nonnegative integers. Also, for all $(r,s)\in[k]\times[\ell]$, 
\begin{equation\*}
 m\_1+n\_s=m\_1+N\_{1s}-m\_1=N\_{1s}
\end{equation\*}
and, if $r\ge2$, 
\begin{equation\*}
 m\_r+n\_s=N\_{r1}-N\_{11}+m\_1+N\_{1s}-m\_1=N\_{rs}. 
\end{equation\*}
Thus, all the necessary proofs are completed. 


