Is true that every finite set has a positively linearly independent subset that positively spans itself? Since it seems that there is no definition for what is positively linearly independent on this site I will try to be concise defining it here. Let $P$ be $n$ by $m$ a matrix and $V$ be a $n$ by $o$ matrix. We say that that $P \cup V$ is positively independent if and only if for every pair of vectors $({\boldsymbol \mu,{\boldsymbol \lambda}})$ with ${ \boldsymbol \mu}$ with non-negative coordinates and
$$
P { \boldsymbol \mu} + V { \boldsymbol \lambda} =\textbf{0},
$$ imples ${ \boldsymbol \mu}={ \boldsymbol 0}$ and ${ \boldsymbol \lambda}= { \boldsymbol 0}.$ Furthermore, we say that $P \cup V$ positively spans its elements if every colum ${\boldsymbol u}$ of $P\cup V$ can be written as
$$
P { \boldsymbol \mu} + V { \boldsymbol \lambda} = u,
$$ for some ${ \boldsymbol \mu}$ with non-negative coordinates.
I'm asking if every matrix $P'$ and $V'$ has a positively linearly independent submatrix $P \cup V$ such that positively spans the set of all ${\boldsymbol u'}$ such that
$$
u'= P' { \boldsymbol \mu'} + V' { \boldsymbol \lambda'}
$$ with ${ \boldsymbol \mu'}$ with non-negative coordinates. Is that true? I strongly suppose that it is true. Any hints?
P.S.: If it is needed, assume $V = {\boldsymbol 0}.$

Consider $P = \begin{bmatrix} 1&-1\end{bmatrix}$ and $V = 0$. The columns of $P$ are not positively linearly independent. We have $P\begin{bmatrix} 2&1\end{bmatrix}^T = [1]$ while $P\begin{bmatrix}1&2\end{bmatrix}^T = [-1]$.
If we were to obtain a positively linearly independent submatrix of $P$, then it's positive span will either have nonnegative entries or have nonpositive entries, because such submatrices are $[1]$ and $[-1]$. Therefore, no such positively linearly independent submatrix exists. Does this answer your question? If you want $V \neq 0$, then I think you can take $P = [1], V = [-1]$ and repeat the argument above.
