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

  1. fix an integer $1\le i \le n$
  2. find all $\overrightarrow{c}\in C$ that uphold $c_i=0$
  3. 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 $k-1$?

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  • $\begingroup$ If $C$ is a vector space of dimension $k$, then choose a basis $\left(e_{i_1},\ldots,e_{i_k}\right)$ for $C$. If $i\in\{i_1,\ldots,i_k\}$, say $i=i_1$, then the new vector space will be spanned by $\left(0,e_{i_2},\ldots,e_{i_k}\right)$, that is by $\left(e_{i_2},\ldots,e_{i_k}\right)$ so the dimension is $k-1$. If $i\notin\{i_1,\ldots,i_k\}$, then you do not change anything since the $i$-th coordinate in $C$ was already zero. $\endgroup$ – Nicolas Apr 9 '16 at 9:31
  • $\begingroup$ Excuse me, I can't understand a thing, what will you do with the vectors which don't have any zero component ?? Will they be removed from the newly built vector space ?? They must be, because if they remain there, the elements of the vector space will be different in no. of components. $\endgroup$ – Anik Bhowmick Aug 1 '18 at 10:02
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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:

  1. $C\subseteq\ker g_i$
  2. $C+\ker g_i=F^n$

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 $k$.

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  • $\begingroup$ the subspace I'm defining is not exactly $Im(g)$ because I only uses vectors that their $i$-th coordinate is $0$ $\endgroup$ – user107761 Apr 9 '16 at 10:31
  • $\begingroup$ @user107761 Now it should be OK $\endgroup$ – egreg Apr 9 '16 at 11:03
  • $\begingroup$ is there a way to solve without Grassman's formula? $\endgroup$ – user107761 Apr 9 '16 at 13:08

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