Let $U$ be a subspace of a finite dimensional $F$ - vector space $V$ with $dim U = k$ and $dimV=n$. Prove that if $\mathscr B_U = \{\vec v_1, \dots, \vec v_k \} $ is a basis for $U$ and $\mathscr B_V = \{\vec v_1, \dots, \vec v_k, \vec v_{k+1}, \dots, \vec v_n \} $ is a basis for V, then $$\mathscr B_{V/U} = \{\vec v_{k+1} +U, \dots, \vec v_n + U \}$$ is a basis for $V/U$.
Solution: Assume $$B_{V/U} = \{\vec v_{k+1} +U, \dots, \vec v_n + U \}$$ is a basis. Then $ c_{k+1}(\vec v_{k+1} +U) + \dots + c_n(\vec v_n + U)= 0 +U \implies$
$(c_{k+1}\vec v_{k+1} + \dots + c_n\vec v_n ) + U= 0 +U$. We can rewrite vector $c_{k+1}\vec v_{k+1} + \dots + c_n\vec v_n $ as a linear combination of vectors from $\mathscr B_U$. So we will get:
$c_{k+1}\vec v_{k+1} + \dots + c_n\vec v_n = c_1\vec v_1 + \dots + c_k \vec v_k$, $\implies$ $ c_1\vec v_1 + \dots + c_k \vec v_k - (c_{k+1}\vec v_{k+1} + \dots + c_n\vec v_n ) = 0$ and since vectors $ \vec v_1, \dots, \vec v_k, \vec v_{k+1}, \dots, \vec v_n$ form a basis for the space $V$, they are linearly independent. So $c_1=\dots=c_k=c_{k+1}=\dots=c_n=0$. Thus the equation $(c_{k+1}\vec v_{k+1} + \dots + c_n\vec v_n ) + U= 0 +U$ will have only trivial solution for $c_{k+1}=\dots=c_n=0$ and vectors $\vec v_{k+1} +U, \dots, \vec v_n + U$ will be linearly independent. Therefore, $\mathscr B_{V/U}$ form a basis for $ U/V$.