Q:

dimension of vector space constructed by reducing another vector space

given $C \subset F^n$ is a vector space of dimension $k$, we construct a new vector space like so: 

fix an integer $1\le i \le n$ 
find all $\overrightarrow{c}\in C$ that uphold $c_i=0$
delete that coordinate to get a vector af length $n-1$.

How can I now show that the dimension of this new vector space is either $k$ or $k-1$?

A:

I guess your $C$ is a subspace of $F^n$.
Fix $i$ with $1\le i\le n$ and consider the map $f_i\colon F^n\to F^{n-1}$ that “discards the $i$-th coordinate”. For instance, if $n=4$ and $i=2$, the map is
$$
(x_1,x_2,x_3,x_4)\mapsto (x_1,x_3,x_4)
$$
The map $f_i$ is linear and surjective, so its kernel has dimension $1$ by the rank-nullity theorem. Consider also $g_i\colon F^n\to F$, that “selects the $i$-th coordinate”. Clearly, $\dim\ker g_i=n-1$.
You want to compute $\dim f_i(C')$, where $C'=\{x\in C:x_i=0\}$ (add arrows on top of letters, if you like them). By the rank-nullity theorem, $\dim f_i(C')=\dim C'-\dim(C'\cap\ker f_i)=\dim C'$, because, by definition, $C'\cap\ker f_i=\{0\}$.
Now note that $C'=C\cap\ker g_i$. By Grassman's formula,
$$
\dim C'=\dim C+\dim\ker g_i-\dim(C+\ker g_i)
$$
Since $\dim\ker g_i=n-1$, there are two cases:

$C\subseteq\ker g_i$
$C+\ker g_i=F^n$

In the first case, $C+\ker g_i=\ker g_i$, so we get
$$
\dim C'=\dim C+\dim\ker g_i-\dim(C+\ker g_i)=\dim C
$$
In the second case we get
$$
\dim C'=\dim C+n-1-n=\dim C-1
$$

Alternative solution. Consider a set of equations in $n$ unknowns so that the solution set is $C$. Since $C$ has dimension $k$, the matrix of the system has rank $n-k$. If you add the equation $x_i=0$ to the above equations, there are two cases: either the rank of the matrix increases by $1$ or it remains the same. So the solution set, which is $C'$, either has dimension $k-1$ or $k$.

