Suppose that $1 < k < n$. Does there exist a constant $\beta > 0$, such that for every $k$ orthonormal vectors $f\_1,\ldots,f\_k \in \mathbb R^n$,
there exist $k$ orthonormal vectors with nonnegative elements, $x\_1,\ldots,x\_k\in \mathbb R\_+^n$, such that 


$$\sum\_{i=1}^k \|x\_i - f\_i\|^2\_2 \leq \beta \sum\_{i=1}^k\|f\_i^-\|\_2^2$$


where $f\_i^- := \max\{-f\_i,0\}$ is the negative part of the vectors $f\_i$?




---


In another way, I am interested in the estimation of the distance of a $n\times k$ dimensional matrix $F$ whose columns are orthonormal from the set of $n\times k$ matrices whose columns are nonnegative and orthonormal, i.e., 


$$
\mathrm{dist}(F;St\_+(n,k))
$$


Where $St\_+(n,k)$ is the set of $n\times k$ matrices whose columns are nonnegative and orthonormal. If we drop orthonormal condition and compute $\mathrm{dist}(F;\mathbb{R}^{n\times k }\_+)$, we obtain $\|F^-\|$ as a lower bound for the above distance. In this term, my question is as follows: Is a multiple of $\|F^-\|$ an upper bound for the above distance, i.e., 



> 
> Is there a constant $c>0$, such that
>  $$
> \|F^-\| \leq \mathrm{dist}(F;St\_+(n,k)) \leq c \|F^-\|
> $$
>  for every $n\times k$ dimentional matrix $F$ whose columns are orthonormal?
> 
> 
> 


In the special case $k=1$, the above statement is true, with $c = 2$. 
I'm interested in the special case of small values of $k$, such as $k=2$. Experimentally,
for $k>1$ and random matrices $F$ and by using Frobenius norm, I get an upper bound for $\mathrm{dist}(F; St\_+(n,k))$ by alternating projection to nonnegative matrices and orthonormal matrices. I guess that the above statement is true for $c \approx 2$, $(\beta \approx 4)$.



Distance from nonnegativity of some orthonormal vectors
$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$


The following two simple lemmas are crucial. 



> 
> **Lemma 1.** For any nonnegative numbers $a\_1,\dots,a\_k$,
>  \begin{equation\*}
>  \sum\_1^k a\_i^2-\max\_1^ka\_i^2\le\sum\_{1\le i<i'\le k}a\_i a\_{i'}. 
> \end{equation\*}
> 
> 
> 


Proof. Without loss of generality, $a\_1=\max\_1^ka\_i$. Then $a\_i^2\le a\_1a\_i$ for all $i=2,\dots,k$. So, Lemma 1 follows.


For any $u\in\R^n$, let $u^+ :=\max\{u,0\}$ and $u^- :=\max\{-u,0\}$, so that $u=u^+-u^-$. 



> 
> **Lemma 2.** For any orthonormal vectors $u$ and $v$, 
>  \begin{equation\*}
>  u^+\cdot v^+\le \|u^-\|+\|v^-\|, 
> \end{equation\*}
>  where $\cdot$ denotes the dot product. 
> 
> 
> 


Proof. We have $0=u\cdot v=u^+\cdot v^+ - u^+\cdot v^- -u^-\cdot v^+ + u^-\cdot v^-
\ge u^+\cdot v^+ - \|u^+\|\,\|v^-\| -\|u^-\|\,\| v^+\|$, whence
Lemma 2 follows. 


As in the question, let now $f\_1,\ldots,f\_k$ be any orthonormal vectors in $\R^n$. Write $f\_i=(f\_{ij})\_{j=1}^n$ and $f^+\_i=(f^+\_{ij})\_{j=1}^n$. Let $(J\_1,\dots,J\_k)$ be any partition of the set $[n]:=\{1,\dots,n\}$ such that for all $i\in[k]$ and $j\in[n]$ we have the implication 
\begin{equation\*}
 j\in J\_i\implies f^+\_{ij}=\max\_{q\in[k]}f^+\_{qj}. 
\end{equation\*}
Define $y\_i=(y\_{ij})\_{j=1}^n$ by 
\begin{equation\*}
 y\_{ij}:=f^+\_{ij}\,\ii{j\in J\_i}, 
\end{equation\*}
where $\ii{}$ denotes the indicator; so, $y\_{ij}=\max\_{q\in[k]}f^+\_{qj}$ for $j\in J\_i$ and $y\_{ij}=0$ for $j\in[n]\setminus J\_i$. 
Hence, in view of Lemmas 1 and 2, 
\begin{multline\*}
 \sum\_1^k\|y\_i-f^+\_i\|^2
 =\sum\_{j\in[n]}\Big(\sum\_{i\in[k]}(f^+\_{ij})^2-\max\_{i\in[k]}(f^+\_{ij})^2\Big)
 \le\sum\_j\sum\_{i<i'}f^+\_{ij}f^+\_{i'j} \\ 
 =\sum\_{i<i'}f^+\_i\cdot f^+\_{i'} 
 \le\sum\_{i<i'}(\|f^-\_i\|+\|f^-\_{i'}\|)
 =2(k-1)\sum\_{i\in[k]}\|f^-\_i\|. \tag{1}
\end{multline\*}


Also, $\sum\_1^k\|f\_i-f^+\_i\|^2=\sum\_1^k\|f^-\_i\|^2\le\sum\_1^k\|f^-\_i\|$. So, by (1) and Minkowski's inequality,  

\begin{equation\*}
 \sum\_{i\in[k]}\|y\_i-f\_i\|^2
 \le(\sqrt{2(k-1)}+1)^2\sum\_{i\in[k]}\|f^-\_i\|\le3k\sum\_{i\in[k]}\|f^-\_i\|=:\ep. \tag{2}
\end{equation\*}


Next, 
\begin{equation}
 0\le1-\|y\_i\|=\|f\_i\|-\|y\_i\|\le\|y\_i-f\_i\|, \tag{3}
\end{equation} 
by the triangle inequality. 


Consider now two possible cases: 


*Case 1: $\ep<1$.* (This is hopefully the main case.) Then, by (2), $\|y\_i-f\_i\|<1$ for all $i$, whence, by (3), $y\_i\ne0$ for all $i$, so that we can let  

\begin{equation\*}
 x\_i:=y\_i/\|y\_i\|. 
\end{equation\*}


Then $x\_1,\dots,x\_k$ are orthonormal vectors in $\R\_+^n$, and  

\begin{equation\*}
 \sum\_{i\in[k]}\|x\_i-y\_i\|^2=\sum\_{i\in[k]}(1-\|y\_i\|)^2\le\sum\_{i\in[k]}\|y\_i-f\_i\|^2\le\ep 
\end{equation\*}
by (3) and (2), 
which yields 
\begin{equation\*}
 \sum\_{i\in[k]}\|x\_i-f\_i\|^2
\le4\ep=12k\sum\_{i\in[k]}\|f^-\_i\|. 
\end{equation\*}


*Case 2: $\ep\ge1$.* Here for any orthonormal $x\_1,\dots,x\_k$ we have 
\begin{equation\*}
 \sum\_{i\in[k]}\|x\_i-f\_i\|^2\le 2\sum\_{i\in[k]}(\|x\_i\|^2+\|f\_i\|^2)=4k\le4k\ep
 =12k^2\sum\_{i\in[k]}\|f^-\_i\|. 
\end{equation\*}


Thus, 
\begin{equation\*}
 \sum\_{i\in[k]}\|x\_i-f\_i\|^2\le
 \left\{
 \begin{aligned}
 12k\sum\_{i\in[k]}\|f^-\_i\|&\text{ if }\ep<1,\\
 12k^2\sum\_{i\in[k]}\|f^-\_i\|&\text{ if }\ep\ge1. 
 \end{aligned}
 \right.
\end{equation\*}


(As follows from the comment by user fedja, here $\|f^-\_i\|$ cannot be replaced by $\|f^-\_i\|^{1+\ep}$, for any real $\ep>0$.) 


