I'm not sure how to approach this question. We let $U$ be a subspace of a vector space $V$ over a field $F$ and also let $(v_{1},\dotsc, v_{n})$ be a basis for $V$ . We must show that there are $1 \le k_{1} \lt \dotsc \lt k_{m} \le n$ such that $(v_{k_{1}} + U, \dotsc, v_{k_{m}} + U)$ is a basis for the quotient space $V/U$. I'm quite confused about what this question is asking so if someone could explain what we are trying to show and give an outline on how to show it that would help greatly. Thanks, Lauren!
[Math] Proof regarding the basis of a quotient space (of a vector space and a subspace)
linear algebra
Best Answer
Well,think about what it means to have a basis for a vector space V over a field F. A basis B is a set of vectors in V for which for every vector v in V, there exists S= ${v_1,v_2,.....v_n}\subseteq B$ and ${a_1,a_2,...,a_n}\in F$ such that $\sum_{i=1}^n a_iv_i = v$ and $\sum_{i=1}^n a_iv_i = 0$ iff for every $i$, $a_i=0$. So let's check. Let U be a subspace of V and consider the quotient space V\U. Consider $S'= S + U\subseteq V/U $ and let's see if this is a basis for V/U. Let's see if it spans V/U. For every $u\in V/U$, u = v + w where v is an arbitrary vector in V where $u-v=w\in U$. Since B is a basis for V, there exists ${v_1,v_2,.....v_i}\in B$ and ${a_1,a_2,...,a_i}\in F$ where $1\leq i \leq n$ such that u = $\sum_{j=1}^i a_jv_j$. Also, since B is a basis for V and U is a subspace of V, there exists ${v_1,v_2,.....v_k}\in B$ and ${b_1,b_2,...,b_k}\in F$ such that for every $w\in U$, w = $\sum_{l=1}^k b_lv_l$ where $1\leq k \leq n$. But this means u = v + w = $\sum_{j=1}^i a_jv_j$ + $\sum_{l=1}^k b_lv_l$ = $\sum_{m=1}^{j+l}(a_j + b_l)v_m$ where $1\leq j+l \leq n$ . But this means S' spans V/U. Since for every m where $1\leq m\leq n$, $v_m\in B$, then $\sum_{i=1}^m a_iv_i = 0$ iff for every $i\leq m$, $a_i=0$. But that means S' is a linearly independent set of vectors in V and that means S' is a basis for V/U. Q.E.D.
The notation of my proof in the indices may be a little sloppy. I'll go over it later, but the basic logic is correct.