Define the $\text{sign}(.)$ function as $\text{sign}:\mathbb{R}^n \to \{-1,0,1\}^n$ which gives you the sign of each coordinate of a given vector.
It is clear that $\text{sign}$ is surjective. Now if we want to restrict this function to a given subspace $V$ of $\mathbb{R}^n,$ things get a bit more complicated.
My question is: What is the cardinality of $\text{sign}(V)$ in such a case?
At first, I thought I would be able to make use of the structure of $\mathbb{F}^n_3$ but that did not work, $\text{sign}(V)$ need not be a vector subspace of $\mathbb{F}_3^n$ in general.
For a hyper-plane, all we need is to count how many times we can have at least one negative sign and at least one positive sign, plus the obvious $(0,\cdots, 0)$.
Now I know that every subspace is an intersection of hyperplanes, but I have no idea if this would even be useful.
The cardinality of $sign(V)$ is at least $3^{\dim V}$, and this lower bound can be achieved (say, if $V$ is generated by vectors of the form $(0,\ldots, 0,1,0,\ldots,0)$).
To see this, simply choose a basis of $V$, and write its vectors as rows of a matrix. Applying row reduction to this matrix (and reordering its columns if necessary), we get a $(\dim V \times n)$-matrix of the form $$ \begin{pmatrix} I & | & *\end{pmatrix} $$ where I is the identity $(\dim V\times \dim V)$-matrix, and $*$ can be any $(\dim V\times (n-\dim V))$-matrix.
The rows of this matrix form a basis of $V$, up to the same reordering of the columns as above. From there, it is easy to see that vectors in $V$ can have any combination of signs for their $\dim V$ first entries. Thus there are at least $3^{\dim V}$ elements in $sign(V)$.
The exact number of elements in $sign(V)$ will depend on the matrix $*$; describing all possibilities seems a more difficult problem.