Different definitions of codimension Let $X$ be a vector space, and $V$ a subspace of $X$. How can I prove that $\dim(X/V)=n \iff$ a basis of V can be extend to a basis of $X$ using just $n$ vectors.
 A: $\Rightarrow$ Let $B$ be a basis for $V$ and let $([w_1],\ldots,[w_n])$ be a basis of $X/V$, for each $i$, $[w_i] = \{x + w_i: x \in X\}$. 
Let $B' = \{w_1,\ldots,w_n\} \subset V$ be a set of representatives of $[w_1],\ldots,[w_n]$. If $\sum_i \alpha_i w_i = 0$ then $\sum_i [\alpha_i w_i] = [0]$ and then $\sum_i \alpha_i [w_i] = [0]$, but since $[w_i]$ are linearly independent, $\alpha_i = 0$ for $1\leq i \leq n$ and $w_1,\ldots,w_n$ are also linearly independent. Now, if $v \in V$ we know that $[v] = \sum_i \alpha_i [w_i]$ for scalars $\alpha_i$ but then $v - \sum_i \alpha_i w_i \in X$ and there are $x_1,\ldots,x_m$ vectors of $B$ such that $v-\sum_{i=1}^n \alpha_i w_i = \sum_{j=1}^m \beta_j x_j$ and then $x$ is a linear combination of elements of $B$ and $B'$. Since $v$ was generic, every element of $V$ is a linear combination of elementos of $B \cup B'$. Showing that $B\cup B'$ is linearly independent can be accomplished passing to the quotient (like we did to B').
$\Leftarrow$ The hypothesis can be translated as: There is a basis $B$ of $V$ such that $B = B' \cup B''$ where $B'$ is a basis for $X$ and $B''$ has $n$ elements. Now, it's a simple matter of passing to the quotient to show that $\dim (X/V) = n$
